# Finding the normal modes for a oscillating system

1. Apr 21, 2015

### skeer

1. The problem statement, all variables and given/known data
The system is conformed by two blocks with masses m (on the left) and M (on the right), and two springs on the left/right has the spring constant of k. The middle spring has a spring constant of 4k. Friction and air resistance can be ignored. All springs are massless.
Find the normal modes.
Diagram:
|~m~~~~M~|
2. Relevant equations
$L = T-V$

$T = \frac{1}{2}(m\dot{x}_1^2 + M\dot{x}_2^2)$
$V = \frac{1}{2}[(x_1^2 + x_2^2) + 4(x_1-x_2)^2]$
$\frac{\partial{L}}{\partial{x_k}} - \frac{d}{dt} \frac{\partial{L}}{\partial{\dot{x_k}}} = 0$
$[A_{ij} - \omega^2 m_{ij}]=0$
3. The attempt at a solution
I have tried to guess a solution for the normal modes but of the for $\eta_1 = x_1 - x_2$ and $\eta_2= x_1+x_2$ but I does not works. I have tried to add some arbitrary coefficient to $\eta_1$ & $\eta_2$ unsuccessfully. Trying to find the eigenvectors is a pain in the neck since the eigenfrequencies are:$\omega^2 = \frac{5k(M+m) \pm k\sqrt{25(M^2+m^2) -14Mm}}{2Mm}$.
I read in a textbook that one could find the coefficient for the etas by knowing that the ratios $\frac{M_{11}}{M_{22}}=\frac{A_{11}}{A_{22}}=\alpha^2$ but for this case the first ratio is $\frac{m}{M}$ and the second is 1 .Therefore, this method doesn't help me :/.

I would appreciate any contribution. Thank you.

2. Apr 21, 2015

### paisiello2

Do you have to solve this problem using the Lagrangian?

3. Apr 22, 2015

### skeer

The Lagragian is not necessary, but is the only method I know. I believe that if I use forces the problem would complicate more.

4. Apr 22, 2015

### paisiello2

I think the force method is easier but the results are the same.

I get a slightly different answer for the Eigen values but even then I think you could probably simplify it a little bit:
ω2 = k(1/M+1/m) [5/2± √(25/4+16/(M/m+m/M+2))]

I am not aware of any other method except plugging the Eigen values into the equations of motion and solving for the mode shapes.

5. Apr 22, 2015

### paisiello2

Duplicate post.