The potential energy of a simple harmonic oscillator,when the particle is halfway to its endpoint,is

$A$ particle performs $S.H.M.$ of amplitude $A$ with angular frequency $\omega$ along a straight line. When it is at a distance $\frac{\sqrt{3}}{2}A$ from the mean position,its kinetic energy is increased by an amount $\frac{1}{2}m\omega^2A^2$ due to an impulsive force. What is its new amplitude?

For a particle performing $S.H.M.$,the total energy is '$n$' times the kinetic energy,when the displacement of a particle from the mean position is $\frac{\sqrt{3}}{2} A$,where $A$ is the amplitude of $S.H.M.$ The value of '$n$' 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:

The amplitude of an $SHM$ particle is $4 \, cm$. At what distance from the mean position will the potential energy and kinetic energy be equal?

The displacements of two particles of same mass executing $SHM$ are represented by the equations $x_1=4 \sin \left(10 t+\frac{\pi}{6}\right)$ and $x_2=5 \cos (\omega t)$. The value of $\omega$ for which the energies of both the particles remain same is (in $\text{ unit}$)