91Ó°ÊÓ

Any item or system that has stored energy as a result of its location or component arrangement is said to have potential energy. Nevertheless, it is unaffected by external factors like air pressure or altitude. On the other hand, kinetic energy is the force that propels a moving object or group of particles.

Calculate the total energy

Consider the given data as below.

The potential energy, P.E. =0.14 J

The kinetic energy, K.E.=0.06 J

Total energy will be

$$\begin{gathered} \mathrm{E}=\mathrm{K}.\mathrm{E}.+\mathrm{P}.\mathrm{E}. \\ =0.06+0.14 \\ =0.2\mathrm{J} \end{gathered}$$

The range of the graph for displacement will be $x∈[-0.3m,0.3m]$.

The total mechanical energy in a simple harmonic motion is given by

$$\begin{gathered} \mathrm{E}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{x}}^2+\frac{1}{2}\mathrm{kx} \\ =\frac{1}{2}{\mathrm{kA}}^2 \\ =\mathrm{CONSTANT} \end{gathered}$$

Now, use the energy equation to find the amplitude

$$\begin{gathered} \mathrm{E}=\frac{1}{2}{\mathrm{kA}}^2 \\ \mathrm{A}=\sqrt{\frac{2×\mathrm{E}}{\mathrm{k}}} \\ =\sqrt{\frac{2×0.2}{10}} \\ =0.2\mathrm{m} \end{gathered}$$

Hence, the amplitude of oscillation, A =0.2 m

The potential energy is given by

$\mathrm{P}.\mathrm{E}.=\frac{1}{2}{\mathrm{kx}}^2$

The displacement is half the amplitude. So,

$$\begin{gathered} \mathrm{x}=\frac{\mathrm{A}}{2} \\ P.E.=\frac{1}{2}\mathrm{k}{\left(\frac{\mathrm{A}}{2}\right)}^2 \\ =\frac{1}{8}{\mathrm{kA}}^2 \\ =\frac{1}{8}×10×{0.2}^2 \\ =0.2\mathrm{J} \end{gathered}$$

Hence, the potential energy is PE=0.2 J.

If the kinetic energy and the potential energy become equal, then the total mechanical energy is given by

$$\begin{gathered} \mathrm{E}=2×\mathrm{K}.\mathrm{E}. \\ =2×\frac{1}{2}{\mathrm{kx}}^2 \end{gathered}$$

Rearranging, we will get $\mathrm{x}=\sqrt{\frac{\mathrm{E}}{\mathrm{k}}}$.

$$\begin{gathered} \mathrm{x}=\sqrt{\frac{0.2}{10}} \\ =0.141\mathrm{m} \end{gathered}$$

Hence, the displacement when potential and kinetic energy is same is x=0.141 m.

The phase angle is given by

$\mathrm{Ï•}={\tan }^{−1}\left(−\frac{{\mathrm{v}}_0}{\mathrm{Ó\neg }{\mathrm{x}}_0}\right)$

Here,

The initial velocity, ${\mathrm{v}}_0=\sqrt{\frac{2{\mathrm{KE}}_0}{\mathrm{m}}}$

The initial displacement, $x_0=-\sqrt{\frac{2P{E}_0}{\mathrm{k}}}$

The angular frequency,$\mathrm{Ó\neg }=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$

Therefore, the phase angle will be,

$$\begin{gathered} \mathrm{ϕ}={\tan }^{−1}−\left(\frac{\sqrt{\frac{2K{E}_0}{m}}}{\left(\sqrt{\frac{k}{m}}\right)−\left(\sqrt{\frac{2P{E}_0}{k}}\right)}\right) \\ ={\tan }^{−1}\left(\sqrt{\frac{K{E}_0}{P{E}_0}}\right) \end{gathered}$$

Substitute known values in the above equation.

$$\begin{gathered} \mathrm{Ï•}={\tan }^{-1}\left(\sqrt{\frac{0.06}{0.14}}\mathrm{Ï€}\right) \\ =0.58\mathrm{rad} \end{gathered}$$

Hence, the phase angle is $\mathrm{Ï•}=0.58\mathrm{rad}$.