Spring-Mass System: Eigenvalues and Eigenvectors

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The discussion centers on finding the eigenvalues and eigenvectors of a spring-mass system represented by a specific matrix. The determinant of the matrix must equal zero to obtain non-trivial solutions for the eigenvalue equation. The user calculated the determinant as a quadratic equation in terms of ω², but expressed concerns about the complexity of the roots and the subsequent difficulty in finding corresponding eigenvectors. Suggestions include treating the equation as a standard quadratic and considering substitutions to simplify the process. The user seeks guidance on whether there are easier methods to solve for eigenfrequencies and eigenvalues.
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?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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