High School Significance of the solution of the Euler-Lagrange equation

[Hamiltonian]

TL;DR

on applying the Euler Lagrange equation onto a system what exactly are you solving for and what is the significance of the solution you get?

I am new to Lagrangian mechanics and I have gone through basic examples of solving the Euler Lagrange equation for simple pendulums or projectiles and things like that. But I am unable to understand what we are exactly solving the equation for or what is the significance of the differential equation you end up with.

for example, if you consider a simple pendulum of mass m and length l
you get the lagrangian as $$L = (1/2)(mgl)\dot \theta^2 - mgl + mglcos\theta$$
and finally solving the Euler Lagrange equation gives you $$ 0 = \ddot \theta + (g\theta)/l$$
what is the significance of this solution?

 

[etotheipi]

You obtain the equations of motion, but I don't think your Lagrangian for this example is correct (even though you somehow got the right answer )? Shouldn't it be, up to an additive constant,$$\mathcal{L} = T - V = \frac{1}{2}ml^2 \dot{\theta}^2 + mgl\cos{\theta}$$Then for the single generalised coordinate $\theta$,$$\frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{\theta}} = \frac{d}{dt} ml^2 \dot{\theta} = ml^2 \ddot{\theta}$$ $$\frac{\partial \mathcal{L}}{\partial \theta} = -mgl \sin{\theta}$$So the equation of motion is$$ml^2 \ddot{\theta} + mgl\sin{\theta} = 0$$If you take the small angle approximation,$$\ddot{\theta} = -\frac{g}{l} \theta$$That's SHM with $\omega = \sqrt{\frac{g}{l}}$ which is exactly what you'd get with forces.

 

[Hamiltonian]

You obtain the equations of motion, but I don't think your Lagrangian for this example is correct (even though you somehow got the right answer )?

my bad shouldn't it be $$L = (1/2)(ml^2)\dot \theta^2 - mgl(1 - cos\theta)$$

 

[vanhees71]

Sure, but it's equivalent to @etotheipi 's, because it just differs by a constant.

 

[etotheipi]

It doesn't matter whether you take the zero of potential energy to be at the bottom of the swing or at the hinge, because that amounts to an additive constant, which doesn't change the equations of motion.

Your kinetic energy is now correct; notice that in polar coordinates the velocity is $\vec{v} = r\dot{\theta} \hat{\theta} + \dot{r}\hat{r}$, and since the motion is constrained to be circular, $\dot{r} = 0$ and $r = l$, so $\vec{v} \cdot \vec{v} = l^2 \dot{\theta}^2$