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e72f0578-1e36-4f81-af6a-109a1ee804b7 | # THERMAL PROPERTIES OF MATTER
bodies like, the moon, Sun and other stars. Light from the moon is found to have a maximum intensity near the wavelength 14 μm. By Wien’s law, the surface of the moon is estimated to have a temperature of 200 K. Solar radiation has a maximum at λₘ = 4753 Å. This corresponds to T = 6060 K. Remember, this is the temperature of the surface of the sun, not its interior.
The most significant feature of the blackbody radiation curves in Fig. 10.18 is that they are universal. They depend only on the temperature and not on the size, shape or material of the blackbody. Attempts to explain blackbody radiation theoretically, at the beginning of the twentieth century, spurred the quantum revolution in physics, as you will learn in later courses.
Energy can be transferred by radiation over large distances, without a medium (i.e., in vacuum). The total electromagnetic energy radiated by a body at absolute temperature T is proportional to its size, its ability to radiate (called emissivity) and most importantly to its temperature. For a body, which is a perfect radiator, the energy emitted per unit time (H) is given by:
H = AσT4 (10.16)
where A is the area and T is the absolute temperature of the body. This relation obtained experimentally by Stefan and later proved theoretically by Boltzmann is known as Stefan-Boltzmann law and the constant σ is called Stefan-Boltzmann constant. Its value in SI units is 5.67 × 10–8 W m–2 K–4. Most bodies emit only a fraction of the rate given by Eq. 10.16. A substance like lamp black comes close to the limit. One, therefore, defines a dimensionless fraction e and writes:
H = AeσT4 (10.17)
Here, e = 1 for a perfect radiator. For a tungsten lamp, for example, e is about 0.4. Thus, a tungsten lamp at a temperature of 3000 K and a surface area of 0.3 cm² radiates at the rate H = 0.3 × 10–4 × 0.4 × 5.67 × 10–8 × (3000)4 = 60 W.
A body at temperature T, with surroundings at temperatures Tₛ, emits, as well as, receives energy. For a perfect radiator, the net rate of loss of radiant energy is:
H = σA (T4– Tₛ4)
# 10.10 NEWTON’S LAW OF COOLING
We all know that hot water or milk when left on a table begins to cool, gradually. Ultimately it attains the temperature of the surroundings. To study how slow or fast a given body can cool on exchanging heat with its surroundings, let us perform the following activity.
Take some water, say 300 mL, in a calorimeter with a stirrer and cover it with a two-holed lid. Fix the stirrer through one hole and fix a thermometer through another hole in the lid and make sure that the bulb of thermometer is immersed in the water. Note the reading of the thermometer. This reading T₁ is the temperature of the surroundings.
Heat the water kept in the calorimeter till it attains a temperature, say 40 °C above room temperature (i.e., temperature of the surroundings). Then, stop heating the water by removing the heat source. Start the stop-watch and note the reading of the thermometer after a fixed interval of time, say after every one minute of stirring gently with the stirrer. Continue to note the temperature (T₂) of water till it attains a temperature about 5 °C above that of the surroundings. Then, plot. | 17 | 11 | Physics | 203 |
d2f808cf-bbeb-4b54-8cd1-40d224fbe8aa | # PHYSICS
a graph by taking each value of temperature ∆T = T₂ – T₁ along y-axis and the corresponding value of t along x-axis (Fig. 10.19).
From Eqs. (10.15) and (10.16) we have
–m s dT = k T dt² (2 – T₁)
TdT² = – k dt = – K dt (10.21)
where K = k/m s
On integrating,
logₑ (T₂ – T₁) = – K t + c (10.22)
or T₂ = T₁ + C′ e–Kt; where C′ = eᶜ (10.23)
Equation (10.23) enables you to calculate the time of cooling of a body through a particular range of temperature.
For small temperature differences, the rate of cooling, due to conduction, convection, and radiation combined, is proportional to the difference in temperature. It is a valid approximation in the transfer of heat from a radiator to a room, the loss of heat through the wall of a room, or the cooling of a cup of tea on the table.
The above activity shows that a hot body loses heat to its surroundings in the form of heat radiation. The rate of loss of heat depends on the difference in temperature between the body and its surroundings. Newton was the first to study, in a systematic manner, the relation between the heat lost by a body in a given enclosure and its temperature.
According to Newton’s law of cooling, the rate of loss of heat, – dQ/dt of the body is directly proportional to the difference of temperature ∆T = (T₂–T₁) of the body and the surroundings. The law holds good only for small difference of temperature. Also, the loss of heat by radiation depends upon the nature of the surface of the body and the area of the exposed surface. We can write
– (10.19)
where k is a positive constant depending upon the area and nature of the surface of the body. Suppose a body of mass m and specific heat capacity s is at temperature T₂. Let T₁ be the temperature of the surroundings. If the temperature falls by a small amount dT₂ in time dt, then the amount of heat lost is
dQ = ms dT
∴ Rate of loss of heat is given by
dQ = ms dT₂ (10.20)
dt dt
Newton’s law of cooling can be verified with the help of the experimental set-up shown in Fig. 10.20(a). The set-up consists of a double-walled vessel (V) containing water between the two walls. A copper calorimeter (C) containing hot water is placed inside the double-walled vessel. Two thermometers through the corks are used to note the temperatures T₂ of water in calorimeter and T₁ of hot water in between the double walls, respectively. Temperature of hot water in the calorimeter is noted after equal intervals of time. A graph is plotted between logₑ (T₂–T₁) [or ln(T₂–T₁)] and time (t). The nature of the | 18 | 11 | Physics | 203 |
b515cf16-5855-4b80-892c-035abb77a46c | # THERMAL PROPERTIES OF MATTER
graph is observed to be a straight line having 8 °C = K (70 °C) ⊳ a negative slope as shown in Fig. 10.20(b). This is in support of Eq. 10.22.
# Example 10.8
A pan filled with hot food cools from 94 °C to 86 °C in 2 minutes when the room temperature is at 20 °C. How long will it take to cool from 71 °C to 69 °C?
Answer: The average temperature of 94 °C and 86 °C is 90 °C, which is 70 °C above the room temperature. K is the same for this situation as for the original.
When we divide above two equations, we have:
8 °C/2 min = K (70 °C)
2 °C/time = K (50 °C)
Time = 0.7 min = 42 s ⊳
# SUMMARY
1. Heat is a form of energy that flows between a body and its surrounding medium by virtue of temperature difference between them. The degree of hotness of the body is quantitatively represented by temperature.
2. A temperature-measuring device (thermometer) makes use of some measurable property (called thermometric property) that changes with temperature. Different thermometers lead to different temperature scales. To construct a temperature scale, two fixed points are chosen and assigned some arbitrary values of temperature. The two numbers fix the origin of the scale and the size of its unit.
3. The Celsius temperature (tC) and the Fahrenheit temperature (tF) are related by:
tF = (9/5) tC + 32
4. The ideal gas equation connecting pressure (P), volume (V) and absolute temperature (T) is:
PV = μRT
where μ is the number of moles and R is the universal gas constant.
5. In the absolute temperature scale, the zero of the scale corresponds to the temperature where every substance in nature has the least possible molecular activity. The Kelvin absolute temperature scale (T) has the same unit size as the Celsius scale (TC), but differs in the origin:
TC = T – 273.15
6. The coefficient of linear expansion (αl) and volume expansion (αv) are defined by the relations:
Δl = αlΔT
ΔV = αvΔT | 19 | 11 | Physics | 203 |
171692e9-3cb2-4db2-8932-fe6805f74f7f | # PHYSICS
where ∆l and ∆V denote the change in length l and volume V for a change of temperature ∆T. The relation between them is :
αv = 3 αl
# 7.
The specific heat capacity of a substance is defined by
s = 1 ∆Q
m ∆T
where m is the mass of the substance and ∆Q is the heat required to change its temperature by ∆T. The molar specific heat capacity of a substance is defined by
C = 1 ΔQ
μ ∆T
where μ is the number of moles of the substance.
# 8.
The latent heat of fusion (Lf) is the heat per unit mass required to change a substance from solid into liquid at the same temperature and pressure. The latent heat of vaporisation (Lv) is the heat per unit mass required to change a substance from liquid to the vapour state without change in the temperature and pressure.
# 9.
The three modes of heat transfer are conduction, convection and radiation.
# 10.
In conduction, heat is transferred between neighbouring parts of a body through molecular collisions, without any flow of matter. For a bar of length L and uniform cross section A with its ends maintained at temperatures TC and TD, the rate of flow of heat H is :
H = K A (TC − TD)
L
where K is the thermal conductivity of the material of the bar.
# 11.
Newton’s Law of Cooling says that the rate of cooling of a body is proportional to the excess temperature of the body over the surroundings :
dQ = – k (T2 – T1)
dt
Where T1 is the temperature of the surrounding medium and T2 is the temperature of the body.
2024-25 | 20 | 11 | Physics | 203 |
79e41b5c-1abe-4eba-b100-06c9b251f01a | # THERMAL PROPERTIES OF MATTER
# POINTS TO PONDER
1. The relation connecting Kelvin temperature (T) and the Celsius temperature (tc)
T = tc + 273.15
and the assignment T = 273.16 K for the triple point of water are exact relations (by choice). With this choice, the Celsius temperature of the melting point of water and boiling point of water (both at 1 atm pressure) are very close to, but not exactly equal to 0 °C and 100 °C respectively. In the original Celsius scale, these latter fixed points were exactly at 0 °C and 100 °C (by choice), but now the triple point of water is the preferred choice for the fixed point, because it has a unique temperature.
2. A liquid in equilibrium with vapour has the same pressure and temperature throughout the system; the two phases in equilibrium differ in their molar volume (i.e. density). This is true for a system with any number of phases in equilibrium.
3. Heat transfer always involves temperature difference between two systems or two parts of the same system. Any energy transfer that does not involve temperature difference in some way is not heat.
4. Convection involves flow of matter within a fluid due to unequal temperatures of its parts. A hot bar placed under a running tap loses heat by conduction between the surface of the bar and water and not by convection within water.
# EXERCISES
1. The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales.
2. Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between TA and TB?
3. The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law:
R = Rₒ [1 + α (T – Tₒ)]
The resistance is 101.6 Ω at the triple-point of water 273.16 K, and 165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω?
4. Answer the following:
1. The triple-point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale)?
2. There were two fixed points in the original Celsius scale as mentioned above which were assigned the number 0 °C and 100 °C respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number 273.16 K. What is the other fixed point on this (Kelvin) scale?
3. The absolute temperature (Kelvin scale) T is related to the temperature tc on the Celsius scale by
tc = T – 273.15
Why do we have 273.15 in this relation, and not 273.16?
4. What is the temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale?
5. Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made: | 21 | 11 | Physics | 203 |
fd0ad155-9d72-4d68-a16e-b305d4e7536a | # PHYSICS
# Temperature
# Pressure
|Triple-point of water|1.250 × 105 Pa|0.200 × 105 Pa|
|---|---|---|
|Normal melting point of sulphur|1.797 × 105 Pa|0.287 × 105 Pa|
(a) What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B?
(b) What do you think is the reason behind the slight difference in answers of thermometers A and B? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings?
# 10.6
A steel tape 1m long is correctly calibrated for a temperature of 27.0 °C. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0 °C. What is the actual length of the steel rod on that day? What is the length of the same steel rod on a day when the temperature is 27.0 °C? Coefficient of linear expansion of steel = 1.20 × 10–5 K–1.
# 10.7
A large steel wheel is to be fitted on to a shaft of the same material. At 27 °C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft? Assume coefficient of linear expansion of the steel to be constant over the required temperature range: αsteel = 1.20 × 10–5 K–1.
# 10.8
A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0 °C. What is the change in the diameter of the hole when the sheet is heated to 227 °C? Coefficient of linear expansion of copper = 1.70 × 10–5 K–1.
# 10.9
A brass wire 1.8 m long at 27 °C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of –39 °C, what is the tension developed in the wire, if its diameter is 2.0 mm? Co-efficient of linear expansion of brass = 2.0 × 10–5 K–1; Young’s modulus of brass = 0.91 × 1011 Pa.
# 10.10
A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250 °C, if the original lengths are at 40.0 °C? Is there a ‘thermal stress’ developed at the junction? The ends of the rod are free to expand (Co-efficient of linear expansion of brass = 2.0 × 10–5 K–1, steel = 1.2 × 10–5 K–1).
# 10.11
The coefficient of volume expansion of glycerine is 49 × 10–5 K–1. What is the fractional change in its density for a 30 °C rise in temperature?
# 10.12
A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings. Specific heat of aluminium = 0.91 J g–1 K–1.
# 10.13
A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500 °C and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper = 0.39 J g–1 K–1; heat of fusion of water = 335 J g–1).
# 10.14
In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150 °C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm³ of water at 27 °C. The final temperature is 40°C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal?
# 10.15
Given below are observations on molar specific heats at room temperature of some common gases. | 22 | 11 | Physics | 203 |
74a8a893-393d-4188-9934-95f858bf8811 | # THERMAL PROPERTIES OF MATTER
# 225
|Gas|Molar specific heat (Cv)|
|---|---|
|Hydrogen|4.87 (cal mo1–1 K–1)|
|Nitrogen|4.97|
|Oxygen|5.02|
|Nitric oxide|4.99|
|Carbon monoxide|5.01|
|Chlorine|6.17|
The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic gas is 2.92 cal/mol K. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine?
# 10.16
A child running a temperature of 101°F is given an antipyrin (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98 °F in 20 minutes, what is the average rate of extra evaporation caused by the drug? Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about 580 cal g–1.
# 10.17
A ‘thermacole’ icebox is a cheap and an efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h. The outside temperature is 45 °C, and co-efficient of thermal conductivity of thermacole is 0.01 J s–1 m–1 K–1. [Heat of fusion of water = 335 × 10³ J kg–1]
# 10.18
A brass boiler has a base area of 0.15 m² and thickness 1.0 cm. It boils water at the rate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 J s–1 m–1 K–1; Heat of vaporisation of water = 2256 × 10³ J kg–1.
# 10.19
Explain why:
1. a body with large reflectivity is a poor emitter
2. a brass tumbler feels much colder than a wooden tray on a chilly day
3. an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace
4. the earth without its atmosphere would be inhospitably cold
5. heating systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water
# 10.20
A body cools from 80 °C to 50 °C in 5 minutes. Calculate the time it takes to cool from 60 °C to 30 °C. The temperature of the surroundings is 20 °C. | 23 | 11 | Physics | 203 |
9485503e-5924-40cf-9ce4-beca3728d481 | # CHAPTER TWO
# MOTION IN A STRAIGHT LINE
# 2.1 INTRODUCTION
Motion is common to everything in the universe. We walk, run and ride a bicycle. Even when we are sleeping, air moves into and out of our lungs and blood flows in arteries and veins. We see leaves falling from trees and water flowing down a dam. Automobiles and planes carry people from one place to the other. The earth rotates once every twenty-four hours and revolves round the sun once in a year. The sun itself is in motion in the Milky Way, which is again moving within its local group of galaxies.
Motion is change in position of an object with time. How does the position change with time? In this chapter, we shall learn how to describe motion. For this, we develop the concepts of velocity and acceleration. We shall confine ourselves to the study of motion of objects along a straight line, also known as rectilinear motion. For the case of rectilinear motion with uniform acceleration, a set of simple equations can be obtained. Finally, to understand the relative nature of motion, we introduce the concept of relative velocity.
In our discussions, we shall treat the objects in motion as point objects. This approximation is valid so far as the size of the object is much smaller than the distance it moves in a reasonable duration of time. In a good number of situations in real-life, the size of objects can be neglected and they can be considered as point-like objects without much error.
In Kinematics, we study ways to describe motion without going into the causes of motion. What causes motion described in this chapter and the next chapter forms the subject matter of Chapter 4.
2024-25 | 0 | 11 | Physics | 102 |
c17dd90b-5a73-4a0f-8989-a2ec26490d83 | # 2.2 INSTANTANEOUS VELOCITY AND SPEED
The average velocity tells us how fast an object has been moving over a given time interval but does not tell us how fast it moves at different instants of time during that interval. For this, we define instantaneous velocity or simply velocity v at an instant t.
The velocity at an instant is defined as the limit of the average velocity as the time interval Δt becomes infinitesimally small. In other words,
v = lim Δx
Δt → 0 Δt
dx
=
dt
Fig. 2.1 Determining velocity from position-time graph. Velocity at t = 4 s is the slope of the tangent to the graph at that instant.
where the symbol lim stands for the operation of taking limit as Δt→0 of the quantity on its right. In the language of calculus, the quantity on the right hand side of Eq. (2.1a) is the differential coefficient of x with respect to t and is denoted by dx/dt (see Appendix 2.1). It is the rate of change of position with respect to time, at that instant.
We can use Eq. (2.1a) for obtaining the value of velocity at an instant either graphically or numerically. Suppose that we want to obtain graphically the value of velocity at time t = 4 s (point P) for the motion of the car represented in Fig. 2.1 calculation.
Let us take Δt = 2 s centred at t = 4 s. Then, by the definition of the average velocity, the slope of line P₁P₂ (Fig. 2.1) gives the value of average velocity over the interval 3 s to 5 s.
**Table 2.1 Limiting value of Δx/Δt at t = 4 s**
|Δt|t = ⌈Δt⌉|t = ⌊t - Δt/2⌋|t = ⌊t + Δt/2⌋|Δx|
|---|---|---|---|---|
|2.0 s|4.0 s| | | |
|1.0 s|4.0 s| | | |
|0.5 s|4.0 s| | | |
|0.1 s|4.0 s| | | |
|0.01 s|4.0 s| | | | | 1 | 11 | Physics | 102 |
e1a9ad28-f0a6-4dca-9d83-632b9b072dd2 | # MOTION IN A STRAIGHT LINE
corresponding values of x, i.e. x (t ) = 0.08 t3 ⊳ = a + 16b – a – 4b = 6.0 × b and x (t ) = 0.08 t3
2. The sixth column lists the difference Δx = x (t2) – x (t1) and the last column gives the ratio of Δx and Δt, i.e. the average velocity corresponding to the value of Δt listed in the first column.
We see from Table 2.1 that as we decrease the value of Δt from 2.0 s to 0.010 s, the value of the average velocity approaches the limiting value 3.84 m s–1 which is the value of velocity at t = 4.0 s, i.e. the value of dx at t = 4.0 s. In this manner, we can calculate velocity at each instant for motion of the car.
Note that for uniform motion, velocity is the same as the average velocity at all instants. Instantaneous speed or simply speed is the magnitude of velocity. For example, a velocity of + 24.0 m s–1 and a velocity of – 24.0 m s–1 — both have an associated speed of 24.0 m s–1. It should be noted that though average speed over a finite interval of time is greater or equal to the magnitude of the average velocity, instantaneous speed at an instant is equal to the magnitude of the instantaneous velocity at that instant. Why so?
# 2.3 ACCELERATION
The velocity of an object, in general, changes during its course of motion. How to describe this change? Should it be described as the rate of change in velocity with distance or with time? This was a problem even in Galileo’s time. It was first thought that this change could be described by the rate of change of velocity with distance. But, through his studies of motion of freely falling objects and motion of objects on an inclined plane, Galileo concluded that the rate of change of velocity with time is a constant of motion for all objects in free fall. On the other hand, the change in velocity with distance is not constant – it decreases with the increasing distance of fall. This led to the concept of acceleration as the rate of change of velocity with time.
The average acceleration over a time interval is defined as the change of velocity divided by the time interval:
a = v2 – v1 = Δv
t2 – t1 Δt
where v2 and v1 are the instantaneous velocities or simply velocities at time t2 and t1. It is the average change of velocity per unit time. The SI unit of acceleration is m s–2.
On a plot of velocity versus time, the average acceleration is the slope of the straight line connecting the points corresponding to (v2, t2) and (v1, t1). | 2 | 11 | Physics | 102 |
2bd69ab9-d0dc-4413-a5c6-69c215fcbf64 | # PHYSICS
Instantaneous acceleration is defined in the same way as the instantaneous velocity:
∆v dv
a = lim ∆t = dt (2.3)
∆ → t 0
The acceleration at an instant is the slope of the tangent to the v–t curve at that instant.
Since velocity is a quantity having both magnitude and direction, a change in velocity may involve either or both of these factors. Acceleration, therefore, may result from a change in speed (magnitude), change in direction or changes in both. Like velocity, acceleration can also be positive, negative or zero.
Position-time graphs for motion with positive, negative and zero acceleration are shown in Figs. 2.4 (a), (b) and (c), respectively. Note that the graph curves upward for positive acceleration; downward for negative acceleration and it is a straight line for zero acceleration.
Although acceleration can vary with time, our study in this chapter will be restricted to motion with constant acceleration. In this case, the average acceleration equals the constant value of acceleration during the interval. If the velocity of an object is vₒ at t = 0 and v at time t, we have:
a = v − vₒ or, v = vₒ + a t (2.4)
t − 0
Fig. 2.2 Position-time graph for motion with (a) positive acceleration; (b) negative acceleration, and (c) zero acceleration.
Let us see how velocity-time graph looks like for some simple cases. Fig. 2.3 shows velocity-time graph for motion with constant acceleration for the following cases:
- (a) An object is moving in a positive direction with a positive acceleration.
- (b) An object is moving in positive direction with a negative acceleration.
Fig. 2.3 Velocity–time graph for motions with constant acceleration. (a) Motion in positive direction with positive acceleration, (b) Motion in positive direction with negative acceleration, (c) Motion in negative direction with negative acceleration, (d) Motion of an object with negative acceleration that changes direction at time t₁. Between times 0 to t₁, it moves in positive x-direction and between t₁ and t₂ it moves in the opposite direction.
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3d34028b-91a0-44b0-97b2-24c5961b1571 | # MOTION IN A STRAIGHT LINE
Fig. 2.4 Area under v–t curve equals displacement of the object over a given time interval.
The v-t curve is a straight line parallel to the time axis and the area under it between t = 0 and t = T is the area of the rectangle of height u and base T. Therefore, area = u × T = uT which is the displacement in this time interval. How come in this case an area is equal to a distance? Think! Note the dimensions of quantities on the two coordinate axes, and you will arrive at the answer.
As explained in the previous section, the area under v-t curve represents the displacement. Therefore, the displacement x of the object is:
x = 1 (v–v₀)t + v₀t
But v − v₀ = at
Therefore, x = 1 at² + v₀t
or, x = v₀t + 1 at²
Equation (2.5) can also be written as x = v + v₀t = v₀t
# 2.4 KINEMATIC EQUATIONS FOR UNIFORMLY ACCELERATED MOTION
For uniformly accelerated motion, we can derive some simple equations that relate displacement (x), time taken (t), initial velocity (v₀), final velocity (v) and acceleration (a). Equation (2.4) already obtained gives a relation between final and initial velocities with uniform acceleration a:
v = v₀ + at
This relation is graphically represented in Fig. 2. The area under this curve is:
Area between instants 0 and t = Area of triangle ABC + Area of rectangle OACD
= 1 (v – v₀)t + v₀t
= ⌈ v + v₀ ⌉ ⌈v − v₀ ⌉ v − v₀
= ⎜ 2 ⎝ ⎜ a ⎝ 2a
v² = v² + 2ax
2024-25 | 4 | 11 | Physics | 102 |
5da7a33d-1689-4cb0-888e-b0a5e1399078 | # PHYSICS
This equation can also be obtained by substituting the value of t from Eq. (2.4) into Eq. (2.6). Thus, we have obtained three important equations:
- v = v₀ + at
- x – x₀ = v₀ t + ½ a t²
- x = x₀ + v₀ t + ½ a t²
- x = v₀ t + ½ a t²
We can write:
v² = v₀² + 2ax (2.9a)
connecting five quantities v, v₀, a, t, and x. These are kinematic equations of rectilinear motion for constant acceleration.
The set of Eq. (2.9a) were obtained by assuming that at t = 0, the position of the particle, x is 0. We can obtain a more general equation if we take the position coordinate at zero, say x₀ at t = 0 as non-zero (replacing x₀). Then Eqs. (2.9a) are modified by x – x₀ to:
v = v₀ + at
x = x₀ + v₀ t + ½ at² (2.9b)
v² = v₀² + 2a (x – x₀) (2.9c)
Example 2.2 Obtain equations of motion for constant acceleration using method of calculus.
Answer: By definition
a = dv/dt
dv = a dt
Integrating both sides:
∫ v dv = ∫ t a dt
v₀ = a ∫ 0 dt (a is constant)
v – v₀ = at
v = v₀ + at
Further, v = dx/dt
Integrating both sides:
∫ x dx = ∫ t v dt
Example 2.3 A ball is thrown vertically upwards with a velocity of 20 m s⁻¹ from the top of a multistorey building. The height of the point from where the ball is thrown is 25.0 m from the ground. (a) How high will the ball rise? and (b) how long will it be before the ball hits the ground? Take g = 10 m s⁻².
Answer: (a) Let us take the y-axis in the vertically upward direction with zero at the ground, as shown in Fig. 2.6. Now vₒ = + 20 m s⁻¹, a = – g = –10 m s⁻², v = 0 m s⁻¹.
If the ball rises to height y from the point of launch, then using the equation:
v² = v₀² + 2a (y – y₀)
we get 0 = (20)² + 2(–10)(y – y₀)
Solving, we get (y – y₀) = 20 m.
(b) We can solve this part of the problem in two ways. Note carefully the methods used. | 5 | 11 | Physics | 102 |
b0e3b721-951d-4960-a33c-70c1e2659f3a | # MOTION IN A STRAIGHT LINE
0 = 25 + 20t + (½)(-10)t²
Or, 5t² – 20t – 25 = 0
Solving this quadratic equation for t, we get t = 5s
Note that the second method is better since we do not have to worry about the path of the motion as the motion is under constant acceleration.
Example 2.4 Free-fall: Discuss the motion of an object under free fall. Neglect air resistance.
# Answer
An object released near the surface of the Earth is accelerated downward under the influence of the force of gravity. The magnitude of acceleration due to gravity is represented by g. If air resistance is neglected, the object is said to be in free fall. If the height through which the object falls is small compared to the earth’s radius, g can be taken to be constant, equal to 9.8 m s–2. Free fall is thus a case of motion with uniform acceleration.
We assume that the motion is in y-direction, more correctly in –y-direction because we choose upward direction as positive. Since the acceleration due to gravity is always downward, it is in the negative direction and we have a = –g = –9.8 m s–2.
The object is released from rest at yv = 0. Therefore, 0 = 0 and the equations of motion become:
v = 0 – gt = –9.8t m
y = 0 – (½)gt² = –4.9t² m
We have, y0 = 45 m, y = 0, v0 = 0, a = –g = –10 m s–2
0 = 45 + (½)(–10)t2²
Solving, we get t2 = 3 s
Therefore, the total time taken by the ball before it hits the ground = t1 + t2 = 2 s + 3 s = 5 s.
# SECOND METHOD
The total time taken can also be calculated by noting the coordinates of initial and final positions of the ball with respect to the origin chosen and using equation:
y = y0 + vt + (1/2)at²
Now y0 = 25 m, y = 0 m, v = 20 m s–1, a = –10 m s–2, t = ? | 6 | 11 | Physics | 102 |
ad580824-184d-49dc-8e9e-38fbb64572ed | # PHYSICS
traversed during successive intervals of time. Since initial velocity is zero, we have
y = − 1 gt²
Using this equation, we can calculate the position of the object after different time intervals, 0, τ, 2 τ, 3 τ… which are given in second column of Table 2.2. If we take (–1/2) gτ² as y₀ — the position coordinate after first time interval τ, then third column gives the positions in the unit of y₀. The fourth column gives the distances traversed in successive τs. We find that the distances are in the simple ratio 1: 3: 5: 7: 9: 11… as shown in the last column. This law was established by Galileo Galilei (1564-1642) who was the first to make quantitative studies of free fall.
# Example 2.6 Stopping distance of vehicles
When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance. It is an important factor for road safety and depends on the initial velocity (v₀) and the braking capacity, or deceleration, –a that is caused by the braking. Derive an expression for stopping distance of a vehicle in terms of v₀ and a.
# Example 2.5 Galileo’s law of odd numbers
“The distances traversed, during equal intervals of time, by a body falling from rest, stand to one another in the same ratio as the odd numbers beginning with unity [namely, 1: 3: 5: 7…...].” Prove it.
Answer: Let us divide the time interval of motion of an object under free fall into many equal intervals τ and find out the distances.
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Thus, the stopping distance is proportional to the square of the initial velocity. Doubling the | 7 | 11 | Physics | 102 |
d4152d93-b826-4f40-9c67-5c6bc21a8ad9 | # MOTION IN A STRAIGHT LINE
initial velocity increases the stopping distance by a factor of 4 (for the same deceleration).
For the car of a particular make, the braking distance was found to be 10 m, 20 m, 34 m and 50 m corresponding to velocities of 11, 15, 20 and 25 m/s which are nearly consistent with the above formula.
Stopping distance is an important factor considered in setting speed limits, for example, in school zones.
# Example 2.7 Reaction time:
When a situation demands our immediate action, it takes some time before we really respond. Reaction time is the time a person takes to observe, think and act. For example, if a person is driving and suddenly a boy appears on the road, then the time elapsed before he slams the brakes of the car is the reaction time. Reaction time depends on complexity of the situation and on an individual.
You can measure your reaction time by a simple experiment. Take a ruler and ask your friend to drop it vertically through the gap between your thumb and forefinger. After you catch it, find the distance d travelled by the ruler. In a particular case, d was found to be 21.0 cm.
Estimate reaction time.
# SUMMARY
1. An object is said to be in motion if its position changes with time. The position of the object can be specified with reference to a conveniently chosen origin. For motion in a straight line, position to the right of the origin is taken as positive and to the left as negative.
2. The average speed of an object is greater or equal to the magnitude of the average velocity over a given time interval.
3. Instantaneous velocity or simply velocity is defined as the limit of the average velocity as the time interval Δt becomes infinitesimally small:
v = lim (Δx/Δt) as Δt → 0
The velocity at a particular instant is equal to the slope of the tangent drawn on position-time graph at that instant. | 8 | 11 | Physics | 102 |
a4edd0bd-b280-47ec-99de-a37c01b62640 | # PHYSICS
1. Average acceleration is the change in velocity divided by the time interval during which the change occurs:
<math>
a = \(\frac{Δv}{Δt}\)
</math>
2. Instantaneous acceleration is defined as the limit of the average acceleration as the time interval Δt goes to zero:
<math>
a = \(\lim_{Δt \to 0} a = \lim_{Δv} = \frac{dv}{dt}\)
</math>
The acceleration of an object at a particular time is the slope of the velocity-time graph at that instant of time. For uniform motion, acceleration is zero and the x-t graph is a straight line inclined to the time axis and the v-t graph is a straight line parallel to the time axis. For motion with uniform acceleration, x-t graph is a parabola while the v-t graph is a straight line inclined to the time axis.
3. The area under the velocity-time curve between times t₁ and t₂ is equal to the displacement of the object during that interval of time.
4. For objects in uniformly accelerated rectilinear motion, the five quantities, displacement x, time taken t, initial velocity v₀, final velocity v, and acceleration a are related by a set of simple equations called kinematic equations of motion:
<math>
v = v₀ + at
</math>
<math>
x = v₀t + \frac{1}{2}at²
</math>
<math>
v² = v₀² + 2ax
</math>
if the position of the object at time t = 0 is 0. If the particle starts at x = x₀, x in the above equations is replaced by (x – x₀).
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3a3e851f-4ae3-4dea-bd35-9caf6efed68e | # MOTION IN A STRAIGHT LINE
# POINTS TO PONDER
1. The origin and the positive direction of an axis are a matter of choice. You should first specify this choice before you assign signs to quantities like displacement, velocity and acceleration.
2. If a particle is speeding up, acceleration is in the direction of velocity; if its speed is decreasing, acceleration is in the direction opposite to that of the velocity. This statement is independent of the choice of the origin and the axis.
3. The sign of acceleration does not tell us whether the particle’s speed is increasing or decreasing. The sign of acceleration (as mentioned in point 3) depends on the choice of the positive direction of the axis. For example, if the vertically upward direction is chosen to be the positive direction of the axis, the acceleration due to gravity is negative. If a particle is falling under gravity, this acceleration, though negative, results in increase in speed. For a particle thrown upward, the same negative acceleration (of gravity) results in decrease in speed.
4. The zero velocity of a particle at any instant does not necessarily imply zero acceleration at that instant. A particle may be momentarily at rest and yet have non-zero acceleration. For example, a particle thrown up has zero velocity at its uppermost point but the acceleration at that instant continues to be the acceleration due to gravity.
5. In the kinematic equations of motion [Eq. (2.9)], the various quantities are algebraic, i.e. they may be positive or negative. The equations are applicable in all situations (for one dimensional motion with constant acceleration) provided the values of different quantities are substituted in the equations with proper signs.
6. The definitions of instantaneous velocity and acceleration (Eqs. (2.1) and (2.3)) are exact and are always correct while the kinematic equations (Eq. (2.9)) are true only for motion in which the magnitude and the direction of acceleration are constant during the course of motion.
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ff16ecdf-059a-41c5-bcbe-7ee97983bd29 | # PHYSICS
# EXERCISES
# 2.1
In which of the following examples of motion, can the body be considered approximately a point object:
- (a) a railway carriage moving without jerks between two stations.
- (b) a monkey sitting on top of a man cycling smoothly on a circular track.
- (c) a spinning cricket ball that turns sharply on hitting the ground.
- (d) a tumbling beaker that has slipped off the edge of a table.
# 2.2
The position-time (x-t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in Fig. 2.9. Choose the correct entries in the brackets below:
- (a) (A/B) lives closer to the school than (B/A)
- (b) (A/B) starts from the school earlier than (B/A)
- (c) (A/B) walks faster than (B/A)
- (d) A and B reach home at the (same/different) time
- (e) (A/B) overtakes (B/A) on the road (once/twice).
Fig. 2.9
# 2.3
A woman starts from her home at 9.00 am, walks with a speed of 5 km h–1 on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 25 km h–1. Choose suitable scales and plot the x-t graph of her motion.
# 2.4
A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.
# 2.5
A car moving along a straight highway with speed of 126 km h–1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?
# 2.6
A player throws a ball upwards with an initial speed of 29.4 m s–1.
- (a) What is the direction of acceleration during the upward motion of the ball?
- (b) What are the velocity and acceleration of the ball at the highest point of its motion?
- (c) Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion.
- (d) To what height does the ball rise and after how long does the ball return to the player’s hands? (Take g = 9.8 m s–2 and neglect air resistance).
# 2.7
Read each statement below carefully and state with reasons and examples, if it is true or false:
A particle in one-dimensional motion
- (a) with zero speed at an instant may have non-zero acceleration at that instant
- (b) with zero speed may have non-zero velocity
- (c) with constant speed must have zero acceleration
- (d) with positive value of acceleration must be speeding up.
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2d1febbc-f059-487a-9c76-152b7a275484 | # MOTION IN A STRAIGHT LINE
# 2.8
A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s.
# 2.9
Explain clearly, with examples, the distinction between:
- (a) magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval;
- (b) magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true? [For simplicity, consider one-dimensional motion only].
# 2.10
A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h–1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h–1. What is the:
- (a) magnitude of average velocity, and
- (b) average speed of the man over the interval of time:
|Interval of Time|Average Velocity|Average Speed|
|---|---|---|
|(i) 0 to 30 min| | |
|(ii) 0 to 50 min| | |
|(iii) 0 to 40 min| | |
[Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero!]
# 2.11
In Exercises 2.9 and 2.10, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?
# 2.12
Look at the graphs (a) to (d) (Fig. 2.10) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.
# 2.13
Figure 2.11 shows the x-t plot of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t > 0? If not, suggest a suitable physical context for this graph.
# 2.14
A police van moving on a highway with a speed of 30 km h–1 fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 km h–1. If the muzzle speed of the bullet is 150 m s–1, with what speed does the bullet hit the thief’s car? (Note: Obtain that speed which is relevant for damaging the thief’s car).
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f104acf6-0f3c-4a97-a9c6-830e9d25779b | # PHYSICS
# 2.15
Suggest a suitable physical situation for each of the following graphs (Fig 2.12):
# 2.16
Figure 2.13 gives the x-t plot of a particle executing one-dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter 13). Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, – 1.2 s.
# 2.17
Figure 2.14 gives the x-t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least? Give the sign of average velocity for each interval.
# 2.18
Figure 2.15 gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest? Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals. What are the accelerations at the points A, B, C and D? | 13 | 11 | Physics | 102 |
f897dd7a-09a2-4b11-95d7-f7fee513bf1e | # CHAPTER THREE
# MOTION IN A PLANE
# 3.1 INTRODUCTION
In the last chapter we developed the concepts of position, displacement, velocity and acceleration that are needed to describe the motion of an object along a straight line. We found that the directional aspect of these quantities can be taken care of by + and – signs, as in one dimension only two directions are possible. But in order to describe motion of an object in two dimensions (a plane) or three dimensions (space), we need to use vectors to describe the above-mentioned physical quantities. Therefore, it is first necessary to learn the language of vectors. What is a vector? How to add, subtract and multiply vectors? What is the result of multiplying a vector by a real number? We shall learn this to enable us to use vectors for defining velocity and acceleration in a plane. We then discuss motion of an object in a plane. As a simple case of motion in a plane, we shall discuss motion with constant acceleration and treat in detail the projectile motion. Circular motion is a familiar class of motion that has a special significance in daily-life situations. We shall discuss uniform circular motion in some detail. The equations developed in this chapter for motion in a plane can be easily extended to the case of three dimensions.
# 3.2 SCALARS AND VECTORS
In physics, we can classify quantities as scalars or vectors. Basically, the difference is that a direction is associated with a vector but not with a scalar. A scalar quantity is a quantity with magnitude only. It is specified completely by a single number, along with the proper unit. Examples are: the distance between two points, mass of an object, the temperature of a body and the time at which a certain event happened. The rules for combining scalars are the rules of ordinary algebra. Scalars can be added, subtracted, multiplied and divided. | 0 | 11 | Physics | 103 |
cbcb275d-be4a-4fa8-9881-8ec423f33ea2 | # PHYSICS
just as the ordinary numbers*. For example, if the length and breadth of a rectangle are 1.0 m and 0.5 m respectively, then its perimeter is the sum of the lengths of the four sides, 1.0 m + 0.5 m + 1.0 m + 0.5 m = 3.0 m. The length of each side is a scalar and the perimeter is also a scalar. Take another example: the maximum and minimum temperatures on a particular day are 35.6 °C and 24.2 °C respectively. Then, the difference between the two temperatures is 11.4 °C. Similarly, if a uniform solid cube of aluminium of side 10 cm has a mass of 2.7 kg, then its volume is 10–3 m³ (a scalar) and its density is 2.7×103 kg m–3 (a scalar).
A vector quantity is a quantity that has both a magnitude and a direction and obeys the triangle law of addition or equivalently the parallelogram law of addition. So, a vector is specified by giving its magnitude by a number and its direction. Some physical quantities that are represented by vectors are displacement, velocity, acceleration and force.
To represent a vector, we use a bold face type in this book. Thus, a velocity vector can be represented by a symbol v. Since bold face is difficult to produce, when written by hand, a vector is often represented by an arrow placed over a letter, say v. Thus, both v and v represent the velocity vector. The magnitude of a vector is often called its absolute value, indicated by |v| = v. Thus, a vector is represented by a bold face, e.g. by A, a, p, q, r, ... x, y, with respective magnitudes denoted by light face A, a, p, q, r, ... x, y.
# 3.2.1 Position and Displacement Vectors
To describe the position of an object moving in a plane, we need to choose a convenient point, say O as origin. Let P and P′ be the positions of the object at time t and t′, respectively [Fig. 3.1(a)]. We join O and P by a straight line. Then, OP is the position vector of the object at time t. An arrow is marked at the head of this line. It is represented by a symbol r, i.e. OP = r. Point P′ is represented by another position vector, OP′ denoted by r′. The length of the vector r represents the magnitude of the vector and its direction is the direction in which P lies as seen from O. If the object moves from P to P′, the vector PP′ (with tail at P and tip at P′) is called the displacement vector corresponding to motion from point P (at time t) to point P′ (at time t′).
It is important to note that the displacement vector is the straight line joining the initial and final positions and does not depend on the actual path undertaken by the object between the two positions. For example, in Fig. 3.1(b), given the initial and final positions as P and Q, the displacement vector is the same PQ for different paths of journey, say PABCQ, PDQ, and PBEFQ. Therefore, the magnitude of displacement is either less or equal to the path length of an object between two points. This fact was emphasised in the previous chapter also while discussing motion along a straight line.
# 3.2.2 Equality of Vectors
Two vectors A and B are said to be equal if, and only if, they have the same magnitude and the same direction.** Figure 3.2(a) shows two equal vectors A and B. We can easily check their equality. Shift B parallel to itself until its tail Q coincides with that of A, i.e. Q coincides with O. Then, since their tips S and P also coincide, the two vectors are said to be equal. In general, equality is indicated.
* Addition and subtraction of scalars make sense only for quantities with same units. However, you can multiply and divide scalars of different units.
In our study, vectors do not have fixed locations. So displacing a vector parallel to itself leaves the vector unchanged. Such vectors are called free vectors. However, in some physical applications, location or line of application of a vector is important. Such vectors are called localised vectors. | 1 | 11 | Physics | 103 |
8667f831-0a27-4d50-ab49-8833040c2e2e | # MOTION IN A PLANE
The factor λ by which a vector A is multiplied could be a scalar having its own physical dimension. Then, the dimension of λ A is the product of the dimensions of λ and A. For example, if we multiply a constant velocity vector by duration (of time), we get a displacement vector.
# 3.4 ADDITION AND SUBTRACTION OF VECTORS — GRAPHICAL METHOD
As mentioned in section 4.2, vectors, by definition, obey the triangle law or equivalently the parallelogram law of addition. We shall now describe this law of addition using the graphical method. Let us consider two vectors A and B that lie in a plane as shown in Fig. 3.4(a). The lengths of the line segments representing these vectors are proportional to the magnitude of the vectors. To find the sum A + B, we place vector B so that its tail is at the head of the vector A, as in Fig. 3.4(b). Then, we join the tail of A to the head of B. This line OQ represents a vector R, that is, the sum of the vectors A and B. Since, in this procedure of vector addition, vectors are equal because they have different directions.
# 3.3 MULTIPLICATION OF VECTORS BY REAL NUMBERS
Multiplying a vector A with a positive number λ gives a vector whose magnitude is changed by the factor λ but the direction is the same as that of A: λA = λA if λ > 0. For example, if A is multiplied by 2, the resultant vector 2A is in the same direction as A and has a magnitude twice of |A| as shown in Fig. 3.3(a). Multiplying a vector A by a negative number −λ gives another vector whose direction is opposite to the direction of A and whose magnitude is λ times |A|. Multiplying a given vector A by negative numbers, say –1 and –1.5, gives vectors as shown in Fig 3.3(b).
# Figures
Fig. 3.3 (a) Vector A and the resultant vector after multiplying A by a positive number 2.
Fig. 3.4 (a) Vectors A and B. (b) Vectors A and B added graphically. (c) Vectors B and A added graphically. (d) Illustrating the associative law of vector addition.
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7036e986-ccd1-4920-bccd-b1920c03870c | # PHYSICS
arranged head to tail, this graphical method is called the head-to-tail method. The two vectors and their resultant form three sides of a triangle, so this method is also known as triangle method of vector addition. If we find the resultant of B + A as in Fig. 3.4(c), the same vector R is obtained. Thus, vector addition is commutative:
A + B = B + A (3.1)
The addition of vectors also obeys the associative law as illustrated in Fig. 3.4(d). The result of adding vectors A and B first and then adding vector C is the same as the result of adding B and C first and then adding vector A:
(A + B) + C = A + (B + C) (3.2)
What is the result of adding two equal and opposite vectors? Consider two vectors –A shown in Fig. 3.3(b). Their sum is A and A + (–A). Since the magnitudes of the two vectors are the same, but the directions are opposite, the resultant vector has zero magnitude and is represented by 0 called a null vector or a zero vector:
A – A = 0
|0|= 0 (3.3)
Since the magnitude of a null vector is zero, its direction cannot be specified. The null vector also results when we multiply a vector A by the number zero. The main properties of 0 are:
A + 0 = A
λ 0 = 0
0 A = 0 (3.4)
Fig. 3.5 (a) Two vectors A and B, – B is also shown. (b) Subtracting vector B from vector A – the result is R₂. For comparison, addition of vectors A and B, i.e. R₁ is also shown.
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4a5a2739-d64f-4930-8fa2-4a76b5a16514 | # MOTION IN A PLANE
Fig. 3.6 (a) Two vectors A and B with their tails brought to a common origin. (b) The sum A + B obtained using the parallelogram method. (c) The parallelogram method of vector addition is equivalent to the triangle method.
# Example 3.1
Rain is falling vertically with a speed of 35 m s–1. Winds starts blowing after sometime with a speed of 12 m s–1 in east to west direction. In which direction should a boy waiting at a bus stop hold his umbrella?
# 3.5 RESOLUTION OF VECTORS
Let a and b be any two non-zero vectors in a plane with different directions and let A be another vector in the same plane (Fig. 3.8). A can be expressed as a sum of two vectors — one obtained by multiplying a by a real number and the other obtained by multiplying b by another real number. To see this, let O and P be the tail and head of the vector A. Then, through O, draw a straight line parallel to a, and through P, a straight line parallel to b. Let them intersect at Q. Then, we have
A = OP = OQ + QP (3.6)
But since OQ is parallel to a, and QP is parallel to b, we can write:
OQ = λ a, and QP = μ b (3.7)
Therefore, A = λ a + μ b (3.8)
Answer: The velocity of the rain and the wind are represented by the vectors vr and vw in Fig. 3.7 and are in the direction specified by the problem. Using the rule of vector addition, we see that the resultant of vr and vw is R as shown in the figure. The magnitude of R is
R = √(vr2 + vw2) = √(352 + 122) m s−1 = 37 m s−1
The direction θ that R makes with the vertical is given by
tan θ = vw / vr = 12 / 35 = 0.343
Or, θ = tan−1(0.343) = 19°
Therefore, the boy should hold his umbrella in the vertical plane at an angle of about 19° with the vertical towards the east.
⦿
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3f45f547-4695-4935-9e7f-09938ff19cff | # PHYSICS
respectively. Using this method one can resolve a given vector into two component vectors along a set of two vectors – all the three lie in the same plane. It is convenient to resolve a general vector along the axes of a rectangular coordinate system using vectors of unit magnitude. These are called unit vectors that we discuss now. A unit vector is a vector of unit magnitude and points in a particular direction. It has no dimension and unit. It is used to specify a direction only. Unit vectors along the x-, y- and z-axes of a rectangular coordinate system are denoted by *i, j and k*, respectively, as shown in Fig. 3.9(a).
Since these are unit vectors, we have:
i = j = k = 1 (3.9)
These unit vectors are perpendicular to each other. In this text, they are printed in bold face with a cap (^) to distinguish them from other vectors. Since we are dealing with motion in two dimensions in this chapter, we require use of only two unit vectors. If we multiply a unit vector n by a scalar, the result is a vector λ n.
In general, a vector A can be written as:
A = |A| n (3.10)
where n is a unit vector along A. We can now resolve a vector A in terms of component vectors that lie along unit vectors i and j. Consider a vector A that lies in the x-y plane as shown in Fig. 3.9(b). We draw lines from the head of A perpendicular to the coordinate axes as in Fig. 3.9(b), and get vectors A₁ and A₂ such that A = A₁ + A₂. Since A is parallel to i, we have:
A = Aₓ i + Aᵧ j (3.12)
where Aₓ and Aᵧ are real numbers. The quantities Aₓ and Aᵧ are called x- and y- components of the vector A. Note that Aₓ is itself not a vector, but Aₓ i is a vector, and so is Aᵧ j. Using simple trigonometry, we can express Aₓ and Aᵧ in terms of the magnitude of A and the angle θ it makes with the x-axis:
Aₓ = A cos θ (3.13)
Aᵧ = A sin θ (3.13)
As is clear from Eq. (3.13), a component of a vector can be positive, negative or zero depending on the value of θ. Now, we have two ways to specify a vector in a plane. It can be specified by:
1. its magnitude A and the direction θ it makes with the x-axis; or
2. its components Aₓ and Aᵧ.
If A and θ are given, Aₓ and Aᵧ can be obtained using Eq. (3.13). If Aₓ and Aᵧ are given, A and θ can be obtained as follows:
A² = Aₓ² + Aᵧ² (3.14)
And tan θ = Aᵧ / Aₓ, θ = tan−1 Aᵧ / Aₓ (3.15)
Fig. 3.9 (a) Unit vectors i, j and k lie along the x-, y-, and z-axes. (b) A vector A is resolved into its components Aₓ and Aᵧ along x-, and y- axes. (c) A₁ and A₂ expressed in terms of i and j. | 5 | 11 | Physics | 103 |
914f1d6f-1935-407e-ac3b-2c0e06198b53 | # MOTION IN A PLANE
So far we have considered a vector lying in an x-y plane. The same procedure can be used to resolve a general vector A into three components along x-, y-, and z-axes in three dimensions. If α, β, and γ are the angles between A and the x-, y-, and z-axes, respectively [Fig. 3.9(d)], we have:
R = A + B = (Ax i + Ay j) + (Bx i + By j) (3.19a)
Let R be their sum. We have:
R = (A + B)x i + (Ay + By) j (3.19b)
Since R = Rx i + Ry j (3.20), we have:
Rx = Ax + Bx, Ry = Ay + By (3.21)
Thus, each component of the resultant vector R is the sum of the corresponding components of A and B.
In three dimensions, we have:
A = Ax i + Ay j + Az k
B = Bx i + By j + Bz k
R = A + B = Rx i + Ry j + Rz k
with Rx = Ax + Bx
Ax = A cosα, Ay = A cosβ, Az = A cosγ (3.16a)
In general, we have:
A = A2 + Ay2 + Az2 (3.16c)
A position vector r can be expressed as:
r = x i + y j + z k (3.17)
where x, y, and z are the components of r along x-, y-, z-axes, respectively.
# 3.6 VECTOR ADDITION – ANALYTICAL METHOD
Although the graphical method of adding vectors helps us in visualising the vectors and the resultant vector, it is sometimes tedious and has limited accuracy. It is much easier to add vectors by combining their respective components.
Consider two vectors A and B in x-y plane with components Ax, Ay and Bx, By:
A = Ax i + Ay j (3.18)
* Note that angles α, β, and γ are angles in space. They are between pairs of lines, which are not coplanar.
Example 3.2 Find the magnitude and direction of the resultant of two vectors A and B in terms of their magnitudes and angle θ between them. | 6 | 11 | Physics | 103 |
136aa5a6-ed63-4da1-a398-18b9301acea5 | # PHYSICS
# Example 3.3
A motorboat is racing towards north at 25 km/h and the water current in that region is 10 km/h in the direction of 60° east of south. Find the resultant velocity of the boat.
Answer: The vector vb representing the velocity of the motorboat and the vector vc representing the water current are shown in Fig. 3.11 in directions specified by the problem. Using the parallelogram method of addition, the resultant R is obtained in the direction shown in the figure.
Let OP and OQ represent the two vectors A and B making an angle θ (Fig. 3.10). Then, using the parallelogram method of vector addition, OS represents the resultant vector R:
R = A + B
SN is normal to OP and PM is normal to OS.
From the geometry of the figure,
OS = ON + SN²
but ON = OP + PN = A + B cos θ
SN = B sin θ
OS = (A + B cos θ) + (B sin θ)²
or, R = A + B² + 2AB cos θ (3.24a)
In Δ OSN, SN = OS sin α = R sin α, and in Δ PSN, SN = B sin θ
Therefore, R sin α = B sin θ or, R θ = Bα (3.24b)
Similarly, PM = A sin α = B sin β
or, Aβ = Bα (3.24c)
Combining Eqs. (3.24b) and (3.24c), we get
R = A = B (3.24d)
Using Eq. (3.24d), we get:
sin α = B sin θ
R = vc or, sin φ = vc sin θ
sin θ sin φ R
where R is given by Eq. (3.24a).
tan α = OPSN = B sin θ + PN A + B cos (3.24f)
Equation (3.24a) gives the magnitude of the resultant and Eqs. (3.24e) and (3.24f) its direction.
Equation (3.24a) is known as the Law of cosines and Eq. (3.24d) as the Law of sines.
# 3.7 MOTION IN A PLANE
In this section we shall see how to describe motion in two dimensions using vectors. | 7 | 11 | Physics | 103 |
12bd02f8-4286-42d6-af59-5e2a4bcb3ac9 | # MOTION IN A PLANE
# 3.7.1 Position Vector and Displacement
Suppose a particle moves along the curve shown by the thick line and is at P at time t and P′ at time t′ [Fig. 3.12(b)]. Then, the displacement is:
r = r′ – r (3.25)
and is directed from P to P′.
We can write Eq. (3.25) in a component form:
r = (x' i + y' j) – (x i + y j)
= Δx i + Δy j
where Δx = x′ – x, Δy = y′ – y (3.26)
# Velocity
The average velocity (v) of an object is the ratio of the displacement and the corresponding time interval:
v = Δr / Δt = (Δx i + Δy j) / Δt = (Δx / Δt) i + (Δy / Δt) j (3.27)
Or, v = v̂x i + vy j
Since v = Δr / Δt, the direction of the average velocity is the same as that of r (Fig. 3.12). The instantaneous velocity is given by the limiting value of the average velocity as the time interval approaches zero:
v = lim (Δr / Δt) = dr / dt (3.28)
The meaning of the limiting process can be easily understood with the help of Fig 3.13(a) to (d). In these figures, the thick line represents the path of an object, which is at P at time t. P₁, P₂, and P₃ represent the positions of the object after times Δt₁, Δt₂, and Δt₃. Δr₁, Δr₂, and Δr₃ are the displacements of the object in times Δt₁, Δt₂, and Δt₃.
As the time interval Δt approaches zero, the average velocity approaches the velocity v. The direction of v is parallel to the line tangent to the path. | 8 | 11 | Physics | 103 |
9229d1fa-cbae-4a46-9857-565be9fa58a3 | # PHYSICS
¥t₃, respectively. The direction of the average velocity v is shown in figures (a), (b) and (c) for three decreasing values of ¥t, i.e. ¥t₁, ¥t₂, and ¥t₃, ( t > t > t ). As ¥t → 0, r
¥ 1 ¥ 2 ¥ 3 ¥ ¥ ¥ 0 and is along the tangent to the path [Fig. 3.13(d)]. Therefore, the direction of velocity at any point on the path of an object is tangential to the path at that point and is in the direction of motion.
We can express v in a component form:
Fig. 3.14 The components vₓ and vy of velocity v and the angle Ò it makes with x-axis. Note that vₓ = v cos Ò, vy = v sin Ò.
The acceleration (instantaneous acceleration) is the limiting value of the average acceleration as the time interval approaches zero:
a = lim (¥v / ¥t) (3. 32a)
Since ¥v = ¥vₓ i + ¥vᵧ j, we have
v = ɵ dx / dt + ɵ dy / dt = vₓ i + vᵧ j
where vₓ = dx / dt, vᵧ = dy / dt (3.30a)
So, if the expressions for the coordinates x and y are known as functions of time, we can use these equations to find vₓ and vy.
The magnitude of v is then
v = √(vₓ² + vᵧ²) (3.30b)
and the direction of v is given by the angle Ò:
tan Ò = vᵧ / vₓ, Ò = tan⁻¹(vᵧ / vₓ) (3.30c)
vₓ, vᵧ and angle Ò are shown in Fig. 3.14 for a velocity vector v at point p.
# Acceleration
The average acceleration a of an object for a time interval ¥t moving in x-y plane is the change in velocity divided by the time interval:
a = (¥v / ¥t) = (¥vₓ i + ¥vᵧ j) = (¥vₓ / ¥t) ɵ + (¥vᵧ / ¥t) ɵ (3.31a)
Or, a = aₓ i + aᵧ j (3.31b)
* In terms of x and y, aₓ and aᵧ can be expressed as
2024-25 | 9 | 11 | Physics | 103 |
c5f6758c-de47-4240-bcd5-4faa08a0a2ce | # MOTION IN A PLANE
Fig. 3.15 The average acceleration for three time intervals (a) Δt1, (b) Δt2, and (c) Δt3, (Δt1 > Δt2 > Δt3). (d) In the limit Δt → 0, the average acceleration becomes the acceleration.
Note that in one dimension, the velocity and the acceleration of an object are always along the same straight line (either in the same direction or in the opposite direction). However, for motion in two or three dimensions, velocity and acceleration vectors may have any angle between 0° and 180° between them.
# 3.8 MOTION IN A PLANE WITH CONSTANT ACCELERATION
Example 3.4 The position of a particle is given by
r = 3.0 i + 2.0 j + 5.0 k where t is in seconds and the coefficients have the proper units for r to be in metres.
(a) Find v(t) and a(t) of the particle.
(b) Find the magnitude and direction of v(t) at t = 1.0 s.
Then, by definition
a = (v - v0) / (t - 0) or v = v0 + at (3.33a)
In terms of components:
vx = v0x + axt
vy = v0y + ayt (3.33b)
Let us now find how the position r changes with time. We follow the method used in the one-dimensional case. Let r0 and r be the position vectors of the particle at time 0 and velocities at these instants be v0 and v. Then, over this time interval t, the average velocity is (v0 + v) / 2. The displacement is the average velocity multiplied by the time interval:
At t = 1.0 s, v = 3.0 i + 4.0 j. Its magnitude is v = √(3² + 4²) = 5.0 m s-1 and direction is | 10 | 11 | Physics | 103 |
81b076aa-6267-44a9-b06e-aa383a08b192 | # PHYSICS
r − r₀ = 0
t = 0
v + v (v + at) + v = (5.0 t + 1.5 t₂)²
2
Therefore, (x(t))² = 5.0t + 1.5t
Given x y t = +1.0t
(t) = 84 m, t = ?
5.0 t + 1.5 t = 84 ⇒
Or, r = r₀ + v₀t + 1/2 at² (3.34a)
At t = 6 s, y = 2 ⇒ t = 6 s
= 1.0 (6)² = 36.0 m
v = dr/dt = (5.0 + 3.0t) î + 2.0t j
At t = 6 s, v = 23.0i + 12.0j
speed = v = √(23² + 12²) ≅ 26 m s⁻¹.
# 3.9 PROJECTILE MOTION
As an application of the ideas developed in the previous sections, we consider the motion of a projectile. An object that is in flight after being thrown or projected is called a projectile. Such a projectile might be a football, a cricket ball, a baseball or any other object. The motion of a projectile may be thought of as the result of two separate, simultaneously occurring components of motions. One component is along a horizontal direction without any acceleration and the other along the vertical direction with constant acceleration due to the force of gravity. It was Galileo who first stated this independency of the horizontal and the vertical components of projectile motion in his Dialogue on the great world systems (1632).
In our discussion, we shall assume that the air resistance has negligible effect on the motion of the projectile. Suppose that the projectile is launched with velocity vₒ that makes an angle θₒ with the x-axis as shown in Fig. 3.16. After the object has been projected, the acceleration acting on it is that due to gravity which is directed vertically downward:
a = -g j
Or, aₓ = 0, aᵧ = -g (3.35)
The components of initial velocity v are:
v = vₒ cos θ
vₒₓ = vₒ sin θ (3.36)
Answer: From Eq. (3.34a) for r₀ = 0, the position of the particle is given by
r(t) = v₀t + 1/2 at²
= (5.0 î)t + 1/2 (3.0 î + 2.0 j)t² | 11 | 11 | Physics | 103 |
54515267-44e7-47f7-b017-f632043d901b | # MOTION IN A PLANE
Now, since g, θₒ and vₒ are constants, Eq. (3.39) is of the form y = a x + b x², in which a and b are constants. This is the equation of a parabola, i.e. the path of the projectile is a parabola (Fig. 3.17).
Fig 3.16 Motion of an object projected with velocity vₒ at angle θ₀.
If we take the initial position to be the origin of the reference frame as shown in Fig. 3.16, we have:
xₒ = 0, yₒ = 0
Then, Eq.(3.34b) becomes:
x = vₒₓ t = (vₒ cos θₒ ) t
and y = (vₒ sin θₒ ) t – ( ½ )g t² (3.37)
The components of velocity at time t can be obtained using Eq.(3.33b):
vₓ = vₒₓ = vₒ cos θₒ
v = v sin θ – g t (3.38)
Equation (3.37) gives the x-, and y-coordinates of the position of a projectile at time t in terms of two parameters — initial speed vₒ and projection angle θₒ.
Notice that the choice of mutually perpendicular x-, and y-directions for the analysis of the projectile motion has resulted in a simplification. One of the components of velocity, i.e. x-component remains constant throughout the motion and only the y-component changes, like an object in free fall in vertical direction. This is shown graphically at few instants in Fig. 3.17. Note that at the point of maximum height, v = 0 and therefore, θ = tan⁻¹ vʸ = 0.
Fig. 3.17 The path of a projectile is a parabola.
How much time does the projectile take to reach the maximum height? Let this time be denoted by tₘ. Since at this point, vy= 0, we have from Eq. (3.38):
vy = vₒ sinθₒ – g tₘ = 0
Or, tₘ = vₒ sinθₒ /g (3.40a)
The total time T during which the projectile is in flight can be obtained by putting y = 0 in Eq. (3.37). We get:
T = 2 (vₒ sin θₒ)/g (3.40b)
Tf is known as the time of flight of the projectile. We note that Tf = 2 tₘ, which is expected because of the symmetry of the parabolic path.
# Maximum height of a projectile
The maximum height hₘ reached by the projectile can be calculated by substituting t = tₘ in Eq. (3.37):
y = hₘ = (vₒ sinθₒ) tₘ - (½)g tₘ²
Equation of path of a projectile
What is the shape of the path followed by the projectile? This can be seen by eliminating the time between the expressions for x and y as given in Eq. (3.37). We obtain:
y = (tan θₒ) x - (g / (2 (vₒ cos θₒ)²)) x² (3.39)
# Horizontal range of a projectile
The horizontal distance travelled by a projectile from its initial position (x = y = 0) to the position where it passes y = 0 during its fall is called the horizontal range. | 12 | 11 | Physics | 103 |
447dab7f-4215-403d-9bd8-b0e7027d21d5 | # PHYSICS
range, R. It is the distance travelled during the time of flight Tf. Therefore, the range R is
R = (vₒ cos θₒ) (Tf)
= (vₒ cos θₒ) (2 vₒ sin θₒ)/g
R = 0
g
v² sin 2θ
Equation (3.42a) shows that for a given projection velocity vₒ, R is maximum when sin 2θₒ is maximum, i.e., when θₒ = 45⁰. The maximum horizontal range is, therefore,
Rm = 0
g
Example 3.6 Galileo, in his book Two New Sciences, stated that “for elevations which exceed or fall short of 45° by equal amounts, the ranges are equal”. Prove this statement.
Answer: For a projectile launched with velocity vₒ at an angle θₒ, the range is given by
R = v² sin 2θ
Now, for angles, (45° + α) and (45° – α), 2θ is (90° + 2α) and (90° – 2α), respectively. The values of sin (90° + 2α) and sin (90° – 2α) are the same, equal to that of cos 2α. Therefore, ranges are equal for elevations which exceed or fall short of 45° by equal amounts α.
Example 3.7 A hiker stands on the edge of a cliff 490 m above the ground and throws a stone horizontally with an initial speed of 15 m s⁻¹. Neglecting air resistance, find the time taken by the stone to reach the ground, and the speed with which it hits the ground. (Take g = 9.8 m s⁻²).
Answer: We choose the origin of the x-, and y-axis at the edge of the cliff and t = 0 s at the instant the stone is thrown. Choose the positive direction of x-axis to be along the initial velocity and the positive direction of y-axis to be the vertically upward direction. The x-, and y- components of the motion can be treated independently. The equations of motion are:
x(t) = xₒ + vₒₓ t
The stone hits the ground when y(t) = –490 m.
–490 m = –(1/2)(9.8) t².
This gives t = 10 s.
The velocity components are vₓ = vₒₓ and vy = voy – g t so that when the stone hits the ground:
vₒₓ = 15 m s⁻¹
voy = 0 – 9.8 × 10 = –98 m s⁻¹
Therefore, the speed of the stone is
v = √(vₓ² + vy²) = √(15² + (–98)²) = √(99 m s⁻¹)
Example 3.8 A cricket ball is thrown at a speed of 28 m s⁻¹ in a direction 30° above the horizontal. Calculate (a) the maximum height, (b) the time taken by the ball to return to the same level, and (c) the distance from the thrower to the point where the ball returns to the same level.
Answer: (a) The maximum height is given by
h = (vₒ sin θₒ)² / (2g) = (28 sin 30°)² / (2 × 9.8)
= 14 × 14 / (2 × 9.8) = 10.0 m
(b) The time taken to return to the same level is
Tf = (2 vₒ sin θₒ) / g = (2 × 28 × sin 30°) / 9.8 = 28 / 9.8 s = 2.9 s
(c) The distance from the thrower to the point where the ball returns to the same level is
R = (v² sin 2θ) / g = (28 × 28 × sin 60°) / 9.8 = 69 m
# 3.10 UNIFORM CIRCULAR MOTION
When an object follows a circular path at a constant speed, the motion of the object is called uniform circular motion. The word “uniform” refers to the speed, which is uniform (constant) throughout the motion. Suppose an object is moving with uniform speed v in a circle of radius R as shown in Fig. 3.18. Since the velocity of the object is changing continuously in direction, the object undergoes acceleration. Let us find the magnitude and the direction of this acceleration. | 13 | 11 | Physics | 103 |
b3446ece-828a-44d9-ab80-0712aaf1e77c | # MOTION IN A PLANE
Fig. 3.18 Velocity and acceleration of an object in uniform circular motion. The time interval Δt decreases from (a) to (c) where it is zero. The acceleration is directed, at each point of the path, towards the centre of the circle.
Let r and r′ be the position vectors and θ be the angle between them. Since the velocity vectors v and v′ are always perpendicular to the position vectors, the angle between them is also Δθ. Therefore, the velocity at a point is along the tangent at that point in the direction of motion. The velocity vectors v and v′ are as shown in Fig. 3.18(a). Δv is obtained in Fig. 3.18 (a2) using the triangle law of vector addition. Since the path is circular, v is perpendicular to r and so is v′. Therefore, Δv is perpendicular to Δr. Since the average acceleration is along Δv, the average acceleration is perpendicular to r. If we place Δv on the line that bisects the angle between r and r′, we see that it is directed towards the centre of the circle. Figure 3.18(b) shows the same quantities for smaller time interval.
Hence, Δv and Δv are again directed towards the centre. In Fig. 3.18(c), as Δt approaches 0, the average acceleration becomes the instantaneous acceleration. It is directed towards the centre. Thus, we find that the acceleration of an object in uniform circular motion is always directed towards the centre of the circle. Let us now find the magnitude of the acceleration.
The magnitude of a is, by definition, given by:
a = Δv / Δt
Let the angle between position vectors r and r′ be small. Therefore, the centripetal acceleration ac is:
ac = v² / r
* In the limit as Δt approaches 0, Δr becomes perpendicular to r. In this limit, Δv approaches 0 and is consequently also perpendicular to V. Therefore, the acceleration is directed towards the centre, at each point of the circular path.
2024-25 | 14 | 11 | Physics | 103 |
1373d9dc-c349-4b76-a081-03b7146adc3a | # PHYSICS
Thus, the acceleration of an object moving with speed v in a circle of radius R has a magnitude v²/R and is always directed towards the centre. This is why this acceleration is called centripetal acceleration (a term proposed by Newton). A thorough analysis of centripetal acceleration was first published in 1673 by the Dutch scientist Christiaan Huygens (1629-1695) but it was probably known to Newton also some years earlier. “Centripetal” comes from a Greek term which means ‘centre-seeking’. Since v and R are constant, the magnitude of the centripetal acceleration is also constant. However, the direction changes — pointing always towards the centre. Therefore, a centripetal acceleration is not a constant vector.
We have another way of describing the velocity and the acceleration of an object in uniform circular motion. As the object moves from P to P′ in time ∆t (= t′ – t), the line CP (Fig. 3.18) turns through an angle ∆θ as shown in the figure. ∆θ is called angular distance. We define the angular speed ω (Greek letter omega) as the time rate of change of angular displacement:
ω = ∆θ / ∆t (3.44)
Now, if the distance travelled by the object during the time ∆t is ∆s, i.e. PP′ is ∆s, then:
v = ∆s / ∆t
but ∆s = R ∆θ. Therefore:
v = R ∆θ / ∆t = R ω (3.45)
We can express centripetal acceleration ac in terms of angular speed:
ac = ω²R (3.46)
Example 3.9 An insect trapped in a circular groove of radius 12 cm moves along in the groove steadily and completes 7 revolutions in 100 s. (a) What is the angular speed, and the linear speed of the motion? (b) Is the acceleration vector a constant vector? What is its magnitude?
Answer: This is an example of uniform circular motion. Here R = 12 cm. The angular speed ω is given by:
ω = 2π/T = 2π × 7/100 = 0.44 rad/s
The linear speed v is:
v = ω R = 0.44 s⁻¹ × 12 cm = 5.3 cm s⁻¹
The direction of velocity v is along the tangent to the circle at every point. The acceleration is directed towards the centre of the circle. Since this direction changes continuously, acceleration here is not a constant vector. However, the magnitude of acceleration is constant:
a = ω² R = (0.44 s⁻¹)² (12 cm) = 2.3 cm s⁻² | 15 | 11 | Physics | 103 |
97dc10e6-abe0-4bba-847e-21bbced0991e | # MOTION IN A PLANE
# SUMMARY
1. Scalar quantities are quantities with magnitudes only. Examples are distance, speed, mass and temperature.
2. Vector quantities are quantities with magnitude and direction both. Examples are displacement, velocity and acceleration. They obey special rules of vector algebra.
3. A vector A multiplied by a real number λ is also a vector λ, whose magnitude is λ times the magnitude of the vector A and whose direction is the same or opposite depending upon whether λ is positive or negative.
4. Two vectors A and B may be added graphically using head-to-tail method or parallelogram method.
5. Vector addition is commutative:
A + B = B + A
It also obeys the associative law:
(A + B) + C = A + (B + C)
6. A null or zero vector is a vector with zero magnitude. Since the magnitude is zero, we don’t have to specify its direction. It has the properties:
A + 0 = A
λ0 = 0
0 A = 0
7. The subtraction of vector B from A is defined as the sum of A and –B:
A – B = A + (–B)
8. A vector A can be resolved into components along two given vectors a and b lying in the same plane:
A = λ a + μ b
where λ and μ are real numbers.
9. A unit vector associated with a vector A has magnitude 1 and is along the vector A:
n = ˆA
The unit vectors i, j, k are vectors of unit magnitude and point in the direction of the x-, y-, and z-axes, respectively in a right-handed coordinate system.
10. A vector A can be expressed as
A = Aₓ i + Aᵧ j
where Aₓ, Aᵧ are its components along x- and y-axes. If vector A makes an angle θ with the x-axis, then Aₓ = A cos θ, Aᵧ = A sin θ and A = √(Aₓ² + Aᵧ²), tan θ = Aᵧ / Aₓ.
11. Vectors can be conveniently added using analytical method. If the sum of two vectors A and B, that lie in the x-y plane, is R, then:
R = Rₓ i + Rᵧ j, where R = A + B, and Rₓ = Aₓ + Bₓ, Rᵧ = Aᵧ + Bᵧ.
12. The position vector of an object in x-y plane is given by r = x i + y j and the displacement from position r to position r’ is given by
Δr = r′− r
= (x ′ − x) i + (y ′ − y) j
= Δx i + Δy j
13. If an object undergoes a displacement Δr in time Δt, its average velocity is given by
v = Δr / Δt. The velocity of an object at time t is the limiting value of the average velocity as Δt tends to zero. | 16 | 11 | Physics | 103 |
9bad37ab-0598-49df-8d7a-efca1568ad20 | # PHYSICS
v = ∆ lim ∆r = dr. It can be written in unit vector notation as:
v = vx i + vy j + vz k where v = ,
x = dx/dt, vy = dy/dt, vz = dz/dt
When position of an object is plotted on a coordinate system, v is always tangent to the curve representing the path of the object.
# 14.
If the velocity of an object changes from v to v′ in time ∆t, then its average acceleration is given by:
a = (v − v′) / ∆t = ∆v / ∆t
The acceleration a at any time t is the limiting value of a as ∆t → 0:
a = ∆ lim ∆v = dv
t → 0 ∆t
a = ax i + ay j + az k
where, ax = dvx / dt, ay = dvy / dt, az = dvz / dt
# 15.
If an object is moving in a plane with constant acceleration a = ax + ay and its position vector at time t = 0 is r0, then at any other time t, it will be at a point given by:
r = r0 + v0 t + (1/2) a t²
and its velocity is given by:
v = v0 + a t
In component form:
x = x0 + v0x t + (1/2) ax t²
y = y0 + v0y t + (1/2) ay t²
vx = v0x + ax t
vy = v0y + ay t
Motion in a plane can be treated as superposition of two separate simultaneous one-dimensional motions along two perpendicular directions.
# 16.
An object that is in flight after being projected is called a projectile. If an object is projected with initial velocity v0 making an angle θ0 with x-axis and if we assume its initial position to coincide with the origin of the coordinate system, then the position and velocity of the projectile at time t are given by:
x = (v0 cos θ0) t
y = (v0 sin θ0) t − (1/2) g t²
vx = v0x = v0 cos θ0
vy = v0 sin θ0 − g t
The path of a projectile is parabolic and is given by:
y = (tan θ0) x − (g x²) / (2 v0 cos θ0)
The maximum height that a projectile attains is: | 17 | 11 | Physics | 103 |
cc1eb87a-01f2-4142-a8da-56deb327728f | # MOTION IN A PLANE
hₘ = (vₒ sinqₒ)²
2g
The time taken to reach this height is:
tₘ = vₒ sinθₒ
g
The horizontal distance travelled by a projectile from its initial position to the position it passes y = 0 during its fall is called the range, R of the projectile. It is:
R = v²
g
# 17.
When an object follows a circular path at constant speed, the motion of the object is called uniform circular motion. The magnitude of its acceleration is aₐ = v² / R. The direction of aₐ is always towards the centre of the circle.
The angular speed ω, is the rate of change of angular distance. It is related to velocity v by v = ω R. The acceleration is ω.
If aₐ = 2R.
T is the time period of revolution of the object in circular motion and ν is its frequency, we have ω = 2πν, v = 2πνR, aₐ = 4π²ν²R
2024-25 | 18 | 11 | Physics | 103 |
cd503bb1-13d2-4ee8-8c4f-6a915d2dc63b | # PHYSICS
# POINTS TO PONDER
1. The path length traversed by an object between two points is, in general, not the same as the magnitude of displacement. The displacement depends only on the end points; the path length (as the name implies) depends on the actual path. The two quantities are equal only if the object does not change its direction during the course of motion. In all other cases, the path length is greater than the magnitude of displacement.
2. In view of point 1 above, the average speed of an object is greater than or equal to the magnitude of the average velocity over a given time interval. The two are equal only if the path length is equal to the magnitude of displacement.
3. The vector equations (3.33a) and (3.34a) do not involve any choice of axes. Of course, you can always resolve them along any two independent axes.
4. The kinematic equations for uniform acceleration do not apply to the case of uniform circular motion since in this case the magnitude of acceleration is constant but its direction is changing.
5. An object subjected to two velocities v₁ and v₂ has a resultant velocity v = v₁ + v₂. Take care to distinguish it from velocity of object 1 relative to velocity of object 2: Here v₁₂ = v₁ − v₂.
6. The resultant acceleration of an object in circular motion is towards the centre only if the speed is constant.
7. The shape of the trajectory of the motion of an object is not determined by the acceleration alone but also depends on the initial conditions of motion (initial position and initial velocity). For example, the trajectory of an object moving under the same acceleration due to gravity can be a straight line or a parabola depending on the initial conditions.
# EXERCISES
1. State, for each of the following physical quantities, if it is a scalar or a vector: volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity.
2. Pick out the two scalar quantities in the following list: force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity.
3. Pick out the only vector quantity in the following list: Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge.
4. State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful:
1. adding any two scalars,
2. adding a scalar to a vector of the same dimensions,
3. multiplying any vector by any scalar,
4. multiplying any two scalars,
5. adding any two vectors,
6. adding a component of a vector to the same vector.
5. Read each statement below carefully and state with reasons, if it is true or false:
1. The magnitude of a vector is always a scalar,
2. each component of a vector is always a scalar,
3. the total path length is always equal to the magnitude of the displacement vector of a particle,
4. the average speed of a particle (defined as total path length divided by the time taken to cover the path) is either greater or equal to the magnitude of average velocity of the particle over the same interval of time,
5. Three vectors not lying in a plane can never add up to give a null vector.
6. Establish the following vector inequalities geometrically or otherwise:
1. |a+b| < |a| + |b|,
2. |a+b| > ||a| − |b|. | 19 | 11 | Physics | 103 |
252ae071-1d76-4c6c-a0f3-718c43509005 | # MOTION IN A PLANE
(c) |a− − −b| < |a| + |b|
(d) |a− − − − −b| ||a| − − − − > − |b||
When does the equality sign above apply?
# 3.7
Given a + b + c + d = 0, which of the following statements are correct:
- (a) a, b, c, and d must each be a null vector,
- (b) The magnitude of (a + c) equals the magnitude of (b + d),
- (c) The magnitude of a can never be greater than the sum of the magnitudes of b, c, and d,
- (d) b + c must lie in the plane of a and d if a and d are not collinear, and in the line of a and d, if they are collinear?
# 3.8
Three girls skating on a circular ice ground of radius 200 m start from a point P on the edge of the ground and reach a point Q diametrically opposite to P following different paths as shown in Fig. 3.19. What is the magnitude of the displacement vector for each? For which girl is this equal to the actual length of path skate?
Fig. 3.19
# 3.9
A cyclist starts from the centre O of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference, and returns to the centre along QO as shown in Fig. 3.20. If the round trip takes 10 min, what is the (a) net displacement, (b) average velocity, and (c) average speed of the cyclist?
Fig. 3.20
# 3.10
On an open ground, a motorist follows a track that turns to his left by an angle of 60° after every 500 m. Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.
# 3.11
A passenger arriving in a new town wishes to go from the station to a hotel located 10 km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23 km long and reaches the hotel in 28 min. What is (a) the average speed of the taxi, (b) the magnitude of average velocity? Are the two equal?
# 3.12
The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 m s⁻¹ can go without hitting the ceiling of the hall?
# 3.13
A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball?
2024-25 | 20 | 11 | Physics | 103 |
42d90d1b-714c-452a-869f-1b1cb9d71db6 | # PHYSICS
# 3.14
A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone?
# 3.15
An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity.
# 3.16
Read each statement below carefully and state, with reasons, if it is true or false:
- (a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre.
- (b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point.
- (c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector.
# 3.17
The position of a particle is given by r = 3.0 ˆi - 2.0t ˆj + 4.0 ˆk m where t is in seconds and the coefficients have the proper units for r to be in metres.
(a) Find the v and a of the particle? (b) What is the magnitude and direction of velocity of the particle at t = 2.0 s?
# 3.18
A particle starts from the origin at t = 0 s with a velocity of 10.0 ɵ m/s and moves in the x-y plane with a constant acceleration of (8.0i + 2.0j) m/s². (a) At what time is the x-coordinate of the particle 16 m? What is the y-coordinate of the particle at that time? (b) What is the speed of the particle at that time?
# 3.19
ɵ and ɵ are unit vectors along x- and y-axis respectively. What is the magnitude and direction of the vectors ɵi + ɵj, and ɵi - ɵj? What are the components of a vector A = 2 ɵi + 3 ɵj along the directions of i + j and i - j? [You may use graphical method]
# 3.20
For any arbitrary motion in space, which of the following relations are true:
- (a) vaverage = (1/2) (v(t₁) + v(t₂))
- (b) vaverage = [r(t₂) - r(t₁)] /(t₂ – t₁)
- (c) v(t) = v(0) + a t
- (d) r(t) = r(0) + v(0) t + (1/2) a t²
- (e) aaverage = [v(t₂) - v(t₁)] /(t₂ – t₁)
(The ‘average’ stands for average of the quantity over the time interval t₁ to t₂)
# 3.21
Read each statement below carefully and state, with reasons and examples, if it is true or false:
- (a) A scalar quantity is one that is conserved in a process.
- (b) A scalar quantity can never take negative values.
- (c) A scalar quantity must be dimensionless.
- (d) A scalar quantity does not vary from one point to another in space.
- (e) A scalar quantity has the same value for observers with different orientations of axes.
# 3.22
An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s apart is 30°, what is the speed of the aircraft? | 21 | 11 | Physics | 103 |
d91bc16c-09ad-418f-a745-d11fb1986784 | # CHAPTER FOUR
# LAWS OF MOTION
# 4.1 INTRODUCTION
In the preceding Chapter, our concern was to describe the motion of a particle in space quantitatively. We saw that uniform motion needs the concept of velocity alone whereas non-uniform motion requires the concept of acceleration in addition. So far, we have not asked the question as to what governs the motion of bodies. In this chapter, we turn to this basic question.
# 4.2 Aristotle’s fallacy
# 4.3 The law of inertia
# 4.4 Newton’s first law of motion
# 4.5 Newton’s second law of motion
# 4.6 Newton’s third law of motion
# 4.7 Conservation of momentum
# 4.8 Equilibrium of a particle
# 4.9 Common forces in mechanics
Circular motion
# 4.10 Solving problems in mechanics
# Summary
# Points to ponder
# Exercises
Let us first guess the answer based on our common experience. To move a football at rest, someone must kick it. To throw a stone upwards, one has to give it an upward push. A breeze causes the branches of a tree to swing; a strong wind can even move heavy objects. A boat moves in a flowing river without anyone rowing it. Clearly, some external agency is needed to provide force to move a body from rest. Likewise, an external force is needed also to retard or stop motion. You can stop a ball rolling down an inclined plane by applying a force against the direction of its motion.
In these examples, the external agency of force (hands, wind, stream, etc.) is in contact with the object. This is not always necessary. A stone released from the top of a building accelerates downward due to the gravitational pull of the earth. A bar magnet can attract an iron nail from a distance. This shows that external agencies (e.g. gravitational and magnetic forces) can exert force on a body even from a distance.
In short, a force is required to put a stationary body in motion or stop a moving body, and some external agency is needed to provide this force. The external agency may or may not be in contact with the body.
So far so good. But what if a body is moving uniformly (e.g. a skater moving straight with constant speed on a horizontal ice slab)? Is an external force required to keep a body in uniform motion? | 0 | 11 | Physics | 104 |
a0c323bf-1b8b-459c-a522-9c7fb7ee765c | # PHYSICS
# 4.2 ARISTOTLE’S FALLACY
The question posed above appears to be simple. However, it took ages to answer it. Indeed, the correct answer to this question given by Galileo in the seventeenth century was the foundation of Newtonian mechanics, which signalled the birth of modern science.
The Greek thinker, Aristotle (384 B.C– 322 B.C.), held the view that if a body is moving, something external is required to keep it moving. According to this view, for example, an arrow shot from a bow keeps flying since the air behind the arrow keeps pushing it. The view was part of an elaborate framework of ideas developed by Aristotle on the motion of bodies in the universe. Most of the Aristotelian ideas on motion are now known to be wrong and need not concern us. For our purpose here, the Aristotelian law of motion may be phrased thus:
|An external force|(i)|(ii)|(iii)|
|---|---|---|---|
|is required to keep a body in motion| | | |
Aristotelian law of motion is flawed, as we shall see. However, it is a natural view that anyone would hold from common experience. Even a small child playing with a simple (non-electric) toy-car on a floor knows intuitively that it needs to constantly drag the string attached to the toy-car with some force to keep it going. If it releases the string, it comes to rest. This experience is common to most terrestrial motion. External forces seem to be needed to keep bodies in motion. Left to themselves, all bodies eventually come to rest.
What is the flaw in Aristotle’s argument? The answer is: a moving toy car comes to rest because the external force of friction on the car by the floor opposes its motion. To counter this force, the child has to apply an external force on the car in the direction of motion. When the car is in uniform motion, there is no net external force acting on it: the force by the child cancels the force (friction) by the floor. The corollary is: if there were no friction, the child would not be required to apply any force to keep the toy car in uniform motion.
The opposing forces such as friction (solids) and viscous forces (for fluids) are always present in the natural world. This explains why forces by external agencies are necessary to overcome the frictional forces to keep bodies in uniform motion. Now we understand where Aristotle went wrong. He coded this practical experience in the form of a basic argument. To get at the law of inertia, Galileo inferred it from observations of motion of a ball on a double inclined plane.
# 4.3 THE LAW OF INERTIA
Galileo studied motion of objects on an inclined plane. Objects (i) moving down an inclined plane accelerate, while those (ii) moving up retard. (iii) Motion on a horizontal plane is an intermediate situation. Galileo concluded that an object moving on a frictionless horizontal plane must neither have acceleration nor retardation, i.e. it should move with constant velocity (Fig. 4.1(a)).
Another experiment by Galileo leading to the same conclusion involves a double inclined plane. A ball released from rest on one of the planes rolls down and climbs up the other. If the planes are smooth, the final height of the ball is nearly the same as the initial height (a little less but never greater). In the ideal situation, when friction is absent, the final height of the ball is the same as its initial height.
If the slope of the second plane is decreased and the experiment repeated, the ball will still reach the same height, but in doing so, it will travel a longer distance. In the limiting case, when the slope of the second plane is zero (i.e. is horizontal), the ball travels an infinite distance. In other words, its motion never ceases. This is, of course, an idealised situation (Fig. 4.1(b)).
Fig. 4.1(b) The law of inertia was inferred by Galileo from observations of motion of a ball on a double inclined plane. | 1 | 11 | Physics | 104 |
0e4ee0d4-ade4-4dc7-ade9-c352cb71fb97 | # LAWS OF MOTION
In practice, the ball does come to a stop after moving a finite distance on the horizontal plane, because of the opposing force of friction which can never be totally eliminated. However, if there were no friction, the ball would continue to move with a constant velocity on the horizontal plane.
Galileo thus, arrived at a new insight on motion that had eluded Aristotle and those who followed him. The state of rest and the state of uniform linear motion (motion with constant velocity) are equivalent. In both cases, there is no net force acting on the body. It is incorrect to assume that a net force is needed to keep a body in uniform motion. To maintain a body in uniform motion, we need to apply an external force to counter the frictional force, so that the two forces sum up to zero net external force.
# Ideas on Motion in Ancient Indian Science
Ancient Indian thinkers had arrived at an elaborate system of ideas on motion. Force, the cause of motion, was thought to be of different kinds: force due to continuous pressure (nodan), as the force of wind on a sailing vessel; impact (abhighat), as when a potter’s rod strikes the wheel; persistent tendency (sanskara) to move in a straight line (vega) or restoration of shape in an elastic body; transmitted force by a string, rod, etc. The notion of (vega) in the Vaisesika theory of motion perhaps comes closest to the concept of inertia. Vega, the tendency to move in a straight line, was thought to be opposed by contact with objects including atmosphere, a parallel to the ideas of friction and air resistance.
It was correctly summarised that the different kinds of motion (translational, rotational and vibrational) of an extended body arise from only the translational motion of its constituent particles. A falling leaf in the wind may have downward motion as a whole (patan) and also rotational and vibrational motion (bhraman, spandan), but each particle of the leaf at an instant only has a definite (small) displacement. There was considerable focus in Indian thought on measurement of motion and units of length and time. It was known that the position of a particle in space can be indicated by distance measured along three axes. Bhaskara (1150 A.D.) had introduced the concept of ‘instantaneous motion’ (tatkaliki gati), which anticipated the modern notion of instantaneous velocity using Differential Calculus. The difference between a wave and a current (of water) was clearly understood; a current is a motion of particles of water under gravity and fluidity while a wave results from the transmission of vibrations of water particles.
# 4.4 NEWTON’S FIRST LAW OF MOTION
The first law of motion can, therefore, be simply expressed as: If the net external force on a body is zero, its acceleration is zero. Acceleration can be non-zero only if there is a net external force on the body.
To summarise, if the net external force is zero, a body at rest continues to remain at rest and a body in motion continues to move with a uniform velocity. This property of the body is called inertia. Inertia means ‘resistance to change’. A body does not change its state of rest or uniform motion, unless an external force compels it to change that state.
Two kinds of situations are encountered in the application of this law in practice. In some examples, we know that the net external force on the object is zero. In that case we can conclude that the acceleration of the object is zero. For example, a spaceship out in interstellar space, far from all other objects and with all its rockets turned off, has no net external force acting on it. Its acceleration, according to the first law, must be zero. If it is in motion, it must continue to move with a uniform velocity. | 2 | 11 | Physics | 104 |
b0d1d67b-53bf-49fa-95fa-d4a154de82f8 | # PHYSICS
More often, however, we do not know all the forces to begin with. In that case, if we know that an object is unaccelerated (i.e. it is either at rest or in uniform linear motion), we can infer from the first law that the net external force on the object must be zero. Gravity is everywhere. For terrestrial phenomena, in particular, every object experiences gravitational force due to the earth. Also, objects in motion generally experience friction, viscous drag, etc. If then, on earth, an object is at rest or in uniform linear motion, it is not because there are no forces acting on it, but because the various external forces cancel out i.e. add up to zero net external force.
Consider a book at rest on a horizontal surface (Fig. 4.2(a)). It is subject to two external forces: the force due to gravity (i.e. its weight W) acting downward and the upward force on the book by the table, the normal force R. R is a self-adjusting force. This is an example of the kind of situation mentioned above. The forces are not quite known fully but the state of motion is known. We observe the book to be at rest. Therefore, we conclude from the first law that the magnitude of R equals W. A statement often encountered is: W = R, forces cancel and, therefore, the book is at rest. This is incorrect reasoning. The correct statement is: “Since the book is observed to be at rest, the net external force on it must be zero, according to the first law. This implies that the normal force R must be equal and opposite to the weight W.”
The property of inertia contained in the First law is evident in many situations. Suppose we are standing in a stationary bus and the driver starts the bus suddenly. We get thrown backward with a jerk. Why? Our feet are in touch with the floor. If there were no friction, we would remain where we were, while the floor of the bus would simply slip forward under our feet and the back of the bus would hit us. However, fortunately, there is some friction between the feet and the floor. If the start is not too sudden, i.e. if the acceleration is moderate, the frictional force would be enough to accelerate our feet along with the bus. But our body is not strictly a rigid body. It is deformable, i.e. it allows some relative displacement between different parts. What this means is that while our feet go with the bus, the rest of the body remains where it is due to inertia. Relative to the bus, therefore, we are thrown backward. As soon as that happens, however, the muscular forces on the rest of the body (by the feet) come into play to move the body along with the bus. A similar thing happens when the bus suddenly stops. Our feet stop due to the friction which does not allow relative motion between the feet and the floor of the bus. But the rest of the body continues to move forward due to inertia. We are thrown forward. The restoring muscular forces again come into play and bring the body to rest.
# Example 4.1
An astronaut accidentally gets separated out of his small spaceship (Fig. 4.2(b)) accelerating in interstellar space at a constant rate of 100 m/s2. What is the acceleration of the astronaut the instant after he is outside the spaceship? (Assume that there are no nearby stars to exert gravitational force on him.)
# Answer
Since there are no nearby stars to exert gravitational force on him and the small spaceship exerts negligible gravitational attraction on him, the net force acting on the astronaut is zero. | 3 | 11 | Physics | 104 |
a03c8dcb-6284-4c2b-85ca-a703b875bebd | # LAWS OF MOTION
An astronaut, once he is out of the spaceship, is zero. By the first law of motion the acceleration of the astronaut is zero.
# 4.5 NEWTON’S SECOND LAW OF MOTION
The first law refers to the simple case when the net external force on a body is zero. The second law of motion refers to the general situation when there is a net external force acting on the body. It relates the net external force to the acceleration of the body.
# Momentum
Momentum of a body is defined to be the product of its mass m and velocity v, and is denoted by p:
p = mv (4.1)
Momentum is clearly a vector quantity. The following common experiences indicate the importance of this quantity for considering the effect of force on motion.
- Suppose a light-weight vehicle (say a small car) and a heavy weight vehicle (say a loaded truck) are parked on a horizontal road. We all know that a much greater force is needed to push the truck than the car to bring them to the same speed in the same time. Similarly, a greater opposing force is needed to stop a heavy body than a light body in the same time, if they are moving with the same speed.
- If two stones, one light and the other heavy, are dropped from the top of a building, a person on the ground will find it easier to catch the light stone than the heavy stone. The mass of a body is thus an important parameter that determines the effect of force on its motion.
- Speed is another important parameter to consider. A bullet fired by a gun can easily pierce human tissue before it stops, resulting in casualty. The same bullet fired with moderate speed will not cause much damage. Thus for a given mass, the greater the speed, the greater is the opposing force needed to stop the body in a certain time. Taken together, the product of mass and velocity, that is momentum, is evidently a relevant variable of motion. The greater the change in the momentum in a given time, the greater is the force that needs to be applied.
- A seasoned cricketer catches a cricket ball coming in with great speed far more easily than a novice, who can hurt his hands in the act. One reason is that the cricketer allows a longer time for his hands to stop the ball. As you may have noticed, he draws in the hands backward in the act of catching the ball (Fig. 4.3). The novice, on the other hand, keeps his hands fixed and tries to catch the ball almost instantly. He needs to provide a much greater force to stop the ball instantly, and this hurts. The conclusion is clear: force not only depends on the change in momentum, but also on how fast the change is brought about. The same change in momentum brought about in a shorter time needs a greater applied force. In short, the greater the rate of change of momentum, the greater is the force.
Observations confirm that the product of mass and velocity (i.e. momentum) is basic to the effect of force on motion. Suppose a fixed force is applied for a certain interval of time on two bodies of different masses, initially at rest, the lighter body picks up a greater speed than the heavier body. However, at the end of the time interval, observations show that each body acquires the same momentum. Thus the same force for the same time causes the same change in momentum for different bodies. This is a crucial clue to the second law of motion.
Fig. 4.3 Force not only depends on the change in momentum but also on how fast the change is brought about. A seasoned cricketer draws in his hands during a catch, allowing greater time for the ball to stop and hence requires a smaller force. | 4 | 11 | Physics | 104 |
635e5c13-5026-4480-a7d0-63fe072b0c0b | # PHYSICS
The character of momentum has not been evident. F ∝ Δp or F = k Δp where k is a constant of proportionality. Taking the limit Δt → 0, the term Δp becomes the derivative or differential co-efficient of p with respect to t, denoted by dp.
Thus, F = k dp dt (4.2)
For a body of fixed mass m,
dp = d(mv) = m dv = m a (4.3)
i.e the Second Law can also be written as F = k m a (4.4) which shows that force is proportional to the product of mass m and acceleration a.
The unit of force has not been defined so far. In fact, we use Eq. (4.4) to define the unit of force. We, therefore, have the liberty to choose any constant value for k. For simplicity, we choose k = 1. The second law then is F = dp = m a (4.5)
In SI unit force is one that causes an acceleration of 1 m/s2 to a mass of 1 kg. This unit is known as newton: 1 N = 1 kg m/s2.
Fig. 4.4 Force is necessary for changing the direction of momentum, even if its magnitude is constant. We can feel this while rotating a stone in a horizontal circle with uniform speed by means of a string.
Let us note at this stage some important points about the second law:
1. In the second law, F = 0 implies a = 0. The second law is obviously consistent with the first law.
2. The second law of motion is a vector law. It is equivalent to three equations, one for each component of the vectors:
Fx = dpx = max dt
Fy = dpy = may dt
Fz = dpz = m az (4.6) dt
This means that if a force is not parallel to the velocity of the body, but makes some angle with it, it changes only the component of velocity along the direction of force. | 5 | 11 | Physics | 104 |
3fb33dd0-cf4c-4e84-a635-dd8f2bf3add0 | # LAWS OF MOTION
component of velocity normal to the force
Answer The retardation ‘a’ of the bullet (assumed constant) is given by
a = – u² / 2s = – 90 × 90 / (2 × 0.6) = – 6750 m s−2
The retarding force, by the second law of motion, is
F = 0.04 kg × 6750 m s−2 = 270 N
The actual resistive force, and therefore, retardation of the bullet may not be uniform. The answer therefore, only indicates the average resistive force.
Example 4.3 The motion of a particle of mass m is described by y = ut + (1/2)gt². Find the force acting on the particle.
Answer We know
y = ut + (1/2)gt²
Now,
v = dy/dt = u + gt
acceleration, a = dv/dt = g
Then the force is given by Eq. (4.5)
F = ma = mg
Thus the given equation describes the motion of a particle under acceleration due to gravity and y is the position coordinate in the direction of g.
# Impulse
We sometimes encounter examples where a large force acts for a very short duration producing a finite change in momentum of the body. For example, when a ball hits a wall and bounces back, the force on the ball by the wall acts for a very short time when the two are in contact, yet the force is large enough to reverse the momentum of the ball. Often, in these situations, the force and the time duration are difficult to ascertain separately. However, the product of force and time, which is the change in momentum of the body remains a measurable quantity. This product is called impulse:
Impulse = Force × time duration = Change in momentum (4.7) | 6 | 11 | Physics | 104 |
08161de0-9aa3-49e3-b5a1-5720c49ccf84 | # 56 PHYSICS
A large force acting for a short time to produce a finite change in momentum is called an impulsive force. In the history of science, impulsive forces were put in a conceptually different category from ordinary forces. Newtonian mechanics has no such distinction. Impulsive force is like any other force – except that it is large and acts for a short time.
Example 4.4 A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of 12 m s–1. If the mass of the ball is 0.15 kg, determine the impulse imparted to the ball. (Assume linear motion of the ball)
Answer
Change in momentum = 0.15 × 12 – (–0.15 × 12) = 3.6 N s,
Impulse = 3.6 N s, in the direction from the batsman to the bowler.
This is an example where the force on the ball by the batsman and the time of contact of the ball and the bat are difficult to know, but the impulse is readily calculated.
# 4.6 NEWTON’S THIRD LAW OF MOTION
The second law relates the external force on a body to its acceleration. What is the origin of the external force on the body? What agency provides the external force? The simple answer in Newtonian mechanics is that the external force on a body always arises due to some other body. Consider a pair of bodies A and B. B gives rise to an external force on A. A natural question is: Does A in turn give rise to an external force on B? In some examples, the answer seems clear. If you press a coiled spring, the spring is compressed by the force of your hand. The compressed spring in turn exerts a force on your hand and you can feel it. But what if the bodies are not in contact? The earth pulls a stone downwards due to gravity. Does the stone exert a force on the earth? The answer is not obvious since we hardly see the effect of the stone on the earth. The answer according to Newton is: Yes, the stone does exert an equal and opposite force on the earth. We do not notice it since the earth is very massive and the effect of a small force on its motion is negligible.
Thus if we are considering the motion of any one body (A or B), only one of the two forces is relevant. It is an error to add up the two forces and claim that the net force is zero. However, if you are considering the system of two bodies as a whole, FAB and FBA are internal forces of the system (A + B). They add up to give a null force. Internal forces in a body or a system of particles thus cancel away. | 7 | 11 | Physics | 104 |
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in pairs. This is an important fact that enables the second law to be applicable to a body or a system of particles (See Chapter 6).
# Example 4.5
Two identical billiard balls strike a rigid wall with the same speed but at different angles, and get reflected without any change in speed, as shown in Fig. 4.6. What is (i) the direction of the force on the wall due to each ball? (ii) the ratio of the magnitudes of impulses imparted to the balls by the wall?
Case (b)
(ₚ) = m u cos 30, (p) = − m u sin 30
x initial y initial
(ₚ) = – m u cos 30, (p) = − m u sin 30
x final y final
Note, while pₓ changes sign after collision, py does not. Therefore,
x-component of impulse = –2 m u cos 30°
y-component of impulse = 0
The direction of impulse (and force) is the same as in (a) and is normal to the wall along the negative x direction. As before, using Newton’s third law, the force on the wall due to the ball is normal to the wall along the positive x direction.
The ratio of the magnitudes of the impulses imparted to the balls in (a) and (b) is
Answer An instinctive answer to (i) might be that the force on the wall in case (a) is normal to the wall, while that in case (b) is inclined at 30°. This answer is wrong. The force on the wall is normal to the wall in both cases.
How to find the force on the wall? The trick is to consider the force (or impulse) on the ball due to the wall using the second law, and then use the third law to answer (i). Let u be the speed of each ball before and after collision with the wall, and m the mass of each ball. Choose the x and y axes as shown in the figure, and consider the change in momentum of the ball in each case:
Case (a)
(pₓ)initial = mu
py initial = 0
(pₓ) final = −mu
py final = 0
Impulse is the change in momentum vector. Therefore,
x-component of impulse = – 2 m u
y-component of impulse = 0
Impulse and force are in the same direction. Clearly, from above, the force on the ball due to the wall is normal to the wall, along the negative x-direction. Using Newton’s third law of motion, this fact is known as the law of conservation of momentum:
In an isolated system (i.e. a system with no external force), mutual forces between pairs of particles in the system can cause momentum change in individual particles, but since the mutual forces for each pair are equal and opposite, the momentum changes cancel in pairs and the total momentum remains unchanged. | 8 | 11 | Physics | 104 |
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The total momentum of an isolated system of interacting particles is conserved.
An important example of the application of the law of conservation of momentum is the collision of two bodies. Consider two bodies A and B, with initial momenta pA and pB. The bodies collide, get apart, with final momenta pA' and pB' respectively. By the Second Law
FAB ∆t = pA' − pA and FBA ∆t = pB' − pB (where we have taken a common interval of time for both forces i.e. the time for which the two bodies are in contact.)
Since FAB = −FBA by the third law,
pA' − pA = −(pB' − pB)
i.e. pA' + pB' = pA + pB (4.9) which shows that the total final momentum of the isolated system equals its initial momentum.
Notice that this is true whether the collision is elastic or inelastic. In elastic collisions, there is a second condition that the total initial kinetic energy of the system equals the total final kinetic energy (See Chapter 5).
# 4.8 EQUILIBRIUM OF A PARTICLE
Equilibrium of a particle in mechanics refers to the situation when the net external force on the particle is zero. According to the first law, this means that the particle is either at rest or in uniform motion.
If two forces F1 and F2, act on a particle, equilibrium requires
F1 = − F2 (4.10) i.e. the two forces on the particle must be equal and opposite. Equilibrium under three concurrent forces F1, F2 and F3 requires that the vector sum of the three forces is zero.
F1 + F2 + F3 = 0 (4.11)
* Equilibrium of a body requires not only translational equilibrium (zero net external force) but also rotational equilibrium (zero net external torque), as we shall see in Chapter 6.
Example 4.6 See Fig. 4.8. A mass of 6 kg is suspended by a rope of length 2 m from the ceiling. A force of 50 N in the horizontal direction is applied at the midpoint P of the rope, as shown. What is the angle the rope makes with the vertical in equilibrium? (Take g = 10 m s−2). Neglect the mass of the rope.
(a) (b) (c)
Fig. 4.8 | 9 | 11 | Physics | 104 |
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Answer Figures 4.8(b) and 4.8(c) are known as free-body diagrams. Figure 4.8(b) is the free-body diagram of W and Fig. 4.8(c) is the free-body diagram of point P.
Consider the equilibrium of the weight W. Clearly, T2 = 6 × 10 = 60 N.
Consider the equilibrium of the point P under the action of three forces - the tensions T1 and T2, and the horizontal force 50 N. The horizontal and vertical components of the resultant force must vanish separately:
- T1 cos θ = T2 = 60 N
- T1 sin θ = 50 N
which gives that
Note the answer does not depend on the length of the rope (assumed massless) nor on the point at which the horizontal force is applied.
# 4.9 COMMON FORCES IN MECHANICS
In mechanics, we encounter several kinds of forces. The gravitational force is, of course, pervasive. Every object on the earth experiences the force of gravity due to the earth. Gravity also governs the motion of celestial bodies. The gravitational force can act at a distance without the need of any intervening medium.
All the other forces common in mechanics are contact forces. As the name suggests, a contact force on an object arises due to contact with some other object: solid or fluid. When bodies are in contact (e.g. a book resting on a table, a system of rigid bodies connected by rods, hinges and other types of supports), there are mutual contact forces (for each pair of bodies) satisfying the third law. The component of contact force normal to the surfaces in contact is called normal reaction. The component parallel to the surfaces in contact is called friction. Contact forces arise also when solids are in contact with fluids. For example, for a solid immersed in a fluid, there is an upward buoyant force equal to the weight of the fluid displaced. The viscous force, air resistance, etc. are also examples of contact forces (Fig. 4.9).
Two other common forces are tension in a string and the force due to spring. When a spring is compressed or extended by an external force, a restoring force is generated. This force is usually proportional to the compression or elongation (for small displacements). The spring force F is written as F = – k x where x is the displacement and k is the force constant. The negative sign denotes that the force is opposite to the displacement from the unstretched state. For an inextensible string, the force constant is very high. The restoring force in a string is called tension. It is customary to use a constant tension T throughout the string. This assumption is true for a string of negligible mass.
We learn that there are four fundamental forces in nature. Of these, the weak and strong forces appear in domains that do not concern us here. Only the gravitational and electrical forces are relevant in the context of mechanics. The different contact forces of mechanics mentioned above fundamentally arise from electrical forces. This may seem surprising.
Fig. 4.9 Some examples of contact forces in mechanics.
* We are not considering, for simplicity, charged and magnetic bodies. For these, besides gravity, there are electrical and magnetic non-contact forces. | 10 | 11 | Physics | 104 |
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since we are talking of uncharged and non-magnetic bodies in mechanics. A
at the microscopic level, all bodies are made of charged constituents (nuclei and electrons) and the various contact forces arising due to elasticity of bodies, molecular collisions and impacts, etc. can ultimately be traced to the electrical forces between the charged constituents of different bodies. The detailed microscopic origin of these forces is, however, complex and not useful for handling problems in mechanics at the macroscopic scale. This is why they are treated as different types of forces with their characteristic properties determined empirically.
We know from experience that as the applied force exceeds a certain limit, the body begins to move. It is found experimentally that the limiting value of static friction fₛ is independent of the area of contact and varies with the normal force (N) approximately as:
# 4.9.1 Friction
Let us return to the example of a body of mass m at rest on a horizontal table. The force of gravity (mg) is cancelled by the normal reaction force (N) of the table. Now suppose a force F is applied horizontally to the body. We know from experience that a small applied force may not be enough to move the body. But if the applied force F were the only external force on the body, it must move with acceleration F/m, however small. Clearly, the body remains at rest because some other force comes into play in the horizontal direction and opposes the applied force F, resulting in zero net force on the body. This force fₛ parallel to the surface of the body in contact with the table is known as frictional force, or simply friction. The subscript stands for static friction to distinguish it from kinetic friction fₖ that we consider later.
Note that static friction does not exist by itself. When there is no applied force, there is no static friction. It comes into play the moment there is an applied force. As the applied force F increases, fₛ also increases, remaining equal and opposite to the applied force (up to a certain limit), keeping the body at rest. Hence, it is called static friction. Static friction opposes impending motion. The term impending motion means motion that would take place (but does not actually take place) under the applied force, if friction were absent.
The law of static friction may thus be written as:
fₛ ≤ μₛ N (4.14)
where μₛ is a constant of proportionality depending only on the nature of the surfaces in contact. The constant μₛ is called the coefficient of static friction. If the applied force F exceeds fₛ max, the body begins to slide on the surface. It is found experimentally that when relative motion has started, the frictional force decreases from the static maximum value fₛ max. Frictional force that opposes relative motion between surfaces in contact is called kinetic or sliding friction and is denoted by fₖ. Kinetic friction, like static friction, is found to be independent of the area of contact. Further, it is nearly independent of the velocity. It satisfies a law similar to that for static friction:
fₖ = μₖ N (4.15)
where μₖ is the coefficient of kinetic friction, depends only on the surfaces in contact. As mentioned above, experiments show that μₖ is less than μₛ. When relative motion has begun, the acceleration of the body according to the second law is:
(F – fₖ)/m
For a body moving with constant velocity, F = fₖ. If the applied force on the body is removed, its acceleration is – fₖ/m and it eventually comes to a stop.
The laws of friction given above do not have the status of fundamental laws like those for gravitational, electric and magnetic forces. They are empirical relations that are only
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approximately true. Yet they are very useful in practical calculations in mechanics. Thus, when two bodies are in contact, each experiences a contact force by the other. Friction, by definition, is the component of the contact force fₛ opposing the impending motion. In equilibrium, the resultant of these forces must be zero. Resolving the weight mg along the two directions shown, we have:
Consider a box lying in the compartment of a train that is accelerating. If the box is stationary relative to the train, it is in fact accelerating along with the train. What forces cause the acceleration of the box? Clearly, the only conceivable force in the horizontal direction is the force of friction. If there were no friction, the floor of the train would slip by and the box would remain at its initial position due to inertia (and hit the back side of the train). This impending relative motion is opposed by the static friction fₛ. Static friction provides the same acceleration to the box as that of the train, keeping it stationary relative to the train.
When θ becomes just a little more than θₘₐₓ, there is a small net force on the block and it begins to slide. Note that θₘₐₓ depends only on μₛ and is independent of the mass of the block.
# Example 4.7
Determine the maximum acceleration of the train in which a box lying on its floor will remain stationary, given that the co-efficient of static friction between the box and the train’s floor is 0.15.
Answer: Since the acceleration of the box is due to the static friction,
ma = fₛ ≤ μₛ N = μₛ mg
i.e. a ≤ μₛ g
∴ a = μₛ g = 0.15 x 10 m s⁻² = 1.5 m s⁻²
# Example 4.8
See Fig. 4.11. A mass of 4 kg rests on a horizontal plane. The plane is gradually inclined until at an angle θ = 15° with the horizontal, the mass just begins to slide. What is the coefficient of static friction between the block and the surface?
# Example 4.9
What is the acceleration of the block and trolley system shown in Fig. 4.12(a), if the coefficient of kinetic friction between the trolley and the surface is 0.04? What is the tension in the string? (Take g = 10 m s⁻²). Neglect the mass of the string.
Answer: Since the acceleration of the box is due to the static friction,
ma = fₛ ≤ μₛ N = μₛ mg
i.e. a ≤ μₛ g
∴ a = μₛ g = 0.15 x 10 m s⁻² = 1.5 m s⁻²
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Answer: As the string is inextensible, and the pully is smooth, the 3 kg block and the 20 kg trolley both have same magnitude of acceleration. Applying second law to motion of the block (Fig. 4.12(b)),
30 – T = 3a
Apply the second law to motion of the trolley (Fig. 4.12(c)),
T – f = 20 a.
Now fk = μk N,
Here μk = 0.04,
N = 20 x 10 = 200 N.
Thus the equation for the motion of the trolley is T – 0.04 x 200 = 20 a or T – 8 = 20a.
These equations give a = 22 m s–2 = 0.96 m s–2 and T = 27.1 N.
# Rolling friction
A body like a ring or a sphere rolling without slipping over a horizontal plane will suffer no friction, in principle. At every instant, there is just one point of contact between the body and the plane and this point has no motion relative to the plane. In this ideal situation, kinetic or static friction is zero and the body should continue to roll with constant velocity. We know, in practice, this will not happen and some resistance to motion (rolling friction) does occur, i.e. to keep the body rolling, some applied force is needed. For the same weight, rolling friction is much smaller (even by 2 or 3 orders of magnitude) than static or sliding friction. This
Fig. 4.13 Some ways of reducing friction. (a) Ball bearings placed between moving parts of a machine. (b) Compressed cushion of air between surfaces in relative motion.
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# 4.10 CIRCULAR MOTION
We have seen in Chapter 4 that acceleration of a body moving in a circle of radius R with uniform speed v is v²/R directed towards the centre. According to the second law, the force f providing this acceleration is:
fc = mv²
where m is the mass of the body. This force directed forwards the centre is called the centripetal force. For a stone rotated in a circle by a string, the centripetal force is provided by the tension in the string. The centripetal force for motion of a planet around the sun is the gravitational force on the planet due to the sun.
For a car taking a circular turn on a horizontal road, the centripetal force is the force of friction.
The circular motion of a car on a flat and banked road gives interesting application of the laws of motion.
# Motion of a car on a level road
Three forces act on the car:
1. The weight of the car, mg
2. Normal reaction, N
3. Frictional force, f
As there is no acceleration in the vertical direction:
N - mg = 0
N = mg
The centripetal force required for circular motion is along the surface of the road, and is provided by the component of the contact force between road and the car tyres along the surface. This by definition is the frictional force. Note that it:
f ≤ µsN
Thus to obtain vmax we put:
f = µsN
Then Eqs. (4.19a) and (4.19b) become:
N cos θ = mg + µsN sin θ | 14 | 11 | Physics | 104 |
d3279162-3de7-4dbf-aa4e-0f8686e37313 | N sin θ + μ N cos θ = mv²/R (4.20b)
From Eq. (4.20a), we obtain
N = mg cos θ – μₛ sin θ
Substituting value of N in Eq. (4.20b), we get
(mg sin θ + μₛ cos θ) = mv²/R
or vₘₐₓ = √(Rg μₛ + tan θ)² / (1 – μₛ tan θ) (4.21)
Comparing this with Eq. (4.18) we see that maximum possible speed of a car on a banked road is greater than that on a flat road.
For μₛ = 0 in Eq. (4.21), vₒ = (Rg tan θ)½ (4.22)
At this speed, frictional force is not needed at all to provide the necessary centripetal force. Driving at this speed on a banked road will cause little wear and tear of the tyres. The same equation also tells you that for v < vₒ, frictional force will be up the slope and that a car can be parked only if tan θ ≤ μₛ.
# Example 4.10
A cyclist speeding at 18 km/h on a level road takes a sharp circular turn of radius 3 m without reducing the speed. The coefficient of static friction between the tyres and the road is 0.1. Will the cyclist slip while taking the turn?
# Answer
On an unbanked road, frictional force alone can provide the centripetal force needed to keep the cyclist moving on a circular turn without slipping. If the speed is too large, or if the turn is too sharp (i.e. of too small a radius) or both, the frictional force is not sufficient to provide the necessary centripetal force, and the cyclist slips. The condition for the cyclist not to slip is given by Eq. (4.18):
v² ≤ μₛ R g
Now, R = 3 m, g = 9.8 m/s², μ = 0.1. That is, μ R g = 2.94 m/s². v = 18 km/h = 5 m/s; i.e., v² = 25 m²/s². The condition is not obeyed. The cyclist will slip while taking the circular turn.
# Example 4.11
A circular racetrack of radius 300 m is banked at an angle of 15°. If the coefficient of friction between the wheels of a race-car and the road is 0.2, what is the (a) optimum speed of the race-car to avoid wear and tear on its tyres, and (b) maximum permissible speed to avoid slipping?
# Answer
On a banked road, the horizontal component of the normal force and the frictional force contribute to provide centripetal force to keep the car moving on a circular turn without slipping. At the optimum speed, the normal reaction’s component is enough to provide the needed centripetal force, and the frictional force is not needed. The optimum speed vₒ is given by Eq. (4.22):
vₒ = (R g tan θ)½
Here R = 300 m, θ = 15°, g = 9.8 m/s²; we have vₒ = 28.1 m/s.
The maximum permissible speed vₘₐₓ is given by Eq. (4.21):
# 4.11 SOLVING PROBLEMS IN MECHANICS
The three laws of motion that you have learnt in this chapter are the foundation of mechanics. You should now be able to handle a large variety of problems in mechanics. A typical problem in mechanics usually does not merely involve a single body under the action of given forces. More often, we will need to consider an assembly of different bodies exerting forces on each other. Besides, each body in the assembly experiences the force of gravity. When trying to solve a problem of this type, it is useful to remember the fact that we can choose any part of the assembly and apply the laws of motion to that part provided we include all forces on the chosen part due to the remaining parts of the assembly. We may call the chosen part of the assembly as the system and the remaining part of the assembly (plus any other agencies of forces) as the environment. We have followed the same | 15 | 11 | Physics | 104 |
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method in solved examples. To handle a typical problem in mechanics systematically, one should use the following steps:
1. Draw a diagram showing schematically the various parts of the assembly of bodies, the links, supports, etc.
2. Choose a convenient part of the assembly as one system.
3. Draw a separate diagram which shows this system and all the forces on the system by the remaining part of the assembly. Include also the forces on the system by other agencies. Do not include the forces on the environment by the system. A diagram of this type is known as ‘a free-body diagram’. (Note this does not imply that the system under consideration is without a net force).
4. In a free-body diagram, include information about forces (their magnitudes and directions) that are either given or you are sure of (e.g., the direction of tension in a string along its length). The rest should be treated as unknowns to be determined using laws of motion.
5. If necessary, follow the same procedure for another choice of the system. In doing so, employ Newton’s third law. That is, if in the free-body diagram of A, the force on A due to B is shown as F, then in the free-body diagram of B, the force on B due to A should be shown as –F.
The following example illustrates the above procedure:
# Example 4.12
See Fig. 4.15. A wooden block of mass 2 kg rests on a soft horizontal floor. When an iron cylinder of mass 25 kg is placed on top of the block, the floor yields steadily and the block and the cylinder together go down with an acceleration of 0.1 m s–2. What is the action of the block on the floor (a) before and (b) after the floor yields? Take g = 10 m s–2. Identify the action-reaction pairs in the problem.
# Answer
(a) The block is at rest on the floor. Its free-body diagram shows two forces on the block, the force of gravitational attraction by the earth equal to 2 × 10 = 20 N; and the normal force R of the floor on the block. By the First Law, the net force on the block must be zero i.e., R = 20 N. Using third law the action of the block (i.e. the force exerted on the floor by the block) is equal to 20 N and directed vertically downwards.
(b) The system (block + cylinder) accelerates downwards with 0.1 m s–2. The free-body diagram of the system shows two forces on the system: the force of gravity due to the earth (270 N); and the normal force R' by the floor. Note, the free-body diagram of the system does not show the internal forces between the block and the cylinder. Applying the second law to the system, 270 – R' = 27 × 0.1 N, i.e. R' = 267.3 N.
By the third law, the action of the system on the floor is equal to 267.3 N vertically downward.
# Action-reaction pairs
For (a): (i) the force of gravity (20 N) on the block by the earth (say, action); the force of gravity on the earth by the block (reaction) equal to 20 N directed upwards (not shown in the figure).
(ii) the force on the floor by the block (action); the force on the block by the floor (reaction).
For (b): (i) the force of gravity (270 N) on the system by the earth (say, action); the force of gravity on the earth by the system (reaction), equal to 270 N. | 16 | 11 | Physics | 104 |
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directed upwards (not shown in the figure). (ii) the force on the floor by the system (action); the force on the system by the floor (reaction). In addition, for (b), the force on the block by the cylinder and the force on the cylinder by the block also constitute an action-reaction pair. The important thing to remember is that an action-reaction pair consists of mutual forces which are always equal and opposite between two bodies. Two forces on the same body which happen to be equal and opposite can never constitute an action-reaction pair. The force of gravity on the mass in (a) or (b) and the normal force on the mass by the floor are not action-reaction pairs. These forces happen to be equal and opposite for (a) since the mass is at rest. They are not so for case (b), as seen already. The weight of the system is 270 N, while the normal force R′ is 267.3 N.
The practice of drawing free-body diagrams is of great help in solving problems in mechanics. It allows you to clearly define your system and consider all forces on the system due to objects that are not part of the system itself. A number of exercises in this and subsequent chapters will help you cultivate this practice.
# SUMMARY
1. Aristotle’s view that a force is necessary to keep a body in uniform motion is wrong. A force is necessary in practice to counter the opposing force of friction.
2. Galileo extrapolated simple observations on motion of bodies on inclined planes, and arrived at the law of inertia. Newton’s first law of motion is the same law rephrased thus: “Everybody continues to be in its state of rest or of uniform motion in a straight line, unless compelled by some external force to act otherwise”. In simple terms, the First Law is “If external force on a body is zero, its acceleration is zero”.
3. Momentum (p) of a body is the product of its mass (m) and velocity (v):
p = mv
4. Newton’s second law of motion:
The rate of change of momentum of a body is proportional to the applied force and takes place in the direction in which the force acts.
Thus
F = k dp/dt = k ma
where F is the net external force on the body and a its acceleration. We set the constant of proportionality k = 1 in SI units. Then
F = dp/dt = ma
The SI unit of force is newton: 1 N = 1 kg m s⁻².
(a) The second law is consistent with the First Law (F = 0 implies a = 0)
(b) It is a vector equation
(c) It is applicable to a particle, and also to a body or a system of particles, provided F is the total external force on the system and a is the acceleration of the system as a whole.
(d) F at a point at a certain instant determines a at the same point at that instant. That is the Second Law is a local law; a at an instant does not depend on the history of motion.
5. Impulse is the product of force and time which equals change in momentum. The notion of impulse is useful when a large force acts for a short time to produce a measurable change in momentum. Since the time of action of the force is very short, one can assume that there is no appreciable change in the position of the body during the action of the impulsive force.
6. Newton’s third law of motion:
To every action, there is always an equal and opposite reaction. | 17 | 11 | Physics | 104 |
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In simple terms, the law can be stated thus: Forces in nature always occur between pairs of bodies. Force on a body A by body B is equal and opposite to the force on the body B by A. Action and reaction forces are simultaneous forces. There is no cause-effect relation between action and reaction. Any of the two mutual forces can be called action and the other reaction. Action and reaction act on different bodies and so they cannot be cancelled out. The internal action and reaction forces between different parts of a body do, however, sum to zero.
# 7. Law of Conservation of Momentum
The total momentum of an isolated system of particles is conserved. The law follows from the second and third law of motion.
# 8. Friction
Frictional force opposes (impending or actual) relative motion between two surfaces in contact. It is the component of the contact force along the common tangent to the surface in contact. Static friction fₛ opposes impending relative motion; kinetic friction fₖ opposes actual relative motion. They are independent of the area of contact and satisfy the following approximate laws:
fₛ ≤ (fₛ)max = μsRfₖ = μkR
μs (co-efficient of static friction) and μk (co-efficient of kinetic friction) are constants characteristic of the pair of surfaces in contact. It is found experimentally that μk is less than μs.
# POINTS TO PONDER
1. Force is not always in the direction of motion. Depending on the situation, F may be along v, opposite to v, normal to v or may make some other angle with v. In every case, it is parallel to acceleration.
2. If v = 0 at an instant, i.e. if a body is momentarily at rest, it does not mean that force or acceleration are necessarily zero at that instant. For example, when a ball thrown upward reaches its maximum height, v = 0 but the force continues to be its weight mg and the acceleration is not zero but g.
3. Force on a body at a given time is determined by the situation at the location of the body at that time. Force is not ‘carried’ by the body from its earlier history of motion. The moment after a stone is released out of an accelerated train, there is no horizontal force (or acceleration) on the stone, if the effects of the surrounding air are neglected. The stone then has only the vertical force of gravity.
4. In the second law of motion F = ma, F stands for the net force due to all material agencies external to the body. a is the effect of the force. ma should not be regarded as yet another force, besides F. | 18 | 11 | Physics | 104 |
9185ea35-1b9a-4113-8238-34b524174d26 | # PHYSICS
5. The centripetal force should not be regarded as yet another kind of force. It is simply a name given to the force that provides inward radial acceleration to a body in circular motion. We should always look for some material force like tension, gravitational force, electrical force, friction, etc as the centripetal force in any circular motion.
6. Static friction is a self-adjusting force up to its limit μₛ N (fₛ ≤ μₛ N). Do not put fₛ = μₛ N without being sure that the maximum value of static friction is coming into play.
7. The familiar equation mg = R for a body on a table is true only if the body is in equilibrium. The two forces mg and R can be different (e.g. a body in an accelerated lift). The equality of mg and R has no connection with the third law.
8. The terms ‘action’ and ‘reaction’ in the third Law of Motion simply stand for simultaneous mutual forces between a pair of bodies. Unlike their meaning in ordinary language, action does not precede or cause reaction. Action and reaction act on different bodies.
9. The different terms like ‘friction’, ‘normal reaction’, ‘tension’, ‘air resistance’, ‘viscous drag’, ‘thrust’, ‘buoyancy’, ‘weight’, ‘centripetal force’ all stand for ‘force’ in different contexts. For clarity, every force and its equivalent terms encountered in mechanics should be reduced to the phrase ‘force on A by B’.
10. For applying the second law of motion, there is no conceptual distinction between inanimate and animate objects. An animate object such as a human also requires an external force to accelerate. For example, without the external force of friction, we cannot walk on the ground.
11. The objective concept of force in physics should not be confused with the subjective concept of the ‘feeling of force’. On a merry-go-around, all parts of our body are subject to an inward force, but we have a feeling of being pushed outward – the direction of impending motion.
# EXERCISES
(For simplicity in numerical calculations, take g = 10 m s⁻²)
# 4.1
Give the magnitude and direction of the net force acting on
- (a) a drop of rain falling down with a constant speed,
- (b) a cork of mass 10 g floating on water,
- (c) a kite skillfully held stationary in the sky,
- (d) a car moving with a constant velocity of 30 km/h on a rough road,
- (e) a high-speed electron in space far from all material objects, and free of electric and magnetic fields.
# 4.2
A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble,
- (a) during its upward motion,
- (b) during its downward motion,
- (c) at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of 45° with the horizontal direction?
Ignore air resistance.
# 4.3
Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg,
- (a) just after it is dropped from the window of a stationary train,
- (b) just after it is dropped from the window of a train running at a constant velocity of 36 km/h,
- (c) just after it is dropped from the window of a train accelerating with 1 m s⁻²,
- (d) lying on the floor of a train which is accelerating with 1 m s⁻², the stone being at rest relative to the train.
Neglect air resistance throughout.
2024-25 | 19 | 11 | Physics | 104 |
0bd76143-ab2b-4b28-9f9c-e70b2d3469dd | # LAWS OF MOTION
# 4.4
One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle (directed towards the centre) is :
- (i) T
- (ii) T − mv2
- (iii) T + mv2
- (iv) 0
T is the tension in the string. [Choose the correct alternative].
# 4.5
A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 m s−1. How long does the body take to stop?
# 4.6
A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s−1 to 3.5 m s−1 in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?
# 4.7
A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body.
# 4.8
The driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. What is the average retarding force on the vehicle? The mass of the three-wheeler is 400 kg and the mass of the driver is 65 kg.
# 4.9
A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 m s−2. Calculate the initial thrust (force) of the blast.
# 4.10
A body of mass 0.40 kg moving initially with a constant speed of 10 m s−1 to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the body at that time to be x = 0, and predict its position at t = –5 s, 25 s, 100 s.
# 4.11
A truck starts from rest and accelerates uniformly at 2.0 m s−2. At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at t = 11 s? (Neglect air resistance.)
# 4.12
A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 m s−1. What is the trajectory of the bob if the string is cut when the bob is (a) at one of its extreme positions, (b) at its mean position.
# 4.13
A man of mass 70 kg stands on a weighing scale in a lift which is moving (a) upwards with a uniform speed of 10 m s−1, (b) downwards with a uniform acceleration of 5 m s−2, (c) upwards with a uniform acceleration of 5 m s−2. What would be the readings on the scale in each case?
(d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?
# 4.14
Figure 4.16 shows the position-time graph of a particle of mass 4 kg. What is the (a) force on the particle for t < 0, t > 4 s, 0 < t < 4 s? (b) impulse at t = 0 and t = 4 s? (Consider one-dimensional motion only).
# 4.15
Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force F = 600 N is applied to (i) A, (ii) B along the direction of string. What is the tension in the string in each case? | 20 | 11 | Physics | 104 |
5391e47f-bdad-4a6c-b14c-2c8b75c6fa20 | # PHYSICS
# 4.16
Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.
# 4.17
A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.
# 4.18
Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 m s⁻¹ collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?
# 4.19
A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 m s⁻¹, what is the recoil speed of the gun?
# 4.20
A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg.)
# 4.21
A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?
# 4.22
If, in Exercise 4.21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks:
- (a) the stone moves radially outwards,
- (b) the stone flies off tangentially from the instant the string breaks,
- (c) the stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle?
# 4.23
Explain why:
- (a) a horse cannot pull a cart and run in empty space,
- (b) passengers are thrown forward from their seats when a speeding bus stops suddenly,
- (c) it is easier to pull a lawn mower than to push it,
- (d) a cricketer moves his hands backwards while holding a catch. | 21 | 11 | Physics | 104 |
e9afefcf-7c71-4a4e-b8a3-5f59f7c779ed | # CHAPTER FOURTEEN
# WAVES
# 14.1 INTRODUCTION
In the previous Chapter, we studied the motion of objects oscillating in isolation. What happens in a system, which is a collection of such objects? A material medium provides such an example. Here, elastic forces bind the constituents to each other and, therefore, the motion of one affects that of the other. If you drop a little pebble in a pond of still water, the water surface gets disturbed. The disturbance does not remain confined to one place, but propagates outward along a circle. If you continue dropping pebbles in the pond, you see circles rapidly moving outward from the point where the water surface is disturbed. It gives a feeling as if the water is moving outward from the point of disturbance. If you put some cork pieces on the disturbed surface, it is seen that the cork pieces move up and down but do not move away from the centre of disturbance. This shows that the water mass does not flow outward with the circles, but rather a moving disturbance is created. Similarly, when we speak, the sound moves outward from us, without any flow of air from one part of the medium to another. The disturbances produced in air are much less obvious and only our ears or a microphone can detect them. These patterns, which move without the actual physical transfer or flow of matter as a whole, are called waves. In this Chapter, we will study such waves.
Waves transport energy and the pattern of disturbance has information that propagate from one point to another. All our communications essentially depend on transmission of signals through waves. Speech means production of sound waves in air and hearing amounts to their detection. Often, communication involves different kinds of waves. For example, sound waves may be first converted into an electric current signal which in turn may generate an electromagnetic wave that may be transmitted by an optical cable or via a | 0 | 11 | Physics | 207 |
f38969d0-fbde-4ed4-a7fa-b20469daa7da | # WAVES
Detection of the original signal will usually involve these steps in reverse order. We shall illustrate this connection through simple examples.
Not all waves require a medium for their propagation. We know that light waves can travel through vacuum. The light emitted by stars, which are hundreds of light years away, reaches us through inter-stellar space, which is practically a vacuum.
The most familiar type of waves such as waves on a string, water waves, sound waves, seismic waves, etc. is the so-called mechanical waves. These waves require a medium for propagation, they cannot propagate through vacuum. They involve oscillations of constituent particles and depend on the elastic properties of the medium.
The electromagnetic waves that you will learn in Class XII are a different type of wave. Electromagnetic waves do not necessarily require a medium - they can travel through vacuum. Light, radiowaves, X-rays, are all electromagnetic waves. In vacuum, all electromagnetic waves have the same speed c, whose value is:
c = 299, 792, 458 ms–1. (14.1)
A third kind of wave is the so-called Matter waves. They are associated with constituents of matter: electrons, protons, neutrons, atoms and molecules. They arise in quantum mechanical description of nature that you will learn in your later studies. Though conceptually more abstract than mechanical or electro-magnetic waves, they have already found applications in several devices basic to modern technology; matter waves associated with electrons are employed in electron microscopes.
In this chapter we will study mechanical waves, which require a material medium for their propagation.
The aesthetic influence of waves on art and literature is seen from very early times; yet the first scientific analysis of wave motion dates back to the seventeenth century. Some of the famous scientists associated with the physics of wave motion are Christiaan Huygens (1629-1695), Robert Hooke and Isaac Newton. The understanding of physics of waves followed the physics of oscillations of masses tied to springs and physics of the simple pendulum. Waves in elastic media are intimately connected with harmonic oscillations. (Stretched strings, coiled springs, air, etc., are examples of elastic media). | 1 | 11 | Physics | 207 |
41cb2782-ad4c-4450-beea-c59605797934 | # PHYSICS
In solids, similar arguments can be made. In a crystalline solid, atoms or group of atoms are arranged in a periodic lattice. In these, each atom or group of atoms is in equilibrium, due to forces from the surrounding atoms. Displacing one atom, keeping the others fixed, leads to restoring forces, exactly as in a spring. So we can think of atoms in a lattice as end points, with springs between pairs of them.
In the subsequent sections of this chapter we are going to discuss various characteristic properties of waves.
# 14.2 TRANSVERSE AND LONGITUDINAL WAVES
We have seen that motion of mechanical waves involves oscillations of constituents of the medium. If the constituents of the medium oscillate perpendicular to the direction of wave propagation, we call the wave a transverse wave. If they oscillate along the direction of wave propagation, we call the wave a longitudinal wave.
Fig. 14.2 shows the propagation of a single pulse along a string, resulting from a single up and down jerk. If the string is very long compared to the size of the pulse, the pulse will damp out before it reaches the other end and reflection from that end may be ignored. Fig. 14.3 shows a similar situation, but this time the external agent gives a continuous periodic sinusoidal up and down jerk to one end of the string. The resulting disturbance on the string is then a sinusoidal wave. In either case the elements of the string oscillate about their equilibrium mean position as the pulse or wave passes through them. The oscillations are normal to the direction of wave motion along the string, so this is an example of transverse wave.
We can look at a wave in two ways. We can fix an instant of time and picture the wave in space. This will give us the shape of the wave as a whole in space at a given instant. Another way is to fix a location i.e. fix our attention on a particular element of string and see its oscillatory motion in time.
Fig. 14.4 describes the situation for longitudinal waves in the most familiar example of the propagation of sound waves. A long pipe filled with air has a piston at one end. A single sudden push forward and pull back of the piston will generate a pulse of condensations (higher density) and rarefactions (lower density) in the medium (air). If the push-pull of the piston is continuous and periodic (sinusoidal), a volume element of air oscillates in the direction parallel to the direction of wave propagation. | 2 | 11 | Physics | 207 |
547e3ae1-4a8e-4d73-a75b-457ef0dd63c4 | # WAVES
A sinusoidal wave will be generated propagating in air along the length of the pipe. This is clearly an example of longitudinal waves.
Given below are some examples of wave motion. State in each case if the wave motion is transverse, longitudinal or a combination of both:
- (a) Motion of a kink in a longitudinal spring produced by displacing one end of the spring sideways.
- (b) Waves produced in a cylinder containing a liquid by moving its piston back and forth.
- (c) Waves produced by a motorboat sailing in water.
- (d) Ultrasonic waves in air produced by a vibrating quartz crystal.
# Answer
- (a) Transverse and longitudinal
- (b) Longitudinal
- (c) Transverse and longitudinal
- (d) Longitudinal
# 14.3 DISPLACEMENT RELATION IN A PROGRESSIVE WAVE
For mathematical description of a travelling wave, we need a function of both position x and time t. Such a function at every instant should give the shape of the wave at that instant. Also, at every given location, it should describe the motion of the constituent of the medium at that location. If we wish to describe a sinusoidal travelling wave (such as the one shown in Fig. 14.3), the corresponding function must also be sinusoidal. For convenience, we shall take the wave to be transverse so that if the position of the constituents of the medium is denoted by x, the displacement from the equilibrium position may be denoted by y. A sinusoidal travelling wave is then described by:
y(x, t) = a sin(kx − ωt + φ) (14.2)
The term φ in the argument of sine function means equivalently that we are considering a linear combination of sine and cosine functions:
y(x, t) = A sin(kx − ωt) + B cos(kx − ωt) (14.3)
From Equations (14.2) and (14.3),
a = √(A² + B²) and φ = tan⁻¹(B/A)
To understand why Equation (14.2) represents a sinusoidal travelling wave, take a fixed instant, say t = t₀. Then, the argument of the sine function in Equation (14.2) is simply... | 3 | 11 | Physics | 207 |
e5244d29-d334-453d-8fac-ecddab186aed | # PHYSICS
kx + constant. Thus, the shape of the wave (at any fixed instant) as a function of x is a sine wave. Similarly, take a fixed location, say x = x₀. Then, the argument of the sine function in Equation (14.2) is constant -ωt. The displacement y, at a fixed location, thus, varies sinusoidally with time. That is, the constituents of the medium at different positions execute simple harmonic motion. Finally, as t increases, x must increase in the positive direction to keep kx – ωt + φ constant. Thus, Eq. (14.2) represents a sinusoidal (harmonic) wave travelling along the positive direction of the x-axis. On the other hand, a function
y(x,t) = a sin(kx + ωt + φ) (14.4) represents a wave travelling in the negative direction of x-axis. Fig. (14.5) gives the names of the various physical quantities appearing in Eq. (14.2) that we now interpret.
|y(x,t)|displacement as a function of position x and time t|
|---|---|
|a|amplitude of a wave|
|ω|angular frequency of the wave|
|k|angular wave number|
|kx–ωt+φ|initial phase angle (a+x = 0, t = 0)|
Fig. 14.5 The meaning of standard symbols in Eq. (14.2)
Using the plots of Fig. 14.6, we now define the various quantities of Eq. (14.2).
# 14.3.1 Amplitude and Phase
In Eq. (14.2), since the sine function varies between 1 and –1, the displacement y (x,t) varies between a and –a. We can take a to be a positive constant, without any loss of generality. Then, the trough is the point of maximum negative displacement, a represents the maximum displacement of the constituents of the medium from their equilibrium position. Note that the displacement y may be positive or negative, but a is positive. It is called the amplitude of the wave.
The quantity (kx – ωt + φ) appearing as the argument of the sine function in Eq. (14.2) is called the phase of the wave. Given the amplitude a, the phase determines the displacement of the wave at any position and at any instant. Clearly φ is the phase at x = 0 and t = 0. Hence, φ is called the initial phase angle. By suitable choice of origin on the x-axis and the initial time, it is possible to have φ = 0. Thus there is no loss of generality in dropping φ, i.e., in taking Eq. (14.2) with φ = 0. | 4 | 11 | Physics | 207 |
940511e2-6b15-47d3-bb95-fe99e0730050 | # 14.3.2 Wavelength and Angular Wave Number
The minimum distance between two points having the same phase is called the wavelength of the wave, usually denoted by λ. For simplicity, we can choose points of the same phase to be crests or troughs. The wavelength is then the distance between two consecutive crests or troughs in a wave. Taking φ = 0 in Eq. (14.2), the displacement at t = 0 is given by
y(x, 0) = a sin(kx) (14.5)
Since the sine function repeats its value after every 2π change in angle, the period of oscillation of the wave is the time it takes for an element to complete one full oscillation. That is
−a sin(ωt) = −a sin(ω(t + T)) = −a sin(ωt + ωT)
That is the displacements at points x and at x + 2nπ are the same, where n=1,2,3,... The least distance between points with the same displacement (at any given instant of time) is obtained by taking n = 1. λ is then given by
λ = 2π/k or k = 2π/λ (14.6)
k is the angular wave number or propagation constant; its SI unit is radian per metre or rad m−1.
# 14.3.3 Period, Angular Frequency and Frequency
In the discussion above, reference has always been made to a wave travelling along a string or a transverse wave. In a longitudinal wave, the displacement of an element of the medium is parallel to the direction of propagation of the wave. In Eq. (14.2), the displacement function for a longitudinal wave is written as,
s(x, t) = a sin(kx – ωt + φ) (14.9)
where s(x, t) is the displacement of an element of the medium in the direction of propagation of the wave at position x and time t. In Eq. (14.9), a is the displacement amplitude; other quantities have the same meaning as in case of a transverse wave except that the displacement function y(x, t) is to be replaced by the function s(x, t).
* Here again, ‘radian’ could be dropped and the units could be written merely as m−1. Thus, k represents 2π times the number of waves (or the total phase difference) that can be accommodated per unit length, with SI units m−1. | 5 | 11 | Physics | 207 |
3153d7ce-59bd-4ce2-bee9-e2d1a7cdb67e | # PHYSICS
A wave travelling along a string is described by,
y(x, t) = 0.005 sin (80.0 x – 3.0 t),
in which the numerical constants are in SI units (0.005 m, 80.0 rad m–1, and 3.0 rad s–1). Calculate (a) the amplitude, (b) the wavelength, and (c) the period and frequency of the wave. Also, calculate the displacement y of the wave at a distance x = 30.0 cm and time t = 20 s?
Answer On comparing this displacement equation with Eq. (14.2),
y (x, t ) = a sin (kx – ωt),
we find
- (a) the amplitude of the wave is 0.005 m = 5 mm.
- (b) the angular wave number k and angular frequency ω are
We, then, relate the wavelength λ to k through Eq. (14.6),
λ = 2π/k
= 2π
80.0 m–1
= 7.85 cm
(c) Now, we relate T to ω by the relation
T = 2π/ω
= 2π
3.0 s
= 2.09 s
and frequency, v = 1/T = 0.48 Hz
The displacement y at x = 30.0 cm and time t = 20 s is given by
y = (0.005 m) sin (80.0 × 0.3 – 3.0 ×20)
= (0.005 m) sin (–36 + 12π)
= (0.005 m) sin (1.699)
= (0.005 m) sin (97°) j 5 mm
# 14.4 THE SPEED OF A TRAVELLING WAVE
To determine the speed of propagation of a travelling wave, we can fix our attention on any particular point on the wave (characterised by some value of the phase) and see how that point moves in time. It is convenient to look at the motion of the crest of the wave. Fig. 14.8 gives
the speed of a mechanical wave is determined by the inertial (linear mass density for strings, mass density in general) and elastic properties (Young’s modulus for linear media/ shear modulus, bulk modulus) of the medium. The medium determines | 6 | 11 | Physics | 207 |
043aebbe-bd19-413b-ae49-3aa85e15941f | WAVES
the speed; Eq. (14.12) then relates wavelength to frequency for the given speed. Of course, as remarked earlier, the medium can support both transverse and longitudinal waves, which will have different speeds in the same medium. Later in this chapter, we shall obtain specific expressions for the speed of mechanical waves in some media.
# 14.4.1 Speed of a Transverse Wave on Stretched String
The speed of a mechanical wave is determined by the restoring force setup in the medium when it is disturbed and the inertial properties (mass density) of the medium. The speed is expected to be directly related to the former and inversely to the latter. For waves on a string, the restoring force is provided by the tension T in the string. The inertial property will in this case be linear mass density μ, which is mass m of the string divided by its length L. Using Newton’s Laws of Motion, an exact formula for the wave speed on a string can be derived, but this derivation is outside the scope of this book. We shall, therefore, use dimensional analysis. We already know that dimensional analysis alone can never yield the exact formula. The overall dimensionless constant is always left undetermined by dimensional analysis.
The dimension of μ is [ML-1] and that of T is like force, namely [MLT-2]. We need to combine these dimensions to get the dimension of speed v [LT-1]. Simple inspection shows that the quantity T/μ has the relevant dimension:
v = C T (14.13)
μ
where C is the undetermined constant of dimensional analysis. In the exact formula, it turns out, C=1. The speed of transverse waves on a stretched string is given by:
v = T (14.14)
μ
Note the important point that the speed v depends only on the properties of the medium T and μ (T is a property of the stretched string).
# 14.4.2 Speed of a Longitudinal Wave (Speed of Sound)
In a longitudinal wave, the constituents of the medium oscillate forward and backward in the direction of propagation of the wave. We have already seen that the sound waves travel in the form of compressions and rarefactions of small volume elements of air. The elastic property that determines the stress under compressional strain is the bulk modulus of the medium defined by (see Chapter 8):
B = - ∆P (14.16)
∆V/V
Here, the change in pressure ∆P produces a volumetric strain ∆V. B has the same dimension as pressure and is given in SI units in terms of pascal (Pa). The inertial property relevant for the propagation of wave is the mass density ρ, with dimensions [ML-3]. Simple inspection reveals that quantity B/ρ has the relevant dimension: | 7 | 11 | Physics | 207 |
2ff8665f-845d-448f-96fc-873ec113820e | # PHYSICS
Liquids and solids generally have higher speed of sound than gases.
[Note for solids, the speed being referred to is the speed of longitudinal waves in the solid].
This happens because they are much more difficult to compress than gases and so have much higher values of bulk modulus.
Thus, if B and ρ are considered to be the only relevant physical quantities,
v = C B (14.18)
ρ
where, as before, C is the undetermined constant from dimensional analysis. The exact derivation shows that C=1. Thus, the general formula for longitudinal waves in a medium is:
v = B (14.19)
ρ
For a linear medium, like a solid bar, the lateral expansion of the bar is negligible and we may consider it to be only under longitudinal strain. In that case, the relevant modulus of elasticity is Young’s modulus, which has the same dimension as the Bulk modulus. Dimensional analysis for this case is the same as before and yields a relation like Eq. (14.18) with an undetermined C, which the exact derivation shows to be unity. Thus, the speed of longitudinal waves in a solid bar is given by:
v = Y (14.20)
ρ
where Y is the Young’s modulus of the material of the bar.
# Table 14.1 Speed of Sound in some Media
|Medium|Speed of Sound (m/s)|
|---|---|
|Air|343|
|Water|1482|
|Steel|5960|
This relation was first given by Newton and is known as Newton’s formula.
# Example 14.4
Estimate the speed of sound in air at standard temperature and pressure. The mass of 1 mole of air is 29.0 ×10–3 kg.
Answer: We know that 1 mole of any gas occupies 22.4 litres at STP. Therefore, density of air at STP is:
ρₒ = (mass of one mole of air) / (volume of one mole of air at STP)
= 29.0 ×10–3 kg / 22.4 × 10–3 m³
= 1.29 kg m–3
2024-25 | 8 | 11 | Physics | 207 |
a179c22c-10be-42a8-bb69-022d147b77e4 | WAVES
According to Newton’s formula for the speed of sound in a medium, we get for the speed of sound in air at STP,
v = 280 m s–1 (14.23)
The result shown in Eq.(14.23) is about 15% smaller as compared to the experimental value of 331 m s–1 as given in Table 14.1. Where did we go wrong? If we examine the basic assumption made by Newton that the pressure variations in a medium during propagation of sound are isothermal, we find that this is not correct. It was pointed out by Laplace that the pressure variations in the propagation of sound waves are so fast that there is little time for the heat flow to maintain constant temperature. These variations, therefore, are adiabatic and not isothermal. For adiabatic processes the ideal gas satisfies the relation (see Section 11.8),
PVγ = constant
i.e. ∆(PVγ) = 0 or γ Pγ Vγ–1 ∆V + Vγ∆P = 0 where γ is the ratio of two specific heats, Cₚ/Cv.
Thus, for an ideal gas the adiabatic bulk modulus is given by,
Bad = − ∆P = γ ∆V/VP
The speed of sound is, therefore, from Eq. (14.19), given by,
v = √(γ P/ρ) (14.24)
This modification of Newton’s formula is referred to as the Laplace correction. For air γ = 7/5. Now using Eq. (14.24) to estimate the speed of sound in air at STP, we get a value 331.3 m s–1, which agrees with the measured speed.
To put the principle of superposition mathematically, let y₁(x,t) and y₂(x,t) be the displacements due to two wave disturbances in the medium. If the waves arrive in a region simultaneously, and therefore, overlap, the net displacement y(x,t) is given by
y(x, t) = y₁(x, t) + y₂(x, t) (14.25)
# 14.5 THE PRINCIPLE OF SUPERPOSITION OF WAVES
What happens when two wave pulses travelling in opposite directions cross each other (Fig. 14.9)? It turns out that wave pulses continue to retain their identities after they have crossed. However, during the time they overlap, the wave pattern is different from either of the individual waves. If we have two or more waves moving in the medium the resultant waveform is the sum of wave functions of individual waves. That is, if the wave functions of the moving waves are... | 9 | 11 | Physics | 207 |
47313231-0704-4acc-94d7-736a15052c22 | # PHYSICS
y1 = f1(x–vt),
y2 = f2(x–vt),
..........
..........
yn = fn(x–vt)
then the wave function describing the disturbance in the medium is
y = f1(x – vt) + f2(x – vt) + ... + fn(x – vt)
= ∑i=1n fi(x – vt) (14.26)
The principle of superposition is basic to the phenomenon of interference.
For simplicity, consider two harmonic travelling waves on a stretched string, both with the same ω (angular frequency) and k (wave number), and, therefore, the same wavelength λ. Their wave speed will be identical. Let us further assume that their amplitudes are equal and they are both travelling in the positive direction of x-axis. The waves only differ in their initial phase. According to Eq. (14.2), the two waves are described by the functions:
y1(x, t) = a sin(kx – ωt) (14.27)
y2(x, t) = a sin(kx – ωt + φ) (14.28)
The net displacement is then, by the principle of superposition, given by
y(x, t) = a sin(kx – ωt) + a sin(kx – ωt + φ) (14.29)
= a × 2 sin(½(kx – ωt) + ½(kx – ωt + φ)) cos(½φ) (14.30)
where we have used the familiar trigonometric identity for sin A + sin B. We then have
y(x, t) = 2a cos(½φ) sin(kx – ωt + ½φ) (14.31)
Eq. (14.31) is also a harmonic travelling wave in the positive direction of x-axis, with the same frequency and wavelength. However, its initial phase angle is φ. The significant thing is that its amplitude is a function of the phase difference φ.
# 14.6 REFLECTION OF WAVES
So far we considered waves propagating in an unbounded medium. What happens if a pulse or a wave meets a boundary? If the boundary is rigid, the pulse or wave gets reflected. | 10 | 11 | Physics | 207 |
1e0e8220-ad71-4a5a-a6c0-405bdec87cf2 | # WAVES
The phenomenon of echo is an example of reflection by a rigid boundary. If the boundary is not completely rigid or is an interface between two different elastic media, the situation is somewhat complicated. A part of the incident wave is reflected and a part is transmitted into the second medium. If a wave is incident obliquely on the boundary between two different media the transmitted wave is called the refracted wave. The incident and refracted waves obey Snell’s law of refraction, and the incident and reflected waves obey the usual laws of reflection.
Fig. 14.11 shows a pulse travelling along a stretched string and being reflected by the boundary. Assuming there is no absorption of energy by the boundary, the reflected wave has the same shape as the incident pulse but it suffers a phase change of π or 180° on reflection. This is because the boundary is rigid and the disturbance must have zero displacement at all times at the boundary. By the principle of superposition, this is possible only if the reflected and incident waves differ by a phase of π, so that the resultant displacement is zero. This reasoning is based on boundary condition on a rigid wall. We can arrive at the same conclusion dynamically also. As the pulse arrives at the wall, it exerts a force on the wall. By Newton’s Third Law, the wall exerts an equal and opposite force on the string generating a reflected pulse that differs by a phase of π.
# 14.6.1 Standing Waves and Normal Modes
We considered above reflection at one boundary. But there are familiar situations (a string fixed at either end or an air column in a pipe with either end closed) in which reflection takes place at two or more boundaries. In a string, for example, a wave travelling in one direction will get reflected at one end, which in turn will travel and get reflected from the other end. This will go on until there is a steady wave pattern set up on the string. Such wave patterns are called standing waves or stationary waves. To see this mathematically, consider a wave travelling along the positive direction of x-axis and a reflected wave of the same amplitude and wavelength in the negative direction of x-axis. From Eqs. (14.2) and (14.4), with φ = 0, we get:
y₁(x, t) = a sin (kx – ωt)
y₂(x, t) = a sin (kx + ωt)
The resultant wave on the string is, according to the principle of superposition:
y(x, t) = y₁(x, t) + y₂(x, t) | 11 | 11 | Physics | 207 |
e4d96f53-32fe-47ee-9f2b-e1d71b18178c | # PHYSICS
= a [sin (kx – ωt) + sin (kx + ωt)]
Using the familiar trigonometric identity
Sin (A+B) + Sin (A–B) = 2 sin A cos B we get,
y (x, t) = 2a sin kx cos ωt (14.37)
Note the important difference in the wave pattern described by Eq. (14.37) from that described by Eq. (14.2) or Eq. (14.4). The terms kx and ωt appear separately, not in the combination kx - ωt. The amplitude of this wave is 2a sin kx. Thus, in this wave pattern, the amplitude varies from point-to-point, but each element of the string oscillates with the same angular frequency ω or time period. There is no phase difference between oscillations of different elements of the wave. The string as a whole vibrates in phase with differing amplitudes at different points. The wave pattern is neither moving to the right nor to the left. Hence, they are called standing or stationary waves. The amplitude is fixed at a given location but, as remarked earlier, it is different at different locations. The points at which the amplitude is zero (i.e., where there is no motion at all) are given by
sin kx = 0.
which implies
kx = nπ; n = 0, 1, 2, 3, ...
Since, k = 2π/λ, we get
x = nλ; n = 0, 1, 2, 3, ... (14.38)
Fig. 14.12 Stationary waves arising from superposition of two harmonic waves travelling in opposite directions.
Note that the positions of zero displacement (nodes) remain fixed at all times.
2024-25 | 12 | 11 | Physics | 207 |
d1c3fa2d-3c3d-4bcc-bcba-2da59bea3836 | # WAVES
Clearly, the distance between any two successive nodes is λ. In the same way, the speed of wave is determined by the properties of the medium. The n = 2 frequency is called the second harmonic; n = 3 is the third harmonic and so on. We can label the various harmonics by the symbol νₙ ( n = 1, 2, ...).
Fig. 14.13 shows the first six harmonics of a stretched string fixed at either end. A string need not vibrate in one of these modes only. Generally, the vibration of a string will be a superposition of different modes; some modes may be more strongly excited and some less. Musical instruments like sitar or violin are based on this principle. Where the string is plucked or bowed, determines which modes are more prominent than others.
Let us next consider normal modes of oscillation of an air column with one end closed. Again the distance between any two consecutive antinodes is λ/2. Eq. (14.38) can be applied to the case of a stretched string of length L fixed at both ends. Taking one end to be at x = 0, the boundary conditions are that x = 0 and x = L are positions of nodes. The x = 0 condition is already satisfied. The x = L node condition requires that the length L is related to λ by:
L = n λ ; n = 1, 2, 3, ... (14.40)
Thus, the possible wavelengths of stationary waves are constrained by the relation:
λ = 2L ; n = 1, 2, 3, … (14.41)
with corresponding frequencies:
v = nv , for n = 1, 2, 3, (14.42)
2L
We have thus obtained the natural frequencies - the normal modes of oscillation of the system. The lowest possible natural frequency of a system is called its fundamental mode or the first harmonic. For the stretched string fixed at either end it is given by:
v = 2L, corresponding to n = 1 of Eq. (14.42). Here v is the speed of the wave.
Fig. 14.13 The first six harmonics of vibrations of a stretched string fixed at both ends.
2024-25 | 13 | 11 | Physics | 207 |
f7b2c988-db0b-4bc8-8542-d6a13d6f52fd | # PHYSICS
and the other open. A glass tube partially filled with water illustrates this system. The end in contact with water is a node, while the open end is an antinode. At the node the pressure changes are the largest, while the displacement is minimum (zero). At the open end - the antinode, it is just the other way - least pressure change and maximum amplitude of displacement. Taking the end in contact with water to be x = 0, the node condition (Eq. 14.38) is already satisfied. If the other end x = L is an antinode, Eq. (14.39) gives
L = n + 1 λ , for n = 0, 1, 2, 3, …
2 2
The possible wavelengths are then restricted by the relation:
λ = 2L (n + 1/2), for n = 0, 1, 2, 3,... (14.43)
The normal modes – the natural frequencies – of the system are
ν = n + 1 v ; n = 0, 1, 2, 3 ... (14.44)
The fundamental frequency corresponds to n= 0, and is given by v.
The higher frequencies are
4L odd harmonics, i.e., odd multiples of the fundamental frequency: 3 v , 5 v , etc.
Fig. 14.14 shows the first six odd harmonics of air column with one end closed and the other open. For a pipe open at both ends, each end is an antinode. It is then easily seen that an open air column at both ends generates all harmonics (See Fig. 14.15).
The systems above, strings and air columns, can also undergo forced oscillations (Chapter 13). If the external frequency is close to one of the natural frequencies, the system shows resonance.
Normal modes of a circular membrane rigidly clamped to the circumference as in a tabla are determined by the boundary condition that no point on the circumference of the membrane vibrates. Estimation of the frequencies of normal
|Fundamental|or|third|fifth|
|---|---|---|---|
|first harmonic| | | | | 14 | 11 | Physics | 207 |
6d47d11b-e8ad-4d79-ab08-d7414421852d | # WAVES
Fig. 14.15 Standing waves in an open pipe, first four harmonics are depicted.
while tuning their instruments with each other. They go on tuning until their sensitive ears do not detect any beats.
To see this mathematically, let us consider two harmonic sound waves of nearly equal angular frequency ω₁ and ω₂ and fix the location to be x = 0 for convenience. Eq. (14.2) with a suitable choice of phase (φ = π/2 for each) and, assuming equal amplitudes, gives
s₁ = a cos ω₁t and s₂ = a cos ω₂t (14.45)
Here we have replaced the symbol y by s, since we are referring to longitudinal not transverse displacement. Let ω₁ be the (slightly) greater of the two frequencies. The resultant displacement is, by the principle of superposition,
s = s₁ + s₂ = a (cosω₁t + cos ω₂ t)
Using the familiar trigonometric identity for cos A + cos B, we get
(ω₁ - ω₂)t (ω₁ + ω₂)t
= 2 a cos (1/2 (ω₁ - ω₂)t) cos (1/2 (ω₁ + ω₂)t) (14.46)
which may be written as :
s = [2 a cos (ωₐ t)] cos (ωb t) (14.47)
If |ω₁ - ω₂| << ω₁, ω₂ >> ωₐ >> ωb, then
ω = (ω₁ − ω₂) and ω = (ω₁ + ω₂)
Now if we assume |ω₁ - ω₂| << ω₁, which means ωₐ >> ωb, we can interpret Eq. (14.47) as follows. The resultant wave is oscillating with the average angular frequency ωₐ; however its amplitude is not constant in time, unlike a pure harmonic wave. The amplitude is the largest when the term cos ωb t takes its limit +1 or –1. In other words, the intensity of the resultant wave waxes and wanes with a frequency which is 2ωb = ω₁ – ω₂.
# 14.7 BEATS
‘Beats’ is an interesting phenomenon arising from interference of waves. When two harmonic sound waves of close (but not equal) frequencies are heard at the same time, we hear a sound of similar frequency (the average of two close frequencies), but we hear something else also. We hear audibly distinct waxing and waning of the intensity of the sound, with a frequency equal to the difference in the two close frequencies. Artists use this phenomenon often. | 15 | 11 | Physics | 207 |
114851d5-3ed3-4dc7-a9ee-a68949980402 | # PHYSICS
ω₂. Since ω = 2πν, the beat frequency νbeat, is given by
νbeat = ν1 – ν2 (14.48)
Fig. 14.16 illustrates the phenomenon of beats for two harmonic waves of frequencies 11 Hz and 9 Hz. The amplitude of the resultant wave shows beats at a frequency of 2 Hz.
# Musical Pillars
Temples often have some pillars portraying human figures playing musical instruments, but seldom do these pillars themselves produce music. At the Nellaiappar temple in Tamil Nadu, gentle taps on a cluster of pillars carved out of a single piece of rock produce the basic notes of Indian classical music, viz. Sa, Re, Ga, Ma, Pa, Dha, Ni, Sa. Vibrations of these pillars depend on elasticity of the stone used, its density and shape.
Musical pillars are categorised into three types: The first is called the Shruti Pillar, as it can produce the basic notes — the “swaras”. The second type is the Gana Thoongal, which generates the basic tunes that make up the “ragas”. The third variety is the Laya Thoongal pillars that produce “taal” (beats) when tapped. The pillars at the Nellaiappar temple are a combination of the Shruti and Laya types.
Archaeologists date the Nelliappar temple to the 7th century and claim it was built by successive rulers of the Pandyan dynasty. The musical pillars of Nelliappar and several other temples in southern India like those at Hampi, Kanyakumari, and Thiruvananthapuram are unique to the country and have no parallel in any other part of the world.
# Example 14.6
Two sitar strings A and B playing the note ‘Dha’ are slightly out of tune and produce beats of frequency 5 Hz. The tension of the string B is slightly increased and the beat frequency is found to decrease to 3 Hz. What is the original frequency of B if the frequency of A is 427 Hz?
Answer: Increase in the tension of a string increases its frequency. If the original frequency of B (νB) were greater than that of A (νA), further increase in νB should have resulted in an increase in the beat frequency. But the beat frequency is found to decrease. This shows that νB < νA. Since νA – νB = 5 Hz, and νA = 427 Hz, we get νB = 422 Hz. | 16 | 11 | Physics | 207 |
ec7665b0-b58a-4f83-afb5-532549d60af7 | # WAVES
# SUMMARY
1. Mechanical waves can exist in material media and are governed by Newton’s Laws.
2. Transverse waves are waves in which the particles of the medium oscillate perpendicular to the direction of wave propagation.
3. Longitudinal waves are waves in which the particles of the medium oscillate along the direction of wave propagation.
4. Progressive wave is a wave that moves from one point of medium to another.
5. The displacement in a sinusoidal wave propagating in the positive x direction is given by
y (x, t) = a sin (kx – ωt + φ)
where a is the amplitude of the wave, k is the angular wave number, ω is the angular frequency, (kx –ωt + φ) is the phase, and φ is the phase constant or phase angle.
6. Wavelength λ of a progressive wave is the distance between two consecutive points of the same phase at a given time. In a stationary wave, it is twice the distance between two consecutive nodes or antinodes.
7. Period T of oscillation of a wave is defined as the time any element of the medium takes to move through one complete oscillation. It is related to the angular frequency ω through the relation
T = 2π / ω
8. Frequency v of a wave is defined as 1/T and is related to angular frequency by
ν = ω / 2π
9. Speed of a progressive wave is given by
v = ω = λ = λν / kT
10. The speed of a transverse wave on a stretched string is set by the properties of the string. The speed on a string with tension T and linear mass density μ is
v = T / μ
11. Sound waves are longitudinal mechanical waves that can travel through solids, liquids, or gases. The speed v of sound wave in a fluid having bulk modulus B and density ρ is
v = B / ρ
The speed of longitudinal waves in a metallic bar is
v = Y / ρ
For gases, since B = γP, the speed of sound is
v = γP / ρ | 17 | 11 | Physics | 207 |
0eacd2cb-ba04-4a5b-80b6-29773b288fa4 | # PHYSICS
# 12.
When two or more waves traverse simultaneously in the same medium, the displacement of any element of the medium is the algebraic sum of the displacements due to each wave. This is known as the principle of superposition of waves
n
y = ∑ fi(x - vt)
i=1
# 13.
Two sinusoidal waves on the same string exhibit interference, adding or cancelling according to the principle of superposition. If the two are travelling in the same direction and have the same amplitude a and frequency but differ in phase by a phase constant φ, the result is a single wave with the same frequency ω:
y (x, t) = 2a cos(1/2 φ) sin(kx - ωt + 1/2 φ)
If φ = 0 or an integral multiple of 2π, the waves are exactly in phase and the interference is constructive; if φ = π, they are exactly out of phase and the interference is destructive.
# 14.
A travelling wave, at a rigid boundary or a closed end, is reflected with a phase reversal but the reflection at an open boundary takes place without any phase change. For an incident wave
yi (x, t) = a sin (kx – ωt)
the reflected wave at a rigid boundary is
yᵣ (x, t) = – a sin (kx + ωt)
For reflection at an open boundary
yᵣ (x,t) = a sin (kx + ωt)
# 15.
The interference of two identical waves moving in opposite directions produces standing waves. For a string with fixed ends, the standing wave is given by
y (x, t) = [2a sin kx] cos ωt
Standing waves are characterised by fixed locations of zero displacement called nodes and fixed locations of maximum displacements called antinodes. The separation between two consecutive nodes or antinodes is λ/2.
A stretched string of length L fixed at both the ends vibrates with frequencies given by
v = n v / 2L, n = 1, 2, 3, ...
The set of frequencies given by the above relation are called the normal modes of oscillation of the system. The oscillation mode with lowest frequency is called the fundamental mode or the first harmonic. The second harmonic is the oscillation mode with n = 2 and so on.
A pipe of length L with one end closed and other end open (such as air columns) vibrates with frequencies given by
v = (n + 1/2) v / 2L, n = 0, 1, 2, 3, ...
The set of frequencies represented by the above relation are the normal modes of oscillation of such a system. The lowest frequency given by v/4L is the fundamental mode or the first harmonic.
# 16.
A string of length L fixed at both ends or an air column closed at one end and open at the other end or open at both the ends, vibrates with certain frequencies called their normal modes. Each of these frequencies is a resonant frequency of the system.
# 17.
Beats arise when two waves having slightly different frequencies, ν₁ and ν₂ and comparable amplitudes, are superposed. The beat frequency is
νbeat = ν₁ - ν₂ | 18 | 11 | Physics | 207 |
a60c1d3c-46ea-401b-8ecf-5a556ce42955 | # WAVES
# POINTS TO PONDER
1. A wave is not motion of matter as a whole in a medium. A wind is different from the sound wave in air. The former involves motion of air from one place to the other. The latter involves compressions and rarefactions of layers of air.
2. In a wave, energy and not the matter is transferred from one point to the other.
3. In a mechanical wave, energy transfer takes place because of the coupling through elastic forces between neighbouring oscillating parts of the medium.
4. Transverse waves can propagate only in medium with shear modulus of elasticity, Longitudinal waves need bulk modulus of elasticity and are therefore, possible in all media, solids, liquids and gases.
5. In a harmonic progressive wave of a given frequency, all particles have the same amplitude but different phases at a given instant of time. In a stationary wave, all particles between two nodes have the same phase at a given instant but have different amplitudes.
6. Relative to an observer at rest in a medium the speed of a mechanical wave in that medium (v) depends only on elastic and other properties (such as mass density) of the medium. It does not depend on the velocity of the source.
# EXERCISES
1. A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?
2. A stone dropped from the top of a tower of height 300 m splashes into the water of a pond near the base of the tower. When is the splash heard at the top given that the speed of sound in air is 340 m s–1? (g = 9.8 m s–2)
3. A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension in the wire so that speed of a transverse wave on the wire equals the speed of sound in dry air at 20°C = 343 m s–1.
4. Use the formula v = γP to explain why the speed of sound in air
- (a) is independent of pressure,
- (b) increases with temperature,
- (c) increases with humidity.
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f8123b3b-980e-4180-9284-0bbd75effa69 | # PHYSICS
# 14.5
You have learnt that a travelling wave in one dimension is represented by a function y = f (x, t) where x and t must appear in the combination x – v t or x + v t, i.e. y = f (x ± v t). Is the converse true? Examine if the following functions for y can possibly represent a travelling wave:
- (a) (x – vt)²
- (b) log [(x + vt)/x₀]
- (c) 1/(x + vt)
# 14.6
A bat emits ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of (a) the reflected sound, (b) the transmitted sound? Speed of sound in air is 340 m s –1 and in water 1486 m s–1.
# 14.7
A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is 1.7 km s–1? The operating frequency of the scanner is 4.2 MHz.
# 14.8
A transverse harmonic wave on a string is described by y(x, t) = 3.0 sin (36 t + 0.018 x + π/4) where x and y are in cm and t in s. The positive direction of x is from left to right.
- (a) Is this a travelling wave or a stationary wave? If it is travelling, what are the speed and direction of its propagation?
- (b) What are its amplitude and frequency?
- (c) What is the initial phase at the origin?
- (d) What is the least distance between two successive crests in the wave?
# 14.9
For the wave described in Exercise 14.8, plot the displacement (y) versus (t) graphs for x = 0, 2 and 4 cm. What are the shapes of these graphs? In which aspects does the oscillatory motion in travelling wave differ from one point to another: amplitude, frequency or phase?
# 14.10
For the travelling harmonic wave y(x, t) = 2.0 cos 2π (10t – 0.0080 x + 0.35) where x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of:
- (a) 4 m,
- (b) 0.5 m,
- (c) λ/2,
- (d) 3λ/4
# 14.11
The transverse displacement of a string (clamped at its both ends) is given by y(x, t) = 0.06 sin 2π x cos (120 πt) where x and y are in m and t in s. The length of the string is 1.5 m and its mass is 3.0 ×10–2 kg.
- (a) Does the function represent a travelling wave or a stationary wave?
- (b) Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency, and speed of each wave? | 20 | 11 | Physics | 207 |
84328559-48ea-4c11-979e-86cad31e5412 | # WAVES
# 14.12
(i) For the wave on a string described in Exercise 15.11, do all the points on the string oscillate with the same (a) frequency, (b) phase, (c) amplitude? Explain your answers. (ii) What is the amplitude of a point 0.375 m away from one end?
# 14.13
Given below are some functions of x and t to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent (i) a travelling wave, (ii) a stationary wave or (iii) none at all:
- (a) y = 2 cos (3x) sin (10t)
- (b) y = 2 x − vt
- (c) y = 3 sin (5x – 0.5t) + 4 cos (5x – 0.5t)
- (d) y = cos x sin t + cos 2x sin 2t
# 14.14
A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5× 10–2 kg and its linear mass density is 4.0 × 10–2 kg m–1. What is (a) the speed of a transverse wave on the string, and (b) the tension in the string?
# 14.15
A metre-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency 340 Hz) when the tube length is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected.
# 14.16
A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod are given to be 2.53 kHz. What is the speed of sound in steel?
# 14.17
A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe if both ends are open? (speed of sound in air is 340 m s–1).
# 14.18
Two sitar strings A and B playing the note ‘Ga’ are slightly out of tune and produce beats of frequency 6 Hz. The tension in the string A is slightly reduced and the beat frequency is found to reduce to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency of B?
# 14.19
Explain why (or how):
- (a) in a sound wave, a displacement node is a pressure antinode and vice versa,
- (b) bats can ascertain distances, directions, nature, and sizes of the obstacles without any “eyes”,
- (c) a violin note and sitar note may have the same frequency, yet we can distinguish between the two notes,
- (d) solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases, and
- (e) the shape of a pulse gets distorted during propagation in a dispersive medium.
2024-25 | 21 | 11 | Physics | 207 |
90402baf-ac4e-4ba6-a778-d4a56c0daadd | # CHAPTER THIRTEEN
# OSCILLATIONS
# 13.1 INTRODUCTION
In our daily life we come across various kinds of motions. You have already learnt about some of them, e.g., rectilinear motion and motion of a projectile. Both these motions are non-repetitive. We have also learnt about uniform circular motions and orbital motion of planets in the solar system. In these cases, the motion is repeated after a certain interval of time, that is, it is periodic. In your childhood, you must have enjoyed rocking in a cradle or swinging on a swing. Both these motions are repetitive in nature but different from the periodic motion of a planet. Here, the object moves to and fro about a mean position. The pendulum of a wall clock executes a similar motion. Examples of such periodic to and fro motion abound: a boat tossing up and down in a river, the piston in a steam engine going back and forth, etc. Such a motion is termed as oscillatory motion. In this chapter we study this motion.
The study of oscillatory motion is basic to physics; its concepts are required for the understanding of many physical phenomena. In musical instruments, like the sitar, the guitar or the violin, we come across vibrating strings that produce pleasing sounds. The membranes in drums and diaphragms in telephone and speaker systems vibrate to and fro about their mean positions. The vibrations of air molecules make the propagation of sound possible. In a solid, the atoms vibrate about their equilibrium positions, the average energy of vibrations being proportional to temperature. AC power supply give voltage that oscillates alternately going positive and negative about the mean value (zero).
The description of a periodic motion, in general, and oscillatory motion, in particular, requires some fundamental concepts, like period, frequency, displacement, amplitude and phase. These concepts are developed in the next section.
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d4eed45b-f3be-4b0a-a41b-f1a81b26febd | # PHYSICS
# 13.2 PERIODIC AND OSCILLATORY MOTIONS
Very often, the body undergoing periodic motion has an equilibrium position somewhere inside its path. When the body is at this position no net external force acts on it. Therefore, if it is left there at rest, it remains there forever. If the body is given a small displacement from the position, a force comes into play which tries to bring the body back to the equilibrium point, giving rise to oscillations or vibrations. For example, a ball placed in a bowl will be in equilibrium at the bottom. If displaced a little from the point, it will perform oscillations in the bowl. Every oscillatory motion is periodic, but every periodic motion need not be oscillatory. Circular motion is a periodic motion, but it is not oscillatory.
There is no significant difference between oscillations and vibrations. It seems that when the frequency is small, we call it oscillation (like, the oscillation of a branch of a tree), while when the frequency is high, we call it vibration (like, the vibration of a string of a musical instrument).
Simple harmonic motion is the simplest form of oscillatory motion. This motion arises when the force on the oscillating body is directly proportional to its displacement from the mean position, which is also the equilibrium position. Further, at any point in its oscillation, this force is directed towards the mean position.
In practice, oscillating bodies eventually come to rest at their equilibrium positions because of the damping due to friction and other dissipative causes. However, they can be forced to remain oscillating by means of some external periodic agency. We discuss the phenomena of damped and forced oscillations later in the chapter.
Any material medium can be pictured as a collection of a large number of coupled oscillators. The collective oscillations of the constituents of a medium manifest themselves as waves. Examples of waves include water waves, seismic waves, electromagnetic waves. We shall study the wave phenomenon in the next chapter.
# 13.2.1 Period and frequency
We have seen that any motion that repeats itself at regular intervals of time is called periodic motion. The smallest interval of time after which the motion is repeated is called its period. Let us denote the period by the symbol T. Its SI unit is second. For periodic motions,
Fig. 13.1 shows some periodic motions. Suppose an insect climbs up a ramp and falls down, it comes back to the initial point and repeats the process identically. If you draw a graph of its height above the ground versus time, it would look something like Fig. 13.1 (a). If a child climbs up a step, comes down, and repeats the process identically, its height above the ground would look like that in Fig. 13.1 (b). When you play the game of bouncing a ball off the ground, between your palm and the ground, its height versus time graph would look like the one in Fig. 13.1 (c).
Note that both the curved parts in Fig. 13.1 (c) are sections of a parabola given by the Newton’s equation of motion (see section 2.6),
h = ut + 1/2 gt² for downward motion, and
h = ut - 1/2 gt² for upward motion, with different values of u in each case. These are examples of periodic motion. Thus, a motion that repeats itself at regular intervals of time is called periodic motion. | 1 | 11 | Physics | 206 |
73378c34-3734-4d82-b311-c568d119d93f | # OSCILLATIONS
which are either too fast or too slow on the scale of seconds, other convenient units of time are used. The period of vibrations of a quartz crystal is expressed in units of microseconds (10–6 s) abbreviated as μs. On the other hand, the orbital period of the planet Mercury is 88 earth days. The Halley’s comet appears after every 76 years.
The reciprocal of T gives the number of repetitions that occur per unit time. This quantity is called the frequency of the periodic motion. It is represented by the symbol ν. The relation between ν and T is
ν = 1/T (13.1)
The unit of ν is thus s–1. After the discoverer of radio waves, Heinrich Rudolph Hertz (1857–1894), a special name has been given to the unit of frequency. It is called hertz (abbreviated as Hz). Thus,
1 hertz = 1 Hz = 1 oscillation per second = 1 s–1 (13.2)
Note, that the frequency, ν, is not necessarily an integer.
# Example 13.1
On an average, a human heart is found to beat 75 times in a minute. Calculate its frequency and period.
Answer: The beat frequency of heart = 75/(1 min) = 75/(60 s) = 1.25 s–1 = 1.25 Hz
The time period T = 1/(1.25 s–1) = 0.8 s ⊳
# 13.2.2 Displacement
In section 3.2, we defined displacement of a particle as the change in its position vector. In this chapter, we use the term displacement in a more general sense. It refers to change with time of any physical property under consideration. For example, in case of rectilinear motion of a steel ball on a surface, the distance from the starting point as a function of time is its position displacement. The choice of origin is a matter of convenience.
Consider a block attached to a spring, the other end of the spring is fixed to a rigid wall [see Fig.13.2(a)]. Generally, it is convenient to measure displacement of the body from its equilibrium position. For an oscillating simple pendulum, the angle from the vertical as a function of time may be regarded in the context of position only. There can be many other kinds of displacement variables. The voltage across a capacitor, changing with time in an A C circuit, is also a displacement variable. In the same way, pressure variations in time in the propagation of sound wave, the changing electric and magnetic fields in a light wave are examples of displacement in different contexts.
The displacement variable may take both positive and negative values. In experiments on oscillations, the displacement is measured for different times. The displacement can be represented by a mathematical function of time. In case of periodic motion, this function is periodic in time. One of the simplest periodic functions is given by
f(t) = A cos ωt (13.3a)
If the argument of this function, ωt, is increased by an integral multiple of 2π radians, the value of the function remains the same. | 2 | 11 | Physics | 206 |
c26a8ac6-fd7f-4f4f-8c3d-c918055a08ab | # PHYSICS
function f (t) is then periodic and its period, T, (ii) This is an example of a periodic motion. It is given by
T = 2π
ω (13.3b)
Thus, the function f (t) is periodic with period T,
f (t) = f (t+T )
The same result is obviously correct if we consider a sine function, f (t ) = A sin ωt. Further, a linear combination of sine and cosine functions like,
f (t) = A sin ωt + B cos ωt (13.3c)
is also a periodic function with the same period T. Taking,
A = D cos φ and B = D sin φ
Eq. (13.3c) can be written as,
f (t) = D sin (ωt + φ ) , (13.3d)
Here D and φ are constant given by
D = A² + B² and φ = tan–1 A
The great importance of periodic sine and cosine functions is due to a remarkable result proved by the French mathematician, Jean Baptiste Joseph Fourier (1768–1830): Any periodic function can be expressed as a superposition of sine and cosine functions of different time periods with suitable coefficients.
# 13.3 SIMPLE HARMONIC MOTION
Consider a particle oscillating back and forth about the origin of an x-axis between the limits +A and –A as shown in Fig. 13.3. This oscillatory motion is said to be simple harmonic if the displacement x of the particle from the origin varies with time as :
x (t) = A cos (ω t + φ ) (13.4)
u Example 13.2 Which of the following functions of time represent (a) periodic and (b) non-periodic motion? Give the period for each case of periodic motion [ω is any positive constant].
(i) sin ωt + cos ωt
(ii) sin ωt + cos 2 ωt + sin 4 ωt
(iii) e–ωᵗ
(iv) log (ωt)
where A, ω and φ are constants.
Thus, simple harmonic motion (SHM) is not any periodic motion but one in which displacement is a sinusoidal function of time.
Answer
(i) sin ωt + cos ωt is a periodic function, it can also be written as 2 sin (ωt + π/4).
Now 2 sin (ωt + π/4)= 2 sin (ωt + π/4+2π) = 2 sin [ω (t + 2π/ω) + π/4]
The periodic time of the function is 2π/ω. | 3 | 11 | Physics | 206 |
0bb073f7-79ba-4941-98cf-1a5e1fa0fb28 | # OSCILLATIONS
any loss of generality]. As the cosine function of time varies from +1 to –1, the displacement varies between the extremes A and – A. Two simple harmonic motions may have same ω and φ but different amplitudes A and B, as shown in Fig. 13.7 (a).
While the amplitude A is fixed for a given SHM, the state of motion (position and velocity) of the particle at any time t is determined by the
ω and φ which characterize a given SHM have standard names, as summarised in Fig. 13.6. Let us understand these quantities.
The amplitude A of SHM is the magnitude of maximum displacement of the particle. [Note, A can be taken to be positive without any loss of generality].
If the amplitude is known, φ can be determined from the displacement at t = 0. Two simple harmonic motions may have the same A and ω but different phase angle φ, as shown in Fig. 13.7 (b).
Finally, the quantity ω can be seen to be related to the period of motion T. Taking, for simplicity, φ = 0 in Eq. (13.4), we have
|x (t)|A|ω|ωt + φ|φ|
|---|---|---|---|---|
|displacement x as a function of time t|amplitude|angular frequency|phase (time-dependent)|phase constant|
2024-25 | 4 | 11 | Physics | 206 |
6ecd67bb-b682-48dd-b489-f9a96d261cae | # PHYSICS
x(t) = A cos ωt (13.5) This function represents a simple harmonic motion having a period T = 2π/ω and a phase angle (–π/4) or (7π/4)
Since the motion has a period T, x(t) is equal to x(t + T). That is,
A cos ωt = A cos ω(t + T) (13.6)
Now the cosine function is periodic with period 2π, i.e., it first repeats itself when the argument changes by 2π. Therefore,
ω(t + T) = ωt + 2π 13.4
that is ω = 2π/T (13.7) ω is called the angular frequency of SHM. Its S.I. unit is radians per second. Since the frequency of oscillations is simply 1/T, ω is 2π times the frequency of oscillation. Two simple harmonic motions may have the same A and φ, but different ω, as seen in Fig. 13.8. In this plot the curve (b) has half the period and twice the frequency of the curve (a).
In this section, we show that the projection of uniform circular motion on a diameter of the circle follows simple harmonic motion. A simple experiment (Fig. 13.9) helps us visualise this connection. Tie a ball to the end of a string and make it move in a horizontal plane about a fixed point with a constant angular speed. The ball would then perform a uniform circular motion in the horizontal plane. Observe the ball sideways or from the front, fixing your attention in the plane of motion. The ball will appear to execute to and fro motion along a horizontal line with the point of rotation as the midpoint. You could alternatively observe the shadow of the ball on a wall which is perpendicular to the plane of the circle. In this process what we are observing is the motion of the ball on a diameter of the circle normal to the direction of viewing.
Fig. 13.8 Plots of Eq. (13.4) for φ = 0 for two different periods.
# Example 13.3
Which of the following functions of time represent (a) simple harmonic motion and (b) periodic but not simple harmonic? Give the period for each case.
|ω|ω|
|---|---|
|(1) sin t – cos t|(2) sin² ωt|
Answer ω ω
(a) sin t – cos t = sin ωt – sin (π/2 – ωt) = 2 cos (π/4) sin (ωt – π/4) = √2 sin (ωt – π/4)
2024-25 | 5 | 11 | Physics | 206 |
139b2049-9c66-405d-bba3-62a9ee398486 | # OSCILLATIONS
# Example 13.4
The figure given below depicts two circular motions. The radius of the circle, the period of revolution, the initial position and the sense of revolution are indicated in the figures. Obtain the simple harmonic motions of the x-projection of the radius vector of the rotating particle P in each case.
Fig. 13.10 will make an angle of ωt + φ with the +ve x-axis. Next, consider the projection of the position vector OP on the x-axis. This will be OP′. The position of P′ on the x-axis, as the particle P moves on the circle, is given by
x(t) = A cos (ωt + φ)
which is the defining equation of SHM. This shows that if P moves uniformly on a circle, its projection P′ on a diameter of the circle executes SHM. The particle P and the circle on which it moves are sometimes referred to as the reference particle and the reference circle, respectively.
We can take projection of the motion of P on any diameter, say the y-axis. In that case, the displacement y(t) of P′ on the y-axis is given by
y = A sin (ωt + φ)
which is also an SHM of the same amplitude as that of the projection on x-axis, but differing by a phase of π/2.
In spite of this connection between circular motion and SHM, the force acting on a particle in linear simple harmonic motion is very different from the centripetal force needed to keep a particle in uniform circular motion.
* The natural unit of angle is radian, defined through the ratio of arc to radius. Angle is a dimensionless quantity. Therefore it is not always necessary to mention the unit ‘radian’ when we use π, its multiples or submultiples. The conversion between radian and degree is not similar to that between metre and centimetre or mile. If the argument of a trigonometric function is stated without units, it is understood that the unit is radian. On the other hand, if degree is to be used as the unit of angle, then it must be shown explicitly. For example, sin(15⁰) means sine of 15 degree, but sin(15) means sine of 15 radians. Hereafter, we will often drop ‘rad’ as the unit, and it should be understood that whenever angle is mentioned as a numerical value, without units, it is to be taken as radians.
2024-25 | 6 | 11 | Physics | 206 |
ea309878-6ed7-4c0f-a6d4-0a3c68f0de24 | # PHYSICS
(b) In this case at t = 0, OP makes an angle of 90ᵒ = π with the x-axis. After a time t, it covers an angle of 2π t in the clockwise sense and makes an angle of π with the x-axis. The projection of OP on the x-axis at time t is given by
x(t) = B cos π 2π t
= B sin 2π t
For T = 30 s,
x(t) = B sin π t
= B sin t
Writing this as x(t) = B cos π t π, and comparing with Eq. (13.4). We find that this represents a SHM of amplitude B, period 30 s, and an initial phase of −π 2.
# 13.5 VELOCITY AND ACCELERATION IN SIMPLE HARMONIC MOTION
The speed of a particle v in uniform circular motion is its angular speed ω times the radius of the circle A.
v = ω A (13.8)
The direction of velocity v at a time t is along the tangent to the circle at the point where the particle is located at that instant. From the geometry of Fig. 13.11, it is clear that the velocity of the projection particle P′ at time t is
v(t) = –ωA sin (ωt + φ) (13.9)
Eq. (13.11) gives the acceleration of a particle in SHM. The same equation can again be obtained directly by differentiating velocity v(t) given by Eq. (13.9) with respect to time:
a(t) = d v(t)
dt (13.12)
We note from Eq. (13.11) the important property that acceleration of a particle in SHM is proportional to displacement. For x(t) > 0, a(t) < 0 and for x(t) < 0, a(t) > 0. Thus, whatever
Fig. 13.11 The velocity, v(t), of the particle P′ is the projection of the velocity v of the reference particle, P. | 7 | 11 | Physics | 206 |
07a0e0a7-5ffa-4414-ae64-0c7e427da971 | # OSCILLATIONS
the value of x between –A and A, the acceleration a(t) is always directed towards the centre. For simplicity, let us put φ = 0 and write the expression for x(t), v(t) and a(t)
x(t) = A cos ωt, v(t) = – ω Asin ωt, a(t)=–ω² A cos ωt
= 10π × 0.707 m s–1 = 22 m s–1
(b) Using Eq. (13.9), the speed of the body
= – (5.0 m)(2π s–1) sin [(2π s–1) ×1.5 s + π/4]
= – (5.0 m)(2π s–1) sin [(3π + π/4)]
(c) Using Eq.(13.10), the acceleration of the body
= –(2π s–1)² ×displacement
= – (2π s–1)² × (–3.535 m)
= 140 m s–2 ⊳
# 13.6 FORCE LAW FOR SIMPLE HARMONIC MOTION
Using Newton’s second law of motion, and the expression for acceleration of a particle undergoing SHM (Eq. 13.11), the force acting on a particle of mass m in SHM is
F (t ) = ma
= –mω² x (t)
i.e., F (t) = –k x (t ) (13.13)
where k = mω² (13.14a)
or ω = k (13.14b)
m
Like acceleration, force is always directed towards the mean position—hence it is sometimes called the restoring force in SHM. To summarise the discussion so far, simple harmonic motion can be defined in two equivalent ways, either by Eq. (13.4) for displacement or by Eq. (13.13) that gives its force law. Going from Eq. (13.4) to Eq. (13.13) required us to differentiate two times. Likewise, by integrating the force law Eq. (13.13) two times, we can get back Eq. (13.4).
u Example 13.5 A body oscillates with SHM according to the equation (in SI units),
x = 5 cos [2π t + π/4]. At t = 1.5 s, calculate the (a) displacement, (b) speed and (c) acceleration of the body.
Answer The angular frequency ω of the body = 2π s–1 and its time period T = 1 s.
At t = 1.5 s
(a) displacement = (5.0 m) cos [(2π s–1)× 1.5 s + π/4]
= (5.0 m) cos [(3π + π/4)]
= –5.0 × 0.707 m
= –3.535 m
u Example 13.6 Two identical springs of spring constant k are attached to a block of mass m and to fixed supports as shown in Fig. 13.14. Show that when the mass is displaced from its equilibrium position on either side, it executes a simple harmonic motion. Find the period of oscillations. | 8 | 11 | Physics | 206 |
458bcf77-b76c-4651-9df6-168a0e3e9cc1 | # 13.7 ENERGY IN SIMPLE HARMONIC MOTION
Both kinetic and potential energies of a particle in SHM vary between zero and their maximum values.
In section 13.5 we have seen that the velocity of a particle executing SHM, is a periodic function of time. It is zero at the extreme positions of displacement. Therefore, the kinetic energy (K) of such a particle, which is defined as
K = 1 mv2
= 1 m ω2 A2 sin2(ωt + φ) (13.15)
is also a periodic function of time, being zero when the displacement is maximum and maximum when the particle is at the mean position. Note, since the sign of v is immaterial in K, the period of K is T/2.
Let the mass be displaced by a small distance x to the right side of the equilibrium position, as shown in Fig. 13.15. Under this situation the spring on the left side gets elongated by a length equal to x and that on the right side gets compressed by the same length. The forces acting on the mass are then,
F1 = -kx (force exerted by the spring on the left side, trying to pull the mass towards the mean position)
F2 = -kx (force exerted by the spring on the right side, trying to push the mass towards the mean position)
The net force, F, acting on the mass is then given by,
F = -2kx
Hence the force acting on the mass is proportional to the displacement and is directed towards the mean position; therefore, the motion executed by the mass is simple harmonic. The time period of oscillations is,
T = 2π√(m/k)
What is the potential energy (U) of a particle executing simple harmonic motion? In Chapter 6, we have seen that the concept of potential energy is possible only for conservative forces. The spring force F = -kx is a conservative force, with associated potential energy
U = 1 kx2 (13.16)
Hence the potential energy of a particle executing simple harmonic motion is,
U(x) = 1 kA2 cos2(ωt + φ) (13.17)
Thus, the potential energy of a particle executing simple harmonic motion is also periodic, with period T/2, being zero at the mean position and maximum at the extreme displacements. | 9 | 11 | Physics | 206 |
3ca41dee-bbc9-47b0-ac75-3879367bfa65 | # OSCILLATIONS
It follows from Eqs. (13.15) and (13.17) that observe that both kinetic energy and the total energy, E, of the system is,
E = U + K
= 1 k A² 2 ω φ 2 ω φ
= 1 k A² [ 2 ω φ cos ( t + φ ) + 2 k A sin ( t + φ ) ]
= 1 k A² [ cos ( t + φ ) + sin ( t + φ ) ]
Thus,
E = 1 k A² (13.18)
The total mechanical energy of a harmonic oscillator is thus independent of time as expected for motion under any conservative force. The time and displacement dependence of the potential and kinetic energies of a linear simple harmonic oscillator are shown in Fig. 13.16.
Using the familiar trigonometric identity, the value of the expression in the brackets is unity.
Both kinetic energy and potential energy peak twice during each period of SHM. For x = 0, the energy is kinetic; at the extremes x = ±A, it is all potential energy. In the course of motion between these limits, kinetic energy increases at the expense of potential energy or vice-versa.
# Example 13.7
A block whose mass is 1 kg is fastened to a spring. The spring has a spring constant of 50 N m–1. The block is pulled to a distance x = 10 cm from its equilibrium position at x = 0 on a frictionless surface from rest at t = 0. Calculate the kinetic, potential and total energies of the block when it is 5 cm away from the mean position.
Answer: The block executes SHM, its angular frequency, as given by Eq. (13.14b), is
ω = k m = 50 N m–1 1 kg = 7.07 rad s–1
Its displacement at any time t is then given by,
x(t) = 0.1 cos (7.07t)
Therefore, when the particle is 5 cm away from the mean position, we have
0.05 = 0.1 cos (7.07t)
Fig. 13.16 Kinetic energy, potential energy and total energy as a function of time [shown in (a)] and displacement [shown in (b)] of a particle in SHM. The kinetic energy and potential energy both repeat after a period T/2. The total energy remains constant at all t or x. | 10 | 11 | Physics | 206 |
339bb63a-e8d2-4cfd-a8b1-0cc97969c138 | # PHYSICS
Or cos (7.07t) = 0.5 and hence let it go. The stone executes a to and fro motion, it is periodic with a period of about two seconds.
sin (7.07t) = 0.866
Then, the velocity of the block at x = 5 cm is
= 0.1 × 7.07 × 0.866 m s–1
= 0.61 m s–1
Hence the K.E. of the block,
= 1 m v2
= ½[1kg × (0.612 m s–1) ]
= 0.19 J (a)
The P.E. of the block,
= 1 k x2
= ½(50 N m–1 × 0.05 m × 0.05 m)
= 0.0625 J
The total energy of the block at x = 5 cm,
= K.E. + P.E.
= 0.25 J
we also know that at maximum displacement, K.E. is zero and hence the total energy of the system is equal to the P.E. Therefore, the total energy of the system,
= ½(50 N m–1 × 0.1 m × 0.1 m)
= 0.25 J
which is same as the sum of the two energies at a displacement of 5 cm. This is in conformity with the principle of conservation of energy.
# 13.8 The Simple Pendulum
It is said that Galileo measured the periods of a swinging chandelier in a church by his pulse beats. He observed that the motion of the chandelier was periodic. The system is a kind of pendulum. You can also make your own pendulum by tying a piece of stone to a long unstretchable thread, approximately 100 cm long. Suspend your pendulum from a suitable support so that it is free to oscillate. Displace the stone to one side by a small distance and let it go.
There are only two forces acting on the bob; the tension T along the string and the vertical.
Fig. 13.17 (a) A bob oscillating about its mean position. (b) The radial force T - mg cosθ provides centripetal force but no torque about the support. The tangential force mg sinθ provides the restoring torque.
Let θ be the angle made by the string with the vertical. When the bob is at the mean position, θ = 0. | 11 | 11 | Physics | 206 |
72e4b364-a663-4453-acc0-fac4d96c32e7 | # OSCILLATIONS
The force due to gravity (=mg). The force mg can be resolved into the component mg cosθ along the string and mg sinθ perpendicular to it. Since the motion of the bob is along a circle of length L and centre at the support point, the bob has a radial acceleration (ω²L) and also a tangential acceleration; the latter arises since motion along the arc of the circle is not uniform. The radial acceleration is provided by the net radial force T –mg cosθ, while the tangential acceleration is provided by mg sinθ. It is more convenient to work with torque about the support since the radial force gives zero torque. Torque τ about the support is entirely provided by the tangential component of force.
τ = –L (mg sinθ) (13.19)
This is the restoring torque that tends to reduce angular displacement — hence the negative sign. By Newton’s law of rotational motion,
τ = I α (13.20)
where I is the moment of inertia of the system about the support and α is the angular acceleration. Thus,
I α = –m g sin θ L (13.21)
Or,
α = − m g L sin θ (13.22)
We can simplify Eq. (13.22) if we assume that the displacement θ is small. We know that sin θ can be expressed as,
sinθ = θ − θ³/3! + θ⁵/5! ± ... (13.23)
where θ is in radians. Now if θ is small, sin θ can be approximated by θ and Eq. (13.22) can then be written as,
α = − mgL θ (13.24)
In Table 13.1, we have listed the angle θ in degrees, its equivalent in radians, and the value of the function sin θ.
|(degrees)|(radians)|sin|
|---|---|---|
|0|0|0|
|10|0.1745|0.1736|
|20|0.3491|0.3420|
|30|0.5236|0.5000|
|40|0.6981|0.6428|
|50|0.8727|0.7660|
|60|1.0472|0.8660|
|70|1.2217|0.9397|
|80|1.3963|0.9848|
|90|1.5708|1.0000|
From this table it can be seen that for θ as large as 20 degrees, sin θ is nearly the same as θ expressed in radians.
Now since the string of the simple pendulum is massless, the moment of inertia I is simply mL². Eq. (13.25) then gives the well-known formula for time period of a simple pendulum.
T = 2π√(L/g) (13.26)
Example 13.8: What is the length of a simple pendulum, which ticks seconds?
Answer: From Eq. (13.26), the time period of a simple pendulum is given by,
T = 2π√(L/g)
From this relation one gets,
L = gT²/4π²
The time period of a simple pendulum, which ticks seconds, is 2 s. Therefore, for g = 9.8 m s–2 and T = 2 s, L is
L = (9.8 m s–2) × (2 s)²/4π² = 1 m
⊳ 2024-25 | 12 | 11 | Physics | 206 |
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