JEE Main 2024 — Oscillations Question with Solution

The total energy of the harmonic oscillator is given by the formula

$E=\frac{1}{2}k{A}^2 ...\left(1\right)$

where, $k$ is the force constant and $A$ is the amplitude.

The potential energy of the oscillator at the given displacement can be written as

$\begin{array}{rcl}U& =& \frac{1}{2}k{\left(\frac{A}{3}\right)}^2\\ & =& \frac{k{A}^2}{2\times 9}\\ & =& \frac{E}{9}\end{array}$

Hence, its kinetic energy is given by

$\begin{array}{rcl}KE& =& E-\frac{E}{9}\\ & =& \frac{8E}{9}\end{array}$

Thus, the required ratio is

$\begin{array}{rcl}\frac{E}{KE}& =& \frac{E}{\frac{8E}{9}}\\ & =& \frac{9}{8}\end{array}$

Hence, $x=9$.