$A$ particle undergoing simple harmonic motion has time-dependent displacement given by $x(t) = A \sin \left( \frac{\pi t}{90} \right)$. The ratio of kinetic energy to potential energy of the particle at $t = 210 \ s$ will be:

$A$ block kept on a frictionless horizontal surface is connected to one end of a horizontal spring of constant $100 \text{ Nm}^{-1}$ whose other end is fixed to a rigid vertical wall. Initially, the block is at its equilibrium position. The block is pulled to a distance of $8 \text{ cm}$ and released. The kinetic energy of the block when it is at a distance of $3 \text{ cm}$ from the mean position is (in $\text{ J}$)

The total energy of a particle performing $S.H.M.$ depends on:

The potential energy of a particle executing $S.H.M.$ is $2.5 \ J$,when its displacement is half of its amplitude. The total energy of the particle is .... $J$.

In $SHM$,the restoring force is $F = -kx$,where $k$ is the force constant,$x$ is the displacement,and $A$ is the amplitude of motion. Then,the total energy depends upon:

$A$ 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 $\left[\sin 30^{\circ} = \cos 60^{\circ} = 0.5, \cos 30^{\circ} = \sin 60^{\circ} = \sqrt{3}/2\right]$