$A$ body executes simple harmonic motion with an amplitude $A$. At what displacement,from the mean position,is the potential energy of the body one fourth of its total energy?

$A$ body is executing simple harmonic motion. At a displacement $x$,its potential energy is $E_1$ and at a displacement $y$,its potential energy is $E_2$. The potential energy $E$ at a displacement $(x + y)$ is

For a particle executing $S.H.M.$, its potential energy is $8$ times its kinetic energy at a certain displacement '$x$' from the mean position. If '$A$' is the amplitude of $S.H.M.$, the value of '$x$' 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

Consider a simple harmonic motion $(SHM)$. Let $K$ and $U$ be kinetic energy and potential energy when the displacement in $SHM$ is one-half $\left(\frac{1}{2}\right)$ the amplitude. Which of the following statements is correct?

For a particle executing $S.H.M.$,the displacement $x$ is given by $x = A \cos \omega t$. Identify the graphs which represent the variation of potential energy $(P.E.)$ as a function of time $t$ and displacement $x$.