MATH 2280 - 2 THIRD MAPLE ASSIGNMENT Treibergs Due Tuesday, March 19, 2002 In this lab we study the forced vibrations and resonance in the damped oscillator based on computing project 3.6 from the text. We begin by studying how the forcing frequency affects the solution to the undamped equation and see the phenomenon of resonance: as the forcing frequency w approaches the natural frequency w0, the amplitude of the steady wave increases. When w=w0 the system is in resonance and the solution blows up. When w is close but not equal to w0 the solution will exhibit beats. Then we consider the damped equation. This time the amplitude of the steady solution will be greatest when the forcing frequency is the practical resonant frequency of the system which will be less for other forcing frequencies. First we illustrate with particular values of the physical constants m, k and c. > with(DEtools): (1.) Consider the case without damping first. Let m=k=1 and c=0. > ode1:=diff(y(t),t,t)+y(t)=0; / 2 \ |d | ode1 := |--- y(t)| + y(t) = 0 | 2 | \dt / > dsolve(ode1,y(t)); y(t) = _C1 cos(t) + _C2 sin(t) The system has a natural frequency w0=1. Now consider the inhomogeneous equation with a forcing term of fixed amplitude F0 and frequency w. > f:=t->F0*cos(w*t); f := t -> F0 cos(w t) > ode2:=diff(y(t),t,t)+y(t)=f(t); > sol2:=simplify(dsolve({ode2,y(0)=0,D(y)(0)=0},y(t))); / 2 \ |d | ode2 := |--- y(t)| + y(t) = F0 cos(w t) | 2 | \dt / F0 (cos(t) - cos(w t)) sol2 := y(t) = ---------------------- 2 -1 + w Using cos(A+B)=cos A cos B - sin A sin B and cos(A-B)=cos A cos B + sin A sin B where A+B=wt and A-B = t we find the trig substitution reduces the difference to a product of sinusoids > subs(cos(t)-cos(w*t)=2*sin((1+w)*t/2) > *sin((w-1)*t/2),sol2); F0 sin(1/2 (1 + w) t) sin(1/2 (-1 + w) t) y(t) = 2 ----------------------------------------- 2 -1 + w Thus the slowly varying function env(t) may by regarded as the enveloping amplitude of the quickly varying function sin((1+w)*t/2). The solution shows beats. > env:=t->2*F0*sin((w-1)*t/2)/(w^2-1); F0 sin(1/2 (-1 + w) t) env := t -> 2 ---------------------- 2 -1 + w > with(plots): Warning, the name changecoords has been redefined > psol1:=subs({w=1.1,F0=1},sol2); > peng:=subs({w=1.1,F0=1},env(t)); psol1 := y(t) = 4.761904762 cos(t) - 4.761904762 cos(1.1 t) peng := 9.523809524 sin(.05000000000 t) > plot([rhs(psol1),peng,-peng],t=0..200); Note that the amplitude of the beats is 2 F0 / |w^2-1| which blows up as w-> w_0=1 the resonant frequency. What happens if w=w0? The forcing term adds energy to each cycle so the soloution grows without bound. > ode3:=diff(y(t),t,t)+y(t)=F0 *cos(t); > sol3:=simplify(dsolve({ode3,y(0)=0,D(y)(0)=0},y(t))); / 2 \ |d | ode3 := |--- y(t)| + y(t) = F0 cos(t) | 2 | \dt / sol3 := y(t) = 1/2 F0 sin(t) t > plot([rhs(subs(F0=1,sol3)),t/2,-t/2],t=0..200); > restart:with(DEtools):with(plots): Warning, the name changecoords has been redefined (2.) The damped case. This time, lets stiffen and damp the dashpot m=1,c=2, k=5. The characteristic equation has roots > solve(r^2+2*r+5=0,r); -1 + 2 I, -1 - 2 I so the pseudofrequency is w1^2=k/m-c^2/4m^2=4. > ode3:=diff(y(t),t,t)+2*diff(y(t),t)+5*y(t)=0; / 2 \ |d | /d \ ode3 := |--- y(t)| + 2 |-- y(t)| + 5 y(t) = 0 | 2 | \dt / \dt / > dsolve(ode3,y(t)); y(t) = _C1 exp(-t) sin(2 t) + _C2 exp(-t) cos(2 t) The system has a natural frequency w1=1. Now consider the inhomogeneous equation with a forcing term of fixed amplitude F0 and frequency w. > f:=t->F0*cos(w*t); f := t -> F0 cos(w t) > ode4:=diff(y(t),t,t)+2*diff(y(t),t)+5*y(t)=f(t); > sol4:=dsolve({ode4,y(0)=0,D(y)(0)=1},y(t)); / 2 \ |d | /d \ ode4 := |--- y(t)| + 2 |-- y(t)| + 5 y(t) = F0 cos(w t) | 2 | \dt / \dt / sol4 := y(t) = 2 4 2 (-25 + 6 w - w + F0 w + 5 F0) exp(-t) sin(2 t) - 1/2 ------------------------------------------------- %1 2 F0 (-5 + w ) exp(-t) cos(2 t) 2 F0 w sin(w t) + ----------------------------- + --------------- %1 %1 2 F0 (-5 + w ) cos(w t) - --------------------- %1 2 4 %1 := 25 - 6 w + w The exponential terms die away and the remaining particular solution has an amplitude > A:=F0*(5-w^2)/(25-6*w^2+w^4); > B:=F0*2*w/(25-6*w^2+w^4); > C:=simplify(sqrt(A^2+B^2),sqrt,symbolic); 2 F0 (5 - w ) A := -------------- 2 4 25 - 6 w + w F0 w B := 2 -------------- 2 4 25 - 6 w + w F0 C := -------------------- 2 4 sqrt(25 - 6 w + w ) > amp:= subs(F0=1,C); 1 amp := -------------------- 2 4 sqrt(25 - 6 w + w ) > plot([rhs(subs({F0=1,w=sqrt(3)},sol4)), subs({F0=1,w=sqrt(3)},amp), -subs({F0=1,w=sqrt(3)},amp)], t=0..20); > plot(amp,w=0..10); The maximum amplitude of the resonance occurs at the maximum of the function amp(w). This practical resonance point 1.732 is somewhat less than the natural frequency w1=2. > solve(diff(amp,w)=0,w); 0, sqrt(3), -sqrt(3) > evalf(%); 0., 1.732050808, -1.732050808 PROBLEMS 1. Choose some other values of F0, k, m and w in the undamped equation (as in ode1). Show that when w is not equal to the natural frequency, then some solutions exhibit beats. Show that as the driving frequency approaches the resonant frequency, the amplitude of the beats become large. Show that if the driving frequency equals the resonant frequerncy then the solution blows up. Use both symbolic computation and plots of your functions to demonstrate these phenomena. 2. Choose some other values of F0, c, k, m and w and initial values (not those in either handout!) for the damped equation (as in ode3). Find the amplitude of the forced steady solution as a function of the driving frequency. Find the practical resonant frequency -- the w for which the amplitide is max. (Be careful to choose m, c, and k so that the practical resonant frequency is positive.) Show an example where the initial data are large compared to the driving force and the natural frequency is much less than the driving frequency such that the natural oscillations dominate for a while until the slaved forced oscillations dominate. Use both symbolic computation and plots of your functions to demonstrate these phenomena. 3. Choose a driving force F(t) = g(t)cos(wt) whose amplitude g(t) is not constant. (e.g. see p. 219-20) Solve and plot an example of damped vibrations that forced by your F(t).