$A$ particle of mass $10 \, g$ is describing $S.H.M.$ along a straight line with a period of $2 \, s$ and an amplitude of $10 \, cm$. Its kinetic energy when it is at $5 \, cm$ from its equilibrium position 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 displacement of a particle of mass $2 \,g$ executing $SHM$ is given by $y=5 \sin \left(4 t+\frac{\pi}{3}\right)$. Here,$y$ is in metres and $t$ is in seconds. The kinetic energy of the particle,when $t=\frac{T}{4}$ is (in $\,J$)

$A$ loaded vertical spring executes $S.H.M.$ with a time period of $4\; sec$. The difference between the kinetic energy and potential energy of this system varies with a period of ........ $sec$.

The potential energy of a simple harmonic oscillator when the particle is half way to its end point is (where $E$ is the total energy)

$A$ body starting at $t=0$ from the origin oscillates simple harmonically with a period of $4 \ s$. After what time will its kinetic energy be $75 \%$ of its total energy?