If a body is executing simple harmonic motion and its current displacement is $\sqrt{3}/2$ times the amplitude from its mean position, then the ratio between potential energy and kinetic energy is

The current displacement is $\frac{\sqrt{3}}{2}$times the amplitude,

i.e.         $\mathrm{y}=\frac{\sqrt{3}}{2}\mathrm{A}$

We know that, potential energy of 11 body in SHM is given by

              $\mathrm{U}=\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{y}}^2$

where, m and $\mathrm{\omega }$ are constants.

            $\mathrm{U}=\frac{1}{2}{\mathrm{m\omega }}^2{\left(\frac{\sqrt{3}}{2}\mathrm{A}\right)}^2=\frac{1}{2}{\mathrm{m\omega }}^2\times \frac{3{\mathrm{A}}^2}{4}$                      …(i)

Similarly, kinetic energy of a body it SHM is given by

            $\mathrm{K}=\frac{1}{2}{\mathrm{m\omega }}^2\left({\mathrm{A}}^2-{\mathrm{y}}^2\right)$

               $=\frac{1}{2}{\mathrm{m\omega }}^2\left({\mathrm{A}}^2-\frac{3{\mathrm{A}}^2}{4}\right)=\frac{1}{2}{\mathrm{m\omega }}^2\frac{{\mathrm{A}}^2}{4}$                       …(ii)

So, the ratio of PE and KE,

       $\mathrm{PE} : \mathrm{KE}=\frac{1}{2}{\mathrm{m\omega }}^2\times \frac{3{\mathrm{A}}^2}{4} : \frac{1}{2}{\mathrm{m\omega }}^2\frac{{\mathrm{A}}^2}{4}=3 : 1$