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Hello,I know solution method for solving differentialequations.

  1. Mar 13, 2012 #1
    Hello,

    I know solution method for solving differentialequations. But last I was asked to solve the equation that describes a mathematical pendulum.

    [tex]\frac{d^{2}\theta}{dt^{2}}+\frac{g}{l}.sin(\theta)=0[/tex]

    How can I solve this the with the most common method?

    thank you
     
  2. jcsd
  3. Mar 13, 2012 #2
    Re: pendulum

    Usually you would assume a small angle to get rid of the sin() term so the problem becomes linear.
    Formally, you could express the solution in terms of the Jacobi elliptic function.
     
  4. Mar 13, 2012 #3
    Re: pendulum

    I'am familiar with the fact that I can assume the sin component equal to the angle, only for small angles..But the questions was related to angles great dan 45°, so I may not neglect the sin component.

    Is it also possible to express the sinus as a power series with n = 5 terms, to get a good result?
    I don't know much about Jacobi elliptic function...

    thanks
     
  5. Mar 13, 2012 #4

    HallsofIvy

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    Science Advisor

    Re: pendulum

    I an concerned about you saying "I know solution method for solving differentialequations". One of the first things you should have learned is that there is NO one method of solving differential equations
    In particularly, this is a badly non-linear equation and there is no "method" for solving it. There are a number of ways of either getting an approximate solution or getting qualitative information about the solution.

    1) Linearization. The set up of your equation implies that you are taking "0" as the bottom of the swing. As long as you do not swing the pendulum "too hard" you have the angle staying small and you can approximate sin(x) by x. Solve [itex]\theta''+ \theta=0[/itex].

    2) "Quadrature". By the chain rule, if we let [itex]\omega= d\theta/dt[/itex],
    [tex]\frac{d^2\theta}{dt^2}= \frac{d\omega}{dt}= \frac{d\omega}{d\theta}\frac{d\theta}{dt}= \omega\frac{d\omega}{dt}[/tex]
    The equation becomes
    [tex]\omega\frac{d\omega}{d\theta}+ (g/l)sin(\theta)= 0[/tex]
    which is a separable first order equation for [itex]\omega[/itex]. Of course, after solving for [itex]\omega[/itex] you have another first order equation for [itex]\theta[/itex]. That equation can be reduced to an integral but, an "elliptic integral, one that has no elementary anti-derivative. That means that it must be done numerically which is why this is an approximation method.

    3) "perturbation". [itex]sin(\theta)= \theta- (1/3!)\theta^3+ (1/5!)\theta^5- ...[/itex]. Dropping all except the [itex]\theta[/itex] gives the linear equation of (1). Then subtract that solution from [itex]\theta[/itex] and put the difference into the equation including the [itex]\theta^2[/itex] term. Continue as long as you wish.

    4) "phase plane". Go ahead and integrate [itex]\int d\omega= -(g/l)sin(\theta)d\theta[/itex] to get [itex]\omega= (g/l)cos(\theta)+ C[/itex] and graph [itex]\omega[/itex] against [itex]\theta[/itex] for different values of C. That gives information about the behavior of solutions.
     
  6. Mar 14, 2012 #5
    Re: pendulum

    Well, I meant the methods I learned for solving (high) second order lineair (non) homogenous differential equations.

    I know methode of annihilation: by wich the differential eqaution is transformed to a differentialpolynoom. For example: [tex](D-a)y(t)=0[/tex] has solution: [tex]Ce^{-at}[/tex]

    Also possible for non homogenous equations..

    I assumed that this a method for solving ODE..

    Back to topic: you proposed a good list of possible way's to solve the equation. I migth be able tot handle the problem now.

    Thank you!
     
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