美国麻省理工大学：《动力工程学》（英文版）Lecture D34 Coupled Oscillators

Lecture D34 Coupled oscillators Spring-Mass System(Undamped/ Unforced) Wmi Mm2 orce on m1 1=-k11-k2(x1-x2) Force on m2 F2= k2(1-2) Newton's second Law 01 k11-k2(x1-x2) 02 2 k2(x1-x2)

Lecture D34 : Coupled Oscillators Spring-Mass System (Undamped/Unforced) Force on m1 : F1 = −k1x1 − k2(x1 − x2) Force on m2 : F2 = k2(x1 − x2) Newton’s Second Law m1 x¨ 1 = −k1x1 − k2(x1 − x2) m2x¨2 = k2(x1 − x2) 1

Solution Try solution of the form 1(t)=X1 sin(wt+o, 2(t=X2 sin(at+o) and see if we can find x1, X2, w and such that the equation is satisfied m010 k1-k2)X1+k2X2 k2X1+( 0,2 k2)X2 or 4) 2 where LA 2 2 2 2

� � � � � � Solution Try solution of the form: x1(t) = X1 sin(ωt+φ), x2(t) = X2 sin(ωt+φ) and see if we can find X1, X2, ω and φ such that the equation is satisfied (m1ω2 − k1 − k2)X1 + k2X2 = 0 k2X1 + (m2ω2 − k2)X2 = 0 or, X1 0 [A] = X2 0 where, m1ω2 − k1 − k2 k2 [A] = k2 m2ω2 − k2 2

Natural Frequencies For non-trivial solutions we need D(u)=de[4]=0 D(u):=m1m2u4-(m1k2+m2(k1+k2))u2+k1k2 We can solve for w2 m1k2+m2(k1+k2)士 2m1m2 Where, △=(m1k2+m2(k1+k2)2-4m1m2k1k2 We can show △<m1k2+m2(k1+k2) 士 are rea

Natural Frequencies For non-trivial solutions we need D(ω) = det[A] = 0 D(ω) := m1m2ω4 − (m1k2 + m2(k1 + k2))ω2 + k1k2 We can solve for ω2 , m1k2 + m2(k1 + k2) ± √Δ ω± 2 = 2m1m2 where, Δ = (m1k2 + m2(k1 + k2))2 − 4m1m2k1k2 We can show • Δ √ > 0 • Δ < m1k2 + m2(k1 + k2) . . . ω± are real 3

Normal modes X1 and x2 will satisfy (m1圣-k1-k2)X+k2X k2X1+(m2 kDxi the same equation!! 2 1 k1-k2 士6 k2 7sm1k+k2.6=1(2m22k2 m1 k1+k2 m21 2 m 2 2k2 X

� � � Normal Modes X1 and X2 will satisfy (m1ω2 ± − k1 − k2)X1 + k2X± = 0 ± 2 k2X1 + (m2ω2 ± − k2)X2 = 0 ± . . . the same equation!! X2 ± m1ω2 ± − k1 − k2 − = = γ ± δ X1 k2 � �2 m1 m1 k1 + k2 γ = 2m2 − k1 + k2 , δ = 2m2 − 2k2 + m1 k2 2k2 m2 4 2

Genera solution Unforced problem X1 sin(w-t+o+)+xl sin (w+t+o+) X2sin(u-t+φ+)+X2sin(u+t+φ+ Constants X,φ± to be determined as a function of initial conditions 10), x2 0) 1(0) and r2(0) Note that xo are determined once x are known

General Solution Unforced Problem 1 sin(ω−t + φ+) + X+ x1(t) = X− 1 sin(ω+t + φ+) 2 sin(ω−t + φ+) + X+ x2(t) = X− 2 sin(ω+t + φ+) Constants X± , φ to be determined as a 1 ± function of initial conditions x1(0), x2(0), x˙1(0) and ˙x2(0) Note that X± 2 1 are determined once X± are known 5

Forced oscillation Spring-Mass System (Undamped) Fosin wt ∧A F1=-k11-k2(x1-2)+F(t) 1 2 Newton's second Law 11 k1C1-k2(e1-22)t Fo sin wt 7722 k 2(1 2

Forced Oscillation Spring-Mass System (Undamped) F1 = −k1x1 − k2(x1 − x2) + F(t) F2 = k2(x1 − x2) Newton’s Second Law m1 x¨ 1 = −k1x1 − k2(x1 − x2) + F0 sin ωt m2x¨2 = k2(x1 − x2) 6

Solution x1(t)=xi(t)+x1(t),x2(t)=x2(t)+a2( af (t) and as(t) are solutions of the unforced problem and we already know We try a particular solution, i(t)and a2(t) of the form i(t)=xi sin wt, aP(t)=X, sin wt k1-k2)X1+k2 k2X1+(m2u2-k2)X=0 We can solve for Xi and x2 if D(w)+o

Solution p c p x1(t) = x1 c (t) + x1(t), x2(t) = x2(t) + x2(t) c x1(t) and x2 c (t) are solutions of the unforced problem and we already know We try a particular solution, p p x1(t) and x2(t) of the form p p 1 sin ωt, x2(t) = Xp x1(t) = Xp 2 sin ωt (m1ω2 − k1 − k2)Xp 1 + k2Xp 2 = F0 1 + (m2ω2 − k2)Xp k2X = 0 p 2 1 and X We can solve for p Xp 2 if D(ω) =6 0 7

Solution(cont'd) =(m202+k2)Fox02m2) 20 D Reca D(u)=m1m2(u2-u2)(u2-2) D(u)≠0 implies that w≠u士 Response for m1=m2,k1=k2 Xiv/E /k/

6 6 ± Solution (cont’d) Xp = (−m2ω2 + k2)F0, Xp = k2F0 1 D(ω) 2 D(ω) Recall −)(ω2 − ω D 2 (ω) = m1m2(ω2 − ω2 +) D(ω) = 0 implies that ω = ω Response for m1 = m2, k1 = k2 5 4 3 2 1 0 1 2 3 4 5 0 0.5 1 1.5 2 2.5 p p 8

COUPLED_OSCILLATORS This lecture serves a dual purpose: it introduces the concept of independent superimposed harmonic motions, and it allows consideration of simple vibration damping Simple Example With Two Coupled Oscillators Consider two masses sliding without friction on a horizontal surface, and connected by springs with no extemal forces applied. The points E1 and E2 are the equilibrium position of Mi and M2 k respectively, and displacements x1 and x2 are are measured to the right The force to the right on Fig M, in the given position is-kr-k(*, -x2), and the force on M2 is k2(x1-x2). Hence mx=一kx1一k(x-x2) (1) m2文2=k2(x1-x2) This pair of 2nd order, coupled, constant coefficient differential equations admits oscillatory solutions, which can be represented as x=Re(元e =Re(元 2e Here Re represents the real part of a complex quantity, and x,, i2 are complex numbers. The unknown frequency a is a real number for this problem, as will be proven, but in more complicated situations, it may also have an imaginary part. If the quantities 元,名 have amplitudes2 and phase angles, 元=e”;=F2 Then, from the definitions (3), Courtesy of Prof. Martinez-SancheZ. Used with permission

Courtesy of Prof. Martinez-Sanchez. Used with permission

x,=, cos(t+o, );x2=i2Icos(ar+2) and we could instead have assumed these sinusoidal functions for the solutions. However the exponential forms in(3)have definite advantages in terms of algebraic cleanliness Substituting (3)into(1)and(2), we obtain Re-m03+k元+k(元-划"}=0(6 R-m202+(一划}= Clearly, these conditions are satisfied of the two square brackets are zero (ma2-k-k2)+k元2=0 k+(mn202-k2)2=0 12_The_Natural_Frequencies An obvious solution is i=i2=0, but we are after a solution where something ens. The only other possibility is that( 8)and (9)are really the same equation, or, in erent words, that the two linear homogeneous equations (8 )and (9)be compatible. The first viewpoint would imply that the coefficients of x,, 2 are proportional in both equations k2 and the second viewpoint(compatibility) would amount to equating to zero the determinant of the coefficients, which, clearly, leads to the same eq(Eq. 10). In future, more involved problems, this second approach will be generally followed, because it is the more general of the Eq. (10) can be reduced to a 4th order Eq. in a, or, luckily in this case, a 2nd equation in @. The absence of terms in o and a can be traced back to the absence of damping terms, i. e, there is no force proportional to i, or i2. Cases with damping can also be handled by the same methods, and we will later see some examples. The equation this case is m,m,a-m, k2+m2(k+k)]o'+k, k The explicit solutions for @*are m+m1(+)士+m1(+人)-4mk m,m2 Courtesy of Prof. Martinez-Sanchez. Used with permission

Courtesy of Prof. Martinez-Sanchez. Used with permission