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. ±A
• B. Zero
• C.
• D.

Answer 1

The correct option is 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 = ± A