$A$ particle is vibrating in $S.H.M.$ with an amplitude of $4 \,cm$. At what displacement from the equilibrium position is its energy half potential and half kinetic?

For a body performing simple harmonic motion,its potential energy is $E_x$ at displacement $x$ and $E_y$ at displacement $y$ from the mean position. The potential energy $E_0$ at displacement $(x+y)$ is

For a particle executing simple harmonic motion,the kinetic energy $K$ is given by $K = K_0 \cos^2 \omega t$. The maximum value of potential energy is

The total mechanical energy of a particle executing simple harmonic motion is $E$. When the displacement is half the amplitude,its kinetic energy will be

The total energy of a simple harmonic oscillator is $E$. What is its kinetic energy at half the amplitude?

$A$ particle of mass $m$ is executing $S.H.M.$ about the origin on the $x$-axis with frequency $f = \frac{\sqrt{Ka}}{\pi m}$, where $K$ is a constant and $a$ is the amplitude of $S.H.M.$ If $x$ is the displacement of the particle at time $t$, the potential energy of the particle will be: