$A$ particle is executing $S.H.M.$ with time period $T^{\prime}$. If the time period of its total mechanical energy is $T$,then $\frac{T^{\prime}}{T}$ is ........

When the potential energy of a particle executing simple harmonic motion is one-fourth of its maximum value during the oscillation,the displacement of the particle from the equilibrium position in terms of its amplitude $a$ is

$A$ particle executing simple harmonic motion along a straight line with an amplitude $A$,attains maximum potential energy when its displacement from the mean position equals

Write the expression of kinetic energy and potential energy of $SHM$ particle $(i)$ as a function of displacement $(ii)$ as a function of time.

The displacement of a particle executing $SHM$ is given by $y = 5 \sin \left(4t + \frac{\pi}{3}\right)$. If $T$ is the time period and the mass of the particle is $2 \text{ g}$, the kinetic energy of the particle when $t = \frac{T}{4}$ is given by (in $\text{ J}$)

The total energy of a particle executing $SHM$ at an amplitude of $4 \, cm$ is $20 \, J$. What will be its total energy at a displacement of $x = 2 \, cm$ (in $J$)?