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
B
A22
C
A2
D
A

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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