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Eigen vals/vecs in MDOF Dynamics Problems

  1. Jun 12, 2008 #1
    Say I have the following simple situation


    fixed walled and two masses connected by springs

    When you write newton's 2nd for each mass you can assume a sinusoid solution for the position as a function of time and plugging back into your equations you can solve for the two fundamental frequencies of the system. My question is this, If i write both equations in a matrix form

    \frac{d^2 \, \vec{x}}{dt^2} = A \, \vec{x}

    with A being a matrix of combinations of spring constants divided by appropriate masses, do the eigenvalues of the matrix correspond to the squares of the natural angular frequencies? If so, why physically is this true? Additionally, what do the eigenvectors mean? What is the physical connection there?

  2. jcsd
  3. Jun 12, 2008 #2
    So, for this situation, if you write down the eigenvalue equation it looks like this:

    \left(\begin{array}{cc}m_1^{-1} & 0 \\ 0 & m_2^{-1}\end{array}\right)
    \left(\begin{array}{cc}-k_1-k_2 & k_2 \\ k_2 & -k_2\end{array}\right)
    \left(\begin{array}{c}x_1\\x_2\end{array}\right) =
    \left(\begin{array}{c}\ddot x_1\\\ddot x_2\end{array}\right)

    (that's what I got, anyway. If it's not quite right, it doesn't really matter because the rest of this discussion doesn't focust on the form of the matrix.)

    Now we assume there are harmonic solutions similar to a single harmonic oscillator, so that our trial solution would be [tex]\ddot x_1 = \omega^2 x_1[/tex] and [tex]\ddot x_2 = \omega^2 x_2[/tex]. (This is done without loss of generality; [tex]\omega[/tex] may be complex, so picking [tex]\omega^2[/tex] is really just handy for looking at the end result. But it's based on the fact that we've taken two derivatives wrt time, so we already know that omega has dimensions of frequency.) If you put that in:

    \left(\begin{array}{cc}-(k_1+k_2)/m_1 & k_2/m_1 \\ k_2/m_2 & -k_2/m_2\end{array}\right)
    \left(\begin{array}{c}x_1\\x_2\end{array}\right) =
    \left(\begin{array}{cc}\omega^2 & 0 \\ 0 & \omega^2\end{array}\right)
    \left(\begin{array}{c} x_1\\ x_2\end{array}\right)

    which is exactly the eigenvalue problem. Solving it will give omega in terms of the k's and m's. Then we go back to our differential equation and look at our trial solution where we assumed [tex]\ddot x_i = \omega^2 x_i[/tex] and solve that, knowing that we have a solution for omega. Of course that will be a harmonic solution; we don't need to know omega for that. But the omegas that we got out of the eigenvalue equation are the ones that go into our trial solution.

    The eigenvectors have significance; a matrix made out of the eigenvectors can be used to diagonalize our k/m matrix. So this tells us what kind of x_1 and x_2 displacements give us the harmonic motion. If m1 = m2 and k1 = k2 then the eigenvectors, ignoring normalization, will look something like (1 1) and (1 -1). This tells us that one type of motion is where both masses move the same direction at the same time (1 1) and another type where they move opposite directions at the same time (1 -1). If the parameters are different, it gets more complicated, but my understanding is that it will be true that the ratio of the components of an eigenvector indicate the ratio of the amplitude of the motions of each mass. The eigenvectors could also be complex, which should indicate a phase difference in the motion of the two masses. This probably won't be the case for two masses, but more complicated systems could give this type of result I suppose.
  4. Jun 12, 2008 #3
    So if i have this right

    Assuming an SHM solution for each component, [tex] \omega_i^2 = \lambda_i[/tex]

    and the ratios of components in any eigenvector [tex]x_i[/tex] give the ratios of he amplitudes of oscillation while in normal mode [tex]i[/tex]
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