Question

The displacement of the 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

• A. 12${\omega }^2$
• B. 36${\omega }^2$
• C. 144${\omega }^2$
• D. $\sqrt{192}$${\omega }^2$

Answer 1

The correct option is B 36${\omega }^2$

$x=12\sin \omega t-16{\sin }^3\omega t=4[3\sin \omega t-4{\sin }^3\omega t]=4\left[\sin 3\omega t\right](by\mathrm{using}\sin 3\theta =3\sin \theta -4{\sin }^3\theta )$

So maximum acceleration
$A_{max}={\left(3\omega \right)}^2\times 4=36{\omega }^2$