91Ó°ÊÓ

Displacement is one-half of the amplitude,$\mathrm{x}=\frac{1}{2}{\mathrm{x}}_{\mathrm{m}}$

We can find the fraction of total energy which is potential energy (P.E) by taking the ratio of P.E and E for the given condition. Then, using this ratio, we can find the fraction of total energy which is kinetic energy (K.E) from the ratio of P.E. Thus, we can find the displacement at which the energy of the system is half K.E and half P.E

Formulae:

The elastic potential energy of a system,$PE=\frac{1}{2}k{x}^2.....\left(1\right)$

The fraction of energy which is K.E using P.E is given as:

$$\begin{gathered} \frac{\mathrm{K}.\mathrm{E}}{\mathrm{E}}=\frac{\mathrm{E}-\mathrm{P}.\mathrm{E}}{\mathrm{E}}(âˆ\mu \mathrm{E}(\mathrm{total}\mathrm{energy})=\mathrm{PE}+\mathrm{KE}) \\ =\frac{\mathrm{E}-{\displaystyle \frac{1}{4}}\mathrm{E}}{\mathrm{E}}(âˆ\mu \mathrm{from}\mathrm{equation}(\mathrm{a})\mathrm{andE}=\frac{1}{2}{\mathrm{kx}}_{\mathrm{m}}^2) \\ =\frac{3}{4} \\ =0.75 \end{gathered}$$

Using equation (i), the elastic potential energy can be given as:

$P.E=\frac{1}{8}k{x_m}^2$

So, the fraction of potential energy to that of the total energy is given as:

$$\begin{gathered} \frac{P.E}{E}=\frac{\left(\begin{array}{c}{\displaystyle \frac{1}{8}k{x_m}^2}\end{array}\right)}{{\displaystyle \frac{1}{2}k{x}_m^2}}(âˆ\mu Totalenergyofthetermsofpotential,E=\frac{1}{2}k{x}_m^2) \\ =\frac{1}{4} \\ =0.25 \end{gathered}$$

Hence, the required value of potential energy ratio is 0.25

We have been given that,

$$\begin{gathered} \frac{P.E}{E}=\frac{\left(\begin{array}{c}{\displaystyle \frac{1}{2}}k{x}^2\end{array}\right)}{{\displaystyle \frac{1}{2}}k{x}_m^2} \\ \frac{x^2}{x_m^2}=\frac{1}{2} \\ x=\frac{x_m}{\sqrt{2}} \end{gathered}$$

Therefore, the displacement at which the energy of the system is half K.E and half P.E is$\raisebox{1ex}{$x_m$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.$