# Lagrangian Mechanics Problem

1. Jun 1, 2016

### Yosty22

1. The problem statement, all variables and given/known data

We have a particle of mass m moving in a plane described by the following Lagrangian:
\frac{1}{2}m((\dot{x}^2)+(\dot{y}^2)+2(\alpha)(\dot{x})(\dot{y}))-\frac{1}{2}k(x^2+y^2+(\beta)xy) for k>0 is a spring constant and \alpha and \beta are time-independent.

Find the normal mode frequencies, \omega_1,2

2. Relevant equations

Euler-Lagrange Equation

3. The attempt at a solution

I think I'm just missing here. There was a lot of math, so I won't explicitly write out all of it, but I will have my final answers. I used the Euler-Lagrange Equation twice: once for x, once for y. This yielded:

m*ddot{x}+m\alpha(\ddot{y}=-kx+\beta(y)
and
m*\ddot{y}+m\alpha(\ddot{x}=-ky+\beta(x)

I solved each for \ddot{x} and equated them, giving me:

\ddot{y} = \frac{\beta+k/(\alpha)}{m(\alpha-1/(\alpha))}*y + \frac{-k-\beta/(\alpha)}{m(\alpha-1/(\alpha))}*x

Am I approaching this the right way to find the frequencies? I know usually in 1D for example, you solve the Euler-Lagrange equation to yield something of the form: \ddot{x}=\omega^2*x, but it is a little more unclear to me as to what to do here. Would I find two seperate frequencies, once in x and once in y and they are two separate answers?

2. Jun 1, 2016

### Yosty22

Update: I should probably transform to coordinates like (r,theta) and find theta dot, right?

3. Jun 2, 2016

### Ray Vickson

Perhaps if you presented the TeX/LaTeX expressions properly you would receive more responses. I edited the first of your equations above. The first one below is exactly what you wrote, but inserting the appropriate controls to make the PF processor understand you want to use LaTeX. As you can see, it looks ugly, and defeats the whole purpose of using LaTeX in the first place. The second one is the properly-edited version, in which all needed control characters are used. Right-click on it to see the TeX commands.

$$m*ddot{x}+m\alpha(\ddot{y}=-kx+\beta(y)$$

$$m \ddot{x}+m\alpha \ddot{y}=-kx+\beta y$$

4. Jun 2, 2016

### ehild

Check your equations. I corrected your TeX codes. Is not a factor of 2 in front of beta in the Lagrangian?
Find the solution in form $x=x_0 e^{i\omega t}$, $y=y_0 e^{i\omega t}$. You get a system of linear homogeneous equations for xo and yo that has nonzero solution for certain ω-as only.

Last edited: Jun 2, 2016