This was very straightforward and remarkably similar to No. 6 part (i) of the same year. No SHM for this part. Forces up = forces down and forces left = forces right. You also needed a little trigonometry to figure out the angle at the bottom. Answer: T = 13W/10 Straightforward. Answer: k = 125W/6 1996 (a) All straightforward, just pick the correct equation. No. of oscillations per second = 5/2π Amplitude = 0.89 m Max acceleration = 258 rads/sec 1996 (b) Straightforward once you establish that the particle is undergoing SHM on its way back until it reaches the natural length (a) away from p, at which stage it travels at constant velocity (this velocity corresponds to the max velocity of the SHM stage). Summary of key points The of a spring or a string is the distance between the and the end of extension Current Position object’s of the string (in other words, how much the string is extended beyond its natural length). the natural length The is the distance between the (R.P.) and the (E.P.). Amplitude Release Point Equilibrium Position When do you first need to note the release point? proving that the motion of the particle is simple harmonic. After Equilibrium Position How do you find equilibrium position for a string in the horizontal direction? Equilibrium position is at the end of the string. In general, what concept do we use to find equilibrium position for a string in the vertical direction? Force down equals force up. What will the equation look like (for a string in the vertical direction)? mg = k {ext} Why should we use the letter to represent extension for this stage? ‘d’ Because we will use to represent distance from E.P. for the next stage. ‘x’ To show SHM Complete the show SHM, extend ___________________________ following: “To the object a distance beyond E.P. and use Fsmall Fbig = x “Extend – ma” The first line of the solution to show SHM in the horizontal direction (one string, no friction) will be: k{ext} = ma –
This was very straightforward and remarkably similar to No. 6 part (i) of the same year. No SHM for this part. Forces up = forces down and forces left = forces right. You also needed a little trigonometry to figure out the angle at the bottom. Answer: T = 13W/10 Straightforward. Answer: k = 125W/6 1996 (a) All straightforward, just pick the correct equation. No. of oscillations per second = 5/2π Amplitude = 0.89 m Max acceleration = 258 rads/sec 1996 (b) Straightforward once you establish that the particle is undergoing SHM on its way back until it reaches the natural length (a) away from p, at which stage it travels at constant velocity (this velocity corresponds to the max velocity of the SHM stage). Summary of key points The extension of a spring or a string is the distance between the object’s Current Position and the end of the natural length of the string (in other words, how much the string is extended beyond its natural length). The Amplitude is the distance between the Release Point (R.P.) and the Equilibrium Position (E.P.). When do you first need to note the release point? After proving that the motion of the particle is simple harmonic. Equilibrium Position How do you find equilibrium position for a string in the horizontal direction? Equilibrium position is at the end of the string. In general, what concept do we use to find equilibrium position for a string in the vertical direction? Force down equals force up. What will the equation look like (for a string in the vertical direction)? mg = k {ext} Why should we use the letter ‘d’ to represent extension for this stage? Because we will use ‘x’ to represent distance from E.P. for the next stage. To show SHM Complete the following: “To show SHM, extend ___________________________ “Extend the object a distance x beyond E.P. and use Fsmall – Fbig = ma” The first line of the solution to show SHM in the horizontal direction (one string, no friction) will be: – k{ext} = ma