The amplitude of a particle executing simple harmonic motion is $6 \ cm$. The distance of the point from the mean position at which the ratio of the potential and kinetic energies of the particle becomes $4:5$ is (in $cm$)

$A$ body is executing simple harmonic motion. At a displacement $x$,its potential energy is $E_1$ and at a displacement $y$,its potential energy is $E_2$. The potential energy $E$ at a displacement $(x + y)$ is

The total energy of a particle executing $S.H.M.$ is $80 \, J$. What is the potential energy when the particle is at a distance of $\frac{3}{4}$ of amplitude from the mean position?

For what value of displacement the kinetic energy and potential energy of a simple harmonic oscillation become equal?

$A$ particle starts oscillating simple harmonically from its equilibrium position with time period $T$. At time $t = \frac{T}{12}$,the ratio of its kinetic energy to potential energy is $\left[\sin \frac{\pi}{3} = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}, \sin \frac{\pi}{6} = \cos \frac{\pi}{3} = \frac{1}{2}\right]$.

$A$ particle undergoing simple harmonic motion has an amplitude of $10 \ cm$. When the particle is at a displacement of $6 \ cm$ from the centre,then the ratio of its kinetic energy to potential energy is