$A$ block is in simple harmonic motion $(S.H.M)$ at the end of a spring with position given by $x = 5 \cos \left(\omega t + \frac{\pi}{4}\right)$. If the total mechanical energy is $100 \ J$, then the potential energy at time $t = 0$ is: (in $J$)

$A$ vertical mass-spring system executes simple harmonic oscillations with a period of $2\, s$. $A$ quantity of this system which exhibits simple harmonic variation with a period of $1\, s$ is

For a particle performing $S.H.M.$,the total energy is '$n$' times the kinetic energy,when the displacement of a particle from the mean position is $\frac{\sqrt{3}}{2} A$,where $A$ is the amplitude of $S.H.M.$ The value of '$n$' is

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

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 starts oscillating simple harmonically from its mean position with time period $T$. At time $t = T/6$,the ratio of the potential energy to kinetic energy of the particle is $\left[\sin 30^{\circ} = \cos 60^{\circ} = 0.5, \cos 30^{\circ} = \sin 60^{\circ} = \sqrt{3}/2\right]$