A) \[\frac{T}{8}\]
B) \[\frac{T}{6}\]
C) \[\frac{T}{3}\]
D) \[\frac{T}{12}\]
Correct Answer: D
Solution :
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.
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