Spring-Mass System: Eigenvalues and Eigenvectors

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SUMMARY

The discussion focuses on solving the eigenvalue problem for a spring-mass system represented by the determinant of a specific matrix. The determinant equation $$ω^{4}m_{1}m_{2}-k(m_{1}+2m_{2})ω^{2}+k^{2}=0$$ is derived, indicating that the eigenvalues (ω²) can be found by solving this quadratic equation. Participants suggest that while the roots may be complex, a substitution (letting u = ω²) can simplify the problem. The goal is to find eigenfrequencies and corresponding eigenvectors for the system's motion.

PREREQUISITES
  • Understanding of eigenvalues and eigenvectors in linear algebra.
  • Familiarity with matrix determinants and quadratic equations.
  • Knowledge of spring-mass systems in mechanical engineering.
  • Proficiency in mathematical notation and manipulation of matrices.
NEXT STEPS
  • Learn how to apply the quadratic formula to solve for eigenvalues in matrix equations.
  • Research methods for simplifying determinant calculations in mechanical systems.
  • Explore the derivation of eigenfrequencies and eigenvectors in spring-mass systems.
  • Study numerical methods for approximating roots of complex polynomial equations.
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Mechanical engineers, physicists, and students studying dynamics who need to analyze spring-mass systems and solve for eigenvalues and eigenvectors.

Valeron21
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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?
 
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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?
 

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