Question

A 200 g mass hangs from a massless spring (k = 10 N/m). At t = 0.0 s, the mass is 20 cm below the equilibrium point and moving upward with a speed of 100 cm/s. What is the a. oscillation frequency? b. distance from equilibrium when the speed is 50 cm/s? c. distance from equilibrium at t = 1.0 s?

Answer 1

Given : m = 200 g (0.2 kg) ; k = 10 N/m ; y0 = -20cm ; v0 = 100 cm/s (1 m/s) ; v1 = 50 cm/s (0.50 m/s)

Solution :

(a) Oscillation frequency

we know that ,frequency of oscillation is given by :

By putting values ,we have :

= 1.125 Hz

Answwe : 1.125 Hz

(b) Distance from equilibrium at v1 = 50 cm/s

According to conservation of Energy,we have :

By rearranging ,we get :

By putting values ,we have :

= 0.2345 m or 23.45 cm

Answer : 23.45 cm

(c) Distance from equilibrium at t = 1 s

The displacement at time t is given by :

Where ,A is the amplitude , is the angular frequency and is the phase constant.

Angular frequency is given by:

= 7.065 rad/s

Amplitude is given by:

= 0.245 m

The phase constant is given by:

= 2.526 rad/s

Here ,negative displacement inducates that the displacement is below the equilibrium point.

i.e. at t = 1 s

= -0.0409m

= -4.09 cm

Answer : x = -4.09 cm (The block is 4.09 cm below the equilibrium point)