If the amplitude of linear $S.H.M.$ is decreased,then:

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 ratio of kinetic energy to the potential energy of a particle executing $SHM$ at a distance equal to half its amplitude,the distance being measured from its equilibrium position,is:

$A$ particle executing simple harmonic motion has a kinetic energy $K = K_0 \cos^2(\omega t)$. The maximum values of the potential energy and the total energy are respectively:

For the minimum time period condition,find the average potential energy between $t = 0$ to $t = 0.05 \ s$ (take $g = 10 \ m/s^2$).

Find the displacement of a simple harmonic oscillator at which its $PE$ is half of the maximum energy of the oscillator.