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I need the parametric equations for a simple pendulum

  1. May 5, 2014 #1
    This is for a personal engineering project. I need the parametric equations y(t) and x(t) for a very simple pendulum. Assume no friction, no forcing, no variation in gravity, a point mass, and the tether angle is significantly less than 30 degrees. It has been a while since I did differential equations, so please link me to some algebraic parametric solutions to this most simple pendulum.

    Thank you.
  2. jcsd
  3. May 6, 2014 #2

    Simon Bridge

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    Its pretty straight forward starting from polar coordinates and converting - but the DE is non-linear for large angles.

    Online resources vary with you eduation level and need:
    http://www.codee.org/library/reviews/summaries/gudermann-and-the-simple-pendulum [Broken]
    Last edited by a moderator: May 6, 2017
  4. May 6, 2014 #3
  5. May 6, 2014 #4
    Thank you for the responses! That is what I needed.

    When I woke up this morning, not having checked here again yet, I was going to tackle this with some algebra. I was going to assume that one of them, say x(t), being an oscillating function, was likely a sine function, and that knowing the point weight is stuck on a circular path, I could use basic trig and algebra to find y(t). I had heard that oscillators make a sine function, but needed to convince myself some way that it was indeed a simple sine function, and not one with another function inside the sine function. The above answers look very believable and will tell me what I need to know.
    Thank you again.

    Oh, maybe x(t) is a more complicated sine function with respect to time, and it is just x(theta) and y(theta) that are simple sine functions. I need to graph this out and get a better look. It is sad that I don't know the graph of sin(sin(t)) off hand, but I'm rusty.
    Last edited: May 6, 2014
  6. May 6, 2014 #5
    Your above link has a derivation I can follow, which shows that x(theta) and y(theta) are simple sine functions. I do not see a x(t), but it is believable that theta(t) is a sine function, so the second respondent's equations for x(t) and y(t) look very believable. That is for small theta, which is close enough for what I'm working with.

    Your other link is for very large theta, giving solutions for it. Not what I needed, but interesting to know they have tackled that too.

    Thanks again.

    I still plan to graph x(t) = sin(sin(t)) in an Open Office spreadsheet to get a better grasp on the shape of the graph.
    Last edited by a moderator: May 6, 2017
  7. May 6, 2014 #6

    Sin(sin(z)) looks like sin(z) but with a more leveled off peak maxing at 0.82 instead of 1.0, at least for z under 180 degrees. Under 40 degrees, the two look extremely similar. Sin can probably replace sin(sin) in my calculations at lower values.
  8. May 6, 2014 #7
    I don't believe that spreadsheet. At 90 degrees, sin(90) = 1, and sin(1) = a bit over zero. Yet it shows it at 0.82.

    The spreadsheet uses radians. I thought I corrected for them by using b1 = sin(sin(3.14*a1/180)). I now realize I should have done b1 = sin(180*(sin(3.14*a1/180))/3.14).
    Last edited: May 6, 2014
  9. May 6, 2014 #8
    I seem to be making errors converting between radians and degrees. Maybe I should just do the x axis in radians to start with, to keep it simple. I just graphed my above equation, and the result looked very high frequency. Not what I expected.
  10. May 6, 2014 #9
    I just repeated the graphs, this time with the a column going from 3.1, 3.0, etc, to -1.6. Column B is sin(sin(a1)), and column c is sin(a1). I graphed it, and I got the same exact graph as the original above, but different scale.

    I just don't understand it. Sin(90 degrees) = 1. Sin(1) = .... Oh. In radians, it is probably about 0.82.

    Well, I'm glad I'm working out these problems now, since I'm going back to school soon. I'm just amazed at the simple mistakes I'm making.

    So y= sin(sin(x)) looks very similar to y = sin(x) for all x, but is flatter and a bit lower in the peaks. Nice to know.
  11. May 6, 2014 #10


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    Do note that the solution given in post #3 is only valid for small θmax. As was mentioned already in post #2, the full equations of motion are non-linear and generally more difficult to solve. Since θmax is small you can approximate

    x = L sin θ ≈ Lθ
    y = - L cos θ ≈ - L + L θ2/2

    inserting θ = θmax sin(√(g/L)t) into this will give you a good approximation of the solution for θmax << 1.

    Also note that what you are plotting has θmax = 1, which is not a small amplitude.
  12. May 7, 2014 #11

    Simon Bridge

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    If you have access to open office you probably have access to GNU/Octave ... which is a better tool than a spreadsheet, if you plan to do more of this sort of thing.

    note: in general, the small angle oscillations will be:
    $$x(t)=x_{max}\sin(\omega t + \delta)$$ ... from that you should be able to figure out y(t) through basic geometry.
    $$\omega = \frac{2\pi}{T}$$ ... where T is the period of the horizontal oscillations.

    This also means that ##\theta (t) = \omega t + \delta## ... because the thing inside a trig function has to be an angle.

    Your ##\sin t## is the special case where ##\omega = 2\pi## radiens per second, and ##\delta=0##radiens.

    In the spreadsheet - be careful to use radiens and not degrees for angles.

    In the end, what you use will depend on what you need the calculation for.
    This information has not been forthcoming.
    Last edited: May 7, 2014
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