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A particle executes simple harmonic motion along a straight line with an amplitude A. The potential energy is maximum when the displacement is

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

$\pm A$

Zero

$\pm \frac{A}{2}$

answer is A.

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Detailed Solution

P.E. $= \frac{1}{2} {mω}^{2} x^{2}$

It is clear P.E. will be maximum when x will be maximum i.e., at $x = \pm A$

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