• # question_answer A point performs simple harmonic oscillation of period T and the equation of motion is given by $x=a\,\sin \,\left( \omega t+\frac{\pi }{6} \right)$.After the elapse of what fraction of the time period, the velocity of the point will be equal to half of its maximum velocity? A)  $\frac{T}{8}$                                   B)  $\frac{T}{6}$ C)  $\frac{T}{3}$                                   D)  $\frac{T}{12}$

Writing the given equation of a point performing SHM $x=a\,\,\sin \,\,\left( \omega t+\frac{\pi }{6} \right)$             ?(i) Differentiating Eq. (i), w.r.t. time, we obtain $v=\frac{dx}{dt}=a\omega \,\,\cos \,\,\left( \omega t+\frac{\pi }{6} \right)$ It is given that $v=\frac{2\omega }{2},$ so that $\frac{a\omega }{2}=a\omega \,\,\cos \,\,\left( \omega t+\frac{g}{6} \right)$ or         $\frac{1}{2}=\cos \,\left( \omega t+\frac{\pi }{6} \right)$ or         $\cos \,\frac{\pi }{3}\,=\cos \,\left( \omega t+\frac{\pi }{6} \right)$ or         $\frac{\pi }{3}=\omega t+\frac{\pi }{6}\,\,\Rightarrow \,\,\omega t\,=\frac{\pi }{6}$ or         $t=\frac{\pi }{6\omega }=\frac{\pi \times T}{6\times 2\pi }=\frac{T}{12}$ Thus, at $\frac{T}{12}$ velocity of the point will be equal to half of its maximum velocity.