Question

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

Answer 1

Answer: (A)

$\pm A$

Answer 2

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 $x = \pm A$