A particle executes simple harmonic motion along a straight line with an amplitude $A$. The potential energy is maximum when the displacement is
A$ \pm A$
B
Zero
C$ \pm \frac{A}{2}$
D$ \pm \frac{A}{{\sqrt 2 }}$
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A$ \pm A$
a (a) $P.E. = \frac{1}{2}m{\omega ^2}{x^2}$
It is clear $P.E.$ will be maximum when $x$ will be maximum i.e., at $x = \pm A$
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