Question

1. A 500 g block is attached to a spring on a frictionless horizontal surface. The block is pulled to stretch the spring by 10 cm, then is gently released. A short time later, as the block passes through the equilibrium position, its speed is 1.0 m/s.

a) What is the amplitude of the oscillation?

b) What is the phase constant?

c) What is the block’s period of oscillation?

d) What is the spring constant of the spring?

e) What is the block’s speed at the point where the spring is compressed by 5.0 cm?

Answer 1

Given Data,

Amplitude of oscillation of spring , A = 10 cm = 0.1 m -- >Answer a

mass of the block = 500 g = 0.5 kg ,

speed at the equilibrium position = 1 m/s.

We know that, Velocity at equilibrium position is given by, V = A*w

Hence, angular frequency of the oscillation is given by, w = V / A

= 1 / 0.1

= 10 rad/s  

Hence, block's period of oscillation is given by, T = 2*pi / w

= 2*pi / 10

= 0.628 s --> Answer C

when the block is fully stretched , the spring will have only Potential energy and no KE

when the block passes the equilibrium position, it will have only KE

by the conservation of Energy , we have

PE = KE

0.5kx^2 = 0.5mv^2

k*(0.1)^2 = 0.5*1^2

k = 50 N/m --> Answer d

The block's speed at the point where the spring is compressed by 5 cm = 0.05 m is given by,

V' = w * sqrt (A^2 - x^2)

= 10 *sqrt (0.1^2 - 0.05^2)

= 0.866 m/s ---> Answer (e)