Question

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

• A. $x=+a$ only
• B. $x=-a$ only
• C. $x=-\frac{a}{2}$ and $+\frac{a}{2}$
• D. $x=-a$ and $+a$

Answer 1

Answer: (B)

$x=-a$ only

At equilibrium position

$F=-\frac{d\vee }{dx}=0$
$\frac{dV}{dx}=\frac{C(a^2-x^2)}{(x^2+a^2{)}^2}=0$
$\therefore$ There are two equilibrium positions
$x_1=a$
$x_2=-a$
Consider $\frac{d^{2}V}{d{x}^2}=\frac{2Cx(x^2-3a^2)}{(x^2+a^2{)}^3}$
$\frac{d^{2}V}{d{x}^2}{|}_{x,}<0$
$\frac{d^{2}V}{d{x}^2}{|}_{x2}>0$

$\because$There is a maxima at $x=a$ and minima at $x=-a$
$\Rightarrow x_1$ is a position of unstable equilibrium and $x_2$ is a position of stable equilibrium.