Physics - SHM Question with Solution | TestHub

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

The total energy of the particle with amplitude $r$ is given by $E=\frac{1}{2}m{\omega }^2{r}^2$.

According to the given condition, $U=\lambda E$.

Substituting the expressions for $U$ and $E$, we get:

$\frac{1}{2}m{\omega }^2{x}^2=\lambda \left(\frac{1}{2}m{\omega }^2{r}^2\right)$

$x^2=\lambda r^2$

$x=r\sqrt{\lambda }$

The displacement of the particle is $r\sqrt{\lambda }$.

Answer: $r\sqrt{\lambda }$