Q. The potential energy of a particle $(U_{x})$ executing SHM is given by
A) $U_{x} = \frac{k}{2}(x - a)^{2}$
B) $U_{x} = k_{1}x + k_{2}x^{2} + k_{3}x^{3}$
C) $U_{x} = Ae^{- bx}$
D) $U_{x} = a\ constant\ $
Explanation
(a) PE of body in SHM at an instant,
$U = \frac{1}{2}k(a - x)^{2} = \frac{1}{2}k(x - a)^{2}$
If the displacement, $y = (a - x)$ then
$U = \frac{1}{2}k(a - x)^{2} = \frac{1}{2}k(x - a)^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.