How Do You Determine Normal Modes in a Coupled Spring System?

[ckelly94]

Homework Statement

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

Homework Equations

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

[itex]\frac{d^{2}x_{1}}{dt^{2}}[/itex]+([itex]\omega_{1}^{2}[/itex]+[itex]\omega_{2}^{2}[/itex])x[itex]_{1}[/itex]-[itex]\omega_{2}^{2}[/itex]x[itex]_{2}[/itex]=0

and

[itex]\frac{d^{2}x_{2}}{dt^{2}}[/itex]-([itex]\omega_{2}^{2}[/itex])x[itex]_{1}[/itex]+([itex]\omega_{2}^{2}[/itex])x[itex]_{2}[/itex]=0

The Attempt at a Solution

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

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

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

( [itex]\stackrel{\omega_{2}^{2}}{3\omega_{2}^{2}+2\omega_{2}^{4}+\omega_{2}^{2}\sqrt{8\omega_{2}^{2}+2}}[/itex] ) ?

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

Thanks in advance!

 

[MisterX]

how to make a matrix

Or do you mean just make it bold?

 

[ckelly94]

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

 

[TSny]

([itex]\omega_{1}^{2}+\omega_{2}^{2}-\lambda[/itex])([itex]\omega_{2}^{2}-\lambda[/itex])-[itex]\omega_{2}^{4}[/itex]=0

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

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

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?

 

[ckelly94]

So I solved for [itex]\lambda[/itex] by multiplying that first equation in your reply:

[itex]\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[/itex]

Which simplifies to

[itex]\lambda^{2}+\lambda(2\omega_{2}^{2}+\omega_{1}^{2})-\omega_{1}^{2}\omega_{2}^{2}=0[/itex]

Using the quadratic formula, I arrived at

[itex]\lambda=CRAP[/itex]

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

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

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

...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 [itex]\omega_{2}^{2}[/itex] in my notes.

 

[ckelly94]

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

I got [itex]\lambda=-2\omega_{2}^{2}\pm\sqrt{\omega_{2}^{2}+3\omega_{2}^{2}}[/itex]

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

 

[ckelly94]

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

Solution should read [itex]\lambda_{1,2}=-2\omega_{2}^{2}\pm\sqrt{10}\omega_{2}^{2}[/itex]

 

[ckelly94]

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

 

[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 [itex]\lambda_{2}[/itex] if [itex]\lambda_{1}[/itex] 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 [itex]\lambda_{1}[/itex] and [itex]\lambda_{2}[/itex] 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 [itex]\lambda_{2}[/itex] independently to describe another phase of motion?

 

[TSny]

You have the equation

([itex]\omega_{1}^{2}+\omega_{2}^{2}-\lambda[/itex])([itex]\omega_{2}^{2}-\lambda[/itex])-[itex]\omega_{2}^{4}[/itex]=0

which I believe is correct.

So I solved for [itex]\lambda[/itex] by multiplying that first equation in your reply:

[itex]\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[/itex]

Check the sign of the λ² term.

 

[ckelly94]

Check the sign of the λ² term.

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

 

[TSny]

([itex]\omega_{1}^{2}+\omega_{2}^{2}-\lambda[/itex])([itex]\omega_{2}^{2}-\lambda[/itex])-[itex]\omega_{2}^{4}[/itex]=0


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

 

[ckelly94]

Goddamn.