A particle executes simple harmonic oscillation with an amplitude a...

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 $\sin \omega t=\frac{1}{2}=\sin \frac{\pi}{6} or \omega t=\frac{\pi}{6} or t=\frac{\pi}{6 \omega} or t=\frac{\pi}{6\left(\frac{2 \pi}{T}\right)} \quad\left(\text{ as } \omega=\frac{2 \pi}{T}\right) or \quad t=\frac{T}{12}$ Hence, the particle travels half of the amplitude from the equilibrium in $\frac{T}{12} \mathrm{~s}$.