A) \[\frac{T}{8}\]
B) \[\frac{T}{6}\]
C) \[\frac{T}{3}\]
D) \[\frac{T}{12}\]
Correct Answer: D
Solution :
Key Idea: Velocity is the time derivative of displacement. Writing .the given equation of a point performing SHM\[x=a\sin \left( \omega t+\frac{\pi }{6} \right)\] ...(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 \[v=\frac{a\omega }{2},\]so that \[\frac{a\omega }{2}=a\omega \cos \left( \omega t+\frac{\pi }{6} \right)\]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\[\omega t+\frac{\pi }{6}=\frac{\pi }{3}\] \[\Rightarrow \]\[\omega t=\frac{\pi }{6}\]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.
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