First slide

Simple hormonic motion

Question

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

Difficult

A

$2 π$
B

$π$
C

$\frac{π}{2}$
D

$\frac{π}{4}$

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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