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Spring-Mass System: Eigenvalues and Eigenvectors

  1. Oct 25, 2013 #1
    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?
     
  2. jcsd
  3. Oct 25, 2013 #2

    Mark44

    Staff: Mentor

    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.
     
  4. Oct 25, 2013 #3
    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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