Question

The amplitude of oscillation of particle in S.H.M. is $\sqrt{3}$ cm. The displacement from mean position at which its potential and kinetic energies are in the ratio $1:2$ is

• A. $1cm$
• B. $0.866cm$
• C. $31cm$
• D. $11cm$

Answer 1

The correct option is A $1cm$

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

$KE=\frac{1}{2}m{\omega }^2(A^2-x^2)$
$\therefore \frac{PE}{KE}=\frac{1}{2}$
$\therefore \frac{A^2-x^2}{x^2}=2$
$\frac{A^2}{x^2}=3$
$\frac{3}{x^2}=3$
$x^2=1$
$x=1$