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

For a body executing $S.H.M. :$
$(a)$ Potential energy is always equal to its $K.E.$
$(b)$ Average potential and kinetic energy over any given time interval are always equal.
$(c)$ Sum of the kinetic and potential energy at any point of time is constant.
$(d)$ Average $K.E.$ in one time period is equal to average potential energy in one time period.
Choose the most appropriate option from the options given below:

$A$ particle is executing Simple Harmonic Motion $(SHM)$. The ratio of potential energy and kinetic energy of the particle when its displacement is half of its amplitude will be

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

Two independent harmonic oscillators of equal mass are oscillating about the origin with angular frequencies $\omega_1$ and $\omega_2$ and have total energies $E_1$ and $E_2$,respectively. The variations of their momenta $p$ with positions $x$ are shown in the figures. If $\frac{a}{b}= n^2$ and $\frac{a}{R}= n$,then the correct equation$(s)$ is(are):
$(A) E_1 \omega_1 = E_2 \omega_2$
$(B) \frac{\omega_2}{\omega_1} = n^2$
$(C) \omega_1 \omega_2 = n^2$
$(D) \frac{E_1}{\omega_1} = \frac{E_2}{\omega_2}$

The potential energy of a simple harmonic oscillator of mass $2\, kg$ in its mean position is $5\, J.$ If its total energy is $9\, J$ and its amplitude is $0.01\, m,$ its time period would be