Having solved the appropriate PDE, we know that the equation for the hanging chain is of the form:
Note that N determines the accuracy of our approximation, and A and B are arbitrary constants corresponding to the amplitude of the solution. Recall also that the most significant variables here are the eigenvalues L (lambda). For the solution to satisfy the boundary condition y(1,t)=0, L must be a root of f(1). Letting N=40, we can obtain these roots as follows:
So the first four real roots are:
Here we graph the solutions corresponding to different eigenvalues. N=40 seems
to be a good approximation for the first four values. Also, for simplicity we are going to set
A=0, and B=0.5. By choosing some different values, you can convince yourselves that
A and B do not effect the qualitative behavior of the solution.
The reason I have chosen "ParametricPlot" instead of "Plot" is because I wanted to rotate the graph so that the chain would be hanging down. The options which appear after the "PlotRange" are just to make the picture look pretty, and you may omit some or all of them. It is important, however, to fix the "PlotRange". This prevents Mathematica from rescaling the graph during the animation phase.
Now you can draw your pictures by choosing different values for time and eigenvalue.
In order to make some animations all we need is to generate a number of frames corresponding to different times for a given eigenvalue. The following command generates 10. You may increase the number to make your movie rum more smoothly.
Now you can generate the frames for your favorite eigenvalue (say L1) by entering the following command. Once all the frames have been drawn, you can animate them by selecting any one of the frames and double clicking on it. Have fun!