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Normal Modes

  1. Dec 19, 2005 #1
    I have just been solving some systems of ODEs to find the nomal mode solutions. Something about it has been bugging me though.

    In a simple case where we have a system of two linear ODEs representing a two-mass spring system, we assume that the solution is a normal mode and so find a general solution using eigenvalues and eigenvectors and get a vector solution involving a linear combination of four terms each multiplied by an arbitary constant. No problem there.

    The part that is worrying me though is this: We assumed that the solution was a normal mode; so what if it wasn't? I.e. do non-normal-mode solutions exist? If they do then why have we got four arbitary constants in our normal-mode solution - surely if a soulution of this system involves four constants then it describes the most general solution.

    Also, is it technically still called a normal mode if we combine two normal modes to get another solution?

    Any guidance on this matter would be great thanks.
  2. jcsd
  3. Dec 19, 2005 #2


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    Normal modes of a system can be written in a special way: the entire solution has the same frequency (all the degrees of freedom are oscillating with the same frequency, although different amplitude and phase).

    A general solution (which is a superposition of normal modes) does not have this property in general. Sometimes there is however, ambiguity: one says that several normal modes are degenerate.
  4. Dec 20, 2005 #3


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    The approach to the problem is a bit different. You do not assume the solution is a normal mode. You set out to FIND the normal modes and for that purpose made the frequency the same for all oscillators.
    If the equation is linear and homogeneous any linear combination of solutions will also be a solution. In fact, ALL solutions can be written as a linear combination of normal modes, so the normal modes form a BASIS for the solution space. A basis consisting of eigenmodes (=normal modes) of the system.
    The question is: Does a basis of normal modes always exist? Fortunately, this is almost always the case (in spring-mass systems it always is). If you write down the system of ODE's in matrix form, you'll see the matrix is symmetric (due to Newton's 3rd law). A theorem in linear algebra will tell you that you can then find such a nice basis of eigenmodes.
  5. Dec 20, 2005 #4
    Thanks guys, I see now. The general solution is a superposition of the normal mode solutions and generally such a solution is not a normal mode itself (unless the constants are chosen VERY carefully).
    So as long as the normal mode solutions are independent functions on the solution space then they will form a basis. I'm interested to see what happens if the eigenvalues are repeated (I realise that this would not happen in a symmetric matrix)? Do we then have to invoke something like the variation of parameter method to get another independent solution and will this be a normal mode?
  6. Dec 21, 2005 #5


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    That's what I called "degenerate normal modes".

    It can of course happen in a symmetric matrix, but you have the guarantee that you will be able to find as many independent solutions as there are eigenvalues.

    As a (trivial) example of such a case, consider 2 equal masses and 2 equal springs, and the two systems are not connected. Then of course you have two times the same frequency, for the two systems (they are identical). But there are two independent solutions:
    x1 oscillates and x2 = 0, and x1 = 0 and x2 oscillates.
    From these two independent solutions, you can build the entire 2-dim solution space where x1 oscillates and x2 oscillates.
  7. Dec 21, 2005 #6


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    The equation:
    [tex]\frac{d^2y}{dt^2}+ \frac{dy}{dt} + y = 0[/tex] has two equal normal modes -1 and -1.
    The general solution is [tex]y = k_1e^{-t} + k_2te^{-t}[/tex]
  8. Dec 22, 2005 #7
    Thanks I see what you are saying. I think the solution, however, should be:

    [tex]y = k_1e^{-t/2}cos(sqrt(3)/2)t + k_2e^{-t/2}sin(sqrt{3}/2)t[/tex]
    Last edited: Dec 22, 2005
  9. Dec 22, 2005 #8


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    My mistake. The equation I wanted to write was:
    [tex]\frac{d^2y}{dt^2} + 2\frac{dy}{dt} + y = 0[/tex]
    that has a double eigenvalue -1 and whose general solution is the one I wrote.
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