Q. The potential energy of a particle with displacement $X$ is $U(X).$ The motion is simple harmonic, when ($K$ is a positive constant)
A) $U = \frac{KX^{2}}{2}$
B) $U = KX^{2}$
C) $U = K$
D) $U = KX$
Explanation
(a) $F = - kx \Rightarrow dW - Fdx = - kxdx$
So$\ \int_{0}^{W}{dW} = \int_{0}^{x}{- kx\ dx \Rightarrow}W = U = \frac{1}{2}kx^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.