# Normal modes of oscillation

1. Dec 13, 2013

### ckelly94

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

So I'm given two horizontal masses coupled by two springs; on the left there is a wall, then a spring with k$_{1}$, then a mass, then a spring with k$_{2}$, and finally another mass, not attached to anything on the right. The masses are equal and move to the right with x$_{1}$ and x$_{2}$, respectively. I'm trying to find the normal modes of oscillation where k$_{1}$=2k$_{2}$.

2. Relevant equations

As usual, we write the equations of motion for each of the masses, i.e.

$\frac{d^{2}x_{1}}{dt^{2}}$+($\omega_{1}^{2}$+$\omega_{2}^{2}$)x$_{1}$-$\omega_{2}^{2}$x$_{2}$=0

and

$\frac{d^{2}x_{2}}{dt^{2}}$-($\omega_{2}^{2}$)x$_{1}$+($\omega_{2}^{2}$)x$_{2}$=0

3. The attempt at a solution

The eigenvalues for this matrix are given by
($\omega_{1}^{2}+\omega_{2}^{2}-\lambda$)($\omega_{2}^{2}-\lambda$)-$\omega_{2}^{4}$=0

At this point, I plugged k$_{1}$=2k$_{2}$ into this mess and determined that $\lambda_{1,2}$=-2$\omega_{2}^{4}$$\pm(\omega_{2}^{2}\sqrt{8\omega_{2}^{2}+2}$)

So did I do something wrong algebraically, or are the eigenvectors, and thus the normal modes of oscillation simply $\lambda$ $\propto$

( $\stackrel{\omega_{2}^{2}}{3\omega_{2}^{2}+2\omega_{2}^{4}+\omega_{2}^{2}\sqrt{8\omega_{2}^{2}+2}}$ ) ?

PS: Sorry about the formatting. I wasn't sure how to make a matrix, but the last line should be a matrix.

2. Dec 13, 2013

### MisterX

3. Dec 13, 2013

### ckelly94

Yeah, but it's just 2 X 1 anyway, so it's not *too* important, I don't think.

4. Dec 13, 2013

### TSny

I believe this equation is correct. Note that you can determine the dimensions of λ from this equation.

This expression cannot be correct because the dimensions are off. For example, the two terms inside the square root have different dimensions.

Can you show the steps you took to get to this result?

5. Dec 13, 2013

### ckelly94

So I solved for $\lambda$ by multiplying that first equation in your reply:

$\omega_{1}^{2}\omega_{2}^{2}+\omega_{2}^{4}-\omega_{2}^{2}\lambda-\omega_{1}^{2}\lambda-\omega_{2}^{2}\lambda-\lambda^{2}-\omega_{2}^{4}=0$

Which simplifies to

$\lambda^{2}+\lambda(2\omega_{2}^{2}+\omega_{1}^{2})-\omega_{1}^{2}\omega_{2}^{2}=0$

Using the quadratic formula, I arrived at

$\lambda=CRAP$

^This is the point where I realized my error. I'm gonna keep going though, just for closure.

$\lambda=-2\omega_{2}^{2}-\omega_{1}^{2}±√(8\omega_{2}^{4}(4\omega_{2}^{2}+1))/2$

Which becomes $-\omega_{2}^{2}-(\omega_{1}^{2}/2)±\omega_{2}^{2}√(8\omega_{2}^{2}+2)$

...which is still really gross and has the wrong dimensions, I believe.

EDIT: No, that's actually not where I ran into problems, because I forgot that I wrote it in terms of $\omega_{2}^{2}$ in my notes.

6. Dec 13, 2013

### ckelly94

Okay so I tried it over, first by doing the quadratic equation, then writing the result in terms of $\omega_{2}^{2}$

I got $\lambda=-2\omega_{2}^{2}\pm\sqrt{\omega_{2}^{2}+3\omega_{2}^{2}}$

Still though, not sure why the dimensions won't work.

7. Dec 13, 2013

### ckelly94

Got it! For some reason I forgot to properly square the first-order term.

Solution should read $\lambda_{1,2}=-2\omega_{2}^{2}\pm\sqrt{10}\omega_{2}^{2}$

8. Dec 13, 2013

### ckelly94

Normal modes of oscillation are therefore $-\omega_{2}^{2}$ and $(5+\sqrt{10})\omega_{2}^{2}$

9. Dec 13, 2013

### ckelly94

I got that last result by just plugging the first (positive) eigenvalue into the first row of the matrix, setting it equal to zero.

Just curious, how would you go about using, say $\lambda_{2}$ if $\lambda_{1}$ is positive? Do you plug that into the second row, or can you plug it into the first row? I guess if it's a 3 X 3 matrix (i.e. there are more masses), you would have to use multiple rows, so $\lambda_{1}$ and $\lambda_{2}$ are both used (and the row doesn't matter/you'd have to use all or multiple rows). In that case, would you have to use $\lambda_{2}$ independently to describe another phase of motion?

10. Dec 13, 2013

### TSny

You have the equation

($\omega_{1}^{2}+\omega_{2}^{2}-\lambda$)($\omega_{2}^{2}-\lambda$)-$\omega_{2}^{4}$=0

which I believe is correct.

Check the sign of the λ2 term.

11. Dec 13, 2013

### ckelly94

...Yes... That's why the last term is negative.

12. Dec 13, 2013

### TSny

($\omega_{1}^{2}+\omega_{2}^{2}-\lambda$)($\omega_{2}^{2}-\lambda$)-$\omega_{2}^{4}$=0

When you expand this out, what will be the sign in front of $\lambda^2$?

13. Dec 13, 2013

Goddamn.