For what value of displacement the kinetic energy and potential energy of a simple harmonic oscillator become equal?

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$x = \pm A$

$x = \frac{A}{2}$

$x = 0$

$x = \pm \frac{A}{\sqrt{2}}$

answer is B.

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Detailed Solution

Given KE and PE of SHM oscillator are equal

$\frac{1}{2} m ω^{2} (A^{2} - x^{2}) = \frac{1}{2} m ω^{2} x^{2} \Rightarrow A^{2} - x^{2} = x^{2} \Rightarrow A^{2} = 2 x^{2} \Rightarrow x = \pm \frac{A}{\sqrt{2}}$

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