$A$ body of mass $m$ is executing $SHM$ with amplitude $a$. When its displacement $x = 1$ unit,the force is $b$. What will be its maximum kinetic energy?

$A$ bob of a simple pendulum of mass $m$ performs $SHM$ with amplitude $A$ and period $T$. The kinetic energy of the pendulum at displacement $x = \frac{A}{2}$ will be:

For what value of displacement the kinetic energy and potential energy of a simple harmonic oscillation become equal?

Assertion $(A)$: In $S.H.M$,kinetic and potential energy become equal when the distance is $1/\sqrt{2}$ times its amplitude. Reason $(R)$: The potential energy of a particle executing $S.H.M$ is periodic with time period being maximum at the extreme displacement.

$A$ particle executes simple harmonic motion with a frequency $f$. The frequency with which its kinetic energy oscillates is

$A$ linear harmonic oscillator has a total mechanical energy of $300 \ J$. If its potential energy at the mean position is $100 \ J$,find its kinetic energy at $x = +\frac{A}{\sqrt{2}}$. (in $J$)