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).

(b) Let x0 = 1, v0 = 0, and w0 = 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, eg plot(Limit(x(t),beta = 1),t = 0..4*Pi);. Note that this will take Maple some time to do. Alternatively, you can just let beta = 0.99999 You might use X:= unapply(x(t), t,beta); and then plot(X(t,0.25), t = 0..4*Pi);

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-bt by defining w12 = w02 - β 2. Obtain the simple expression for the case when omega1 = 0.9375, beta = 0.25, x0=1.0, v0=0.0

Hints 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)): x2(t) :=simplify(evalc(subs({omega1 = 0.9375, beta = 0.25, x0=1.0, v0=0.0},%))); If asked, choose table. You may have to use x2 and then unapply to get x2(t).

(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