A body is executing S.H.M. At a displacement | vedclass

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(C) The potential energy $U$ of a body executing $S.H.M.$ at displacement $x$ is given by $U = \frac{1}{2} k x^2$,where $k$ is the force constant.Give

Vedclass · Accepted Answer

$49$

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(C) The potential energy $U$ of a body executing $S.H.M.$ at displacement $x$ is given by $U = \frac{1}{2} k x^2$,where $k$ is the force constant.Given:$U_x = \frac{1}{2} k x^2 = 9 \ J$  --- $(1)$$U_y = \frac{1}{2} k y^2 = 16 \ J$ --- $(2)$We need to find the potential energy at displacement $(x+y)$,which is $U_{(x+y)} = \frac{1}{2} k (x+y)^2$.Expanding this,we get:$U_{(x+y)} = \frac{1}{2} k (x^2 + y^2 + 2xy) = \frac{1}{2} k x^2 + \frac{1}{2} k y^2 + 2 \left( \sqrt{\frac{1}{2} k x^2} ight) \left( \sqrt{\frac{1}{2} k y^2} ight)$.Substituting the given values:$U_{(x+y)} = 9 + 16 + 2 	imes \sqrt{9} 	imes \sqrt{16}$$U_{(x+y)} = 25 + 2 	imes 3 	imes 4$$U_{(x+y)} = 25 + 24 = 49 \ J$.

$A$ body is executing $S.H.M$. At a displacement $x$ its potential energy is $9 \ J$ and at a displacement $y$ its potential energy is $16 \ J$. The potential energy at displacement $(x+y)$ is (in $J$)

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