$A$ loaded vertical spring executes $S.H.M.$ with a time period of $4\; sec$. The difference between the kinetic energy and potential energy of this system varies with a period of ........ $sec$.

For a particle executing $S.H.M.$,the potential energy $V$ versus displacement $x$ graph is given. The value of the restoring force constant is ..... $N/m$.

$A$ 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 $J$)

As a body performs $S.H.M.$,its potential energy $U$ varies with time $t$ as indicated in:

For any $S.H.M.$,the amplitude is $6\, cm$. If the instantaneous potential energy is half the total energy,then the distance of the particle from its mean position is .... $cm$.

$A$ particle starts oscillating simple harmonically from its equilibrium position with time period $T$. What is the ratio of potential energy to kinetic energy of the particle at time $t = \frac{T}{12}$? (Given: $\sin(\frac{\pi}{6}) = \frac{1}{2}$)