A) T/4
B) T/8
C) T/12
D) 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 \[\,\omega t=\frac{\pi }{6}\] or \[t=\frac{\pi }{6\omega }\] or \[t=\frac{\pi }{6\left( \frac{2\pi }{T} \right)}\] \[\left( as\,\,\omega =\frac{2\pi }{T} \right)\] or \[t=\frac{T}{12}\] Hence, the particle travels half of the amplitude from the equilibrium in \[\frac{T}{12}s\].You need to login to perform this action.
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