# Q. 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 :

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Solution:

## 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 $\therefore \dfrac{1}{2} kx^{2} = \dfrac{1}{2} kA^{2} - \dfrac{1}{2} kx^{2}$ $x = \pm \dfrac{A}{\sqrt{2}}$ $\therefore$ Correct answer is (3)

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