Question

In Exercise 13.9, 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

x = 2sin 20t

x = 2cos 20t

x = –2cos 20t

The functions have the same frequency and amplitude, but different initial phases.

Distance travelled by the mass sideways, A = 2.0 cm

Force constant of the spring, k = 1200 N m–1

Mass, m = 3 kg

Angular frequency of oscillation:

$w=\sqrt\frac{k}{m}$

$=\sqrt\frac{1200}{3}=\sqrt400=20\,rad\,s^{-1}$

When the mass is at the mean position, initial phase is 0.

Displacement, x = Asin ωt

= 2sin 20t

At the maximum stretched position, the mass is toward the extreme right. Hence, the

initial phase is $\frac{\pi}{2}.$

Displacement, $x= A \,sin(wt+\frac{\pi}{2})$

$=2 \,sin(20t+\frac{\pi}{2})$

= 2cos 20t

At the maximum compressed position, the mass is toward the extreme left. Hence, the

initial phase is $\frac{3\pi}{2}$

Displacement, $x= A \,sin(wt+\frac{\pi}{2})$

$=2\,sin(20t+\frac{3\pi}{2})=-2\,cos\,20t$

The functions have the same frequency $(\frac{20}{2\pi}Hz)$ and amplitude (2 cm), but different initial phases (initial phases $(0,\frac{\pi}{2},\frac{3\pi}{2})$