Physics - SHM Question with Solution | TestHub

The potential energy of a particle executing S.H.M. at a displacement $x$ from the mean position is given by $U=\frac{1}{2}m{\omega }^2{x}^2$

The kinetic energy of the particle at displacement $x$ is given by $K=\frac{1}{2}m{\omega }^2(A^2-x^2)$

Given that the potential energy is 8 times the kinetic energy:

$U=8K$

$\frac{1}{2}m{\omega }^2{x}^2=8\times \frac{1}{2}m{\omega }^2(A^2-x^2)$

$x^2=8(A^2-x^2)$

$x^2=8A^2-8x^2$

$9x^2=8A^2$

$x^2=\frac{8}{9}{A}^2$

$x=\frac{2\sqrt{2}}{3}A$

Answer: $\frac{2\sqrt{2}}{3}A$