Question

Solve the foregoing problem if the potential energy has the form $U\left(x\right)=a/x^2-b/x$ where a and b are positive constants.

Answer 1

If $U\left(x\right)=\frac{a}{x^2}-\frac{b}{x}$
then the equilibrium position is $x=x_0$ when $U^{{}^{\prime }}\left(x_0\right)=0$
or $-\frac{2a}{x_0^3}+\frac{b}{x_0^2}=0\Rightarrow =\frac{2a}{b}$
Now write : $x=x_0+y$
Then $U\left(x\right)=\frac{a}{x_0^2}-\frac{b}{x_0}+(x-x_0)U^{{}^{\prime }}\left(x_0\right)+\frac{1}{2}(x-x_0{)}^2{U}^{{}^{\prime \prime }}(x_0)$
But $U^{{}^{\prime \prime }}\left(x_0\right)=\frac{6a}{x_0^4}-\frac{2b}{x_0^3}=\left(2a/b{)}^{-3}\right(3b-2b)=b^4/8a^3$
So finally : $U\left(x\right)=U\left(x_0\right)+\frac{1}{2}\left(\frac{b^4}{8a^3}\right)y^2+...$
We neglect remaining terms for small oscillations and compare with the P.E. for a harmonic, oscillator :
$\frac{1}{2}m{\omega }^2{y}^2=\frac{1}{2}\left(\frac{b^4}{8a^3}\right)y^2,$ so $\omega =\frac{b^2}{\sqrt{8a^{3}m}}$
Thus $T=2\pi \frac{\sqrt{t}8m{a}^3}{b^2}$
Note : Equilibrium position is generally a minimum of the potential energy. Then $U^{{}^{\prime }}\left(x_0\right)=0,U^{{}^{\prime \prime }}\left(x_0\right)>0$. The equilibrium position can in principle be a maximum but then $U^{{}^{\prime \prime }}\left(x_0\right)<0$ and the frequency of oscillations about this equilibrium position will be dimensionally.