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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$\begingroup$ 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: $\endgroup$– Strong LizardCommented Mar 17, 2022 at 14:32
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$\begingroup$ 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 $T\omega= 2\pi$. $\endgroup$– Strong LizardCommented Mar 17, 2022 at 14:42
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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 $$2\pi\sqrt{\frac{l}{g}}$$
