Computer Algebra Recipes

3.2. SECOND-ORDER MODELS 125
0.4
0.2
0
–0.2
–0.4
10 t 20
1
0.8
0.6
0.4
0.2
0
10 t 20
Figure 3.4: Steady-state (left) and transient (right) contributions.
In the steady-state picture, the shorter of the two oscillatory curves corresponds
to x1(t), the taller one to x2(t). The motion of mass m1 is approximately 180
degrees out of phase with m2. The period of oscillation of both masses is
identical with that of the driving force.
In the transient picture, the curve that rises above 1 is x2(t). Both curves
decay to essentially zero in less than 20 time units.
To animate the solution, the complete (steady-state plus transient) timedependent
displacements are added to the given equilibrium coordinates.
> X1:=2+x1ss+x1tr: X2:=5+x2ss+x2tr:
The motion of the two masses is now animated, each mass being represented
by a size-10 (default color red) circle. To produce motion along the horizontal
axis, the horizontal coordinate of each mass is entered in a list with the vertical
coordinate set equal to zero. The time range is such that the transient parts
vanish, revealing the steady-state motion. You can take a larger time interval
if you desire. To obtain a smooth animation, 200 frames are used.
> animate(f[X1,0],[X2,0]g,t=0..20,frames=200,style=point,
symbol=circle,symbolsize=20,tickmarks=[3,2]);
Execute the command on your computer to see the animated motion.
PROBLEMS:
Problem 3-10: Hand calculation
Making use of the integral transform package, mimic a hand calculation to
derive the solution for the second example.
Problem 3-11: Transform package approach
Use the integral transform package and the Laplace transform method to solve
the following ODEs. Con¯rm the solutions using the method= laplace option
in the dsolve command. Plot each solution over a suitable time range that
includes the steady-state regime. Extract the transient part of the solution and