A) anon-periodic
B) periodic but not SHM
C) SHM with period 0.2 s
D) SHM with period 0.1 s
Correct Answer: D
Solution :
Given, \[y=0.2\sin \,(10\pi t+1.5\pi )\cos (10\pi t+1.5\pi )\] \[=0.1\sin 2(10\pi t+1.5\pi )\] \[[\,\,\,\,\sin 2A=2\sin A\cos A]\] \[=0.1\sin (20\pi t+3.0\pi )\] So, the motion of particle is SHM Time period, \[T=\frac{2\pi }{\omega }=\frac{2\pi }{20\pi }=\frac{1}{10}=0.1\,\,s\]You need to login to perform this action.
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