A particle of mass $m$ moves in a potential field $U(x)=\frac{a}{x^2}-\frac{b}{x}$, where $a$ and $b$ are positive constants. What is the period of small oscillations about the stable equilibrium position?

Solution

1. The equilibrium position $x_0$ is found by setting the first derivative of the potential $U(x)=a{x}^{-2}-b{x}^{-1}$ to zero: $$\begin{array}{rl}{U}^{\prime }(x)& =-2a{x}^{-3}+b{x}^{-2}\\ & =0\end{array}$$, which yields $x_0=2a/b$.
2. The effective spring constant $k$ is the second derivative evaluated at $x_0$: $U^{\prime \prime }(x)=6a{x}^{-4}-2b{x}^{-3}$, so $$\begin{array}{rl}k& =U^{\prime \prime }(2a/b)\\ & =6a(b/2a{)}^4-2b(b/2a{)}^3\\ & =\frac{6a{b}^4}{16a^4}-\frac{2b^4}{8a^3}\\ & =\frac{3b^4}{8a^3}-\frac{2b^4}{8a^3}\\ & =\frac{b^4}{8a^3}\end{array}$$.
3. The period of small oscillations is given by $$\begin{array}{rl}T& =2\pi \sqrt{m/k}\\ & =2\pi \sqrt{m/(b^4/8a^3)}\\ & =2\pi \sqrt{8m{a}^3/b^4}\\ & =4\pi \sqrt{2m{a}^3/b^4}\end{array}$$.
4. Re-evaluating the derivative $U^{\prime \prime }(x)=6a{x}^{-4}-2b{x}^{-3}$ at $x_0=2a/b$ gives $$\begin{array}{rl}k& =6a(b^4/16a^4)-2b(b^3/8a^3)\\ & =\frac{3b^4}{8a^3}-\frac{2b^4}{8a^3}\\ & =\frac{b^4}{8a^3}\end{array}$$, leading to $$\begin{array}{rl}T& =2\pi \sqrt{m\cdot 8a^3/b^4}\\ & =2\pi \sqrt{8}\sqrt{m{a}^3/b^4}\end{array}$$.
5. Hence the answer is (A).