$A$ body of mass $1 \ kg$ is executing simple harmonic motion $(SHM)$. Its displacement $y$ (in $cm$) at time $t$ is given by $y = 6 \sin (100 t + \pi/4) \ cm$. Its maximum kinetic energy is (in $J$)

$A$ 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?

$A$ body executes simple harmonic motion with an amplitude $A$. At what displacement,from the mean position,is the potential energy of the body one fourth of its total energy?

Find the displacement of a simple harmonic oscillator at which its $PE$ is half of the maximum energy of the oscillator.

Which graph represents the difference between total energy and potential energy of a particle executing $SHM$ versus its distance from the mean position?

The total energy of a particle executing $SHM$ at an amplitude of $4 \, cm$ is $20 \, J$. What will be its total energy at a displacement of $x = 2 \, cm$ (in $J$)?