$A$ particle executes simple harmonic motion along a straight line with an amplitude $A$. The potential energy is maximum when the displacement is

An object of mass $3 \,kg$ is executing simple harmonic motion with an amplitude $\frac{2}{\pi} \,m$. If the kinetic energy of the object when it crosses the mean position is $6 \,J$, the time period of oscillation of the object is (in $\,s$)

The displacements of two particles of same mass executing $SHM$ are represented by the equations $x_1=4 \sin \left(10 t+\frac{\pi}{6}\right)$ and $x_2=5 \cos (\omega t)$. The value of $\omega$ for which the energies of both the particles remain same is (in $\text{ unit}$)

The amplitude of a particle executing $SHM$ is made three-fourth keeping its time period constant. Its total energy will be

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

$A$ 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