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?

Asked by Pragya Singh | 1 year ago |  56

1 Answer

Solution :-

Distance travelled by the mass sideways, a = 2.0 cm

Angular frequency of oscillation:

ω = \( \dfrac{\sqrt{k}}{m}\)

\( \dfrac{\sqrt{1200}}{3}\)


We get,

= 20 rad s-1


(a) As time is noted from the mean position,

Hence, using

x = a sin ωt

We have,

x = 2 sin 20 t


(b) At maximum stretched position, the body is at the extreme right position,

with an initial phase of \( \dfrac{π}{2}\) rad. Then,

x = \( \dfrac{ a sin (ωt + π) }{2}\)

= a cos ωt

= 2 cos 20 t


(c) At maximum compressed position, the body is at left position,

with an initial phase of \( \dfrac{ 3π}{2}\) rad.


x =\( \dfrac{ a sin (ωt + 3π) }{2}\)

= – a cos ωt

= – 2 cos 20 t


The functions neither differ in amplitude nor in frequency. They differ in initial phase.

Answered by Abhisek | 1 year ago

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