For a body performing simple harmonic motion, its potential energy is $E_x$ at displacement $x$ and…
For a body performing simple harmonic motion, its potential energy is $E_x$ at displacement $x$ and $\mathrm{E}_{\mathrm{y}}$ at displacement y from mean position. The potential energy $\mathrm{E}_0$ at displacement $(\mathrm{x}+\mathrm{y})$ is
$\sqrt{\mathrm{E}_{\mathrm{x}}^2+\mathrm{E}_y^2}$
$\sqrt{E_x-E_y}$
$\quad E_x+E_y$
$E_x+E_y+2 \sqrt{E_x E_y}$
Solution
Potential energy at $\mathrm{x}=\mathrm{E}_{\mathrm{x}}=\frac{1}{2} \mathrm{kx}^2$
$\therefore \quad \mathrm{x}=\sqrt{\frac{2 \mathrm{E}_{\mathrm{x}}}{\mathrm{k}}}$
Similarly,
Potential energy at $\mathrm{y}=\mathrm{E}_{\mathrm{y}}=\frac{1}{2} \mathrm{ky}^2$
$\therefore \quad y=\sqrt{\frac{2 E_y}{k}}$
P.E. at displacement $(\mathrm{x}+\mathrm{y})=\mathrm{E}=\frac{1}{2} \mathrm{k}(\mathrm{x}+\mathrm{y})^2$
$\begin{array}{ll}
\therefore & E=\frac{1}{2} k\left(x^2+y^2+2 x y\right) \\
\therefore & E=\frac{k}{2}\left[\frac{2 E_1}{k}+\frac{2 E_2}{k}+\frac{2 \times 2 \sqrt{E_1 E_2}}{k}\right] \\
\therefore & E=E_x+E_y+2 \sqrt{E_x E_y}
\end{array}$