$A$ 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 ........

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 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?

$A$ block is in simple harmonic motion $(S.H.M)$ at the end of a spring with position given by $x = 5 \cos \left(\omega t + \frac{\pi}{4}\right)$. If the total mechanical energy is $100 \ J$, then the potential energy at time $t = 0$ is: (in $J$)

The potential energy of a simple harmonic oscillator,when the particle is halfway to its endpoint,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 $t = \frac{T}{6}$.