Confused about the solution to the pendulum differential equation

Question

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!

Answer 1

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 $\theta_0 = 0$.

Answer 2

A constant function has any period. i.e. you cannot write a lowest period and so it is not a meaningful quantity.