A body is executing Simple Harmonic Motion. A | vedclass

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(B) The potential energy of a body executing Simple Harmonic Motion at displacement $x$ is given by $E_1 = \frac{1}{2}kx^2$.From this,we get $x = \sqr

Vedclass · Accepted Answer

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

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(B) The potential energy of a body executing Simple Harmonic Motion at displacement $x$ is given by $E_1 = \frac{1}{2}kx^2$.From this,we get $x = \sqrt{\frac{2E_1}{k}}$.Similarly,at displacement $y$,the potential energy is $E_2 = \frac{1}{2}ky^2$,which gives $y = \sqrt{\frac{2E_2}{k}}$.The potential energy $E$ at displacement $(x + y)$ is $E = \frac{1}{2}k(x + y)^2$.Taking the square root,we get $\sqrt{E} = \sqrt{\frac{1}{2}k}(x + y) = \sqrt{\frac{k}{2}}(x + y)$.Substituting the values of $x$ and $y$,we get $\sqrt{E} = \sqrt{\frac{k}{2}} \left( \sqrt{\frac{2E_1}{k}} + \sqrt{\frac{2E_2}{k}} \right)$.Simplifying this,we get $\sqrt{E} = \sqrt{\frac{k}{2}} \cdot \sqrt{\frac{2}{k}} (\sqrt{E_1} + \sqrt{E_2})$.Therefore,$\sqrt{E} = \sqrt{E_1} + \sqrt{E_2}$.

$A$ 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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