Lagrangian function of a mass-spring-system with deflections in 2D [closed]

Question

I’m looking for the lagrangian function of the following problem (as seen in the picture). We have a mass connected to two springs. We can deglect the mass in two dimensions.

My main problems are:

1. What coordinates should I use?

2. How can I define the potential energy of the springs in regard of the mass?

In Newtonian mechanics, the approach is clear. I use polar-like coordinates and obtain a 2D-vector field resulting in two coupled ode’s.

I assume the potential energy is way easier than I think. Probably something like (in Cartesian coordinates):

$$ V = \frac{1}{2} k \left( x^2 + y^2 \right) $$

But isn’t that the potential energy for a setup with two additional vertical springs? Somehow I can't wrap my head around this. I would very much appreciate your help!

Answer 1

The potential energy of a spring is $V(\mathbf{r})=\frac{1}{2}k\|\mathbf{r}-\mathbf{r}_0\|^2$ where $\mathbf{r}_0$ is the equilibrium position. You haven't provided the equilibrium distance or how the springs are allowed to rotate, so this problem is inherently ambiguous.

Let's assume that they are stretched far enough that any "competition" between the two is far enough past their equilibrium distance, so it just looks like they both want to pull it towards where they are anchored to the wall. Then the equilibrium points are the left wall, which I set as the origin $(0,0)$, and the right wall $(2a,0)$. Thus we have something like: $$V(x,y)=\frac{1}{2}k_\text{left}(x^2+y^2)+\frac{1}{2}k_\text{right}\left((x-2a)^2+y^2\right)$$

You can check the limits that when the point is on the right wall the right spring doesn't pull and likewise for the left.

Also, I wouldn't use polar coordinates and the mass is needed for the equations of motion.

Correction: If the origin is the center of the two walls we actually get: $$V(x,y)=\frac{1}{2}k_\text{left}\left((x+a)^2+y^2\right)+\frac{1}{2}k_\text{right}\left((x-a)^2+y^2\right)$$

Answer 2

Assuming the springs are identical, the potential energy associated with each spring will be $$ U = \frac{1}{2} k (\ell(x,y) - \ell_0)^2 $$ where $\ell_0$ is the uncompressed length of the springs and $\ell(x,y)$ is the length of the spring when the mass is at the position $(x,y)$. This latter function can be found via straightforward geometry.

Note that the resulting functional form for the potential energy is not particularly "clean" unless $\ell_0 = 0$.