Physics - SHM Question with Solution | TestHub

Let displacement equation of particle executing SHM is $y=a\sin \omega t$ As particle travels half of the amplitude from the equilibrium position, so $y=\frac{a}{2}$ Therefore, $\frac{a}{2}=a\sin \omega t$ or $\sin \omega t=\frac{1}{2}=\sin \frac{\pi }{6}$ or $\omega t=\frac{\pi }{6}$ $or$ $t=\frac{\pi }{6\omega }$ or $t=\frac{\pi }{6\left(\frac{2\pi }{T}\right)}\left(\text{ as }\omega =\frac{2\pi }{T}\right)$ or $t=\frac{T}{12}$ Hence, the particle travels half of the amplitude from the equilibrium in $\frac{T}{12}\mathrm{s}$.