Analysis of Cases

For convenience of analysis, m is taken to be 1.   The initial conditions used are y(0) = -1, and y¢(0) = 0.

Case 1:                    Ideal mass-spring system

The solution with the given initial conditions is 

The solution to Case 1 oscillates with the same amplitude and period for all time.

 

    Case 2:                Mass-spring system with damping but no rubber band

 

 

There is an important bifurcation at the point where b equals the square root of 50.  When b takes on this value the system is said to be critically damped and takes on the solution:

                       
                                                                                                            This graph shows the equilibrium position at y = 4/5.

 

When b is greater than the square root of 50 (eg. b=9), the system is said to be overdamped, and has solutions very similar to the critically damped case.  The interesting cases are where b is less than the square root of 50 (eg. b=2).  This system has a solution:           

 

The solution is composed of sines and cosines with exponentially decreasing amplitudes

 

The value of b in our experiments was significantly lower than the value necessary for the critically damped case.  More specifically, we found a reasonable value to be 1.5.

 

Rubber band Equations

   

    Case 3:                 Mass-spring system with rubber band but no damping

 

Maple did not take well to a piecewise function being in this differential equation so it was noted that h(y) could also be expressed as:   

                                             

Unfortunately, this insight did not change Maple’s disposition about this particular differential equation.  So we decided to solve this method in a brute force fashion.  We considered the differential equation to be two equations, one where y(t) is positive, and one where y(t) is negative.  Then, we would alternate between these equations, only needing to know the velocity from one to create the next part of the curve.  The solution to this equation looks like this:   

   
The curve in blue is the actual solution to the above equation.   Then to see the effects of the rubber band component more completely we placed the red and light blue curves onto the graph.  

 

The red graph is the solution to the equation:   

and the light blue curve is the solution to the equation:   

 

These are both of the same form of Case 1.  The solutions to these equations mainly just vary on period and slightly on amplitude.  The effects of this can been seen clearly when comparing the solution to the original equation to the graphs of components extended.

 

To get a better look at what this system should actually do a damping term is added to this equation.  The equation becomes:

Case 4:                    
                                         Damped mass-spring system with rubber band

 

Having to use the same brute force tactics as before we obtained the following graph:

The blue line is the solution for the system with the piecewise rubber band function and the dotted red line the solution the system with no rubber band.

 

 

Unfortunately with the initial conditions that were used in all the above analysis only allowed for one cross over from the region of the system where the rubber band is taut and where it is slack.  To compensate for this, we explored another set of initial conditions. 

To give more information about the interaction between the equations where the rubber band is forcing and where it is not we analyzed the solution to the equation where damping is 1,(i.e. b=1) and let the initial starting position to be –10 (i.e. y(0)=-10 and y¢(0) = 0.)

The graph of the solution with these initial conditions is:

The blue line is the solution to the equation, the red line the spring-only system.

 

 

back to main page