Question

A massless spring hangs from the ceiling with a small object attached to its lower end. The object is initially held at rest in a position y_i such that the spring is at its rest length. The object is then released from y_i and oscillates up and down, with its lowest position being 14 cm below y_i. (a) What is the frequency of the oscillation? (b) What is the speed of the object when it is 12 cm below the initial position? (c) An object of mass 240 g is attached to the first object, after which the system oscillates with half the original frequency. What is the mass of the first object? (d) How far below y_i is the new equilibrium (rest) position with both objects attached to the sping?

Answer 1

For simple harmonic oscillation, F=kx mg = kx Assume that when the spring-block system is at rest, it is having twice the amplitude 14 cm -0.07 m m × 9.8 m/2-k× 0.07m k 9.8 m 0.07s -=140 1 1/ -140 1 =11.83 So, frequency of the oscillation is,

2π ㄧㄧㄨ ]- 11.83 2x3.14 f=1.88 Hz Maximum potential energy is, PD, xkx0.072 = 2.45 x 10-3 k When the displacement is, x=12cm =0.12m Potential energy is,

(PE),--kỷ --xkx0. 122 = 7.2 x10-3k So, the change in potential energy is, APE) (7.2- 2.45)x10-3 k 4.75x10-3 k This change in energv is equal to the kinetic energv, KEm

4.75x10-3 k v2x4.75x103 x -2x4.75x10 x140 v-1.33 v 1.15m Therefore the speed of the object when it is 12 cm below the initial position is1.15 f is the new frequency after the mass of 240 gm is attached to the object.

(m+0.240 kg) 2m ㄧㄧㄨㄧ- (m+0.240 kg) 4 m 4m (m+0.240kg) 3m = 0.240 m 0.08 kg m-80 gm Therefore the mass of the first object is S0 gm For the equilibrium position, we have quadrupled the force, for which we need to quadruple the displacement 4x7 cm -28 cm

Therefore the new equilibrium position is 28 cm below the yi position