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I Large oscillations of pendulum

  1. Sep 17, 2016 #1
    When I solve this problem i have the equation:
    $$x''+Asinx=0$$
    How solve this equation if x large?
    I think we use some approximations
     
  2. jcsd
  3. Sep 17, 2016 #2

    epenguin

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    It's horrible. You cannot solve it exactly with familiar functions, and it involves things called elliptic integrals. Specialised stuff :eek:I think.
    If you like that sort of thing :oldwink: I can tell you there is a treatment in a book so old it is probably available as free e-book somewhere "The Dynamics Of Particles" by A G Webster, pp 45-48. He does give a formula for the period, a series in squared signs of the amplitude.
    Maybe you could follow the argument up to the point where Webster gets
    t = something multiplied by ∫ between limits one of which is essentially the pendulum angle of something of form
    dx/√(1 - k2 sin2(x/2)) . That is the elliptic integral.
    I guess you could plot some solutions if you can understand what he is talking about and use Wolfram app or Mathematica to give you the elliptic integrals numerically.

    Noting your age I would say this is surely not in your syllabus for a long time, if ever.
     
    Last edited: Sep 17, 2016
  4. Sep 18, 2016 #3
  5. Sep 18, 2016 #4
  6. Sep 18, 2016 #5

    epenguin

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    If I am not mistaken this form has different subject on the left hand side than the form I gave which was t = ...
    If you want to go further I am not the person to ask - but it seems he has these last minutes arrived!. :oldsmile:

    What I think it is useful for you to be familiar with is that an equation x'' + f(x) = 0 , which is the equation of a particle moving under a force that depends only on position in one dimension, can be reduced to t = an integral of a function of x by multiplying the equation by 2x' and then the first term is (x2)' . So you can get an equation with x' as LHS, and on the RHS you'll get through just an integral with respect to t of something involving a square root. By the way before square rooting this is just the energy equation. Just if you're lucky this will be an integral that corresponds to functions you know the integral of.

    If you're interested you could work this out for simple harmonic motion - second term kx. You should find you get the same results as what is usually done in a more direct way. You can also solve it with 'elementary' integration when the one-dimensional force is inverse square. But in other cases you may also find that the integrals are not any of the 'elementary' ones, as here.
     
    Last edited: Sep 19, 2016
  7. Sep 18, 2016 #6
    thanks for all helping ^^
     
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