$A$ particle executing simple harmonic motion along a straight line with an amplitude $A$,attains maximum potential energy when its displacement from the mean position equals

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 performs $S.H.M.$ from the mean position. Its amplitude is $A$ and total energy is $E$. At a particular instant,its kinetic energy is $\frac{3E}{4}$. The displacement of the particle at that instant is:

$A$ particle executes simple harmonic motion along a straight line with an amplitude $A$. The potential energy is maximum when the displacement is

$A$ particle starts its oscillation from the equilibrium position with time period $T$. Find the ratio of kinetic energy to potential energy of the particle at time $t = \frac{T}{6}$.

$A$ simple pendulum of mass $m$ executes $S.H.M.$ with total energy $E$. If at an instant it is at one of the extreme positions,then its linear momentum after a phase shift of $\frac{\pi}{3} \, rad$ will be