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Homework Help: Pendulum's Period

  1. May 1, 2007 #1
    I have never taken a differential equations class. I have this problem that I could really use some help on.


    1. The problem statement, all variables and given/known data
    A pendulum' motion is the following equation [tex] \ddot\theta = (\frac{-g}{l})\sin(\theta))[/tex]

    I need to find the period of oscillations for various starting angles between 1 and 60 degrees. I am not to approximate [tex]\sin(\theta) = (\theta)[/tex] (that is for a later part.)

    2. Relevant equations

    3. The attempt at a solution
    I dont know how to solve the D.E. The Boas book says its not solvable. I thought that the period would be between times where the acceleration is 0... so I set [tex] \ddot\theta=0[/tex], but didnt get very far with that.

    I need to use Maple or some sort of Euler engine? I am clueless about this. Please help me, thank you.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    Last edited: May 1, 2007
  2. jcsd
  3. May 1, 2007 #2
    You can't solve the ODE analytically without the small angle approximation because it is a nonlinear coefficient. I am not familiar with Maple or Euler engine, but I know that Maple has numerical options. Yes, try solving it numerically with Maple or Euler(?).
  4. May 1, 2007 #3
    The other option is to disregard the profs advice and solve for amall angles and by Taylor's expansion account for the error accumulation. My hunch is that the point here is to show up conventional math by numerical solutions.
  5. May 1, 2007 #4
    No, I cant do that... Thats the second part of the problem. Then for the third part I have to state where the small angle approximation breaks down (using the first two parts).

    So, here is where Im at... I need to solve this thing numerically somehow. I messed with Maple a bit and couldnt figure out how to do it. I will try more tomorrow but if somebody has any hints on how I could solve this numerically I would *greatly* appreciate it.

    So, if/when I get this numerical solution, I will have a function of t, [tex]\theta (t) [/tex] = something, right? Then I would take the derivative, set that equal to zero to find the times and thus get the period?

    Sorry if that didnt make sense, im not real versed in the subject. As I said I have never taken Diff Eq, but am wishing I had.
  6. May 2, 2007 #5


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    Once you get your numerical solution, you can find the period by observation. How much time passes between each time [itex] \theta = 0 [/itex].

    You can use a variation of the http://en.wikipedia.org/wiki/Euler_method" [Broken] is much better.
    Last edited by a moderator: May 2, 2017
  7. May 2, 2007 #6
    Ok, I have used the Euler method and found the period to be roughly 2 second.

    Now I need to make the small angle approximation, and compare the results.

    I make the approximation and solve the differential equation to get [tex] \theta(t) = \alpha e^{i\sqrt{\frac{g}{L}}t}+\beta e^{-i\sqrt{\frac{g}{L}}t} [/tex]

    Now, the boundry condition are [tex] \dot\theta(0) = 0 [/tex] and [tex] \theta (0) = \theta _0 [/tex].

    This gives me a the function [tex] \theta (t) = \frac{\theta_0}{2}i\sqrt{\frac{g}{L}}*(e^{i\sqrt{\frac{g}{L}}t} -e^{-i\sqrt{\frac{g}{L}}t} [/tex])

    But there are no zeros for this save the case where t=0 are there?!?

    Where have I gone wrong?
  8. May 2, 2007 #7
    Huh? What are you asking about?

    I think you made a mistake using the boundary conditions.
    [tex] \dot{\theta} = \alpha \omega e^{i \omega t} - B \omega e^{-i \omega t}[/tex]
    [tex] 0 = \alpha i \omega - \beta i \omega[/tex]
    omega is non-zero so
    [tex] \alpha = \beta[/tex]

    and the first condition shows that
    [tex]\alpha + \beta = \theta_0[/tex]
    [tex] 2 \alpha = \theta_0[/tex]

    Also, do you really care about the complex solutions? I would just take the real components and write the equation as [tex]\theta(t) = c_1 sin \omega t + c_2 cos \omega t[/tex] where [tex]\omega = \sqrt{\frac{g}{l}}[/tex].
  9. May 3, 2007 #8
    Ok I find the function to be [tex] \theta (t) = \frac{\theta_0}{2} (exp (i \sqrt\frac{g}{L} t) + (exp (-i \sqrt\frac{g}{L} t) [/tex]

    Now to find the period of oscillation I need to find the times between where [tex] \theta (t) = 0 [/tex], which only occurs at t=0...

    Any help on how to find these periods? Thx.
  10. May 3, 2007 #9
    Are you aware of Euler's formula at all? If not, Euler's formula states that
    [tex]e^{i\theta} = cos\theta + isin\theta[/tex]
    in your case
    [tex]exp{i \sqrt{\frac{g}{l}} t} = cos\sqrt{\frac{g}{l}t} + isin \sqrt{\frac{g}{l}t}[/tex]

    You can write your exponentials in terms of sines and cosines. In your current state, you will have both a complex solution and a real solution, though you can get just the real solutions if needed/wanted. Once you have your solution in terms of sines and cosines I assume, you seem to be at a pretty good level, you know how to find the period.
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