18.03 Topic 9: Applications: forced oscillations.Author: Jeremy OrloffReading: SN §FR; EP §2.6; §2.7 (up to ’Reactance & Impedance’).Applications (See §2.4 -reading and problems and also §2.7 for RLC circuits andpractical resonance)Damped Forced Harmonic OscillatorSolve DE: my00+ by0+ ky = F0cos ωtThink of this as a system with a knob you can use to set the input frequency ω. Weare interested in the response for various values of ω.Finding the homogeneous part:Characteristic roots =−b ±√b2− 4mk2m. Let β =p|b2− 4mk|2m.i) b2− 4mk > 0 (overdamped): yh= c1e(−b/2m+β)t+ c2e(−b/2m−β)t.ii) b2− 4mk < 0 (underdamped): yh= c1e−bt/2mcos βt + c2e−bt/2msin βt.iii) b2− 4mk = 0 (critically damped): yh= c1e−bt/2m+ c2te−bt/2m.Particular solution (will do this many times):Char. polynomial = P (r) = mr2+ br + k.Complexify: m˜y00+ b˜y0+ k˜y = F0eiωt, y = Re(˜y).Exponential Input Theorem: ˜yp=F0eiωtP (iω)=F0eiωtk − mω2+ ibω⇒ yp= Re(˜yp) =F0|P (iω)|cos(ωt − φ) =F0p(k − mω2)2+ b2ω2cos(ωt − φ),where φ = Arg(P (iω)) = tan−1bωk − mω2in first or second quadrants.Let A =F0|P (iω)|=F0p(k − mω2)2+ b2ω2⇒ yp= A cos(ωt − φ).General solution: y = yp+ yh; where yh= transient and yp= periodic solution,yhis called the transient because it goes 0 as t → ∞. This implies all solutions areasymptotically the same as yp, hence periodic.The other quantities (discussed more below) also have names. First we give theirgeneral definition, valid for any linear time invariant system.1. Output amplitude: amplitude of the sinusoidal solution.2. Gain or amplitude response: the amount by which the system scales the inputamplitude, i.e. the ratio of the output to input amplitudes.3. Complex gain: the gain for the complexified equation.4. Phase lag: the angle by which the output maximum trails the input maximum.5. Time lag: the time by which the output maximum trails the input maximum.6. Frequency response: both amplitude response and phase lag.Now we’ll give these quantities for the system discussed above. Pay attention to the118.03 topics 9 2abstract statements involving P (iω), they are more useful to know than the formulaswith square roots etc.Amplitude: A = A(ω) =F0|P (iω)|=F0p(k − mω2)2+ b2ω2.Gain: g = g(ω) =A(ω)F0=1|P (iω)|=1p(k − mω2)2+ b2ω2.Complex gain:1p(iω)=1k − mω2+ ibω.Phase lag: φ = φ(ω) = Arg(P (iω)) = tan−1bωk − mω2.Time lag: φ/ω.Important note: The gain depends on what we designate as the input and output.For example, consider the equation mx00+ bx0+ kx = kB cos(ωt). This has periodicsolution xp=kB|P (iω)|cos(ωt − φ). If we designate B cos(ωt) as the input and x as theoutput then the gain is g(ω) =k|P (iω)|.Phase Lag:In the picture below the dotted line is the input and the solid line is the response.The damping causes a lag between when the input reaches its maximum and whenthe output does. In radians, the angle φ is called the phase lag and in units of timeφ/ω is the time lag. The lag is important, but in this class we will be more interestedin the amplitude response.ty............................................................................................................................•φ/ωtime lagOO18.03 topics 9 3Amplitude response and practical resonanceThe gain is a function of ω. It tells us the size of the system’s response to the giveninput frequency. If the amplitude has a peak at ωrwe call this the practical resonancefrequency.Example: Consider the systemmy00+ by0+ ky = F0cos(ωt)where we consider F0cos(ωt) to be the input. Find the practical resonant frequency.answer: The sinusoidal solution to this equation isyp=F0cos(ωt − φ)|P (iω)|=F0p(k − mω2)2+ b2ω2cos(ωt − φ) (where φ = Arg(P (iω)).The gain (output/input) isg = g(ω) =1p(k − mω2)2+ b2ω2.Here we consider the system parameters m, b, k to be fixed and that the gain dependson the input parameter ω.Finding practical resonance: We want ωrwhere g(w) has a maximum⇒ f(ω) =1g2= (k − mω2)2+ b2ω2has a minimum⇒ f0(ω) = −4mω(k − mω2) + 2b2ω = 0Solving for ω we findωr= 0 or ωr=pk/m − b2/2m2=qω20− b2/2m2.The last expression gives ωrin terms of the natural frequency ω0=pk/m.ωgω0ωrωgω0ωr=pω20− b2/2m2(practical resonance).b/m > ω0√2(no practical resonance).Notice that in this case if the damping gets to large there is no practical resonance.For this example see the mathlet Amplitude Phase Second Order I. Look at E&P §2.7on radio circuits for another example and an application of this.18.03 topics 9 4Undamped forced systemSolve the DE: my00+ ky = F0cos ωt.Homog. part: yh= c1cos ω0t + c2sin ω0t (should be able to do this without work: ω0=pk/m).Particular solution (will do this many times):Complexify: m˜y00+ k˜y = F0eiωt, y = Re(˜y)Characteristic polynonial: P (r) = mr2+ k ⇒ P (iω) = k − mω2.Exponential Input Theorem ⇒ ˜yp=F0eiωtP (iw)=F0eiωtk − mω2if w 6= ω0F0teiωtP0(iw)=F0teiωt2imωif ω = ω0⇒ yp=F0cos ωt|k − mω2|=F0cos ωtm|ω20− ω2|if ω < ω0−F0cos ωt|k − mω2|=−F0cos ωtm|ω20− ω2|if ω > ω0F0t sin ω0t2mω0if ω = ω0General solution (to the inhomogeneous DE): y = yp+ yh.Resonance and amplitude response of the undamped harmonic oscillatorIn ypthe gain = g = g(ω) =1m|ω20− ω2|is a function of ω.The right hand plot below shows g as a function of ω. Note, it is similar to thedamped amplitude response except the peak is infinitely high. As w gets closer to ω0the amplitude increases.When ω = ω0we have yp=F0t sin ω0t2mω0. This is called pure resonance (like a swing).The frequency ω0is called the resonant or natural frequency of the system.In the left-hand plot below notice how the response is oscillatory but not periodic.The amplitude keeps growing in time (caused by the factor of t in yp).Note carefully the different units and different meanings in the plots below.The left hand plot is output vs. time (for a fixed input frequency) and the right handplot is output amplitude vs. input frequency.x and g are in physical units dependent on the system; t is in time; ω is in radians.txResonant response (ω = ω0)ωgω0Undamped amplitude responseEnd of topic 9
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