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