Question

A particle of mass m moves due to a conservative force with potential energy V(x) = $x_{2}+a_{2}Cx $, where C and a are positive constants. The position(s) of stable equilibrium is/are given as

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Updated on : 2022-09-05

Solution

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Correct option is B)

At equilibrium
position

$F=−dxdV =0$

$dxdV =(x_{2}+a_{2})_{2}C(a_{2}−x_{2}) =0$

$∴$ There are two equilibrium
positions

$x_{1}=a$

$x_{2}=−a$

Consider $dx_{2}d_{2}V =(x_{2}+a_{2})_{3}2Cx(x_{2}−3a_{2}) $

$dx_{2}d_{2}V ∣_{x_{1}}<0$

$dx_{2}d_{2}V ∣_{x_{2}}>0$

$∵$There is a maxima at $x=a$ and minima at $x=−a$

$⇒x_{1}$ is a position of unstable
equilibrium and $x_{2}$ is a position of stable equilibrium.

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