Why does $\omega = \sqrt{V''(x_0) / m}$?

Question

I know that in an equation such that $$\ddot{x} + \omega^2x = 0,$$ the angular frequency $ = \omega$. But why is that ever $ \sqrt{V''(x_0) / m}$? (where $x_0$ is the equilibrium point). I just saw that used in the solutions to a problem set without explanation.

Answer 1

Consider an arbitrary potential energy $V(x)$; take $x_0$ to be an equilibrium point, that is, $V'(x_0)=0$. Next, Taylor expand $V(x)$ for $x$ close to $x_0$: $$ V(x)\approx V(x_0)+(x-x_0)V'(x_0)+\frac{1}{2}(x-x_0)^2 V''(x_0) $$

The first term is just a constant (ie, irrelevant for energies), and the second one is, by definition, null; therefore we find $V(x)\approx \frac{1}{2}V''(x_0)(x-x_0)^2$. Note that this resambles a usual harmonic energy, thus we may rewrite it as $V(x)=\frac{1}{2}k(x-x_0)^2$, where $k=V''(x_0)$.

Finally, take the usual formula $\omega=\sqrt{k/m}$ to get the result you seek.