$A$ particle is executing $SHM$ with an amplitude $A$. What is the displacement of the particle when its potential energy is half of its total energy?

$A$ particle is executing simple harmonic motion with a time period $T$. At time $t = 0$,it is at its position of equilibrium. The kinetic energy-time graph of the particle will look like:

An object of mass $0.5\, \text{kg}$ is executing simple harmonic motion. Its amplitude is $5\, \text{cm}$ and time period $T$ is $0.2\, \text{s}$. What will be the potential energy of the object at an instant $t = \frac{T}{4}\, \text{s}$ starting from the mean position? Assume that the initial phase of the oscillation is zero. (In $\text{J}$)

For a particle performing $S.H.M.$,when displacement is $x$,the potential energy and restoring force acting on it are denoted by $E$ and $F$ respectively. The relation between $x, E$ and $F$ is

$A$ body is executing a linear $S.H.M.$ Its potential energies at the displacements $x$ and $y$ are $E_1$ and $E_2$ respectively. Its potential energy at displacement $(x+y)$ will be

$A$ particle starts oscillating simple harmonically from its equilibrium position. The ratio of kinetic energy and potential energy of the particle at time $t = T/12$ is: ($T =$ time period)