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The potential energy of a particle of mass I kg in motion along the x-axis is given by: $U = 4 (1 - \cos 2 x)$ , where x is in metres. The period of small oscillation (in sec) is

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$2 π$

$\frac{π}{2}$

$\frac{π}{4}$

$π$

answer is C.

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Detailed Solution

$F = - \frac{dU}{dx} = - 8 \sin 2 x$

For small oscillations,

$\sin 2 x = 2 x$

i.e., a = -16x

Since a = -x

the oscillations are SH in nature

$T = 2 π \sqrt{| \frac{x}{a} |} = 2 π \sqrt{\frac{1}{16}} = \frac{π}{2} \sec$

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