For a particle executing $S.H.M.$, the potential energy is $n$ times the kinetic energy when its displacement from mean position is $(\frac{2\sqrt{2}}{3})A$, where $A$ is the amplitude of $S.H.M.$ The value of $n$ is

$A$ mass $0.4 \,kg$ performs $S.H.M.$ with a frequency $\frac{16}{\pi} \,Hz$. At a certain displacement, it has kinetic energy $2 \,J$ and potential energy $1.2 \,J$. The amplitude of oscillation is (in $m$)

$A$ particle of mass $m$ executes simple harmonic motion with amplitude $a$ and frequency $v$. The average kinetic energy during its motion from the position of equilibrium to the end is

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 $\text{ J}$)

The total mechanical energy of a particle executing simple harmonic motion is $E$. When the displacement is half the amplitude,its kinetic energy will be

$A$ body is executing simple harmonic motion. At a displacement $x$,its potential energy is $E_1$ and at a displacement $y$,its potential energy is $E_2$. The potential energy $E$ at a displacement $(x+y)$ is