$A$ body is performing $S.H.M.$ of amplitude $A$. The displacement of the body from a point where kinetic energy is maximum to a point where potential energy is maximum,is

For a particle executing simple harmonic motion,the ratio of kinetic and potential energies at a point where displacement is one half of the amplitude is

$A$ 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 $J$)

$A$ particle is executing $S.H.M.$ with total mechanical energy $90 \,J$ and amplitude $6 \,cm$. If its energy is somehow decreased to $40 \,J$,then its amplitude will become ........ $cm$.

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

The ratio between kinetic and potential energies of a body executing simple harmonic motion, when it is at a distance of $\frac{1}{N}$ of its amplitude from the mean position is