MT-1620 al.2002 For each eigenvalue, the homogeneous solution is e sina. t coSO. t homogeneous Still an undetermined constant in each case(An ) which can be determined from the initial conditions Each homogeneous solution physically represents a possible free vibration mode Arrange natural frequencies from lowest (o, to highest (o By superposition, any combinations of these is a valid solution Example: Two mass system(from Unit 19) Figure 22.1 Representation of dual spring-mass system Paul A Lagace @2001 Unit 22-6
MIT - 16.20 Fall, 2002 For each eigenvalue, the homogeneous solution is: r r r q i hom = φi () e iω r t = C1 φi () sinωr t + C2 φi () cosωr t ~ ~ ~ ~ homogeneous Still an undetermined constant in each case (An) which can be determined from the Initial Conditions • Each homogeneous solution physically represents a possible free vibration mode • Arrange natural frequencies from lowest (ω1) to highest (ωn) • By superposition, any combinations of these is a valid solution Example: Two mass system (from Unit 19) Figure 22.1 Representation of dual spring-mass system Paul A. Lagace © 2001 Unit 22 - 6
MT-1620 al.2002 The governing equation was m. 0 +)-k]91F 0m22 Thus, from equation(22-5) (k+k)-02m-k2 k k2-0 This gives: (k+k)-a m, k2 0m,一 k2=0 This leads to a quadratic equation in (2. Solving gives two roots(o 1 2 and 022)and the natural frequencies are o, and O2 Find the associated eigenvectors in terms of A2(i.e normalized by a2) Paul A Lagace @2001 Unit 22-7
q q MIT - 16.20 Fall, 2002 The governing equation was: m1 0 ˙˙1 (k1 + k2 ) −k2 q1 F1 0 m2 ˙˙2 + = −k2 k2 q2 F2 Thus, from equation (22-5): (k1 + k2 ) − ω2 m1 −k2 = 0 2 −k2 k2 − ω m2 This gives: [(k1 + k2 ) − ω2 m1 ][k2 − ω2 m2 ] − k22 = 0 This leads to a quadratic equation in ω2. Solving gives two roots (ω 12 and ω22) and the natural frequencies are ω1 and ω2 Find the associated eigenvectors in terms of A2 (i.e. normalized by A2) Paul A. Lagace © 2001 Unit 22 - 7
MT-1620 a.2002 Go back to equation (22-4)and divide through by A2 421「41 k Normalized constant k for o. mode k+k2-0n2m1 Thus the eigenvectors are k 2 k k2-02m 2)|k+k2-m2mn1 For the case of Initial Conditions of 0, the cos term goes away and are left with sin w. t Physically the modes are Paul A. Lagace @2001 Unit 22-8
MIT - 16.20 Fall, 2002 Go back to equation (22-4) and divide through by A2: (k1 + k2 ) − ωr 2 m1 −k2 A1 = 0 −k2 k2 − ωr 2 m2 1 Normalized constant k ⇒ A1 = k1 + k2 2 − ωr 2 m1 for ωr mode Thus the eigenvectors are: k2 k2 1 k1 + k2 − ω12 m1 φ() k1 + k2 − ω22 2 m1 φi () = i = ~ ~ 1 1 For the case of Initial Conditions of 0, the cos term goes away and are left with… r q t() = φ() sin ωr t ~ ~ Physically the modes are: Paul A. Lagace © 2001 Unit 22 - 8