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Large oscillations of pendulum

  1. Sep 18, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the large oscillation period T of pendulum. Suppose that the amplitude is ##\theta_0##
    We can write oscillation period T by the sum of a series, know that:
    $$\int_0^1 \frac{dt}{\sqrt{(1-t^2)(1-k^2t^2)}}=\frac{\pi}{2} \sum_{n=0}^{∞}(\frac{(2n)!}{2^{2n}(n!)^2})^2$$
    Let ##T_0=2\pi\sqrt{\frac{l}g}## which l is the length of pendulum, we have the graph of ratio ##\frac{T}{T_0}## respects to amplitude ##\theta_0##:
    3226861545_173192223_574_574.jpg

    2. Relevant equations
    The differential equation is:
    $$\ddot{\theta}+\frac{g}{l}sin\theta=0$$

    3. The attempt at a solution
    I think that using the condition and graph to solve this equation. But how?

    Thanks for helping
     
  2. jcsd
  3. Sep 18, 2016 #2

    Simon Bridge

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    At some point in the attempt at a solution, you may get an integral that can be made to look like the one in your problem statement.
    As to the details, look to the coursework you have just recently completed.
     
  4. Sep 18, 2016 #3
    Thanks, but by a point of the graph I always have: ##(\frac{T}{T_0})_i=\theta_i##
    How to get an intergral?
     
  5. Sep 18, 2016 #4

    Simon Bridge

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    That does not make any sense. The graph is for ##(T/T_0) = f(\theta_0)## where ##f## is given by the curve.
    Is that how your most recent coursework has handled things?

    Note: you usually need to integrate in order to solve differential equations.
    I cannot tell you anything more specific because I don't know what you've just done in your course.
     
  6. Sep 18, 2016 #5
    I think the differential equation dont help me anything, so I decided to use the conservation of energy, it seems useful because that have more data.
    The angular velocity of pendulum at time t:
    $$\frac{1}{2}mv^2=mglcos\theta-mglcos\theta_0\Rightarrow \dot{\theta}=\sqrt{\frac{2g}{l}(cos\theta-\cos\theta_0)}=\frac{d\theta}{dt}$$
    So:
    $$dt=\frac{d\theta}{\sqrt{\dfrac{2g}{l}(cos\theta-\cos\theta_0)}}$$
    The interal, let ##t=sin\theta## we have:
    $$I=\int_0^{\frac{\pi}{2}} \frac{d\theta}{\sqrt{1-k^2sin^2{\theta}}}$$
    It seems similar form but I can't make they exactly alike.
     
    Last edited: Sep 18, 2016
  7. Sep 18, 2016 #6

    Simon Bridge

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    ... is that the sort of thing you can just "let"?
    Don't you already have t defined to be time elsewhere?

    How are you choosing the limits of the integration ... notice the integral example goes from 0 to 1?
    Notes: ##\int_0^T t\; dt = T## also ##2g/l = 4\pi/T_0## and if you put a backslash in front of the "sin" etc you get proper typesetting for trig.

    ... basically that is the approach though - keep playing around until you get something promising.
     
  8. Sep 18, 2016 #7
    $$dT=\frac{d\theta}{\sqrt{\dfrac{2g}{l}(cos\theta-\cos\theta_0)}}$$
    Let ##cos\theta=1-2sin^2(\frac{\theta}{2})##
    So I have:
    $$dT=\frac{d\theta}{\sqrt{\dfrac{4g}{l}[sin^2(\frac{\theta_0}{2})-sin^2(\frac{\theta}{2})]}}$$
    $$\int_0^{\frac{T}4}dT=\frac{T_0}{4\pi}\int_0^{\theta_0}\frac{d\theta}{{\sqrt{sin^2(\frac{\theta_0}{2})-sin^2(\frac{\theta}{2})]}}}$$
    Let ##t=\dfrac{sin\frac{\theta}{2}}{sin\frac{\theta_0}{2}}## then:
    $$\int_0^{\frac{T}4}dT=\frac{T_0}{2\pi}\int_0^{1}\frac{dt}{{\sqrt{(1-sin^2(\frac{\theta_0}{2})t^2)(1-t^2)}}}$$
    So
    $$\frac{T}{T_0}=\sum_{n=0}^{∞}(\frac{(2n)!}{2^{2n}(n!)^2})^2$$
    I think it correct :((
     
  9. Sep 18, 2016 #8
    But I dont see how this ratio respectsto ##\theta_0##. As you told, Does ## f(\theta_0)## equals sum?
     
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