The potential energy of a particle $(U_x)$ executing $S.H.M.$ is given by

The amplitude of a particle executing simple harmonic motion is $6 \ cm$. The distance of the point from the mean position at which the ratio of the potential and kinetic energies of the particle becomes $4:5$ is (in $cm$)

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 $\text{ unit}$)

$A$ particle executes $SHM$ of amplitude $A$. The distance from the mean position when its kinetic energy becomes equal to its potential energy is:

The ratio between kinetic and potential energies of a body executing simple harmonic motion,when it is at a distance of $\frac{1}{N}$ of its amplitude from the mean position is

$A$ body performing simple harmonic motion has potential energy $P_{1}$ at displacement $x_{1}$. Its potential energy is $P_{2}$ at displacement $x_{2}$. The potential energy $P$ at displacement $(x_{1}+x_{2})$ is