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Question

A particle is executing simple harmonic motion (SHM) of amplitude A, along the x-axis, about x = 0. When its potential Energy (PE) equals kinetic energy (KE), the position of the particle will be :


A
A2
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B
A22
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C
A2
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D
A
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Solution

The correct option is D $$\dfrac{A}{\sqrt{2}}$$
Potential energy $$(U) = \dfrac{1}{2} kx^2$$
Kinetic energy $$(K) = \dfrac{1}{2} kA^2 - \dfrac{1}{2} kx^2$$
According to the question, $$U = k$$
$$\therefore \dfrac{1}{2} kx^2 = \dfrac{1}{2} kA^2 - \dfrac{1}{2} kx^2$$
$$x = \pm \dfrac{A}{\sqrt{2}}$$

Physics

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