text: Click to Chat
  
  1800-2000-838
  
  +91-120-4616500
  
  CART 0
  
  • 0
  
  MY CART (5)
  
  Use Coupon: CART20 and get 20% off on all online Study Material
  
  ITEM
  DETAILS
  MRP
  DISCOUNT
  FINAL PRICE
  Total Price: Rs.
  
  There are no items in this cart.
  Continue Shopping
  • Complete Physics Course - Class 11
  • OFFERED PRICE: Rs. 2,968
  • View Details
  
  ```Dynamics of Harmonic Oscillations
  
  Harmonic oscillations
  
  The oscillations of a physical system results from two basic properties namely elasticity and inertia. Let us consider a body displaced from a mean position. The restoring force brings the body to the mean position.
  
  (i) At extreme position when the displacement is maximum, velocity is zero. The acceleration becomes maximum and directed towards the mean position.
  
  (ii) Under the influence of restoring force, the body comes back to the mean position and overshoots because of negative velocity gained at the mean position.
  
  (iii) When the displacement is negative maximum, the velocity becomes zero and the acceleration is maximum in the positive direction. Hence the body moves towards the mean position. Again when the displacement is zero in the mean position velocity becomes positive.
  
  (iv) Due to inertia the body overshoots the mean position once again. This process repeats itself periodically. Hence the system oscillates.
  
  The restoring force is directly proportional to the displacement and directed towards the mean position.
  
  That is,
  
  F ∝ y
  
  F = -ky             …... (1)
  
  where k is is the force constant. It is the force required to give unit displacement. It is expressed in N m−1.
  
  From Newton’s second law,   F  = ma             …... (2)
  
  −k y = ma
  
  Or, a = (-k/m) y               …... (3)
  
  From definition of SHM acceleration a = −ω2y
  
  The acceleration is directly proportional to the negative of the displacement.
  
  Comparing the above equations we get,
  
  ω = √k/m            …... (4)
  
  Therefore the period of SHM is
  
  T = 2π/ω = 2π√m/k
  
  So,
  
  T =  2π√inertial factor/spring factor            …... (5)
  
  Angular harmonic oscillator
  
  Simple harmonic motion can also be angular. In this case, the restoring torque required for producing SHM is directly proportional to the angular displacement and is directed towards the mean position.
  
  Consider a wire suspended vertically from a rigid support. Let some weight be suspended from the lower end of the wire. When the wire is twisted through an angle θ from the mean position, a restoring torque acts on it tending to return it to the mean position. Here restoring torque is proportional to angular displacement θ.
  
  Hence τ = − C θ                  …... (1)
  
  where C is called torque constant.
  
  It is equal to the moment of the couple required to produce unit angular displacement. Its unit is N m rad−1.
  
  The negative sign shows that torque is acting in the opposite direction to the angular displacement. This is the case of angular simple harmonic motion.
  
  Examples : Torsional pendulum, balance wheel of a watch.
  
  But, τ = Iα                  …... (2)
  
  where τ is torque, I is the moment of inertia and α is angular acceleration.
  
  Thus, angular acceleration,
  
  α = r/I = – Cθ/I        …... (3)
  
  This is similar to α = – ω2y
  
  Replacing y by θ, and a by α we get,
  
  α = – ω2θ = – (C/I) θ
  
  So, ω = √C/I
  
  Thus, period of SHM, T = 2π√I/C
  
  Therefore, frequency, n = 1/T = 1/(2π√I/C) = (1/2π)√(C/I)
  
  Thus,
  
  Refer this video to know more about on, “Dynamic of Harmonic Oscillations”.
  
  Problem (JEE Main):
  
  Let T1 and T2 be the time periods of two springs A and B when a mass m is suspended from them separately. Now both the springs are connected in parallel and same mass m is suspended with them. Now let T be the time period in this position. Then
  
  (a) T = T1 + T2            (b) T = T1T2/ T1 + T2
  
  (c) T2 = T12 + T22        (d) 1/T2 = 1/T12 + 1/T22
  
  Solution:
  
  We know that,
  
  T1 = 2π√m/k1
  
  Or, k1 = 4π2m/T12
  
  T2 = 2π√m/k2
  
  Or, k2 = 4π2m/T22
  
  In parallel, k = k1 + k2
  
  Substituting the values of k, k1 and k2 we get,
  
  1/T2 = 1/T12 + 1/T22
  
  Thus from the above observation, we conclude that, option (d) is correct.
  
  Many systems that are in stable equilibrium will oscillate with simple harmonic motion when displaced by from equilibrium by a small amount.
  
  Near equilibrium the force acting to restore the system can be approximated by the Hooke’s law no matter how complex the "actual" force.
  
  The force holding atoms together in a solid can be approximated by spring forces when the vibrations of atoms in a solid are small.
  
  The simple harmonic oscillator (SHO) is a mass connected to some elastic object of negligible mass that is fixed at the other end and constrained so that it may only move in one dimension.
  
  The motion is periodic with a frequency that depends on the nature of the mass and the elastic object (here assumed to be a spring).
  
  A more realistic physical model is one that includes dissipative forces: the damped harmonic oscillator.
  
  The motion is sinusoidal in time and demonstrates a single resonant frequency.
  
  In order for simple harmonic motion to take place, the net force of the object at the end of the pendulum must be proportional to the displacement.
  
  Question 1
  
  A particle executes linear simple harmonic motion with an amplitude of 2 cm. When the particle is at 1 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is
  
  (a) 1/ 2π√3         (b) 2π√3
  
  (c) 2π/√3           (d) √3/2π
  
  Question 2
  
  If the initial displacement is doubled what happens to the maximum kinetic energy of the thingy?
  
  (a) It is unchanged.                            (b) It is doubled.
  
  (c) It is increased by a factor of 4.        (d) We can't tell from the information provided.
  
  Question 3
  
  An object swinging on the end of a string forms a simple pendulum. Some students (and some texts) often cite the simple pendulum's motion as an example of SHM. That is not quite accurate because the motion is really
  
  (a) approximately SHM only for small amplitudes.
  
  (b) exactly SHM only for amplitudes that are smaller than a certain value.
  
  (c) approximately SHM for all amplitudes.
  
  Question 4
  
  The amplitude of the oscillation is determined by
  
  (a) the amount of the initial displacement.
  
  (b) the mass of the thingy and the properties of the spring
  
  (c) the local gravitational field, g.
  
  (d) all of the above.
  
  Question 5
  
  Which one of the following statements is not true for a body vibrating in simple harmonic motion when damping is present?
  
  (a) The damping force is always in the opposite direction to the velocity.
  
  (b) The damping force is always in the opposite direction to the acceleration.
  
  (c) The presence of damping gradually reduces the maximum potential energy of the system
  
  (d) The presence of damping gradually reduces the maximum kinetic energy of the system.
  
  Q.1
  Q.2
  Q.3
  Q.4
  Q.5
  
  c
  
  c
  
  a
  
  a
  
  b
  
  Related Resources:-
  
  You might like to simple harmonic motion.
  
  For getting an idea of the type of questions asked, refer the  Previous Year Question Papers.