LECTURE 17: FORCED OSCILLATIONS AND RESONANCE MINGFENG ZHAO October 15, 2014 Mass-Spring System: Figure 1. Mass-Spring System Let x(t) be the displacement of the mass, then mx00 + cx0 + kx = F (t) . We are interested in periodic forcing, that is, F (t) = F0 cos(ωt). So we have mx00 + cx0 + kx = F0 cos(ωt) . Rewrite the equation, we have x00 + 2px0 + ω02 x = F0 cos(ωt) , m where c p= ≥ 0, 2m r and ω0 = 1 k > 0. m 2 MINGFENG ZHAO Undamped forced motion and resonance The differential equation for the undamped forced motion (c = 0) is: x00 + ω02 x = F0 cos(ωt). m The characteristic equation of x00 + ω02 x = 0 is: r2 + ω02 = 0. Solve r2 + ω02 = 0, we get r1 = ω0 i, and r2 = −ω0 i. The general solution to x00 + ω02 x = 0 is: x(t) = A cos(ω0 t) + B sin(ω0 t) . F0 cos(ωt), we have two cases: m F0 I. If ω = 6 ω0 , a particular solution to x00 + ω02 x = cos(ωt) can be: m Since the nonhomogeneous function is xp (t) = F0 cos(ωt) . m[ω02 − ω 2 ] II. If ω = ω0 . Let xp (t) = Dt cos(ω0 t) + Et sin(ω0 t) be a particular solution to x00 + ω02 x = F0 cos(ω0 t), then m x0p (t) = D cos(ω0 t) − Dω0 t sin(ω0 t) + E sin(ω0 t) + Eω0 t cos(ω0 t) x00p (t) = −Dω0 sin(ω0 t) − Dω0 sin(ω0 t) − Dω02 t cos(ω0 t) + Eω0 cos(ω0 t) + Eω0 cos(ω0 t) − Eω02 t sin(ω0 t) = −2Dω0 sin(ω0 t) − Dω02 t cos(ω0 t) + 2Eω0 cos(ω0 t) − Eω02 t sin(ω0 t) x00 + ω02 x = −2Dω0 sin(ω0 t) − Dω02 t cos(ω0 t) + 2Eω0 cos(ω0 t) − Eω02 t sin(ω0 t) +ω02 [Dt cos(ω0 t) + Et sin(ω0 t)] = −2Dω0 sin(ω0 t) + 2Eω0 cos(ω0 t) = F0 cos(ω0 t). m Then we have D = 0, and E = F0 . 2mω0 LECTURE 17: FORCED OSCILLATIONS AND RESONANCE So a particular solution to x00 + ω02 x = F0 cos(ω0 t) can be: m xp (t) = In summary, the general solution to x00 + ω02 x = x(t) = 3 F0 t sin(ω0 t) . 2mω0 F0 cos(ω0 t) is: m A cos(ω0 t) + B sin(ω0 t) + F0 cos(ωt), if ω0 6= ω, m[ω02 − ω 2 ] A cos(ω0 t) + B sin(ω0 t) + F0 t sin(ω0 t), 2mω0 . if ω0 = ω F0 F0 t and t, in particular, the 2mω0 2mω0 amplitude of x(t) will go to infinity. This kind of behavior is called resonance or perhaps pure resonance. In the case that ω0 = ω, when t → ∞, the graph of x(t) oscillates between − Figure 2. Graph of 1 t sin(πt) π Damped forced motion and practical resonance The differential equation for the damped forced motion (c > 0) is: x00 + 2px0 + ω02 x = F0 cos(ωt). m The characteristic equation of x00 + 2px0 + ω02 x = 0 is: r2 + 2pr + ω02 = 0. Solve r2 + 2pr + ω02 = 0, we have r1,2 = −p ± q p2 − ω02 . 4 MINGFENG ZHAO The general solution to x00 + 2px0 + ω02 x = 0 is: Aer1 t + Ber2 t , overdamping, that is, p2 − ω02 > 0 Ae−pt + Bte−pt , critical damping, that is, p2 − ω02 = 0 . xc (t) = q q Ae−pt cos ω02 − p2 · t + Be−pt sin ω02 − p2 · t , underdamping, that is, p2 − ω02 < 0 In any case, since p 6= 0, then xc (t) → 0 as t → ∞. Since ωi is not a solution to r2 + 2pr + ω02 = 0, then we can let F0 xp (t) = D cos(ωt) + E sin(ωt) be a particular solution to x00 + 2px0 + ω02 x = cos(ωt), then m x0p (t) = −Dω sin(ωt) + Eω cos(ωt) x00p (t) = −Dω 2 cos(ωt) − Eω 2 sin(ωt) x00p (t) + 2px0p (t) + ω02 xp (t) = −Dω 2 cos(ωt) − Eω 2 sin(ωt) + 2p [−Dω sin(ωt) + Eω cos(ωt)] +ω02 [D cos(ωt) + E sin(ωt)] = −Dω 2 + 2pEω + Dω02 cos(ωt) + −Eω 2 − 2pDω + Eω02 sin(ωt) = F0 cos(ωt). m Then we have −Dω 2 + 2pEω + Dω02 = F0 , m and − Eω 2 − 2pDω + Eω02 = 0. Solve D and E, we get D= F0 [ω02 − ω 2 ] , m [(2ωp)2 + (ω02 − ω 2 )2 ] So a particular solution to x00 + 2px0 + ω02 x = xp (t) = and E = 2ωpF0 . + (ω02 − ω 2 )2 ] m [(2ωp)2 F0 cos(ωt) is: m F0 [ω02 − ω 2 ] 2ωpF0 · cos(ωt) + sin(ωt) . 2 2 2 2 2 m [(2ωp) + (ω0 − ω ) ] m [(2ωp) + (ω02 − ω 2 )2 ] The general solution xc to x00 + 2px0 + ω02 x = 0 is called the transient solution, denoted by xtr , the particular solution F0 xp found in above to x00 + 2px0 + ω02 x = cos(ωt) is called the steady periodic solution, denoted by xsp , which is a m periodic function with frequency ω. Since xtr (t) → 0 as t → ∞, so for any given initial data, at infinity, the solution will be close the the steady periodic solution xsp (t). LECTURE 17: FORCED OSCILLATIONS AND RESONANCE 5 Figure 3. Solutions with differential initial data for k = m = F0 = 1, c = 0.7 and ω = 1.1 For the steady periodic solution: xsp (t) = F0 [ω02 − ω 2 ] 2ωpF0 · cos(ωt) + sin(ωt). m [(2ωp)2 + (ω02 − ω 2 )2 ] m [(2ωp)2 + (ω02 − ω 2 )2 ] we know that I. xsp (t) is a periodic function II. xsp (t) has frequency ω, and the period 2π . ω III. The amplitude of xsp is: s C(ω) = F0 [ω02 − ω 2 ] m [(2ωp)2 + (ω02 − ω 2 )2 ] = 1 F0 ·p . 2 m (2ωp) + (ω02 − ω 2 )2 2 + 2ωpF0 2 m [(2ωp) + (ω02 − ω 2 )2 ] Notice that C 0 (ω) Then = 1 2(2ωp) · 2p + 2(ω02 − ω 2 ) · (−2ω) F0 · − · 3 m 2 [(2ωp)2 + (ω02 − ω 2 )] 2 = F0 2ω[ω02 − ω 2 − 2p2 ] · m [(2ωp)2 + (ω02 − ω 2 )] 23 = 2F0 −ω[ω 2 − (ω02 − 2p2 )] · . m [(2ωp)2 + (ω02 − ω 2 )] 23 2 6 MINGFENG ZHAO – If ω02 − 2p2 ≤ 0, then C(ω) has the maximum value at ω = 0. But we assume that ω > 0, so C(ω) can not attain its maximum, that is, C(ω) < C(0) = F0 , mω02 ∀ω > 0. p – If ω02 − 2p2 > 0, then C(ω) has the maximum value at ω = ω02 − 2p2 , that is, q 2F0 F0 1 =p max C(ω) = C( ω02 − 2p2 ) = ·p 2 2 2 2 m c [4k 2 − c2 ] 4p [ω0 − p ] In this case, ω = p ω02 − 2p2 is called the practical resonance frequency, max C(ω) = F0 1 ·p 2 m 4p [ω02 − p2 ] is called the practical resonance amplitude. Figure 4. Graph of C(ω) for k = m = F0 = 1 with c = 0.4, 0.8 and 1.6 Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C. Canada V6T 1Z2 E-mail address: mingfeng@math.ubc.ca