Spring-Mass System: Eigenvalues and Eigenvectors

Valeron21
Messages
8
Reaction score
0
The det. of the following matrix:

$$
\begin{matrix}
2k-ω^{2}m_{1} & -k\\ -k & k-ω^{2}m_{2}\\
\end{matrix}
$$

must be equal to 0 for there to be a non-trivial solution to the equation: $$(k - ω^{2}m)x =0$$


Where m is the mass matrix:

$$
\begin{matrix}
m_{1} & 0\\ 0& m_{2}\\
\end{matrix}
$$
k is the stiffness matrix:
$$
\begin{matrix}
2k& -k\\ -k & k\\
\end{matrix}
$$

and ω^2 is the eigenvalue.

I worked out the determinant as $$ω^{4}m_{1}m{2}-k(m_{1}+2m_{2})ω^{2}+k^{2}=0$$ and I could probably solve to find the two positive roots of this, but the roots are going to be pretty horrible and they are going to make finding the corresponding eigenvectors pretty difficult, no?

Am I doing something wrong? Is there an easier way to do this? Perhaps a substitution/alteration to the matrix?
 
Physics news on Phys.org
Valeron21 said:
The det. of the following matrix:

$$
\begin{bmatrix}
2k-ω^{2}m_{1} & -k\\ -k & k-ω^{2}m_{2}\\
\end{bmatrix}
$$

must be equal to 0 for there to be a non-trivial solution to the equation: $$(k - ω^{2}m)x =0$$


Where m is the mass matrix:

$$
\begin{bmatrix}
m_{1} & 0\\ 0& m_{2}\\
\end{bmatrix}
$$
k is the stiffness matrix:
$$
\begin{bmatrix}
2k& -k\\ -k & k\\
\end{bmatrix}
$$

and ω^2 is the eigenvalue.

I worked out the determinant as $$ω^{4}m_{1}m_{2}-k(m_{1}+2m_{2})ω^{2}+k^{2}=0$$ and I could probably solve to find the two positive roots of this, but the roots are going to be pretty horrible and they are going to make finding the corresponding eigenvectors pretty difficult, no?

Am I doing something wrong? Is there an easier way to do this? Perhaps a substitution/alteration to the matrix?
I didn't check your work...

The equation you're trying to solve is a little messy, but not all that bad. It's quadratic in form - Let u = ω2, and then you have a quadratic.

BTW, in what I quoted I changed "matrix" to "bmatrix" to get them to look like matrices.
 
Hmm, I'm not really sure that makes it significantly easier, though.

To give a bit more context - I need to find the eigenfrequencies and corresponding eigenvalues, then give a solution of the form:


$$ \underline{X}=\sum_{i=1}^{}\underline{U_{i}}[A_{i}cos(ω_{i}t)+B_{i}sin(ω_{i}t)]$$

where $$\underline{U_{i}}$$ is the eigenvector associated with each eigenfrequency.

So, unless I'm being stupid here, I really don't see how I'm going to calculate that if I don't get some more manageable roots.


Is this in the right section, btw?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top