The average energy possessed by an oscillator at a temperature $300 \ K$ is (Boltzmann constant $= 1.38 \times 10^{-23} \ J K^{-1}$).

What is the ratio of the kinetic energy at the mean position to the potential energy at $y = A / 2$ for a particle performing $SHM$ (in $: 1$)?

At a given point of time,the displacement of a simple harmonic oscillator is given by $y = A \cos(30^{\circ})$. If the amplitude is $40 \, cm$ and the kinetic energy at that time is $200 \, J$,the value of the force constant is $1.0 \times 10^{x} \, Nm^{-1}$. The value of $x$ 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?

$A$ particle is executing simple harmonic motion with a time period of $3 \,s$. At a position where the displacement of the particle is $60 \%$ of its amplitude, the ratio of the kinetic and potential energies of the particle is

$A$ body performs linear simple harmonic motion of amplitude $A$. At what displacement from the mean position is the potential energy of the body one-fourth of its total energy?