# How to obtain the time period from the Lagrangian equation for a simple pendulum? [closed]

Solving the Lagrangian equation for a simple pendulum we get the following equation: $$\ddot{\theta} + \frac{g \theta}{l} = 0,$$ (when $$\theta$$ is small enough). We already know that time period of a simple pendulum is given by $$T= 2π \sqrt{l/g}.$$ But how can we derive the time period $$T$$ from the equation that is obtained from the Lagrangian equation?

• It's been more than a year since I asked this question and now it seems so trivial to me. If anyone ever search for this question come to this the following might be helpful: Commented Mar 17, 2022 at 14:32
• The equation of SHM in standard form is $\ddot{x}+\omega^2 x=0$ , where $\omega$ is angular frequency of the SHM. Which means, in this case, $\omega=\sqrt{\frac{g}{l}}$. Then time period follows from the relation $T\omega= 2\pi$. Commented Mar 17, 2022 at 14:42

$$\frac{d^2 \theta}{dt^2}=-g\frac{\theta}{l}$$
From the analogies you can just say that the time period would be $$2\pi\sqrt{\frac{l}{g}}$$