The displacement-time equation of a particle | vedclass

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Question

(A) The displacement equation is $x = A \sin \left( \omega t + \frac{\pi}{6} \right)$.At $x = -\frac{A}{2}$:$-\frac{A}{2} = A \sin \left( \omega t_1 +

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$\frac{\pi}{3\omega}$

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(A) The displacement equation is $x = A \sin \left( \omega t + \frac{\pi}{6} \right)$.At $x = -\frac{A}{2}$:$-\frac{A}{2} = A \sin \left( \omega t_1 + \frac{\pi}{6} \right)$$\sin \left( \omega t_1 + \frac{\pi}{6} \right) = -\frac{1}{2} = \sin \left( -\frac{\pi}{6} \right)$$\omega t_1 + \frac{\pi}{6} = -\frac{\pi}{6}$$\omega t_1 = -\frac{\pi}{3} \implies t_1 = -\frac{\pi}{3\omega}$.At $x = +\frac{A}{2}$:$\frac{A}{2} = A \sin \left( \omega t_2 + \frac{\pi}{6} \right)$$\sin \left( \omega t_2 + \frac{\pi}{6} \right) = \frac{1}{2} = \sin \left( \frac{\pi}{6} \right)$$\omega t_2 + \frac{\pi}{6} = \frac{\pi}{6}$$\omega t_2 = 0 \implies t_2 = 0$.The time taken is $\Delta t = t_2 - t_1 = 0 - \left( -\frac{\pi}{3\omega} \right) = \frac{\pi}{3\omega}$.

The displacement-time equation of a particle executing $SHM$ is $x = A \sin \left( \omega t + \frac{\pi}{6} \right)$. The time taken by the particle to go directly from $x = -\frac{A}{2}$ to $x = +\frac{A}{2}$ is

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