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Homework Help: Direction of oscillations

  1. Apr 8, 2005 #1
    how would one go about finding the direction of oscillation for a differntial equation?

    for example [tex] \frac{d^2 y}{dt^2} = -2y [/tex]
    has eigenvalues [itex] \pm \sqrt{2}i [/itex]
    and the corresponding matrix is [tex] \left(\begin{array}{cc}0&2\\-2&0\end{array}\right) [/tex] so the solution curves will form closed loops around the origin. But what about the direction -i.e. counter clockwise or clockwise, how would one go about figuring this out??

    please help!!! I have an exam tomorrow on this stuff and this is where i am snagged!
  2. jcsd
  3. Apr 9, 2005 #2
    the two associated eigenvectors would be computed as follows

    [itex]AX=\lambda X[/itex]

    [tex] \left(\begin{array}{cc}0&2\\-2&0\end{array}\right) \left(\begin{array}{cc}x\\y\end{array}\right) = i \sqrt{2} \left(\begin{array}{cc}x\\y\end{array}\right)[/tex]

    which gives teh equation
    [tex] 2y = i \sqrt{2} x [/tex] and
    [tex] -2x = i \sqrt{2} y [/tex]
    which gives me eigenvector which is [itex] 2y = i \sqrt{2} x[/itex] which gives me [tex] V_{1} = \left(\begin{array}{cc}2\\i \sqrt{2}\end{array}\right)[/tex] and [tex] V_{2} = \left(\begin{array}{cc}2\\-i \sqrt{2}\end{array}\right) [/tex]
    is this correct?? Am i supposed to simplify the eigenvectors??
    Please help!!!!!!!!
  4. Apr 9, 2005 #3


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    In this situation, the eigenvalues are not sufficient to tell you the direction of oscillation (I would say "rotation").
    Consider the general example: [tex] \left(\begin{array}{cc}a&b\\c&d\end{array}\right)[/tex].
    The characteristic equation is [tex] \left|\begin{array}{cc}a-\lambda&b\\c&d-\lambda\end{array}\right|= \lambda^2- (a+d))\lamba+ ad-bc= 0[/tex].
    The solutions to that depend on two numbers, not 3. Specifically, a+ d, the "trace" of the matrix and ad-bc, the "determinant" of the matrix. Any two matrices having the same trace and matrix will have the same eigenvalues. In particular, [tex] \left(\begin{array}{cc}0&-2\\2&0\end{array}\right)[/tex] has exactly the same eigenvalues but opposite rotation. In this case, since the eigenvectors are complex and you are working in the real plane, they don't help.

    You can, easily, determine the direction by looking at any one point. In particular, just because it is easy, suppose x= 1, y= 0. Then multiplication of (1, 0) by the matrix gives dx/dt= 2, dy/dt= -2. That is, x is increasing, y is decreasing- that "y is decreasing" tells us that the rotation is counter-clockwise.
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