The displacement of a particle varies with time as $x = 12\sin \omega t - 16{\sin ^3}\omega t$ (in $cm$). If its motion is $S.H.M.$, then its maximum acceleration is
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(b) $x = 12\sin \omega \,t - 16\,{\sin ^3}\omega \,t = 4[3\sin \omega \,t - 4{\sin ^3}\omega \,t]$
$ = 4[\sin 3\omega \,t]$ (By using $\sin 3\theta = 3\sin \theta - 4{\sin ^3}\theta )$
$\therefore $ maximum acceleration ${A_{\max }} = {(3\omega )^2} \times 4 = 36{\omega ^2}$
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