The potential energy of a particle of mass 4kg in motion along the x-axis is given by U = 4(1-cos4x) J. The time period of the particle for small oscillation ( $\sin θ \approx θ$ ) is $(\frac{π}{k})$ s. The value of K is _____

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answer is 2.

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Detailed Solution

$U = 4 (1 - \cos 4 x)$

$F = - \frac{d U}{d x} = - 16 \sin 4 x$

For small $θ, \sin θ \approx θ$

$∴ F = - 16 \times 4 x$

$m a = - 64 x$

$4 a = - 64 x$

$a = - 16 x$

$w = \sqrt{16} = 4$ rad/sec

$T = \frac{2 π}{w} = \frac{2 π}{4} = \frac{π}{2}$

$∴ K = 2$

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