Potential energy function along positive x-axis is given by
$U (x) = - a x + \frac{b}{x}$ (only for $x > 0$ ) where $a$ and $b$ are constants. It is known that system has only one stable equilibrium configuration. Find the possible values of a and b. Justify your answer.

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Solution

$U (x) = - a x + \frac{b}{x}$
$F = - \frac{d U}{d t} = - (- a - \frac{b}{x^{2}}) = a + \frac{b}{x^{2}}$
$F = 0$ for equilibrium, $a + \frac{b}{x^{2}} = 0$
For stable equilibrium , $- \frac{d U}{d t^{2}} < 0$
$\frac{2 b}{x^{3}} < 0$ here $x > 0$ then,
$b > 0, a = - \frac{b}{x^{2}}$

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Potential energy function along positive x-axis is given byU(x)=−ax+bx (only for x>0) where a and b are constants. It is known that system has only one stable equilibrium configuration. Find the possible values of a and b. Justify your answer.

U(x)=−ax+bxF=−dUdt=−(−a−bx2)=a+bx2F=0 for equilibrium,a+bx2=0For stable equilibrium ,−dUdt2<02bx3<0 here x>0 then,b>0,a=−bx2

Potential energy function along positive x-axis is given by
$U (x) = - a x + \frac{b}{x}$ (only for $x > 0$ ) where $a$ and $b$ are constants. It is known that system has only one stable equilibrium configuration. Find the possible values of a and b. Justify your answer.

$U (x) = - a x + \frac{b}{x}$
$F = - \frac{d U}{d t} = - (- a - \frac{b}{x^{2}}) = a + \frac{b}{x^{2}}$
$F = 0$ for equilibrium, $a + \frac{b}{x^{2}} = 0$
For stable equilibrium , $- \frac{d U}{d t^{2}} < 0$
$\frac{2 b}{x^{3}} < 0$ here $x > 0$ then,
$b > 0, a = - \frac{b}{x^{2}}$