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In Exercise 9, let us take the position...

In Exercise 9, let us take the position of mass when the spring is unstretched 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 at the maximum compressed position.
In what way to these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?

Text Solution

Verified by Experts

Amplitude from exercise : `14.9, A = 2 cm`
Angular frequency `omega = sqrt((k)/(m)) = sqrt((1200)/(3)) = sqrt(400) = 20 rad s^(-1)`
Initial phase in the maximum compressed position `phi = (3pi)/(2) rad`
Displacement of block at time t `x(t) = A sin (omega t+phi)`
` = 2sin (2omega t+(3pi)/(2))= -2cos 20 t cm`.
Hence, in all these three cases, amplitude and frequency are same but differ only in initial phase because motions starts from different positions.
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KUMAR PRAKASHAN-OSCILLATIONS-SECTION-B (NUMERICAL FROM TEXTUAL EXERCISE)
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