$A$ 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:

The $P.E.$ of a particle executing $SHM$ at a distance $x$ from its equilibrium position is

$A$ particle is executing simple harmonic motion with amplitude $A$. The ratio of the kinetic energies of the particle when it is at displacements of $\frac{A}{4}$ and $\frac{A}{2}$ from the mean position is

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

$A$ 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

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