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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