The maximum potential energy of a block executing simple harmonic motion is $25 \ J$. $A$ is the amplitude of oscillation. At $x = A / 2$,the kinetic energy of the block is $...............$ (in $J$)

$A$ body is executing simple harmonic motion with frequency $n$,the frequency of its potential energy is:

$A$ particle is executing simple harmonic motion with a time period of $3 \,s$. At a position where the displacement of the particle is $60 \%$ of its amplitude, the ratio of the kinetic and potential energies of the particle is

$A$ particle starts oscillating simple harmonically from its mean position with time period $T$. At time $t=\frac{T}{12}$,the ratio of the potential energy to kinetic energy of the particle is $\left(\sin 30^{\circ}=\cos 60^{\circ}=0.5, \cos 30^{\circ}=\sin 60^{\circ}=\frac{\sqrt{3}}{2}\right)$

Find the displacement of a simple harmonic oscillator at which its $PE$ is half of the maximum energy of the oscillator.

If $T$ is the period of $SHM$,then write the period of kinetic and potential energy.