A particle is executing $SHM$ with amplitude $A,$ time period $T,$ maximum acceleration $a_o$ and maximum velocity $v_0.$ Its starts from mean position at $t=0$ and at time $t,$ it has the displacement $A/2,$ acceleration $a$ and velocity $v$ then
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Let the displacement at t is $x=A \sin \omega t=A \sin \left(\frac{2 \pi}{T}\right) t$
here, $x=\frac{A}{2}=A \sin \left(\frac{2 \pi}{T}\right) t$
or, $\sin \left(\frac{2 \pi}{T}\right) t=\frac{1}{2}=\sin \frac{\pi}{6} \Rightarrow t=\frac{T}{12}$
$\frac{d x}{d t}=\frac{2 \pi}{T} A \cos \left(\frac{2 \pi}{T}\right) t$
$\frac{d^{2} x}{d t^{2}}=-\left(\frac{2 \pi}{T}\right)^{2} A \sin \left(\frac{2 \pi}{T}\right) t$
so, $v_{\max }=v_{0}=\left(\frac{2 \pi}{T}\right) A, a_{\max }=a_{0}=\left(\frac{2 \pi}{T}\right)^{2} A$
now at $t=T / 12,$ the velocity, $v=\frac{d x}{d t}=\frac{2 \pi}{T} A \cos \left(\frac{2 \pi}{T}\right) \frac{T}{12}$
$=\frac{2 \pi}{T} A \cos \frac{\pi}{6}=\frac{2 \pi}{T} A \frac{\sqrt{3}}{2}$
$\therefore v=\sqrt{3} v_{0}$
at $t=T / 12,$ the acceleration, $a=\frac{d^{2} x}{d t^{2}}=\left(\frac{2 \pi}{T}\right)^{2} A \sin \left(\frac{2 \pi}{T}\right) \frac{T}{12}$
$=\frac{2 \pi}{T} A \sin \frac{\pi}{6}=\frac{2 \pi}{T} A \frac{1}{2}$
$\therefore a=\frac{a_{0}}{2}$
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