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

Question

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?

Answer 1

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: $$ \omega = \sqrt{\frac{g}{l}} $$

Hope this helps.

Answer 2

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 $\sqrt{g/l}$.