A) periodic but not SHM
B) non-periodic
C) simple harmonic motion with period 0.1 s
D) simple harmonic motion with period 0.2 s
Correct Answer: C
Solution :
Given, \[y=0.2\sin (10\pi t+1.5\pi )\cos (10\pi t+1.5\pi )\] \[y=0.1\sin 2(10\pi t+1.5\pi )\] [using 2 sin A cos A = sin2 A] \[y=0.1\sin (20\pi t+3\pi )\] This equation represents simple harmonic motion of angular frequency \[20\pi .\] \[\therefore \]Time period \[T=\frac{2\pi }{\omega }=\frac{2\pi }{20\pi }\] \[=\frac{1}{10}=0.1s\]You need to login to perform this action.
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