A point performs simple harmonic oscillation of period T and the...

Key Idea : Velocity is the time derivative of displacement. Writing the given equation of a point performing SHM $x=a \sin \left(\omega t+\frac{\pi}{6}\right)$ Differentiating Eq. (i), w.r.t. time, we obtain $y=\frac{d x}{d t}=a \omega \cos \left(\omega t+\frac{\pi}{6}\right)$ It is given that $y=\frac{a \omega}{2}$, so that or $\frac{a \omega}{2}=a \omega \cos \left(\omega t+\frac{\pi}{6}\right)$\frac{1}{2}=\cos \left(\omega t+\frac{\pi}{6}\right)$or$\cos \frac{\pi}{3}=\cos \left(\omega t+\frac{\pi}{6}\right)$$\begin{array}{ll} \text{or } & \omega t+\frac{\pi}{6}=\frac{\pi}{3} \\ \Rightarrow & \omega t=\frac{\pi}{6} \\ \text{or } & t=\frac{\pi}{6 \omega}=\frac{\pi \times T}{6 \times 2 \pi}=\frac{T}{12} \end{array}$$ Thus, at $\frac{T}{12}$ velocity of the point will be equal to half of its maximum velocity.