A particle in $S.H.M.$ is described by the displacement function $x(t) = a\cos (\omega t + \theta )$. If the initial $(t = 0)$ position of the particle is $1\, cm $ and its initial velocity is $\pi \,cm/s$. The angular frequency of the particle is $\pi \,rad/s$, then it’s amplitude is
Diffcult
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(b) $x = a\cos (\omega \,t + \theta )$ ….(i)
and $v = \frac{{dx}}{{dt}} = - a\omega \sin (\omega \,t + \theta )$ ….(ii)
Given at $t = 0$, $x = 1\,cm$ and $v = \pi $ and $\omega = \pi $
Putting these values in equation (i) and (ii) we will get $\sin \theta = \frac{{ - 1}}{a}$ and $\cos \theta = \frac{1}{A}$
==> ${\sin ^2}\theta + {\cos ^2}\theta = {\left( { - \frac{1}{a}} \right)^2} + {\left( {\frac{1}{a}} \right)^2}$
==> $a = \sqrt 2 \,cm$
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