A body is executing simple harmonic motion. A | vedclass

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Question

(C) The potential energy $U$ of a body executing simple harmonic motion at displacement $x$ is given by $U = \frac{1}{2} kx^2$.Given:$E_1 = \frac{1}{2

Vedclass · Accepted Answer

$E_1 + E_2 + 2\sqrt{E_1 E_2}$

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(C) The potential energy $U$ of a body executing simple harmonic motion at displacement $x$ is given by $U = \frac{1}{2} kx^2$.Given:$E_1 = \frac{1}{2} kx^2$  --- $(i)$$E_2 = \frac{1}{2} ky^2$  --- $(ii)$We need to find the potential energy $E$ at displacement $(x + y)$:$E = \frac{1}{2} k(x + y)^2$$E = \frac{1}{2} k(x^2 + y^2 + 2xy)$$E = \frac{1}{2} kx^2 + \frac{1}{2} ky^2 + kxy$Substituting values from $(i)$ and $(ii)$:$E = E_1 + E_2 + kxy$Since $\sqrt{E_1} = \sqrt{\frac{1}{2} k} x$ and $\sqrt{E_2} = \sqrt{\frac{1}{2} k} y$,we have:$2\sqrt{E_1 E_2} = 2 \sqrt{\frac{1}{2} k x^2 \cdot \frac{1}{2} k y^2} = 2 \cdot \frac{1}{2} kxy = kxy$Therefore,$E = E_1 + E_2 + 2\sqrt{E_1 E_2}$.

Column-$I$	Column-$II$
$(a)$ $y = \frac{A}{\sqrt{2}}$	$(i)$ $K = \frac{3E}{4}$
$(b)$ $y = \frac{\sqrt{3}A}{2}$	$(ii)$ $K = \frac{E}{4}$
	$(iii)$ $K = \frac{E}{2}$

$A$ 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 a displacement $(x + y)$ is

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