Thank you for reporting, we will resolve it shortly
Q.
A body executes simple harmonic motion with amplitude $a$. At what displacement from the equilibrium position is its energy half kinetic and half potential?
J & K CETJ & K CET 2001
Solution:
The kinetic energy (KE) of a body executing SHM with amplitude a undergoing displacement $y$ is
$K E=\frac{1}{2} m \omega^{2}\left(a^{2}-y^{2}\right) \ldots$ (i)
where $\omega$ is angular velocity, and $m$ the mass.
Also, potential energy (PE) is $P E=\frac{1}{2} m \omega^{2} y^{2} \ldots$ (ii)
Given, $K E=P E $
$\therefore $ $\frac{1}{2} m \omega^{2}\left(a^{2}-y^{2}\right)=\frac{1}{2} m \omega^{2} y^{2}$
$ \Rightarrow a^{2}-y^{2}=y^{2} $
$\Rightarrow y=\frac{a}{\sqrt{2}}$