$A$ 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?

$A$ body is executing simple harmonic motion. At a displacement $x$,its potential energy is $E_1$ and at a displacement $y$,its potential energy is $E_2$. The potential energy $E$ at a displacement $(x + y)$ is

For a particle executing simple harmonic motion,the kinetic energy $K$ is given by $K = K_0 \cos^2 \omega t$. The maximum value of potential energy 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?

The variation of kinetic energy $(KE)$ of a particle executing simple harmonic motion with the displacement $(x)$ starting from mean position to extreme position $(A)$ is given by

$A$ mass $0.4 \,kg$ performs $S.H.M.$ with a frequency $\frac{16}{\pi} \,Hz$. At a certain displacement, it has kinetic energy $2 \,J$ and potential energy $1.2 \,J$. The amplitude of oscillation is (in $m$)