Displacement-time equation of a particle executing $SHM$ is $x\, = \,A\,\sin \,\left( {\omega t\, + \,\frac{\pi }{6}} \right)$ Time taken by the particle to go directly from $x\, = \, - \frac{A}{2}$ to $x\, = \, + \frac{A}{2}$ is
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displacement, $X=A \sin \left(\omega t+\frac{\pi}{6}\right)$
At $X=-\frac{A}{2}$
$-\frac{A}{2}=A \sin \left(\omega t_{1}+\frac{\pi}{6}\right)$
$\omega t_{1}+\frac{\pi}{6}=\sin ^{-1}(-0.5)$
$\omega t_{1}+\frac{\pi}{6}=-\frac{\pi}{6}$
$t_{1}=-\frac{\pi}{3 \omega} \ldots .(1)$
At $X=+\frac{A}{2}$
$\frac{A}{2}=A \sin \left(\omega t_{2}+\frac{\pi}{6}\right)$
$\omega t_{2}+\frac{\pi}{6}=\sin ^{-1}(0.5)$
$\omega t_{2}+\frac{\pi}{6}=\frac{\pi}{6}$
$\omega t_{2}=0$
$t_{2}=0 \ldots \ldots .(2)$
The time taken by the particle,
$T=t_{2}-t_{1}$
$T=0-\left(-\frac{\pi}{3 \omega}\right)$
$T=\frac{\pi}{3 \omega}$
The correct option is $A.$
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