Physics - SHM Question with Solution | TestHub

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 $y=\frac{dx}{dt}=a\omega \cos \left(\omega t+\frac{\pi }{6}\right)$ 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.