125Lecture 14 Tuesday March 10 2015Recap of previous lecture's discussion of steady state response to a harmonic forcingm{d2x/dt2}+c{dx/dt}+kx = Fo cos (ωt-θ)where the ω-dependent complex function %G is defined by %G(ω)= (1/k) / [ 1 - (ω/ωn)2 + 2i ζ ω/ωn ]and the ω-dependent real functions G and φ are byCorollary: If the forcing is the sum of two harmonic forcings F = F1 cos (ω1t-θ1) + F2 cos (ω2t-θ2)then the response is the sum of the two corresponding harmonic responses xsteady-state(t) = F1 G(ω1) cos(ω1t-θ1−φ(ω1)) + F2 G(ω2) cos(ω2t-θ2−φ(ω2)Resonance continued… The response of a mechanical oscillator to a harmonic force isillustrated in several on-line videos. For example, seehttp://www.youtube.com/watch?v=aZNnwQ8HJHU. This video shows a (vertical) mass-springsystem driven by a prescribed harmonic displacement of its support. Schematically, it is G(ω) ≡|%G(ω) |=1 / k(1− (ω/ωn)2)2+ (2ζω/ωn)2φ(ω) ≡ −arg%G(ω) = tan−1[2ζωωn/ (ωn2−ω2)]Main conclusions:=> The steady-state response is non-decaying and sinusoidal at the samefrequency ω as the forcing (not necessarily the natural frequency ωn).=> The steady-state response has an amplitude which is different from theforcing by a factor G(ω) and has a phase delay with respect to the forcing by φ(ω). xsteady-state(t) = Re[Foe−iθ%G(ω)eiωt]= Fo G cos(ωt-θ−φ)126The equation of motion is, (for X defined as an absolute displacement above some fixedreference level, i.e., relative to the lab, not the moving ceiling) neglecting the dissipation terms,may be derived by summing all the upward vertical forces acting on the mass m (L is the spring'snatural length) m&&X = −mg − k{X − y − A + L}We recognize that there is a static equilibrium Xeq at X = Xeq = A-L-mg/k (This is obtained bysolving the above for X when y=0 and &&X = 0.) We define x(t) = X(t) - Xeq and derive m&&x + kx = ky(t)for which the effective force is ky(t). x(t) represents displacement relative to some fixed heightA-L-mg/k in the lab.If y(t) = Yo cosωt such that Feff = kYo cosωt, then, using the above formulas for the steadystate response, we find x(t) = kG(ω) Yo cos( ωt-φ(ω) ). The absolute motion of the mass x(t)is equal to the absolute motion of the wall Yo cosωt …. times a frequency dependentamplification kG … and with a frequency dependent phase delay.The video http://www.youtube.com/watch?v=aZNnwQ8HJHU shows the motion of themass together with the motion of the vertical rod (equivalent to the motion of the ceiling). Itshows1) at low frequency, the ceiling motion and the mass motion are in phase φ=0 Themotion of the mass that they illustrate is about twice that of the ceiling/rod. Thus kG at thisfrequency is about 2, (greater than the zero frequency case in which kG =1) though still muchless than resonance.2) that at resonant frequency, the mass moves much more than the ceiling does. This isresonance. Notice also that the mass's motion is about φ = 90 degrees out of phase from theceiling; when the mass is at the top of its motion, the rod/ceiling is at the midpoint of its motion.1273) At high frequency, the video shows that the mass has little amplitude, (you can't seethe background well, but you can perhaps see the mass is not moving relative to the background(a curtain in the lab?) What little motion it has is φ=180 degrees out of phase with the ceiling.When the rod/ceiling is at the top of its motion, the mass is at the bottom of its.Energy Balance and Power flow in the steady stateOne might suppose it peculiar that the amplitude of the response varies so much with frequencyeven though the force strength Fo is constant. Somehow the system extracts more or less energyfrom the forcing even though the forcing is constant. How can this be? The answer is that thework done by a force is not just up to the force, but depends on the velocity of the responsetoo…. which in turn depends on the force, but also on the parameters m, k, c.Power dissipated is c v2. Power input by the force is Fv. Both of these vary in time. Tocalculate these powers we first recallF(t) = Fo cos(ωt-θ)andx = G Fo cos{ ωt-θ−φ}sov = dx/dt = - Gω Fo sin{ ωt-θ−φ}( By the way, at resonance ω = ωn, φ is π/2 and the velocity is exactly in phase with the force. v= Fo ω G cos(ωt-θ) This is another indication of why the amplitude is high at resonance, theforce is doing maximal work. )The power output of the force is the force times the velocity.F v = Fo cos(ωt-θ) [ -Fo G(ω) ω sin(ωt-θ−φ) ]When not at resonance, the two factors are not in precise phase, so there are times during thecycle when the force is doing negative work on the system. At other times it is doing positivework. The average power output could be obtained by integrating the above expression for Fvover one period and then dividing by the period. ( T is 2π/ω) It is though, more easily obtainedby equating the average power input to the average power dissipation (which is the time averageof cv2, being the velocity times the dissipating force.) This is easy to do because the time-average of sin2 is 1/2.1Tc{−(Fo)G(ω)ωsin(ωt −θ−φ)}2oneperiod∫dt = cFo2ω2G2/ 2This is the average power dissipated <cv2>, and is also the average power input <Fv> . We see itis proportional to G2 and independent of φ.128Definition: Quality Factor "Q" = the peak value of Gkcompared to its quasi static value at ω =0: Gk = 1. The maximumamplification possible. See plot. It happens at ω ∼ ωn. Ourtheory tells us that Q = 1/2ζ inversely proportional to damping.The points ω1 and ω2 are called the "half power points" becausethere the force is doing work on average at a rate half as fast as atresonance, because the amplitude of the motion there is down by afactor of √2.Q is not usually measured by the peak value of Gk, because the lowfrequency limit where Gk is unity is not usually available tonormalize against (equivalently, k is not known) . Far morecommonly, the resonance region is analyzed by measuring its"width", the distance between the half power points. From a scrutiny of the formula for G onecan see, after a bit of algebra and if the damping is small, that Q is also equal to the naturalfrequency divided by the width,width = Δω = ω2 -ω1 = ωn / Q.Correspondences: large Q - low damping – strong resonance -