MATH 286 Sections D1 & X1: Assignment 5Solutions for the graded problems.Section 3.4, #14 [3 points] (a). The IVP associated to the mass-spring system is25x00+ 10x0+ 226x = 0, x(0) = 20, x0(0) = 41.The roots of the characteristic polynomial P(r) = 25r2+ 10r + 226 are −15± 3i, so thesolution isx(t) = c1e−15tcos(3t) + c2e−15tcos(3t) = Ce−15tcos(3t − α).Where C =qc21+ c22, sin α =c2C, cos α =c1C. Plugging in the IC’s we can solve for c1, c2(or C, α), and we getx(t) = 25e−15tcos(3t − α), α = arctan(3/4).From this equation, the solution does indeed have the graph given by Figure 3.4.15. Notethat this system is underdamped, since it has complex roots (and therefore oscillatesabout its steady state equilibrium).(b). The pseudoperiod of the oscillation of x(t) is just the period of the cos(3t − α)term. This is the period that the solution would have if it did not have the decayingamplitude factor of 25e−15t. The pseudoperiod is thereforeT =2πω=2π3.The envelope curves for x(t) are given by the amplitude factor on x(t) : they are ±25e−15t.Section 3.6, #18 [2 points] We havemx00+ cx0+ kx = F0cos(ωt ), (m = 1, c = 10, k = 650, F0= 100).If we apply the method of undetermined coefficients, the particular (steady state solu-tion) of this of the form byxp(t) = C(ω) cos(ωt − α) = A cos(ωt) + B sin(ωt ),where the amplitude factor is (refering to equation (21) in the text):C(ω) =F0p(k − mω2)2+ c2ω2=100p(650 − ω2)2+ 100ω21To investigate practical resonance, we need to try and maximize the amplitude C(ω)of the steady state solution. This maximum will occur when C0(ω) = 0. Solving theequation0 = C0(ω) = −50(−4ω )(650 − ω2) + 200ω((650 − ω2)2+ 100ω2)3/2,we see that ω∗= 0, 10√6 are the only solutions. After plugging in these two values,we see C(ω) must be maximized at ω∗= 10√6, and this is the practical resonancefrequency.Section 3.6, #24 [2 points] If an undamped mass-spring system is acted on by a forceF(t) = F0cos3(ωt), we will have resonance whenever F(t) contains a term of the formA cos(ω0t) or B sin( ω0t) where ω0=qkmis the natural frequency of the unforced sys-tem. Now, using the product formula cos a cos b =12(cos(a + b) + cos(a − b)) twice:cos3(ωt) = cos( ω t)cos(2ωt ) + 12=cos(3ωt ) + cos(ωt))4+cos(ωt )2=14cos(3ωt ) +34cos(ωt ),we see that resonance occurs when ω = ω0OR when 3ω = ω0.Section 3.7, #24 [1 point] We have to prove that if R, L, C are all positive, then everysolution I ofLI00+ RI0+ I/C = 0is transient. If we let m = L, c = R and k =1C, then the above equation is precisely theequation of a free damped mass spring system with mass m, dashpot constant c, andspring constant k. However, in class, we already showed that all possible solutions form, k, c > 0 are transient (i.e., tend to 0 as t → ∞). Thus we are done.Note: One can also prove this directly by considering the roots of the characteristicpolynomial of the above equation and showing that they always have negative real parts.Section 4.1, #8 [2 points] Letx1= x, x2= x0, y1= y, y2= y0.Then the given pair of second order equations becomesx01= x2x02= −3x2−4x1+ 2y1y01= y2y02= −2y2+ 3x1− y1+ cos
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