Question 294: Oscillations

Question

For a body performing simple harmonic motion, its potential energy is $\mathrm{E}_{\mathrm{x}}$ at displacement x and $\mathrm{E}_{\mathrm{y}}$ at displacement y from mean position. The potential energy $E_0$ at displacement $(x+y)$ is

Accepted Answer

Potential energy at $\mathrm{x}=\mathrm{E}_{\mathrm{x}}=\frac{1}{2} \mathrm{kx}^2$
$	herefore\quad\mathrm{x}=\sqrt{\frac{2 \mathrm{E}_{\mathrm{x}}}{\mathrm{k}}}\quad	ext{.... (i)}$
Similarly,
Potential energy at $\mathrm{y}=\mathrm{E}_{\mathrm{y}}=\frac{1}{2} \mathrm{ky}^2$
$	herefore \quad \mathrm{y}=\sqrt{\frac{2 \mathrm{E}_{\mathrm{y}}}{\mathrm{k}}}\quad	ext{.... (ii)}$
P.E. at displacement $(x+y)=E=\frac{1}{2} k(x+y)^2$
$$\begin{array}{ll}
	herefore & E=\frac{1}{2} k\left(x^2+y^2+2 x yight) \
	herefore & E=\frac{k}{2}\left[\frac{2 E_1}{k}+\frac{2 E_2}{k}+\frac{2 	imes 2 \sqrt{E_1 E_2}}{k}ight] \
	herefore & E=E_x+E_y+2 \sqrt{E_x E_y}
\end{array}$$

Answer

$\sqrt{\mathrm{E}_{\mathrm{x}}^2+\mathrm{E}_{\mathrm{y}}^2}$

Answer

$\sqrt{E_x-E_y}$

Answer

$E_x+E_y$

Answer

$E_x+E_y+2 \sqrt{E_x E_y}$

Oscillations

For a body performing simple harmonic motion, its potential energy is \mathrm{E}_{\mathrm{x}} at displacement x and \mathrm{E}_{\mathrm{y}} at displacement y from mean position. The potential energy E_0 at displacement (x+y) is

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