What kind of system does this lagrangian describe?

[brotof]

Homework Statement

Consider the following Lagrangian:
\begin{equation} 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)\end{equation}
Assume that \begin{equation} b^2 - 4ac \ne 0 \end{equation}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 \begin{equation} b^2 - 4ac \ne 0 \end{equation} have?

Homework Equations

\begin{equation} \frac{\partial L}{\partial q_i} -\frac{d}{dt}\frac{\partial L}{\partial \dot{q_i} } = 0 \end{equation}

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
\begin{equation} x(t) = \frac{1}{ac-b^2}(A_1 c \cos(\omega t + \phi_1) - A_2 b \cos(\omega t + \phi_2)) \end{equation}
\begin{equation} y(t) = \frac{1}{ac-b^2}(A_2 a \cos(\omega t + \phi_2) - A_1 b \cos(\omega t + \phi_1)) \end{equation}

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

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 \begin{equation} b^2 - 4ac \ne 0 \end{equation} "means" when I can't figure out what kind of system this is.

Thanks in advance,
Brotof

 

[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?

 

[brotof]

I see that the Lagrangian then can be written:

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

Also, bot my equations of motion reduces to a single one:
\begin{equation}
\frac{b^2}{c}\dot{x} + b\dot{y} + \frac{k}{m}(\frac{b^2}{c} x + by) = 0
\end{equation}

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

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

 

[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
In other terms, a degree of freedom has been lost somewhere. Think about this example:
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?

 

[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
\begin{equation}
b^2 = ac
\end{equation}
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.

\begin{equation}
\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
\end{equation}
and of course the same way for the kinetic energy term.

Then we get

\begin{equation}
bx + cy = A\cos(\omega t + \phi)
\end{equation}

Anyway, the help is appreciated!