homework and exercises - Why does $omega = sqrt{V''(x_0) / m}$?

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.