# Confused about the solution to the pendulum differential equation

So I’ve learned how to derive the exact solution to the pendulum differential equation (in respect to its period), $$\ddot{\theta} + \frac{g}{l}\sin\theta=0$$, where $$g$$ is gravitational acceleration and $$l$$ is the length of the string attached to the pendulum. The solution turns out to be: $$T = 4\sqrt{\frac{l}{g}}K\left(\sin{\left(\frac{\theta_0}2\right)}\right),$$

Where $$\theta_0$$ is the starting angle of the pendulum, and $$K(k)$$ is the Complete Elliptic Integral of the First Kind, which can be expressed as: $$K(k) = \int_0^{\frac{\pi}2}\frac{d\eta}{\sqrt{1-k^2\sin^2\eta}}$$

If you set $$\theta_0 = 0, K\left(\sin{\left(\frac{\theta_0}2\right)}\right) = \frac{\pi}2$$, which would make $$T = 2\pi\sqrt{\frac{l}{g}}$$. But this doesn’t make sense—how could the period be equal to anything other than $$0$$? If $$\theta_0 = 0$$, the pendulum isn’t in motion!

• I have edited your question to include what I think are better tags; feel free to revert this if you don't like them. Apr 28, 2022 at 15:04

In the derivation of the period, it is implicitly assumed that $$\theta_0 \neq 0$$. Specifically, the standard derivation via energy conservation leads to the integral $$T = 2 \sqrt{\frac{l}g} \int_0^{\theta_0} \frac{d \theta}{\sqrt{\sin^2 (\theta_0/2) - \sin^2(\theta/2)}}$$ If $$\theta_0 = 0$$, then this integral is obviously zero. But to put this in the form you're using above for an elliptic integral, we then need to make the substitution $$\sin \eta = \frac{\sin(\theta/2)}{\sin(\theta_0/2)},$$ which is not a valid operation when $$\theta_0 = 0$$.
• @Mailbox: You mean the integral when $\theta_0 = 0$? Probably not (or at least in any way that's interesting); it's equal to zero by definition because of the limits of integration. Apr 28, 2022 at 15:11
• Hello again! I looked into this a little more myself, and I’m not sure this answer fully explains the problem for when $\theta_0 = 0$. Firstly, when $\theta_0 = 0$, the integral above does not equal zero—it becomes undefined. Second, if we take the limit as $\theta \to 0$, the integral converges to $\pi$, leaving us with the same problem as before: $T$ converges to $2\pi\sqrt{\frac{l}{g}}$, which is still nonsensical even when considering limits. How could a pendulum have such a large period when its starting angle is infinitesimally small? Apr 29, 2022 at 13:33
• As to the first point, yeah, I suppose that it's probably better to take that integral as undefined. But if we go back a step or two in the derivation, the period is defined as the amount of time required to go from maximum displacement $\theta = \theta_0$ to the equilibrium position $\theta = 0$. If $\theta_0 = 0$, I don't see how this quantity can be anything other than zero. Apr 29, 2022 at 15:17
• There are varying definitions of the elliptic integral — the integral that some sources call $K(x)$ is called $K(x^2)$ by other sources. So that's not necessarily wrong. Apr 28, 2022 at 14:47