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Class 11
Physics
Oscillations
in exercise 14 9 let us take the position of mass...

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:

Given,

Spring constant, k = 1200 N/m

Mass, m = 6 kg

Displacement, A = 4.0 cm = 0.04 cm

ω = 14.14 s^-1

( i )

Since time is measured from mean position,

x = A sin ω t

x = 4 sin 14.14t

( ii ) At the maximum stretched position, the mass has an initial phase of π/2 rad.

Then, x = A sin( ωt + π/2 ) = A cos ωt

= 4 cos 14.14t

( iii ) At the maximum compressed position, the mass is at its leftmost position with an initial phase of 3π/2 rad.

Then, x = A sin( ωt + 3π/2 )

= -4 cos14.14 t

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