A) A simple harmonic motion with a period \[2\pi /\omega \]
B) A simple harmonic motion with a period \[\pi /\omega \]
C) A periodic but not simple harmonic motion with a period \[2\pi /\omega \]
D) A periodic but not simple harmonic, motion with a period \[\pi /\omega \]
Correct Answer: D
Solution :
\[y={{\sin }^{2}}\omega \,t\]\[=\frac{1-\cos 2\omega t}{2}\]Þ Period,\[T=\frac{2\pi }{2\omega }=\frac{\pi }{\omega }\] The given function is not satisfying the standard differential equation of S.H.M. \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=-\,{{\omega }^{2}}y\]. Hence it represents periodic motion but not S.H.M.You need to login to perform this action.
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