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

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2π

π/2

$\sqrt{2}$ π

answer is C.

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Detailed Solution

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

$\begin{array}{l} a = \frac{F}{m} = - 8 \sin 2 x \\ For small oscillations sin 2x \approx 2x \\ i.e., a=-16x \\ since a α -x \\ ω = \sqrt{16} = 4 \\ Oscillations are simple harmonic in nature \\ T = \frac{2 π}{ω} = \frac{π}{2} s \end{array}$

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