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I Can anyone evaluate this integral?

  1. Oct 18, 2016 #1
    Hello All! I am trying to solve the simple pendulum without using a small angle approximation. But I end up with this integral:
    Is this possible to evaluate? If so, could I get a hint about what methods to use? If not, is it still possible to get an expression for the period of a simple pendulum without using small angle?

  2. jcsd
  3. Oct 18, 2016 #2


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    Any integral is the area under the curve of the function, so one way to evaluate is to take values of the function you have written between pi/4 and theta for some given theta and do a piecewise linear area approximation for the value of the integral. The more points you evaluate the piecewise area for, the more accurate your evaluation will be for that specific theta.

    Not sure if that what you mean by evaluate.
  4. Oct 18, 2016 #3
    While you could do this with a Riemann sum, I think OP might be asking for an analytical answer.

    Now as for the OP, you'd be in for quite a ride, but what makes me wonder is how you got to that equation from a simple pendulum? I once tried to do it, and didn't arrive to something as close as this.
  5. Oct 18, 2016 #4


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    The expression looks weird. At the lower limit, the denominator = 0, while afterwards (assuming [itex]\theta > \frac{\pi}{4}[/itex]) the denominator is imaginary.
  6. Oct 18, 2016 #5
    This is a simple pendulum with amplitude pi/4 so theta is going to be less then pi/4 but yes.
  7. Oct 18, 2016 #6
    So what parameter of the pendulum is supposed to be represented by this integral?
    And if the amplitude is pi/4 what represent the upper limit of the integral? An angle less than pi/4?
  8. Oct 18, 2016 #7
    When you integrate the equation of motion for a simple pendulum, this is what you get. The amplitude is pi/4 so I'm saying that at t=0 the angle is pi/4. Then this integral is from the initial condition to some arbitrary angle theta at time t.

    My end goal is to get an expression for the period.
  9. Oct 18, 2016 #8
    But what if I want an expression with some arbitrary theta? I could do something like this but I don't know what my bounds should be if I want the Time for one period. Because at half a period the angle is pi/4 and at one period the angle is pi/4!
  10. Oct 18, 2016 #9
    I will upload the rest of my work if you want to see it. I just integrated the equation of motion for a simple pendulum.

    A friend was saying this is an elliptic integral, are elliptic integrals analytically solvable? If so I'm down to try it!
  11. Oct 18, 2016 #10
    When you integrate an equation you get another equation. What you show there is not an equation. There is no equal sign between two terms.
    So my question. What is this integral supposed to be (equal to)?
    But if you do the math correctly you end up with some elliptic function or an expression that can be reduced to elliptic functions. There is nothing to "solve" about it.
  12. Oct 19, 2016 #11


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    The integral can be expressed in terms of elliptic integrals, special functions whose values can only be approximated. This is part of the work by Legendre done about 200 years ago.
    Last edited: Oct 19, 2016
  13. Oct 19, 2016 #12
    While you could solve it analytically to get an expression of elliptic function, a pendulum equation for big thetas is non-linear, so the linear solutions can't be applied at least not in the way it is currently written. I see sources that say it involves complex analysis.

    Numerically you can still use a Riemann sum and get a graph that would give you an approximation in the real space, but physically it would be hard to pin point how the variables change the behaviour of the pendulum.

    I am still wanting to know how you got to this equation.
  14. Oct 19, 2016 #13
    Ok cool! I will figure out how to deal with elliptic integrals now! Thanks!
  15. Oct 19, 2016 #14
    First I looked at a pendulum and wrote down $$-gsin\theta=l\ddot{\theta}$$ $$l\ddot{\theta}+\frac{g}{l}sin\theta=0$$
    Chain Rule $$\frac{d\dot{\theta}}{d\theta}\frac{d\theta}{dt}+\frac{g}{l}sin\theta=0$$

    Separate and Integrate with bounds ##0## to ##\dot{\theta}## and ##\pi/4## to ##\theta## to get
    ##\frac{\dot{\theta}^{2}}{2} = \frac{g}{l}(cos\theta-\sqrt{2}/2)## or ##\dot{\theta} = \sqrt{\frac{2g}{l}(cos\theta-\sqrt{2}/2)}##
    You can separate variables:
    $$d\theta = \sqrt{\frac{2g}{l}(cos\theta-\sqrt{2}/2)}dt$$
    And Integrate:
    $$\int_{\pi/4}^{\theta}\frac{d\theta}{\sqrt{\frac{2g}{l}(cos\theta-\sqrt{2}/2)}}=\int_{0}^{t}dt = t$$

    I still don't see any problems with this.
  16. Oct 19, 2016 #15
  17. Oct 19, 2016 #16
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