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Physics-

General

Easy

Question

The P.E. of a particle executing SHM at a distance x from its equilibrium position is

$\frac{1}{2} m ω^{2} x^{2}$
$\frac{1}{2} m ω^{2} a^{2}$
$\frac{1}{2} m ω^{2} (a^{2} - x^{2})$
,Zero

The correct answer is: $1 half m omega squared x squared$

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