Question

A particle that can move along the x-axis is part of a system with potential energy

$U\left(x\right)=\frac{A}{x^2}-\frac{B}{x}$

where A and B are positive constants.

a. Where are the particle’s equilibrium positions?

b. For each, is it a point of stable or unstable equilibrium?

Short answer

(a) The equilibrium position is $x=\frac{2A}{B}$

(b) Since $-{\left.\frac{d^{2}U}{d{x}^2}\right|}_{x=\frac{2A}{B}}<0$, the equilibrium position $x=\frac{2A}{B}$is stable.

Step-by-step solution

Given information (part a)

A system has potential energy$U\left(x\right)=\frac{A}{x^2}-\frac{B}{x}$, whereA and B are positive constants.

Explanation (part a)

To find the equilibrium positions,

$$\begin{gathered} U\left(x\right)=\frac{A}{x^2}-\frac{B}{x} \\ -\frac{dU\left(x\right)}{dx}=0 \\ -\frac{d}{dx}\left(\frac{A}{x^2}-\frac{B}{x}\right)=0 \\ \frac{2A}{x^3}-\frac{B}{x^2}=0 \\ 2A-Bx=0 \\ x=\frac{2A}{B} \end{gathered}$$

Given information (part b)

A system has potential energy$U\left(x\right)=\frac{A}{x^2}-\frac{B}{x}$, whereA and B are positive constants.

Explanation (part b)

To determine the stability,

$$\begin{gathered} -\frac{d^{2}U}{d{x}^2}=-\frac{d^2}{d{x}^2}\left(\frac{A}{x^2}-\frac{B}{x}\right) \\ -\frac{d^{2}U}{d{x}^2}=-\frac{d}{dx}\left(\frac{B}{x^2}-\frac{2A}{x^3}\right) \\ -\frac{d^{2}U}{d{x}^2}=\frac{2B}{x^3}-\frac{6A}{x^4} \\ -{\left.\frac{d^{2}U}{d{x}^2}\right|}_{x=\frac{2A}{B}}=\frac{2B}{{\left(\frac{2A}{B}\right)}^3}-\frac{6A}{{\left(\frac{2A}{B}\right)}^4} \\ -{\left.\frac{d^{2}U}{d{x}^2}\right|}_{x=\frac{2A}{B}}=\frac{B^4}{4A^3}-\frac{3B^4}{8A^3} \\ -{\left.\frac{d^{2}U}{d{x}^2}\right|}_{x=\frac{2A}{B}}=-\frac{B^4}{8A^3}<0\text{, since}A>0,B>0 \end{gathered}$$