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Undamped 2 DOF vibration. What should the eigen vectors be here

  1. Mar 23, 2013 #1
    Normal modes of vibration, two masses, two spring, arranged vertically with m2 at the top, m1 underneath arranged (top to bottom) m2, k2, m1, k1, rigid support

    I have solved the first part of an undamped coupled spring problem to give

    [itex]m_1m_2 \omega ^ 4 + ((m_1+m_2)k_2+m_2k_1)\omega ^2 +k_1k_2=0[/itex] Since this is a show that Q, I know this is correct.

    With [itex]k_1=5,\; k_2=10,\; m_1=20,\; m_2=50 [/itex] I get [itex]\omega_1=0.2365,\;\omega_2=0.9456[/itex]

    This comes from the equation [itex]\begin{pmatrix}
    m_1 \omega^2+k_1+k_2 & -k_2 \\
    -k_2 & m_2 \omega^2+k_2
    \end{pmatrix}
    \begin{pmatrix}
    X_1 \\
    X_2
    \end{pmatrix}=\begin{pmatrix}
    0 \\
    0
    \end{pmatrix}[/itex]

    I have formed the impression (which must be wrong) that my values of [itex]\omega[/itex] should be eigen values with eigen vectors of [itex]\begin{pmatrix}
    1 \\
    1
    \end{pmatrix}[/itex] and [itex]\begin{pmatrix}
    1 \\
    -1
    \end{pmatrix}[/itex] Which describe the first two principle modes of vibration.

    I expect to be able to check by substituting my values of [itex]\omega[/itex] into the matrix equation and get the zero matrix at the RHS when I used the eigen vectors for [itex]\begin{pmatrix}
    X_1 \\
    X_2
    \end{pmatrix}[/itex]
    However when I multiply out the matrix and the eigen vectors with my values of [itex]\omega[/itex] I get nothing like [itex]\begin{pmatrix}
    0 \\
    0
    \end{pmatrix}[/itex]

    Where is my understanding going wrong with this?
     
  2. jcsd
  3. Mar 23, 2013 #2

    AlephZero

    User Avatar
    Science Advisor
    Homework Helper

    Your "impression" about the correct answer is wrong.

    To see why it's wrong, imagine a two springs with equal stiffness, with a large mass at the bottom, and a small mass at the mid point. This is almost the same as a single mass at the end of the spring, and the eigenvector for the lowest mode will be approximately [itex]\begin{pmatrix}
    0.5 \\
    1
    \end{pmatrix}[/itex] not [itex]\begin{pmatrix}
    1 \\
    1
    \end{pmatrix}[/itex]

    The eigenector for the second mode will be close to [itex]\begin{pmatrix}
    1 \\
    0
    \end{pmatrix}[/itex], though that is a bit harder to "see" intuitively.

    In general the eigenvectors depend on all the mass and stiffness properties. To calculate them, substitute the numbers for k m and ##\omega## into your equation [itex]\begin{pmatrix}
    m_1 \omega^2+k_1+k_2 & -k_2 \\
    -k_2 & m_2 \omega^2+k_2
    \end{pmatrix}
    \begin{pmatrix}
    X_1 \\
    X_2
    \end{pmatrix}=\begin{pmatrix}
    0 \\
    0
    \end{pmatrix}[/itex]
    If you calculated ##\omega## correctly, the system of equations will be singular, and you can solve them for the ratio of ##X_1## to ##X_2##.

    Repeat with the other value of ##\omega## to find the other eigenvector.
     
  4. Mar 24, 2013 #3
    Thank you very much for your prompt answer. I might even have ended up understanding eigenvectors in this context. It was well worth ploughing through the latex to get your answer.
     
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