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Exact solution for pendulum

  1. Jan 25, 2006 #1
    Yup. How can I solve:
    [tex]d^2\theta / dt^2 = -(g/L)sin(\theta)[/tex]
     
  2. jcsd
  3. Jan 25, 2006 #2

    arildno

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    There doesn't exist a nice, exact solution of this diff. eq for general initial values.
     
    Last edited: Jan 25, 2006
  4. Jan 25, 2006 #3

    benorin

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    Look here

    The definite integral that gives so little as the period of the pendulum involves a complete elliptic integral of the first kind! and indeed this is not nice either.
     
  5. Jan 25, 2006 #4

    saltydog

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    I disagree. It is very nice. Beautiful in fact. I'll start it for you:

    We obtain the first integral:

    [tex]\frac{1}{2}\left(\frac{d\theta}{dt}\right)^2-\frac{g}{L}Cos(\theta)=C[/tex]

    You know, integrate the second order ODE directly to get this. Try it. :smile:

    If [itex]\theta=\omega[/itex] is the maximum displacement of the pendulum from its equlibrium position, then [itex]\theta^{'}=0[/itex] for this value and we can evaluate C, which is:

    [tex]C=-\frac{gCos(\omega)}{L}[/tex]

    Now we solve for [itex]\frac{d\theta}{dt}[/itex] and obtain:

    [tex]\frac{d\theta}{dt}=\sqrt{\frac{2g}{L}}\sqrt{Cos(\theta)-Cos(\omega)}[/tex]

    or:

    [tex]dt=\sqrt{\frac{L}{2g}}\frac{d\theta}{\sqrt{Cos(\theta)-Cos(\omega)}}[/tex]

    Letting:

    [tex]Cos(\theta)=1-2k^2 Sin^2(\phi)[/tex]

    [tex]k=Sin(\frac{\omega}{2})[/tex]

    we can then reduce the last dif. eq. to an elliptic integral and solve it accordingly . . .

    . . . equal rights for special functions . . .
     
    Last edited: Jan 25, 2006
  6. Jan 26, 2006 #5

    arildno

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    Equal rights for special functions?
    As long as you don't count in the Haenkel function, I'll accept that.
     
  7. Jan 26, 2006 #6

    dextercioby

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    The solution angle(t) is proportional to sinus amplitudinis, one of Jacobi's elliptical functions.


    Daniel.
     
  8. Jan 26, 2006 #7

    saltydog

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    Well I' d like to finish this since I think the analysis is as I saidit was above.

    First some clarification:

    1. I obtained the first integral by writing the ODE as:

    [tex]d\left[\theta^{'}\right]=-\frac{g}{L}Sin[\theta][/tex]

    multiplying both sides by [itex]\theta^{'}[/tex] and writing it as:

    [tex]\theta^{'}d\left[\theta^{'}\right]=-\frac{g}{L}Sin[\theta]d\theta[/tex]

    The variables have been separated and can now be directly integrated to give the first integral (pretty sure that's poke-a-poke, kinda' rough for me too).

    2. I glossed over the fact that I chose the initial velocity to be zero. This makes the algebra easier although we can't model round-and-round motion with this solution.

    I left the problem above as:

    [tex]dt=\sqrt{\frac{L}{2g}}\frac{d\theta}{\sqrt{Cos(\theta)-Cos(\omega)}}[/tex]

    and seek a transformation that will convert the RHS to an elliptic integral:

    [tex]F[\phi,k]=\int_0^{\phi} \frac{dv}{\sqrt{1-k^2 Sin^2(v)}}[/tex]

    Well, it turns out that if:

    [tex]Cos(\theta)=1-2k^2Sin^2(\phi)\quad \text{and}\quad k=Sin(\omega/2)[/tex]

    then the following can be shown to be true:

    [tex]Cos(\theta)-Cos(\omega)=2k^2Cos^2(\phi)[/tex]

    [tex]Sin(\theta)=2k Sin(\phi)\sqrt{1-k^2Sin^2(\phi)}[/tex]

    [tex]Sin(\theta)d\theta=4k^2 Sin(\phi) Cos(\phi)d\phi[/tex]

    Making this substitution, which I think is quite a challenge in itself, we obtain:

    [tex]dt=\sqrt{\frac{L}{g}}\frac{d\phi}{\sqrt{1-k^2 Sin^2(\phi)}}[/tex]

    Integrating from t=0 to some time t:

    [tex]t=\sqrt{\frac{L}{g}}\int_{0}^{\phi}\frac{d\phi}{\sqrt{1-k^2 Sin^2(\phi)}}[/tex]


    Now, this last integral is an elliptic integral. It has an inverse. We can then take the inverse elliptic integral of both sides in order to isolate [itex]\phi[/itex]. A special function called the JacobiSN function expresses the sin of the inverse. That is if:

    [tex]F[\phi,k]=\int_0^{\phi} \frac{dv}{\sqrt{1-k^2 Sin^2(v)}}[/tex]

    then:

    [tex]\text{JacobiSN}\left\{F[\phi,k]\right\}=Sin(\phi)[/tex]

    Doing this to the equation above yields:

    [tex]\text{JacobiSN}\left(t\sqrt{\frac{g}{L}},k\right)=Sin(\phi)=\frac{1}{k}Sin(\theta/2)[/tex]

    Isolating theta:

    [tex]\theta(t)=2\text{ArcSin}\left[k\text{JacobiSN}\left(t\sqrt{g/L},k\right)\right],\;k=sin(\omega/2)[/tex]

    See, told you it was beautiful:smile:

    A sample plot of this function (for a 3 foot pendulum raised 1 radian) is attached.
     

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    Last edited: Jan 26, 2006
  9. Feb 4, 2006 #8

    Hootenanny

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    Saltydog,

    I can follow your derivation of
    [tex]\theta(t)=2\text{ArcSin}\left[k\text{JacobiSN}\left(t\sqrt{g/L},k\right)\right],\;k=sin(\omega/2)[/tex]
    but I was wondering if there was anyway you could find an exact solution for the period of a pendulum when [itex]\sin\theta \not\approx \theta [/itex]

    Thanks
     
  10. Feb 4, 2006 #9

    HallsofIvy

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    No there is not exact formula- just as there is no exact formula for circumference of an ellipse.
     
  11. Feb 4, 2006 #10

    Hootenanny

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    So how would I solve it? Could I do an expansion?
     
  12. Feb 4, 2006 #11

    saltydog

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    Well hey Hootenanny. Do I need to look up what that means so I know who I'm talkin' to? Anyway, not sure what Hall is referring to but my understanding is that there is a way:

    Using the three transformation rules above with the expression:

    [tex]\frac{d\theta}{dt}=\sqrt{\frac{2g}{L}}\sqrt{Cos(\theta)-Cos(\omega)}[/tex]

    we obtain:

    [tex]\frac{d\theta}{dt}=2k\sqrt{\frac{g}{L}} Cos(\phi)[/tex]

    Since the desired value of theta (the positions of maximum displacement in order to calculate the period) is that for which [itex]\frac{d\theta}{dt}=0[/itex], this then corresponds to:

    [tex]\phi=\pi/2[/tex]


    Letting P(k) be the period of the pendulum, we get:

    [tex]P(k)=4\sqrt{\frac{L}{g}}\int_0^{\pi/2} \frac{d\phi}{\sqrt{1-k^2 Sin^2(\phi)}}[/tex]

    or:

    [tex]P(k)=4\sqrt{\frac{L}{g}}K(k)[/tex]

    where K(k) is the complete elliptic integral of the first kind . . . equal rights . . . ok, you guys are getting annoyed by that.

    Anyway, when k=0, this reduces to:

    [tex]P=2\pi\sqrt{\frac{L}{g}}[/tex]
     
    Last edited: Feb 4, 2006
  13. Feb 4, 2006 #12

    Hootenanny

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    Sorry, I'm sure this is gonna sound really stupid but which three transformation rules?
     
  14. Feb 4, 2006 #13

    saltydog

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    These:

    [tex]Cos(\theta)=1-2k^2Sin^2(\phi)\quad \text{and}\quad k=Sin(\omega/2)[/tex]

    [tex]Cos(\theta)-Cos(\omega)=2k^2Cos^2(\phi)[/tex]

    [tex]Sin(\theta)=2k Sin(\phi)\sqrt{1-k^2Sin^2(\phi)}[/tex]

    [tex]Sin(\theta)d\theta=4k^2 Sin(\phi) Cos(\phi)d\phi[/tex]

    Tell you what though, it would be a good idea if you went through the change of variables using these transformations that lead to the elliptic form of the integral as I described above.
     
  15. Feb 4, 2006 #14

    saltydog

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    Alright, I don't understand this. What, I can say that can't I?

    So I plot the period function:

    [tex]P(k)=4\sqrt{\frac{L}{g}}K(k)[/tex]

    with:

    [tex]k=sin[\omega/2][/tex]

    with omega being the maximum displacement in radians. The plot is attached. Well, anyway, when k=0, that corresponds to the pendulum at rest. Why then is the period:

    [tex]P(0)=2\pi \sqrt{L/g}\approx 1.92 \text{sec}[/tex]

    (for a 3 foot pendulum I mean)

    when it's not moving?

    Tell you what though, if I had easy access to a college physics lab you can bet I'd be in there with a rigid pendulum checking all this.
     

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    Last edited: Feb 4, 2006
  16. Feb 4, 2006 #15

    Hootenanny

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    I'm just thinking that because just because k = 0 doesn't have to mean that the pendulum is at rest:

    [tex]k= \sin [\omega/2] [/tex]
    [itex] \sin = 0 [/itex] when [itex]\omega/2 = 0 [/itex].
    [tex] \omega = \frac{v}{r} [/tex]

    So when k = 0 could it just indicate that the pendulum is at its maximum displacement because v = 0??
     
  17. Feb 4, 2006 #16

    HallsofIvy

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    No, [itex]\omega[/itex] is NOT the angle of motion, it is the angle of maximum displacement and is a constant. If that maximum displacement if 0, so that k= 0, then there is no motion.
     
  18. Feb 4, 2006 #17

    saltydog

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    The analysis I gave above is taken from "Introduction to Non-Linear Differential and Integral Equations", by H. Davis., Ch. 7. What, you didn't think I did all that from scratch did you?

    For example, Davis gives this problem at the end of the section:

    A pendulum is displaced through an angle of 45 degrees. Compute the peroid. Answer: [itex]6.53 \sqrt{L/g}[/itex] seconds.

    Ok, so I use:

    [tex]\text{maxdisp}=(45/360) 2\pi[/tex]

    [tex]k[\omega]=Sin[\omega/2][/tex]

    I then calculate:

    [tex]4 \text{EllipticK[(k[maxdisp])}^2]}[/tex]

    Mathematica returns:

    [tex]6.534[/tex]

    so this gives me some confidence the formula is correct.

    I tell you what, if we can't get it resolved here I may risk stepping into the General Physics Forum and presenting this apparent discrepancy to them to explain. It's not just a "limiting" issue is is? I mean even if the max displacement is close to zero, the period will be around 1.9 seconds and approaches the limiting case P(0) as omega goes to zero?
     
    Last edited: Feb 4, 2006
  19. Feb 5, 2006 #18

    Hootenanny

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    So your saying that there may be a minimum displacement before a significant error is induced?
     
  20. Feb 5, 2006 #19

    mathwonk

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    instead of, or perhaps in addition to, these complicated formulas, it is enlightening to draw the phase curves.

    i.e. consider it as a system with y = dx/dt, and then dy/dx = - Csin(x). Then find and plot all the equilibria, i.e. points where both dy/dt = 0 = dx/dt.


    Then linearize the system near each equilibrium, i.e. solve the usual approximate system d^2x/dt^2 = -x, and get an idea of the possible local behavior near the equilibria.

    in fact these linearizations do give a good approximation to that behavior.
    then try to draw in the full phase curves guided by these local pictures.

    this gives a rather enlightening picture of the whole situation, although some further analysis is needed to be sure the linear models do reflect the non linear solutions in this case; i.e. since the linear equilibria are "centers", it is not immediate that the non - linear equilibria are also centers, but here they are.
     
  21. Feb 6, 2006 #20

    saltydog

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    I agree with Mathwonk regarding the importance of qualitative analysis of differential equations and am a big advocator of such. I only presented the analytical approach for it's sheer beauty. Although some here have criticized Bob Devaney and his approach to Differential Equations (emphasis on qualitative analysis), I find such absolutely essential in obtaining an intutitive understanding of dynamics. Tell you what though, I say yes, do the phase-portrait for that one but really one should then proceed to open the throttle wide:

    [tex]\frac{d^2\theta}{dt^2}+\frac{1}{q}\frac{d\theta}{dt}+Sin(\theta)=gCos(\omega t)[/tex]

    Although I've not studied this one in detail, I understand it exhibits chaos and allows for the existence of strange attractors. Thus one should be able to draw both a Feigenbaum diagram and a strange attractor for it. Yep, yep, I think this is a good one for Hootenanny to work on. Please post a complete report of your findings, plots too.:rolleyes:

    Edit: Oh yea, just set it up in Mathematica and solve it numerically then plot the attractor parametrically and the Feigenbaum plot as a Poincare' section.
     
    Last edited: Feb 6, 2006
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