For a body performing simple harmonic motion, its potential energy is E x at displacement x and E y at displacement y from mean position. The potential energy E 0 at displacement ( x + y ) is

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For a body performing simple harmonic motion, its potential energy is E x ​ at displacement x and E y ​ at displacement y from mean position. The potential energy E 0 ​ at displacement ( x + y ) is

TestHub · Accepted Answer

The potential energy of a body performing simple harmonic motion at a displacement $x$ from the mean position is given by $E_{x} = \frac{1}{2} k x^{2}$ .

From this, the displacement $x$ is:

$x = \frac{2 E _{x}}{k}$

Similarly, the potential energy at displacement $y$ is $E_{y} = \frac{1}{2} k y^{2}$ , which gives:

$y = \frac{2 E _{y}}{k}$

The potential energy $E_{0}$ at displacement $(x + y)$ is:

$E_{0} = \frac{1}{2} k (x + y)^{2}$

Expanding the square:

$E_{0} = \frac{1}{2} k (x^{2} + y^{2} + 2 x y)$

$E_{0} = \frac{1}{2} k x^{2} + \frac{1}{2} k y^{2} + k x y$

Substituting the values of $\frac{1}{2} k x^{2}$ , $\frac{1}{2} k y^{2}$ , $x$ , and $y$ :

$E_{0} = E_{x} + E_{y} + k (\frac{2 E _{x}}{k}) (\frac{2 E _{y}}{k})$

$E_{0} = E_{x} + E_{y} + k (\frac{2}{k} E_{x} E_{y})$

$E_{0} = E_{x} + E_{y} + 2 E_{x} E_{y}$

Answer: $E_{x} + E_{y} + 2 E_{x} E_{y}$

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