Explanation:
Velocity is the time dervative of displacement.
Writing the given equation of a point performing SHM
\(x = a\sin \left( {\omega t + \frac{\pi }{6}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\)
Differentiating Eq. (1), w.r.t. time, we obtain
\(v=\dfrac{d x}{d t}=a \omega \cos \left(\omega t+\dfrac{\pi}{6}\right)\)
It is given that \(v=\dfrac{a \omega}{2}\), so that
\(\dfrac{a \omega}{2}=a \omega \cos \left(\omega t+\dfrac{\pi}{6}\right)\)
\(\begin{aligned}\cos \dfrac{\pi}{3} & =\cos \left(\omega t+\dfrac{\pi}{6}\right) \\\Rightarrow \quad t & =\dfrac{\pi}{6 \omega}=\dfrac{\pi \times T}{6 \times 2 \pi}=\dfrac{T}{12}\end{aligned}\)
Thus, at \(T/12\) velocity of the point will be equal to half of its maximum velocity.