A particle is executing simple harmonic motion (SHM). The ratio of potential energy and kinetic energy of the particle when its displacement is half of its amplitude will be

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.

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