Explanation

(c) 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, $\frac{a}{2} = a\sin{\omega t} \Rightarrow \sin{\omega t} = \frac{1}{2} = \sin\frac{\pi}{6}$
$\Rightarrow \omega t = \frac{\pi}{6} \Rightarrow t = \frac{\pi}{6\omega} \Rightarrow t = \frac{\pi}{6\left( \frac{2\pi}{T} \right)}\left( As\ \omega = \frac{2\pi}{T} \right)$
$\Rightarrow t = \frac{T}{12}$
Hence, the particle travels half of the amplitude from the equilibrium in $\frac{T}{12}s$