The potential energy of a particle of mass $1 k g$ in motion along the $x$ -axis is given by $U = 4 (1 - cos 2 x) J$ . Here $x$ is in meter. The period of small oscillations (in sec) is _______.

Question

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Answer 1

$M = 1 k g$
$U = 4 (1 - cos 2 x)$
$F = - \frac{d u}{d x}$
$F = - 4 [0 - (- sin 2 x) (2)]$
$F = - 8 sin 2 x N$
$a = - 8 sin 2 x m / sec^{2} [∵ m = 1 k g]$
$\Rightarrow a ≃ - 8 (2 x)$
$a = - 16 x$ [For small $x, sin x \approx x$ ]
$\Rightarrow ω^{2} = 4^{2}$
$ω = 4$ rad /ac
$T = \frac{2 π}{w} = \frac{π}{2} sec$

Answer 2

M = 1 k g

U = 4 (1 - cos 2 x)

F = - \frac{d u}{d x}

F = - 4 [0 - (- sin 2 x) (2)]

F = - 8 sin 2 x N

a = - 8 sin 2 x m / sec^{2} [∵ m = 1 k g]

\Rightarrow a ≃ - 8 (2 x)

a = - 16 x

[For small

x, sin x \approx x

]

\Rightarrow ω^{2} = 4^{2}

ω = 4

rad /ac

T = \frac{2 π}{w} = \frac{π}{2} sec