91Ó°ÊÓ

1. Position of the block is$x=x_m\cos (\mathrm{Ó\neg }t+\mathrm{Ï•})$
2. $\mathrm{Ï•}=\frac{\mathrm{Ï€}}{5}\mathrm{rad}$

A particle in simple harmonic motion has, at any time, potential energy

$\mathit{p}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{k}{\mathit{x}}^{\mathbf{2}}$

First, we have to find the position and potential energy of the block, then, we have to find the total mechanical energy which equals to maximum potential energy according to the law of conservation of energy. By taking the ratio of potential energy to maximum potential energy (E), we can find the percentage of total mechanical energy which is potential energy at t=0.

Formulae:

The displacement equation of a wave,$x=x_m\cos (\mathrm{Ó\neg }t+\mathrm{Ï•})$ (i)

The total potential energy of a particle, $E=\frac{1}{2}k{x_m}^2$ (ii)

The displacement equation at t = 0 can be given as:

$$\begin{gathered} x_0=x_m\cos \left(\mathrm{Ï•}\right) \\ =x_m\cos \left(\frac{\mathrm{Ï€}}{5}\right) \end{gathered}$$

Here, the total mechanical energy equals to its maximum potential energy. So, total energy is given as:

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

Now its potential energy at t = 0 is,

$$\begin{gathered} P.E_0=\frac{1}{2}k{x_0}^2 \\ =\frac{1}{2}k{\left(x_m\cos \left(\frac{\mathrm{Ï€}}{5}\right)\right)}^2 \\ =\frac{1}{2}k{x_m}^2{\cos }^2\left(\frac{\mathrm{Ï€}}{5}\right) \end{gathered}$$

Percentage of total mechanical energy is potential energy at t = 0 is,

$$\begin{gathered} \frac{P.E_0}{E}×100=\frac{{\displaystyle \frac{1}{2}}k{x_m}^2{\cos }^2\left({\displaystyle \frac{\mathrm{π}}{5}}\right)}{{\displaystyle \frac{1}{2}}k{x_m}^2} \\ ={\cos }^2\left(\frac{\mathrm{π}}{5}\right)×100 \\ =65.5\% \end{gathered}$$

Therefore, percentage of total mechanical energy is potential energy at t = 0 is 65.5 %