# Lagrangian mechanics for two springs

1. Mar 15, 2009

### wglmb

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

A spring of negligible mass and spring constant k, hanging vertically with one end at a fixed point O, supports a mass m, and beneath it as second, identical spring carrying a second, identical mass.
Using a generalised coordinates the vertical displacements x and y of the masses from their positions with the springs unextended, write down the Lagrangian.
Find the position of equilibrium and the normal modes and frequencies of vertical oscillations.

2. Relevant equations

3. The attempt at a solution

KE: $$\frac{1}{2}m(\dot{x}^{2}+\dot{y}^{2})$$
PE: $$mg(l+x)+mg(2l+y)+\frac{1}{2}kx^{2}+\frac{1}{2}k(y-x)^{2}$$ $$=mg(3l+x+y)+kx^{2}+\frac{1}{2}ky^{2}-kxy$$

Lagrangian: $$L=\frac{1}{2}m(\dot{x}^{2}+\dot{y}^{2})-mg(3l+x+y)-kx^{2}-\frac{1}{2}ky^{2}+kxy$$

Equations of motion:
$$\frac{d}{dt}(\frac{\partial L}{\partial \dot{x}})-\frac{\partial L}{\partial x}=0 \Rightarrow m\ddot{x}+mg+2kx-ky=0$$

$$\frac{d}{dt}(\frac{\partial L}{\partial \dot{y}})-\frac{\partial L}{\partial y}=0 \Rightarrow m\ddot{y}+mg+ky-kx=0$$

Know that $$B_{ij}\ddot{q}_{j}+A_{ij}q_{j}=0$$

$$\Rightarrow B=\left( \begin{array}{cc} m & 0 \\ 0 & m \end{array} \right), A=\left( \begin{array}{cc} 2k & -k \\ -k & k \end{array} \right)$$

$$B^{-1}A=\left( \begin{array}{cc} \frac{1}{m} & 0 \\ 0 & \frac{1}{m} \end{array} \right)\left( \begin{array}{cc} 2k & -k \\ -k & k \end{array} \right) =\left( \begin{array}{cc} \frac{2k}{m} & \frac{-k}{m} \\ \frac{-k}{m} & \frac{k}{m} \end{array} \right)$$

Eigenvalues turn out to be:
$$\lambda=\frac{k}{2m}(3+\sqrt{5}) and \lambda=\frac{k}{2m}(3-\sqrt{5})$$

This doesn't look very likely to be right.
Then when I work out the eigenvectors, I get (0,0) in both cases...
... where have I gone wrong?

2. Mar 15, 2009

### Dr.D

When you moved the masses down x and y respectively, the strain energies increased as you have indicated, but the gravitational energies also increase?

3. Mar 15, 2009

### wglmb

Haven't I got that included in the $$mg(3l+x+y)$$ part of the PE? This comes from $$mg(l+x)+mg(2l+y)$$ which is mg times the height of the first particle plus mg times the height of the second.

4. Mar 15, 2009

### Dr.D

The gravitational potential energy is decreasing with increasing values of x and y, so you need some negative signs.

5. Mar 16, 2009

### wglmb

Ah, I see what you mean - of course! I don't have time just now but I'll work that through later and see how I manage :)

Edit:

Hmm, well I still get the same matrices A and B, since they come from the double-dotted terms and the x&y terms. The constant terms (now -mg for both of the equations of motion) aren't used.
I have a feeling those constant terms shouldn't even be there, since none of the examples we've been given have any...

Last edited: Mar 16, 2009
6. Mar 16, 2009

### wglmb

It's OK, I've figured it out now. Thanks :)