The displacement-time equation of a particle | vedclass

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Question

(A) The displacement of the particle is given by $x = A \sin(\omega t + \phi)$.At $x = -\frac{A}{2}$,we have $-\frac{A}{2} = A \sin(	heta_1) \implies

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

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(A) The displacement of the particle is given by $x = A \sin(\omega t + \phi)$.At $x = -\frac{A}{2}$,we have $-\frac{A}{2} = A \sin(	heta_1) \implies \sin(	heta_1) = -\frac{1}{2}$,so $	heta_1 = -\frac{\pi}{6}$ (or $330^{\circ}$).At $x = +\frac{A}{2}$,we have $+\frac{A}{2} = A \sin(	heta_2) \implies \sin(	heta_2) = \frac{1}{2}$,so $	heta_2 = \frac{\pi}{6}$ (or $30^{\circ}$).The phase difference $\Delta 	heta = 	heta_2 - 	heta_1 = \frac{\pi}{6} - (-\frac{\pi}{6}) = \frac{\pi}{3}$ radians (which corresponds to $60^{\circ}$).Since $\Delta 	heta = \omega \Delta t$,the time taken is $\Delta t = \frac{\Delta 	heta}{\omega} = \frac{\pi / 3}{\omega} = \frac{\pi}{3\omega}$.

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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