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
$P.E.=\frac{1}{2}m{\omega }^2{x}^2=\frac{1}{2}k{x}^2$
Where k = $m{\omega }^2$ is constant.