$A$ particle of mass $m$ is hanging vertically by an ideal spring of force constant $K$. If the mass is made to oscillate vertically,its total energy is

$A$ particle performs $S.H.M.$ Its potential energies are $U_{1}$ and $U_{2}$ at displacements $x_{1}$ and $x_{2}$ respectively. At displacement $(x_{1} + x_{2})$,its potential energy $U$ is:

Starting from the mean position,a body oscillates simple harmonically with a period $T$. After what time will its kinetic energy be $75 \%$ of the total energy? $(\sin 30^{\circ} = 0.5)$

The frequency at which kinetic energy changes into potential energy in a simple harmonic motion ($S$.$H$.$M$.) with frequency $f$ is:

$A$ 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: $\sin(\frac{\pi}{6}) = \frac{1}{2}$)

$A$ particle of mass $m$ is executing $S.H.M.$ about the origin on the $x$-axis with frequency $f = \frac{\sqrt{Ka}}{\pi m}$, where $K$ is a constant and $a$ is the amplitude of $S.H.M.$ If $x$ is the displacement of the particle at time $t$, the potential energy of the particle will be: