A) \[f/2\]
B) \[f\]
C) \[2f\]
D) \[4f\]
Correct Answer: C
Solution :
Kinetic energy \[K=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}m{{a}^{2}}{{\omega }^{2}}{{\cos }^{2}}\omega \,t\] \[=\frac{1}{2}m{{\omega }^{2}}{{a}^{2}}(1+\cos 2\omega \,t)\] hence kinetic energy varies periodically with double the frequency of S.H.M. i.e. \[2\omega \].
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