Lab V
Simple Harmonic Motion

Continued

II. (a) Use Maple to solve the differential
equation describing a damped harmonic oscillator with parameters the same as in
I(a) except now there is a damping force characterized by -bv(t) where the damping constant,
β= b/2m.

Remember in part Ia there was a mass, m, attached to a spring of force
constant, k. The initial position of the mass was at x0 and its initial
velocity v0 and the angular frequency was w0 = (k/m)^{1}^{/2}.

**Hint:**You will
solve the equation diff(x(t),t,t) +
2*beta*diff(x(t),t) + omega0^2*x(t) = 0 with x(0)=x0,D(x)(0)=v0 (like last
week's lab);

You can also solve the DFQ using Solve DE Intercatively. Go
ahead and do this too (just for fun).

Assign your solution

(b) Let x0 = 1, v0 = 0, and w_{0} = 1.
Plot the position x(t) for t from 0 to 4p for the case of β = 0.25 (underdamping),
β = 1 (critical damping), β = 4 (overdamping).

**Hints:** In the case of beta = 1, you will need to plot the limit as beta goes
to 1, you may as well do this for all of them, eg
plot(Limit(x(t),beta = 0.25),t = 0..4*Pi);.
Note that this will take Maple some time to do.

You probably should restart before part c

(c) For the underdamping
case of part (b), obtain a simple expression for x(t)
in terms of cosines, sines and e^{-}^{b}^{t} by
defining w_{1}^{2} = w_{0}^{2} - β^{
2}. Obtain the simple expression for the case
when omega1 = 0.9375, beta = 0.25, x0=1.0, v0=0.0

**Hints** This may work better if you input in text rather than math, but it works for the instructor in math too. Solve the same equation for x(t) in terms of omega0 exactly as you did and then assign
the solution and assume omega1, beta are > 0. Then use
subs(omega0=sqrt(beta^2+omega1^2),x(t)): Define x2(t)
:=simplify(evalc(subs({omega1 = 0.9375, beta = 0.25,
x0=1.0, v0=0.0},%))); If asked, choose table.

(d) Make a phase plot for this showing the
decay in the position and velocity.

**Hint:** Define X(t) and V(t): X:= unapply(x2(t),t);
V:= unapply(diff(x2(t),t),t); and plot them like last
time