JEE Main 2023 — Oscillations Question with Solution

The formula to calculate the potential energy of the particle executing SHM is given by

$P=\frac{1}{2}m{\omega }^2{x}^2 ...\left(1\right)$

The formula to calculate the kinetic energy of the same particle is given by

$K=\frac{1}{2}m{\omega }^2\left(A^2-x^2\right) ...\left(2\right)$

Divide equation (1) by equation (2) to obtain the required ratio.

$\begin{array}{rcl}\frac{P}{K}& =& \frac{{\displaystyle \frac{1}{2}}m{\omega }^2{x}^2}{{\displaystyle \frac{1}{2}}m{\omega }^2\left(A^2-x^2\right)}\\ & =& \frac{x^2}{A^2-x^2} ...\left(3\right)\end{array}$

Substitute $\frac{A}{2}$ for $x$ into equation (3) to calculate the required ratio.

$\begin{array}{rcl}\frac{P}{K}& =& \frac{{\displaystyle {\left({\displaystyle \frac{A}{2}}\right)}^2}}{A^2-{\left({\displaystyle \frac{A}{2}}\right)}^2}\\ & =& \frac{1}{3}\end{array}$