First slide

Simple harmonic motion

Question

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

Easy

A

$\pm A$
B

Zero
C

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

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

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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