A) \[2s\]
B) \[3s\]
C) \[8s\]
D) \[12s\]
Correct Answer: A
Solution :
: Here, Amplitude, \[A=0.2\text{ }m\] Time period, \[T=24\text{ }s\] Since time is noted from the mean position, hence displacement x of a particle from its mean position is given by \[x=A\sin \omega t\] Here, \[x=0.1m\] \[\therefore \] \[0.1=0.2\sin \omega t\] \[\frac{1}{2}=\sin \omega t\] \[\sin \left( \frac{\pi }{6} \right)=\sin \omega t\] \[\omega t=\frac{\pi }{6}\] \[t=\frac{\pi }{60}=\frac{\pi }{6}\left( \frac{T}{2\pi } \right)\] \[\left( \because \,\,\omega =\frac{2\pi }{T} \right)\] \[=\frac{\pi \times 24}{6\times 2\pi }=2s\]
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