Two-Degree-Of-Freedom Linear System: Eigenvalue problem

[Valeron21]

I've found the characteristic equation of the system I'm trying to solve:
$$ω^{4}m_{1}m_{2}-k(m_{1}+2m_{2})ω^{2}+k^{2}=0$$

I now need to find the eigenfrequencies, i.e. the two positive roots of this equation, and then find the corresponding eigenvectors. I've been OK with other examples, but because of the different masses here, I don't see any obvious substitution or cancellation that will leave the eigenfrequencies in a manageable form.

I've been at this a while and I'm a bit lost.


Thanks in advance.

 

[Dick]

I've found the characteristic equation of the system I'm trying to solve:
$$ω^{4}m_{1}m_{2}-k(m_{1}+2m_{2})ω^{2}+k^{2}=0$$

I now need to find the eigenfrequencies, i.e. the two positive roots of this equation, and then find the corresponding eigenvectors. I've been OK with other examples, but because of the different masses here, I don't see any obvious substitution or cancellation that will leave the eigenfrequencies in a manageable form.

I've been at this a while and I'm a bit lost.


Thanks in advance.

If you substitute u=ω^2 then it's quadratic in u. Once you know u, you can find ω, right?

 

[Valeron21]

If you substitute u=ω^2 then it's quadratic in u. Once you know u, you can find ω, right?

Sorry. The characteristic equation isn't really the problem - I should have worded that better - it's how I use what will be pretty horrible roots to find the corresponding eigenvectors that I'm having trouble with.

EDIT: $$(k-ω_{i}^{2}m)u_{i}=0$$
is what I would usually sub the values of ω into to find the eigenvectors, where k is the stiffness matrix, m is the mass matrix and u the eigenvector.

 

[nasu]

From your characteristic equation it seems that you have only one k and two masses. Is that correct?

 

[Valeron21]

From your characteristic equation it seems that you have only one k and two masses. Is that correct?

Yes, that is correct. Both springs in the system have the same stiffness constant, k.

 

[nasu]

So how does it look like? You have two masses and two springs but how do you couple them?
I am just curios.
For the solution, it does not have to simplify to a simple expression. The fact that it does not do not implies that the solution is wrong.
Once you find a solution for frequency, you plug it in one of the equations and find the ratio of the amplitudes for that mode. You may want a numerical solution, in order to "see" how does the mode look like.

 

[Valeron21]

So how does it look like? You have two masses and two springs but how do you couple them?
I am just curios.
For the solution, it does not have to simplify to a simple expression. The fact that it does not do not implies that the solution is wrong.
Once you find a solution for frequency, you plug it in one of the equations and find the ratio of the amplitudes for that mode. You may want a numerical solution, in order to "see" how does the mode look like.

Hmm, maybe there's something wrong in how I'm approaching this?

I've always previously, when working with 2.D.O.F. systems, assumed a solution of the form:

$$ \underline{X}={U}[A_{}cos(ω_{}t)+B_{}sin(ω_{}t)]$$

then through a process of differentiation and substitution obtained this equation:
$$( \underline{k}-ω^{2} \underline{m}) \underline{X}=0$$

then, for a non-trivial solution, the det =/=0, etc. And here I am now with the roots of the characteristic equation.

Previously, I've always been able to sub the values of ω into this:
$$(\underline{k}-ω_{i}^{2}\underline{m})\underline{u}_{i}=0$$

there's always been a nice cancellation and I've been able to use an arbitrary value to find the ratio, as you say, between the two elements in the eigenvector. But here, with horrible roots, I'm not sure what to do. And I'm not sure which equations you're referring to.

Sorry if I'm being stupid.

 

[nasu]

The method is OK.
It's just that you don't have to obtain a nice solution. In the textbook examples they pick-up parameters to get a simple result, as an example.

I mean the equations which you wrote in matrix format.
Maybe you are too formal about it. You don't really need the matrix form for two equations.
What stops you from substituting the values for omega back into the equations?