# Why doesn't $\omega = \sqrt{\frac{U''(x_0)}{m}}$ work for a simple pendulum?

In a simple pendulum, we know that the angular frequency of small oscillations is $$\omega = \sqrt{\frac{g}{l}}$$. However $$\sqrt{\frac{U''(x_0)}{m}}$$ gives $$\sqrt{gl}$$ as the angular frequency.

Let $$l$$ be the length of the string, let $$\theta$$ be the angle made by the string with the vertical, and let $$m$$ be the mass of the bob.

We have $$U(\theta) = -mgl\cos\theta$$ and $$U''(\theta) = mgl\cos\theta$$.

$$U$$ has a minimum at $$\theta = 0$$.

So the angular frequency of small oscillations is $$\sqrt{\frac{U''(0)}{m}} = \sqrt{\frac{mgl}{m}} = \sqrt{gl}$$.

What mistake have I made here?

You have to understand where the division by $$m$$ comes from. This is because you are assuming the kinetic energy to be $$K(\dot x) = \frac{m}{2}\dot x^2$$. In general, the relevant quantity is $$K''(0)$$ which corresponds to $$m$$ in the previous example. This is the kinetic equivalent for the harmonic approximation of potential energy. The correct formula is therefore: $$\omega = \sqrt{\frac{U''(x_0)}{K''(0)}}$$ which has the correct dimensions of inverse of time. In the case of the simple pendulum: $$U = -mgl\cos\theta\\ K = \frac{ml^2}{2}\dot\theta^2$$ so you indeed retrieve: $$\omega = \sqrt{\frac{g}{l}}$$
• Thanks, I don't know why I didn't spot my fairly obvious error before. I was implicitly assuming that the force on the bob was $m\ddot{\theta}$ instead of $ml\ddot{\theta}$. Commented Jun 15, 2022 at 12:04
If $$x=l\theta$$, then $$\frac{d}{dx}=\frac{1}{l}\frac{d}{d\theta}$$ so $$\frac{d^2 V(x)}{dx^2}= \frac{1}{l^2}\frac{d^2 V}{d\theta^2}$$ so converting derivatives w/r to $$x$$ to derivatives w/r to $$\theta$$ yields $$\sqrt{g/l}$$.