The total energy of a particle executing $S.H.M.$ is proportional to

$A$ particle executes $SHM$ of amplitude $A$. The distance from the mean position when its kinetic energy becomes equal to its potential energy is:

The displacement-time graph of a particle executing $SHM$ is as shown in the figure. The corresponding graph between potential energy $(PE)$ and time is:

The amplitude of an $SHM$ particle is $4 \, cm$. At what distance from the mean position will the potential energy and kinetic energy be equal?

$A$ particle performs $S.H.M.$ of amplitude $A$ with angular frequency $\omega$ along a straight line. When it is at a distance $\frac{\sqrt{3}}{2}A$ from the mean position,its kinetic energy is increased by an amount $\frac{1}{2}m\omega^2A^2$ due to an impulsive force. What is its new amplitude?

When the displacement of a simple harmonic oscillator is one third of its amplitude,the ratio of total energy to the kinetic energy is $\frac{x}{8}$,where $x=$ . . . . . . .