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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?

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Well you have actually got the equation of motion of the pendulum. Now all you have to do is understand it

It's simple if you can see what is happening in the equation we may write the equation as follows

$$\frac{d^2 \theta}{dt^2}=-g\frac{\theta}{l}$$

This equation must be looking very familiar to you if you have solved for the simple harmonic oscillator.

From the analogies you can just say that the time period would be $$2\pi\sqrt{\frac{l}{g}}$$

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