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Question

A particle of mass 2 kg moves in simple harmonic motion and its potential energy U varies with position x as shown. The period of oscillation of the particle is:

219039.png


A
2π5 s
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B
22π5 s
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C
2π5 s
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D
4π5 s
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Solution

The correct option is D $$\displaystyle \frac{4\pi }{5}$$ s
$$\displaystyle \frac{1}{2}kA^{2}=1.0$$
$$\displaystyle \frac{1}{2}k\left ( 0.4 \right )^{2}=1.0$$
or $$k=12.5$$ or $$\displaystyle \frac{25}{2}$$ N/m
$$\displaystyle T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{2}{\frac{25}{2}}}=\frac{4\pi }{5}$$ s

Physics

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