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Simple Pendulum nonlinear second order differential equation

  1. Nov 22, 2012 #1
    1. The problem statement, all variables and given/known data

    given: [itex]dt=-\frac{1}{2}\sqrt{\frac{l}{g}}\frac{d\theta}{\sqrt{sin^2(\alpha/2)-sin^2(\theta/2)}}[/itex]

    make the change of variables [itex]sin(\theta/2)=sin(\alpha/2)sin(\phi)[/itex]

    to show that: [itex]dt=-\sqrt{\frac{l}{g}}\frac{d\phi}{\sqrt{1-k^2sin^2(\phi)}}[/itex]

    where [itex]k=sin(\alpha/2)[/itex]

    2. Relevant equations

    given in (A)

    3. The attempt at a solution

    Substituting for θ i have gotten to:

    [itex]dt=-\frac{1}{2}\sqrt{\frac{l}{g}}\frac{d\theta}{\sqrt{k^2-k^2sin^2(\phi)}}[/itex]

    I'm not sure how to go any farther, or how to substitute for dθ
     
    Last edited: Nov 22, 2012
  2. jcsd
  3. Nov 22, 2012 #2

    Simon Bridge

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    You have to substitute for ##d\theta## as well.

    if ##\sin(\theta/2)=k\sin(\phi)##
    then ##\theta=?## and ##d\theta/d\phi=?##

    Note: ##\sqrt{k^2-k^2\sin^2(\phi)}=k\cos(\phi)##
     
    Last edited: Nov 22, 2012
  4. Nov 22, 2012 #3
    Differenting both sides and then doing some algebra gives me

    [itex]d\theta=\frac{2sin(\alpha/2)cos(\phi)d\phi}{cos(\theta/2)}[/itex]

    Also thank you for pointing out that my denominator can be rewritten as kcos(θ). That seems potentially helpful.

    If [itex]sin(\theta/2)=ksin(\phi)[/itex], then [itex]\theta=2arcsin(ksin(\phi))=2\phi*arcsin(k)[/itex]?
     
  5. Nov 22, 2012 #4

    Simon Bridge

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    That won't help unless you can find ##\cos(\theta/2)## in terms of ##\phi##.
    (What did you differentiate both sides with respect to?!
    It looks like you differentiated the LHS wrt ##\theta## and the RHS wrt ##\phi## which is a nono.)

    Why not start by solving the equation for ##\theta## and then differentiating that?
     
  6. Nov 22, 2012 #5
    I did differentiate the LHS wrt θ and the RHS wrt phi. Come to think of it, that doesn't make sense at all!

    If I were to solve for [itex]\theta[/itex], I think I'd get:

    [itex]\theta=2arcsin(ksin(\phi))[/itex]

    then differentiating with respect to phi,

    [itex]\frac{d\theta}{d\phi}=2\frac{kcos(\phi)}{\sqrt{1-k^2sin(\phi)}}[/itex]

    so I can substitute [itex]d\theta=2d\phi*\frac{kcos(\phi)}{\sqrt{1-k^2sin(\phi)}}[/itex] in?
     
    Last edited: Nov 22, 2012
  7. Nov 22, 2012 #6

    Simon Bridge

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    That's the idea!
    Notice how that aready looks a lot like the answer you are looking for?
    And you have a ##k\cos(\phi)## in the numerator to cancel out that stuff you already have in the denominator!

    Aside:
    trig function in LaTeX look better if you put a \ in front of them so ##\arcsin(\phi)## instead of ##arcsin(\phi)## see?
     
    Last edited: Nov 23, 2012
  8. Nov 22, 2012 #7
    I see the difference! I'm a little new here for thanks for the tip.

    I believe I was able to solve the problem with your help. It is very much appreciated! Thanks for the expedient responses and have a nice holiday season.
     
  9. Nov 23, 2012 #8

    Simon Bridge

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    No worries - not holidays in NZ yet though ;)
     
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