The total energy of a particle,executing simple harmonic motion 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?

Write the maximum velocity of a $SHM$ oscillator in terms of mechanical energy $E$ and mass of oscillator $m$.

$A$ 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 of mass $10 \ g$ performs simple harmonic motion with an amplitude of $10 \ cm$ and a time period of $2 \ s$. What is its kinetic energy at a displacement of $5 \ cm$?

The displacement of a particle of mass $2 \,g$ executing $SHM$ is given by $y=5 \sin \left(4 t+\frac{\pi}{3}\right)$. Here,$y$ is in metres and $t$ is in seconds. The kinetic energy of the particle,when $t=\frac{T}{4}$ is (in $\,J$)