If a particle is executing simple harmonic motion with time period $T$,then the time period of its total mechanical energy is ...........

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 variations of kinetic energy $K(x)$,potential energy $U(x)$,and total energy $E$ as a function of displacement $x$ of a particle in $SHM$ are as shown in the figure. The value of $|x_0|$ is

The equation of motion of a particle executing simple harmonic motion is given by $x=3 \sin \left(6 t+\frac{\pi}{6}\right)$,where $x$ is in metres and $t$ is in seconds. The ratio of the potential and kinetic energies of the particle at time $t=0$ 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

The amplitude of a particle executing $S.H.M.$ is $3 \,cm$. The displacement at which its kinetic energy will be $25 \%$ more than the potential energy is (in $\,cm$)