$A$ particle starts from the mean position and performs $S.H.M.$ with a period of $4 \ s$. At what time is its kinetic energy $50 \%$ of its total energy (in $s$)?

$A$ particle starting from mean position executes simple harmonic motion with a period $8 \ s$. The minimum time in which its potential energy becomes half of the total energy is . . . . . . . (in $s$)

$A$ particle starts oscillating simple harmonically from its mean position with time period $T$. At time $t = T/6$,the ratio of the potential energy to kinetic energy of the particle is $\left[\sin 30^{\circ} = \cos 60^{\circ} = 0.5, \cos 30^{\circ} = \sin 60^{\circ} = \sqrt{3}/2\right]$

$A$ 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 is performing $S.H.M.$ with energy of vibration $90 \,J$ and amplitude $6 \,cm$. When the particle reaches a distance of $4 \,cm$ from the mean position,it is stopped for a moment and then released. The new energy of vibration will be ........... $J$.

For any $S.H.M.$,the amplitude is $6\, cm$. If the instantaneous potential energy is half the total energy,then the distance of the particle from its mean position is .... $cm$.