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Driven simple pendulum - system of first order ODEs

  1. Nov 24, 2015 #1
    1. The problem statement, all variables and given/known data

    We have a driven pendulum described by the following differential equation:

    [itex] \frac{d^2\theta}{dt^2} = \frac{-g}{l}\sin(\theta) + C\cos(\theta)\sin(\Omega t) [/itex]

    I need to turn this second order differential equation into a system of first order differential equations (then use a computer to solve the first orders, but that's not the problem here).

    2. Relevant equations

    None needed

    3. The attempt at a solution

    We are told to use some numeric values: l = 10cm, g = 9.81m/s^2, capital omega = 5/s, C = 2/s^2, and we are told to turn the equation into a dimensionless equation using the following notation:

    [itex] \omega^2 = g/l [/itex]

    [itex] \beta = \frac{\Omega}{\omega} [/itex]

    [itex] \gamma = \frac{C}{\omega} [/itex]

    [itex] x= \omega t [/itex]

    Now, putting these into the ODE gives

    [itex] \frac{d^2\theta}{dt^2} = \omega^2\sin(\theta) + \omega^2\gamma\cos(\theta)\sin(\beta x) [/itex]

    But, the only way I can think of turning this into a system of first order ODEs is by using some dummy variable, y.

    In other words, let

    [itex] \frac{d\theta}{dt} = y [/itex]

    and

    [itex] \frac{dy}{dt} = \omega^2\sin(\theta) + \omega^2\gamma\cos(\theta)\sin(\beta x) [/itex]

    Is there no way to get it all in terms of theta and x?

    EDIT:
    I accidentally posted this before it was complete because I hit the "enter" key. Is there a way to turn this feature off? I don't want to get into trouble over posting something which doesn't fit with the rules.
     
    Last edited: Nov 24, 2015
  2. jcsd
  3. Nov 24, 2015 #2

    Orodruin

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    What do you mean? You had it as a system including only theta and t and you rewrite it as a system of two first order equations. That necessarily has to involve another dependent variable, one you decided to call y. It is unclear why you introduce x unless you want a dimensionless number, but then you should replace all occurences of t with x and not mix the notation.
     
  4. Nov 24, 2015 #3
    Yeah, my thinking is the same and I think I just need to leave it at that (besides, numerically solving that system results in a graph which looks sensible).

    But there is a reason why I need it in terms of x - a later part of the question is that it asks for a plot of theta against [itex] \frac{d\theta}{dx} [/itex], but the solution is for theta as a function of time. Also, it's a numerical solution - I don't actually know what theta is in analytical form so I can't just differentiate it with respect to x by hand.

    My thinking is this:
    I have the quantity x = omega * t. What If I just do [itex] dx = \omega dt [/itex], giving [itex] \frac{1}{\omega} dx = dt [/itex] and just substitute that into the system? I'm going to try that now.
     
    Last edited: Nov 24, 2015
  5. Nov 25, 2015 #4

    Orodruin

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    Have you heard of the chain rule?
     
  6. Dec 1, 2015 #5
    Yes.

    Also I tried to solve this entirely in terms of theta and x, as I said I would above, but it was pointless since the only way to get Python to numerically solve this is to tell it that x is defined as omega * t, then give t values. So obviously it gave the same thing.
     
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