Question

In Exercise 14.9, let us take the position of mass when the spring is unstreched as x = 0, and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass is (a) at the mean position, (b) at the maximum stretched position, and (c) at the maximum compressed position. In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?

Answer 1

Given: The position of the mass when spring is un-stretched is x=0, the mass of the object is 3 kg, spring constant is 1200 N/m and the amplitude is 2 cm.

The direction from left to right is positive.

The angular frequency of spring-mass system is given as,

ω= k m (1)

Where, k is the spring constant and m is the mass.

By substituting the given values in equation (1), we get

ω= 1200 3 = 400 =20  rad/s

a)

At the mean position the SHM equation is given as,

x=A( sinωt+ϕ )(2)

When the mass is at mean position, the initial phase angle is zero.

By substituting the given values in equation (2), we get

x=2sin20t

b)

At maximum stretched position, the mass is towards extreme right. Thus, the initial phase is π 2 .

By substituting the given values in equation (2), we get

x=2sin( 20t+ π 2 ) x=2cos( 20t )

c)

At maximum compressed position the mass is towards extreme left. The initial phase is 3π 2

By substituting the given values in equation (2), we get

x=2sin( 20t+ 3π 2 ) x=−2cos( 20t )

All the above functions have the same frequency and amplitude.

The amplitude and frequency of resulting function is given by,

A=2 cm ν= 20 2π  Hz

Thus, the function have same amplitude and frequency but different initial phases.