$A$ linear harmonic oscillator of force constant $2 \times 10^6 \, N/m$ and amplitude $0.01 \, m$ has a total mechanical energy of $160 \, J$. Its

The potential energy function for a particle executing linear simple harmonic motion is given by $V(x) = k x^{2} / 2$,where $k$ is the force constant of the oscillator. For $k = 0.5 \; N m^{-1}$,the graph of $V(x)$ versus $x$ is shown in the figure. Show that a particle of total energy $1 \; J$ moving under this potential must 'turn back' when it reaches $x = \pm 2 \; m$.

What is the displacement of a body in $SHM$ when the potential energy becomes three times its kinetic energy?

If the frequency of a simple harmonic motion is $f$,what is the frequency of its kinetic energy?

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

Starting from the mean position,a body oscillates simple harmonically with a period of $2\,s$. After what time will its kinetic energy be $75\%$ of the total energy?