For a body performing simple harmonic motion, | vedclass

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(D) The potential energy of a body in simple harmonic motion at displacement $x$ is given by $E_x = \frac{1}{2} kx^2$. From this,we can write $x = \sq

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The potential energy of a body in simple harmonic motion at displacement $x$ is given by $E_x = \frac{1}{2} kx^2$. From this,we can write $x = \sqrt{\frac{2 E_x}{k}}$. Similarly,the potential energy at displacement $y$ is $E_y = \frac{1}{2} ky^2$,which gives $y = \sqrt{\frac{2 E_y}{k}}$. The potential energy $E_0$ at displacement $(x+y)$ is given by $E_0 = \frac{1}{2} k(x+y)^2$. Expanding this expression,we get $E_0 = \frac{1}{2} k(x^2 + y^2 + 2xy)$. Substituting the values of $x^2$,$y^2$,and $xy$: $E_0 = \frac{1}{2} k \left( \frac{2 E_x}{k} + \frac{2 E_y}{k} + 2 \sqrt{\frac{2 E_x}{k}} \sqrt{\frac{2 E_y}{k}} \right)$. $E_0 = \frac{1}{2} k \left( \frac{2 E_x}{k} + \frac{2 E_y}{k} + \frac{4 \sqrt{E_x E_y}}{k} \right)$. $E_0 = E_x + E_y + 2 \sqrt{E_x E_y}$.

For a body performing simple harmonic motion,its potential energy is $E_x$ at displacement $x$ and $E_y$ at displacement $y$ from the mean position. The potential energy $E_0$ at displacement $(x+y)$ is

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