A) \[2\pi \]
B) \[\pi \]
C) \[\frac{\pi }{2}\]
D) \[\sqrt{2}\pi \]
Correct Answer: C
Solution :
[c] \[U=4\left( 1-cos2x \right)J\] \[F=-\frac{dU}{dx}=-8\sin 2x\] for small oscillation \[sin2x\approx 2x\] \[F=-16x\] Comparing with \[F=-kx\] k=16 \[m{{\omega }^{2}}=16\] \[\omega =\sqrt{\frac{16}{m}}\] \[T=2\pi \sqrt{\frac{m}{16}}\] \[T=2\pi \sqrt{\frac{1}{16}}=\frac{\pi }{2}\sec \]You need to login to perform this action.
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