A) \[\frac{T}{4}\]
B) \[\frac{T}{8}\]
C) \[\frac{T}{12}\]
D) \[\frac{T}{2}\]
Correct Answer: C
Solution :
Let displacement equation of particle executing SHM is \[y=a\sin \omega t\] As particle travels half of the amplitude from the equilibrium position, so \[y=\frac{a}{2}\] Therefore, \[\frac{a}{2}=a\sin \omega t\] \[or\sin \omega t=\frac{1}{2}=\sin \frac{\pi }{6}\] \[or\omega t=\frac{\pi }{6}\] \[ort=\frac{\pi }{6\omega }\] \[ort=\frac{\pi }{6\left( \frac{2\pi }{T} \right)}\] \[\left( as\,\omega =\frac{2\pi }{T} \right)\] \[ort=\frac{T}{12}\] Hence, the particle travels half of the amplitude from the equilibrium in \[\frac{T}{12}\]sec.
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