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Proving the period of a pendulum

  1. Mar 30, 2017 #1
    1. The problem statement, all variables and given/known data
    I'm writing a small project as part of my degree, and I am stuck on proving one (probably simple!) thing towards the end. I've shown that the pendulum is periodic and has a total period of 4t2.

    2. Relevant equations
    oXMcWhE.png

    3SIvxGm.png


    3. The attempt at a solution

    I think I need to rewrite the same ODE with starting point t=0 but I'm having a complete mindblank on how to proceed from here. Any help would be greatly appreciated!
     
  2. jcsd
  3. Mar 30, 2017 #2

    haruspex

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    Not entirely sure about your notation, nor what your starting points are.
    You seem to be drawing on some earlier equation ##\ddot\theta=csin(\theta)##, but what is csin there? Do you just mean a constant c multiplied by sine? "csin" is a standard abbreviation for complex sine in some computer languages.
    You also seem to know that ##\theta(t_2)=\dot\theta(t_2)=\dot\theta(0)=0##.
    Anything else?
     
  4. Mar 30, 2017 #3
    Hi,

    Sorry I should've given more information. The equation I'm drawing on as is as you said is ##\ddot\theta = -csin\theta## and this is just the regular sine function multiplied by a constant, c. No complex sine - sorry for the confusion.


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    ouSsKmX.png

    Here is everything I know that's relevant to periodicity in my paper. Thanks for your reply!
     
    Last edited by a moderator: May 8, 2017
  5. Mar 30, 2017 #4

    haruspex

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    The latex is clearer if you use \sin rather than just sin.

    As far as I can see, nowhere did you define the initial conditions (t=0). But in the second stage (post #1) you use ##\dot\theta(0)=0##.

    You have an equation ##\theta(t)=-\theta(4t_2-t)##.
    What would that imply at t=2t2? Is the minus sign a mistake?
     
  6. Mar 30, 2017 #5
    Noted for the future.

    The minus sign is a typo, thanks for pointing that out!

    At time t=2t2, we have that it's equal to ##\theta(-t)## by symmetry, since it's the furthest "left" position of the mass (taking moving from left to right as being negative).
     
  7. Mar 30, 2017 #6
    I think I need to prove that ##\theta(4t_2) = \theta(0)## but I'm not entirely sure... I think to do that I would need to basically run through a very similar argument again and then use cauchy-lipschitz to show uniqueness
     
  8. Mar 30, 2017 #7

    haruspex

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    You need to show that the initial conditions going forward from 4t2 are the same as those going forward from t=0. So that's the values of ##\theta## and ##\dot\theta##.
     
  9. Mar 30, 2017 #8
    Thanks. It's very late here in the UK so I'll actually write this up tomorrow - the flu really isn't helping my thought process!

    So in short I need to show that ##\theta(0) = \theta(4t_2)## and that ##\dot\theta(0) = \dot\theta(4t_2)##? And then use a similar argument from before, from my second post with:

    $$\begin{cases}

    \ddot y& = -csiny(t)\\

    y(t_2)&=-\theta(2t_2-t_2)=-\theta(t_2)=0 \\

    \dot y(t_2)&=\dot\theta(2t_2-t_2)=\dot\theta(t_2)

    \end{cases}$$

    Or am I way off?

    And then I would need to change this to reflect 0 and ##4t_2##?

    $$-\theta(2t_2-t)=\theta(t), t\in[t_2, 2t_2]$$
     
  10. Mar 30, 2017 #9

    haruspex

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    Yes.
    As you wish, but to me it would be suffice to say that the values of ##\theta## and ##\dot\theta## define the state of the system, so if the value pair recurs then the behaviour is cyclic. Likewise, for your antisymmetric result in post #3, I would just observe the effect of time reversal on the differential equation.
    These are mechanistic arguments rather than purely mathematical ones.
     
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