A) \[2f\]
B) \[\frac{f}{2}\]
C) \[\frac{f}{3}\]
D) none of these
Correct Answer: A
Solution :
Using the relation, \[x=A\sin \omega t,\] \[U=\frac{1}{2}k{{x}^{2}}=\frac{1}{2}k{{A}^{2}}{{\sin }^{2}}\omega t\] \[=\frac{1}{4}k{{A}^{2}}(1-\cos \,2\omega t)\] So, frequency of potential energy is twice of frequency of displacement
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