Question

A particle is performing linear SHM, along x-axis, when the displacement becomes, A/$\sqrt{2}$, (A is the amplitude), the relation that holds between kinetic energy and potential energy is then

• A. KE = 2PE
• B. KE = PE
• C. KE = 3PE
• D. KE = 4PE

Answer 1

The correct option is B KE = PE

In SHM,

$KE=\frac{1}{2}m{\omega }^2(A^2-x^2)$

$PE=\frac{1}{2}m{\omega }^2{x}^2$

At $x=\frac{A}{\sqrt{2}}$

$KE=\frac{1}{4}m{\omega }^2{A}^2$

$PE=\frac{1}{4}m{\omega }^2{A}^2$

Hence $KE=PE$

Option $\text{B}$ is the correct answer.