The problem is

Using the equations of motion for the simple pendulum in cartesian coordinates (Eq.(3.7)), numarically integrate the trajectory for the initial conditions

(θ[0], θ'[0]) = (0.1, 0)

Plot the pendulum angle theta[t] and the pendulum length l[t]=sqrt[x[t]^2+y[t]^2] as a function of time.

The equation (3.7) is

x''[t] == (-x[t] x'[t]^2 - x[t] y'[t]^2 + x[t] y[t] g)/l[t]^2
y''[t] == (-y[t] x'[t]^2 - y[t] y'[t]^2 - x[t]^2 g)/l[t]^2

At first i try to make EOM(equation of motion) only express for x[t] and y[t]. So use l[t]=sqrt[x[t]^2+y[t]^2]]. And in this problem the initial boundary condition expressed polar coordinate, sine I try to change IBC(Initial Boundary Condition) as

θ[t] == ArcTan[y[t]/x[t]]; 
Tan[θ[t]] == Sin[θ[t]]/Cos[θ[t]] == y[t]/x[t]



Actually I don't know that another two IBC correctly.

there is my process by using Mathematica and solutions what I want to plot. Actually I just try to make to plot similarly to solution. So I was assumed the one of IBC.

enter image description here

That is, I want to know method of changing IBC, and I want to plot the graph in solution.

  • 1
    $\begingroup$ Share the pdf file which contains the equations and other related stuff. $\endgroup$ – zhk Apr 9 '17 at 13:59
  • $\begingroup$ Reading your question couple of times, I still didn't know what exactly you want? Do you want to know that why the MMA results are not the same for ll vs tt as shown in the pdf? $\endgroup$ – zhk Apr 9 '17 at 14:39

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