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The potential energy of a particle of mass $4 k g$ in motion along the x – axis is given by $U = 4 (1 - c o s 4 x) J .$ The time period of the particle for small oscillation $(\sin θ ≃ θ)$ is $\frac{π}{K} .$ The value of K is ___

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answer is D.

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Detailed Solution

$U = 4 (1 - \cos 4 x), F = - \frac{d U}{d x} = - (\sin 4 x) 4 = - 64 x$

For small $θ$

$\sin θ \approx θ$

$F = - 64 x, a = - \frac{64 x}{m} = - 16 x$

$ω^{2} = 16, T = \frac{2 π}{ω} = \frac{π}{2}$

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