# In simple pendulum problem, how to change Initial boundary conditions polar into cartesian coordinate

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]


Since,

y[0]=Sin[θ[0]]
x[0]=Cos[θ[0]]


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.

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

• Share the pdf file which contains the equations and other related stuff. – zhk Apr 9 '17 at 13:59
• 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? – zhk Apr 9 '17 at 14:39