A) \[f\]
B) \[f/2\]
C) \[2f\]
D) 0
Correct Answer: C
Solution :
If \[x=A\text{ }sin\text{ }\omega t\] Then, \[PE=\frac{1}{2}m{{A}^{2}}{{\omega }^{2}}{{\sin }^{2}}\omega t\] \[\therefore \] \[PE=\frac{1}{2}m{{A}^{2}}{{\omega }^{2}}\left( \frac{1-\cos 2\omega t}{2} \right)\] \[\therefore \] \[\omega =2\omega \] Or \[2\pi f=2\times 2\pi f\] \[f=2f\]
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