The displacement of a particle of mass $2 \text{ g}$ executing simple harmonic motion is $x = 8 \cos \left(50 t + \frac{\pi}{12}\right) \text{ m}$,where $t$ is time in seconds. The maximum kinetic energy of the particle is (in $\text{ J}$)

$A$ particle of mass $10 \, g$ is describing $S.H.M.$ along a straight line with a period of $2 \, s$ and an amplitude of $10 \, cm$. Its kinetic energy when it is at $5 \, cm$ from its equilibrium position is

The potential energy of a simple harmonic oscillator when the particle is half way to its end point is (where $E$ is the total energy)

In a simple harmonic motion,when the displacement is one-half the amplitude,what fraction of the total energy is kinetic?

$A$ particle starts oscillating simple harmonically from its equilibrium position with time period $T$. At time $t = \frac{T}{12}$,the ratio of its kinetic energy to potential energy is $\left[\sin \frac{\pi}{3} = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}, \sin \frac{\pi}{6} = \cos \frac{\pi}{3} = \frac{1}{2}\right]$.

$A$ body is executing a linear $S.H.M.$ Its potential energies at the displacements $x$ and $y$ are $E_1$ and $E_2$ respectively. Its potential energy at displacement $(x+y)$ will be