$v = \frac{{dx}}{{dt}} = - A\omega \sin \left( {\omega \,t + \frac{\pi }{4}} \right)$
For maximum speed, $\sin \,\left( {\omega \,t + \frac{\pi }{4}} \right) = 1$
==> $\omega \,t + \frac{\pi }{4} = \frac{\pi }{2}$ or $\omega \,t = \frac{\pi }{2} - \frac{\pi }{4}$
==> $t = \frac{\pi }{{4\omega }}$
$1.$ If the total energy of the particle is $E$, it will perform periodic motion only if
$(A)$ $E$ $<0$ $(B)$ $E$ $>0$ $(C)$ $\mathrm{V}_0 > \mathrm{E}>0$ $(D)$ $E > V_0$
$2.$ For periodic motion of small amplitude $\mathrm{A}$, the time period $\mathrm{T}$ of this particle is proportional to
$(A)$ $\mathrm{A} \sqrt{\frac{\mathrm{m}}{\alpha}}$ $(B)$ $\frac{1}{\mathrm{~A}} \sqrt{\frac{\mathrm{m}}{\alpha}}$ $(C)$ $\mathrm{A} \sqrt{\frac{\alpha}{\mathrm{m}}}$ $(D)$ $\mathrm{A} \sqrt{\frac{\alpha}{\mathrm{m}}}$
$3.$ The acceleration of this particle for $|\mathrm{x}|>\mathrm{X}_0$ is
$(A)$ proportional to $\mathrm{V}_0$
$(B)$ proportional to $\frac{\mathrm{V}_0}{\mathrm{mX}_0}$
$(C)$ proportional to $\sqrt{\frac{\mathrm{V}_0}{\mathrm{mX}_0}}$
$(D)$ zero
Give the answer qustion $1,2$ and $3.$

