A body is executing simple harmonic motion. A | vedclass

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(B) The potential energy $U$ of a body executing simple harmonic motion $(SHM)$ at a displacement $d$ is given by $U = \frac{1}{2} k d^2$, where $k =

Vedclass · Accepted Answer

$\sqrt{E}=\sqrt{E_1}+\sqrt{E_2}$

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(B) The potential energy $U$ of a body executing simple harmonic motion $(SHM)$ at a displacement $d$ is given by $U = \frac{1}{2} k d^2$, where $k = m \omega^2$ is the force constant.Given:$E_1 = \frac{1}{2} k x^2 \implies x = \sqrt{\frac{2 E_1}{k}}$$E_2 = \frac{1}{2} k y^2 \implies y = \sqrt{\frac{2 E_2}{k}}$At displacement $(x+y)$, the potential energy $E$ is:$E = \frac{1}{2} k (x+y)^2$Taking the square root of both sides:$\sqrt{E} = \sqrt{\frac{1}{2} k} (x+y)$Substituting the expressions for $x$ and $y$:$\sqrt{E} = \sqrt{\frac{1}{2} k} \left( \sqrt{\frac{2 E_1}{k}} + \sqrt{\frac{2 E_2}{k}} \right)$$\sqrt{E} = \sqrt{\frac{1}{2} k} \cdot \sqrt{\frac{2}{k}} (\sqrt{E_1} + \sqrt{E_2})$$\sqrt{E} = \sqrt{\frac{1}{2} k \cdot \frac{2}{k}} (\sqrt{E_1} + \sqrt{E_2})$$\sqrt{E} = \sqrt{1} (\sqrt{E_1} + \sqrt{E_2})$$\sqrt{E} = \sqrt{E_1} + \sqrt{E_2}$

$A$ body is executing simple harmonic motion. At a displacement $x$ its potential energy is $E_1$ and at a displacement $y$ its potential energy is $E_2$. The potential energy $(E)$ at displacement $(x+y)$ is

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