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Question

The kinetic energy and potential energy of a particle executing simple harmonic motion will be equal, when displacement (amplitude=a) is :


A
a2
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B
a2
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C
a2
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D
a23
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Solution

The correct option is C $$\dfrac { a }{ \sqrt { 2 } } $$
Suppose at displacement  $$y$$ from mean position. 
K.E $$=\dfrac{1}{2}m(a^{2}-y^{2})\omega^{2}$$

P.E $$=\dfrac{1}{2}m\omega^{2}y^{2}$$

potential energy = kinetic energy
$$\Rightarrow \dfrac{1}{2}m(a^{2}-y^{2})\omega^{2}=\dfrac{1}{2}m\omega^{2}y^{2}$$

$$\Rightarrow a^{2}=2y^{2}\Rightarrow y=\dfrac{a}{\sqrt{2}}$$

Physics

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