The displacement between the position of maximum potential energy and the position of maximum kinetic energy for a particle executing $S.H.M.$ is

$A$ particle starts its oscillation from the equilibrium position with time period $T$. Find the ratio of kinetic energy to potential energy of the particle at time $t = \frac{T}{6}$.

Draw plots of mechanical energy,potential energy,and kinetic energy versus displacement for different positions of a block attached to a spring.

The total mechanical energy of a particle executing simple harmonic motion is $E$. When the displacement is half the amplitude,its kinetic energy will be

$A$ 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

Position of a $3 \ kg$ mass moving along the $X$-axis is given by $x = 0.3 \cos (\omega t) \ m$. If $K(t)$ denotes the kinetic energy at time $t$,then the value of $\frac{K\left(\frac{\pi}{6 \omega}\right)}{K\left(\frac{\pi}{3 \omega}\right)}$ is