# What kind of system does this lagrangian describe?

1. Feb 17, 2014

### brotof

1. The problem statement, all variables and given/known data
Consider the following Lagrangian:
$$L = \frac{m}{2}(a\dot{x}^2 + 2b\dot{x}\dot{y} + c\dot{y}^2)- \frac{k}{2}(ax^2 + 2bxy + cy^2)$$
Assume that $$b^2 - 4ac \ne 0$$Find the equations of motion and examine the cases a=b=0 and b=c, c=-a. Which kind of physical system is described by this Lagrangian? What meaning does $$b^2 - 4ac \ne 0$$ have?

2. Relevant equations
$$\frac{\partial L}{\partial q_i} -\frac{d}{dt}\frac{\partial L}{\partial \dot{q_i} } = 0$$

3. The attempt at a solution
I'm able to solve the maths for this problem (assuming no errors in my calculations). In general for a, c not equal to 0 i get
$$x(t) = \frac{1}{ac-b^2}(A_1 c \cos(\omega t + \phi_1) - A_2 b \cos(\omega t + \phi_2))$$
$$y(t) = \frac{1}{ac-b^2}(A_2 a \cos(\omega t + \phi_2) - A_1 b \cos(\omega t + \phi_1))$$

While for the cases a=c=0 and b=0, c=-a i get
$$x(t) = A_1 \cos(\omega t + \phi_1)$$
$$y(t) = A_2 \cos(\omega t + \phi_2)$$

The problem is I can't exactly figure out what kind of system this could be. My first thoughts are some kind of 2D harmonic oscillator or some system of 2 masses coupled with springs. I'm having problem answering what $$b^2 - 4ac \ne 0$$ "means" when I can't figure out what kind of system this is.

Brotof

2. Feb 18, 2014

### Goddar

Hi.
I didn't check your solution but it looks ok... now:
"2D harmonic oscillator or some system of 2 masses coupled with springs" are examples of two coupled oscillators, so you're basically right.
For your next question, how can you express the Lagrangian when b^2= 4ac? What does it say about the relative phases of x and y?

3. Feb 19, 2014

### brotof

I see that the Lagrangian then can be written:

L = \frac{m}{2c}(b\dot{x} + c\dot{y})^2 -\frac{k}{2c}(bx + cy)^2

Also, bot my equations of motion reduces to a single one:

\frac{b^2}{c}\dot{x} + b\dot{y} + \frac{k}{m}(\frac{b^2}{c} x + by) = 0

Hmm. I'm not exactly sure what this says about the relative phases. We get the solution

(\frac{b^2}{c} x + by) = A \cos(\omega t + \phi)
.
Does this necessarily mean they have the same phase? I'm still not able to visualize this system.

Last edited: Feb 19, 2014
4. Feb 19, 2014

### Goddar

The above is wrong. If:
ax2+2bxy+cy2 and b2= 4ac, then you have equivalently:
(x√a + y√b)2
And your Lagrangian can be written in terms of a single variable:
u = x√a + y√b
you have first a system of two masses and three springs, something like that:
|^^O^^O^^|
Now replace the central spring by a rigid rod:
|^^O––O^^|
See the connection?

5. Feb 19, 2014

### brotof

Thanks a lot, that makes sense!

Im really sorry, I was a bit hasty and have made an error. It's supposed to be

b^2 = ac

without the factor 4.

I don't think I'm wrong when taking that into account that I meant b^2 = ac, even though I wrote it a bit more complicated that necessary.

\frac{k}{2}(ax^2 + 2bxy + cy^2) = \frac{k}{2}(\frac{b^2}{c}x^2 + 2bxy + cy^2) = \frac{k}{2c}(b^2 x^2 + 2bcxy + c^2 y^2) = \frac{k}{2c}(bx + cy)^2

and of course the same way for the kinetic energy term.

Then we get

bx + cy = A\cos(\omega t + \phi)

Anyway, the help is appreciated!