The following are the equations of a nonlinear pendulum with mass m and length L under gravity g. The angle with respect to the verical is theta:

Compare the nonlinear solution with the following linearized version around the critical point (2pi,0) (i.e. k=2), with the same initial conditions. In order to obtain a comparison figure, you can execute the nonlinear version and during the execution select the opcion "File, Edit RHS" in order to change ¨sin(x)¨ by ¨(x-2*pi)¨. Then compare the phase diagrams (see spiral figure below).

The linealized pendulum presents a non-realistic behavior far away from the critical point (0,2*pi).

# nonlinear single pendulum

dx/dt = y

dy/dt = -b/M*y-g/L*sin(x)

# initial conditions

x(0)=0

y(0)=6

# initial energy

Ei=M*g*L*(1-cos(0))+.5*M*(L*6)^2

# total energy

Ep=M*g*L*(1-cos(x))

Ec=.5*M*(L*y)^2

Et=Ep+Ec

aux E.P.=Ep

aux E.C.=Ec

aux E.T.=Et

# Parameters

param M=10,b=4,g=9.8,L=4

param scale=0.0083333

@ bounds=50000

@ xplot=x, yplot=y

@ xlo=-10, xhi=10, ylo=-6, yhi=6

@ DT=0.025

done

The equation can be written as a system of two equations by means of two states: the angle x=theta and the angular velocity y=theta’.

Its behavior is shown, as well as a phase diagram and total energy.