# Q. 14.10 (a) In Exercise 14.9, let us take the position of mass when the spring is unstreched as $\inline x=0,$ 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 $\inline (t=0),$ the mass is(a) at the mean position,In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?

S Sayak

Amplitude is A = 0.02 m

Time period is $\\\omega$

$\\\omega =\sqrt{\frac{k}{m}}\\ \omega =\sqrt{\frac{1200}{3}}\\ \omega =20\ rad/s$

(a) At t = 0 the mass is at mean position i.e. at t = 0, x = 0

$\\x(t)=0.02sin\left ( 20t \right )$

Here x is in metres and t is in seconds.

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