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Question

A particle executes simple harmonic motion with a frequency $$'f'$$. The frequency with which its kinetic energy oscillates is?


A
f2
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B
f
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C
2f
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D
4f
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Solution

The correct option is C $$2f$$
Let, $$x=A\sin\omega t$$
$$v=\cfrac {dx}{dt}=A\omega \cos \omega t$$
Kinetic energy, $$K=\cfrac {1}{2}mv^2$$
$$\implies K=\cfrac {1}{2}m\omega ^2A^2\cos^2\omega t$$
$$\implies K=\cfrac {1}{2}m\omega^2A^2\left (\cfrac {1+cos2\omega t}{2}\right)$$
$$\implies K=\cfrac {1}{4}m\omega^2A^2(1+\cos^2\omega t)$$
$$\therefore \omega_K=2\omega$$
$$\therefore$$ Frequency of oscillation of K.E $$=2f\quad \left[f=\cfrac {\omega}{2\pi}\right]$$

Physics

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