$A$ spring-mass system performs $S.H.M.$ If the mass is doubled keeping the amplitude same,then the total energy of $S.H.M.$ will become:

$A$ particle is executing simple harmonic motion with a time period $T$. At time $t = 0$,it is at its position of equilibrium. The kinetic energy-time graph of the particle will look like:

The potential energy of a simple harmonic oscillator when the particle is half way to its end point is (where $E$ is the total energy)

For a particle executing $S.H.M.$,the displacement $x$ is given by $x = A \cos \omega t$. Identify the graphs which represent the variation of potential energy $(P.E.)$ as a function of time $t$ and displacement $x$.

For a particle executing $S.H.M.$, the potential energy is $n$ times the kinetic energy when its displacement from mean position is $(\frac{2\sqrt{2}}{3})A$, where $A$ is the amplitude of $S.H.M.$ The value of $n$ 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 $\left[\sin 30^{\circ} = \cos 60^{\circ} = 0.5, \cos 30^{\circ} = \sin 60^{\circ} = \sqrt{3}/2\right]$