Small Oscillations

First let's find the equilibrium point. This occurs where $\frac{dU}{dx} = 0$ . Computing, $\frac{dU}{dx} = -\frac{2}{x^3} + \frac{1}{x^2} = 0 \implies x_0 =2.$ At the equilibrium point $x_0$ , $U(x_0) = -\frac14$ . The second derivative of the potential is: $\frac{d^2 U}{dx^2} = \frac{6}{x^4} - \frac{2}{x^3}.$ Evaluated at $x_0 = 2$ , this is equal to $\frac18$ . The Taylor expansion to second-order at that point is therefore: $U(x) \approx -\frac14 + \frac{1}{16} (x-2)^2.$ Consequently, the frequency of oscillation is: $\omega_0 =\sqrt{\biggl.\frac{1}{m}\frac{d^2 U}{dx^2}\biggr|_{x=x_0}}= \sqrt{\frac{1}{8m}}.$