$A$ 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 $t = \frac{T}{6}$.

$A$ particle is executing simple harmonic motion starting from its mean position. If the time period of the particle is $1.5 \ s$,then the minimum time at which the ratio of the kinetic and total energies of the particle becomes $3: 4$ is

The amplitude of a particle executing simple harmonic motion is $6 \ cm$. The distance of the point from the mean position at which the ratio of the potential and kinetic energies of the particle becomes $4:5$ is (in $cm$)

$A$ body performs simple harmonic oscillations along the straight line $ABCDE$ with $C$ as the midpoint of $AE.$ Its kinetic energies at $B$ and $D$ are each one-fourth of its maximum value. If $AE = 2R,$ the distance between $B$ and $D$ is

Starting from the origin,a particle oscillates simple harmonically with a time period of $2 \ s$. After what time will its kinetic energy be $75 \%$ of the total energy?

$A$ particle of mass $4 \,kg$ is executing $SHM$. Its displacement is given by the equation $y=8 \cos [100 t+\pi / 4] \,cm$. Its maximum kinetic energy is (in $\,J$)