Question

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

A
a2
B
a2
C
a2
D
a23

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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