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Coupled Pendulums - Possible Solutions and Superposition

  1. May 30, 2012 #1
    I am tasked to answer the following question on the subject of coupled pendulums:

    QUESTION: The Antisymmetric and symmetric normal modes can be superposed to obtain new solutions. Can ANY solutions be written as a superposition of these two modes? Justify your answer.

    Hint: Is a solution completely characterised by its initial positions and velocities?


    My answer: I say yes any solution can be written as a superposition of these two modes as it's a consequence of linearity of the system (It's a linear system because restoring force is linearly proportional to the displacement from equillibrium). If the antisymmetric and symmetric modes are each a solution, then any linear combination of those is also a solution.

    What do you guys think? Is that wrong?
     
  2. jcsd
  3. May 30, 2012 #2

    K^2

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    Transformation to normal coordinates just decouples your equations. If solution for each normal coordinate is general, then overall solution is general.
     
  4. May 30, 2012 #3
    Yeah, the solutions are general for each normal mode, so I guess that does indeed mean their superposition is also a general solution. Thank you.
     
  5. May 30, 2012 #4
    Hi alyosha17
    No, your answer is not correct, you are not answering the question
    The question is, and it is emphasized: can ANY solution be written as a superposition of the two modes
    You are saying that any superposition of the two modes is a solution
    That would be like saying
    here are two linearly independent vectors, is any other vector a linear combination of those ?
    and you would be answering that any combination of those vectors is a new vector, but you didn't answer the real question.
    if you have 2 linearly independent vectors, then yes, any vector of the plane, for a 2D space will be writeable as a combination of those, they form a basis. but what if this is >=3D space ?
    your question came with a hint (is a solution *completely characterized* by its initial positions and velocities ?)
    I think you will have trouble answering the problem without relying on this hint :)

    Cheers...
     
  6. May 30, 2012 #5
    Thanks for input,but I don't really understand what the hint is insinuating though, to be honest. The hint is what threw me off in the first place and provoked me to post this question on this forum. But I still don't really get why it disagrees with my answer. Could you be less cryptic in explaining?

    for simplicity's sake I'm assuming there is two degrees of freedom for my specific pendulum. That is to say, motion of the pendulum in a plane only.

    If I have two normal modes, then I can say any combination of motion of the system can be written as a linear combination of these normal modes.

    If I have solutions for each of the modal frequencies of a coupled pendulum (solutions, say, x1 and x2) for each frequency (both symmetric and asymmetric frequencies), then I can write equations where i have, say:

    X1(t[itex]_{i}[/itex])=A[itex]_{1}[/itex]cos(ω[itex]_{sym}[/itex]t[itex]_{i}[/itex])
    X2(t[itex]_{i}[/itex])=A[itex]_{2}[/itex]cos(ω[itex]_{sym}[/itex]t[itex]_{i}[/itex])

    and

    [itex]\dot{X1}[/itex](t[itex]_{i}[/itex])=-ωX1
    [itex]\dot{X2}[/itex](t[itex]_{i}[/itex])=-ωX2

    Which will give me equations for the initial conditions (position and velocity) of the pendulum which I can then use to solve for A1 and A2 by using the normal mode general solutions, to get a general solution for the motion of the coupled pendulum.

    Beyond the initial position and velocity, what other factors would characterise the motion? A coupling factor maybe?

    Im confused.
     
    Last edited: May 30, 2012
  7. May 31, 2012 #6
    Hi again alyosha17,
    Sorry, I didn't mean to be cryptic, nor did I mean to say that your answer was wrong per se.
    What you said is correct, but it was just not answering the question which is about making sure that all solutions are accounted for.
    That is, you have found some linearly independent solutions, just by luck, say.
    But can you be sure that their linear combinations account for all possible solutions ?
    The hint is about leading you to the dimensionality of the problem
    Is a system entirely specified by its positions and velocities ? (yes)
    In the end, solving the whole system's equations of motion boils down to a 2x2 differential equations matrix for which those two normal modes are the two eigenvectors and are, indeed, enough to account for everything.
    For a triple pendulum, you would need 3 normal modes, etc.
    (At least this is how I understand what the question is about)

    Cheers...
     
  8. May 31, 2012 #7
    Oh thanks. I see what you meant now. I guess I shall answer the question with a caveat explaining such things.
     
  9. May 31, 2012 #8
    You can prove this rigorously by showing that the normal mode solutions are orthogonal (or not!). Take a look at the math of Fourier series if you're a bit rusty on such things.
     
  10. May 31, 2012 #9

    Ken G

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    I'm worried that your use of the word "caveat" suggests you did not completely get the point-- it is not a "caveat" that oli4 is talking about, it is the entire crux of the issue. The question tells you that solutions can be superimposed to yield a solution, so putting that in the answer does not earn you any credit (as someone who has done grading, I can tell you that it is always frustrating when a student writes an answer that seems to imply they expect getting credit for merely repeating something that is given in the question!).

    What the question is all about, as oli4 said, is whether all solutions can be found this way. You showed with an example that the answer is "yes" in a particular case (2 degrees of freedom), which may be all that is needed, but the question might still be looking for a more general answer in any dimensionality. The hint is the key-- I think you are supposed to count the number of constraints, connect it to the number of degrees of freedom in the problem, and match it to the number of normal modes. If superpositions of normal modes constrains the full dimensionality of the possible solutions, then you must account for all possible solutions that way. (To be completely technically correct, you must bring in a generalized concept of orthogonality, as was mentioned, but I'm not sure the teacher is looking for that kind of detail-- it sounds to me that the crux of the matter is counting constraints and matching to the degrees of freedom supplied by the superposition of normal modes. Maybe the whole issue is in 2D, and you don't need to worry about higher D, but you can still give a more concise answer than giving the full general solution in 2D.)

    Also, I wanted to clarify what the "D" means, because I think we have a confusion there. The "D" means counting the degrees of freedom, which means counting the number of independent quantities you need to describe the configuration of the system (what the system "looks like" if you had a photograph of it at t=0). You are right that a pendulum in a plane is thus "2D", but that is not typically an application of the concept of "normal modes." Normal modes usually applies to coupled oscillators, like pendulums with springs between them, so there "2D" means two pendulums moving in one dimension (indeed, normal modes are almost always used with just one spatial dimension). Then there is a symmetric and an antisymmetric normal mode, so 2 normal modes spans all the degrees of freedom of the system, but not because it is in a plane, it's because there are two pendulums.
     
    Last edited: May 31, 2012
  11. May 31, 2012 #10
    Ah, I see now. I suppose that is in fact what it's asking. And thanks for pointing out the degrees-of-freedom/dimensional space confusion I had. I think I was misinterpreting those concepts.
     
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