# Coupled Oscillator

1. Aug 11, 2009

### CNX

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

One mass $m$ constrained to the x-axis, another mass $m$ constrained to the y-axis. Each mass has a spring connecting it to the origin with elastic constant $k$ and they are connected together by elastic constant $c$. I.e. we have a right-angle triangle made from the springs with lengths $b$, $b$, and $\sqrt{2} b$.

Write the Lagrangian, find the normal mode frequencies.

3. The attempt at a solution

Again having trouble with the coupling. For the two springs connected to the origin the potentials are straightforward:

$$V = \frac{1}{2} k x^2 + \frac{1}{2} k y^2$$

Given the geometry wouldn't the coupling spring add the potential,

$$V = \frac{1}{2} c \left [ \sqrt{x^2 + y^2} - \sqrt{2} b \right ]^2 = \frac{1}{2} c \left [ x^2 + y^2 - 2 \sqrt{2 x^2 + 2 y^2} + 2 b^2 \right ]$$

But I don't know how to put this in matrix form...

Last edited: Aug 11, 2009
2. Aug 11, 2009

### arunma

I reworked the problem and got the same potential as you, so it looks like you do indeed have the spring interaction potential correct. It looks a bit pesky because of the presence of the square root. This would make a closed form solution a bit difficult.

I did think of a potential trick you could use (I don't remember if I read this in a textbook or came up with it myself...hopefully the former). You could try converting to polar coordinates with,

$$x = r cos(\theta)$$
$$y = r sin(\theta)$$

Be careful here, because in this context $$r$$ and $$\theta$$ don't have any physical meaning; it's merely a math trick. You can then rewrite the Lagrangian with these new generalized coordinates. If my algebra/calculus are right, you should get the following set of differential equations:

$$m\ddot{r} = -(k + c)r + cb$$
$$\ddot{\theta} = 0$$

This seems comparatively a lot easier than what you would likely get by writing the Lagrangian using the generalized coordinates that you were working with. And the first differential equation looks like it will give you the oscillatory motion (without damping) that you would expect. After you solve for the two coordinates, you can transform back into the coordinates given in the problem.

Disclaimer: I don't know if this method will work, and indeed I see one potential pitfall. When you transform into polar coordinates, you get the weird effect of $$\theta = \theta + 2\pi$$. I normally leave it to the mathematicians to prove that physics math tricks actually work, but in this case I could see this as possibly being the cause of an incorrect solution. But hey, try it out and see what happens

3. Aug 11, 2009

### CNX

Thanks for your insight. I may have misled you into thinking I needed to the differential equations because I asked for the Lagrangian. I'm trying to get the normal mode frequencies by solving the eigenvalue problem.

I was thinking your trick might help still but it seems $\theta$ drops out of the expression for V.

$$T = \frac{1}{2} m \dot{r}^2$$

$$V = \frac{1}{2} c (r^2 - 2 \sqrt{2} r)$$

(constant term dropped in V)