The $P.E.$ of a particle executing $SHM$ at a distance $x$ from its equilibrium position is

$A$ particle is executing $SHM$ with an amplitude $A$. What is the displacement of the particle when its potential energy is half of its total energy?

The graph shows the variation of displacement of a particle executing $S$.$H$.$M$. with time. We infer from this graph that

The potential energy of a particle with displacement $X$ is $U(X)$. The motion is simple harmonic,when ($K$ is a positive constant)

$A$ particle is executing simple harmonic motion $(SHM)$ of amplitude $A$ along the $x$-axis about $x = 0$. When its potential energy $(PE)$ equals kinetic energy $(KE)$,the position of the particle will be

$A$ particle starts oscillating simple harmonically from its mean position with time period $T$. At time $t = T/6$,the ratio of the potential energy to kinetic energy of the particle is $\left[\sin 30^{\circ} = \cos 60^{\circ} = 0.5, \cos 30^{\circ} = \sin 60^{\circ} = \sqrt{3}/2\right]$