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Question

When particle oscillates simple harmonically its potential energy varies periodically. If frequency of the particle is n, the frequency of the potential energy is


A
n/2
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B
n
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C
2n
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D
4n
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Solution

The correct option is B 2n
$$PE=\frac { 1 }{ 2 } m\omega ^{ 2 }x^{ 2 }\\ =\frac { 1 }{ 2 } m\omega ^{ 2 }(x_{ 0 }^{ 2 }\sin ^{ 2 }{ \omega t } )\\ =\frac { 1 }{ 2 } m\omega ^{ 2 }x_{ 0 }^{ 2 }\left( \frac { 1-\cos { 2\omega t }  }{ 2 }  \right) \\ =\frac { 1 }{ 4 } m\omega ^{ 2 }x_{ 0 }^{ 2 }-\frac { 1 }{ 4 } m\omega ^{ 2 }x_{ 0 }^{ 2 }\cos { 2\omega t }  $$
frequency of above equation i.e. frequency of P.E. is $$2\omega$$ or $$2n$$

Physics

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