Question

The potential energy of a particle executing simple harmonic motion at a distance $x$ from the equilibrium position is proportional to

• A. $\sqrt{x}$
• B. $x$
• C. $x^2$
• D. $x^3$

Answer 1

The correct option is C $x^2$

The potential energy of a particle of mass $m$ executing simple harmonic motion of angular frequency $\omega$ at a distance $x$ from the equilibrium position is given by
$\text{P.E.}=\frac{1}{2}m{\omega }^2{x}^2=\frac{1}{2}k{x}^2$
where $k=m{\omega }^2$ is constant.
Hence, the correct choice is (c).