美国麻省理工大学：Dynamics（动力工程学）：d34

Lecture D34 : Coupled Oscillators Spring-Mass System (Undamped/Unforced) Force on m 1 : F 1 = ?k 1 x 1 ?k 2 (x 1 ?x 2 ) Force on m 2 : F 2 = k 2 (x 1 ?x 2 ) Newton’s Second Law ¨ m 1 x 1 = ?k 1 x 1 ?k 2 (x 1 ?x 2 ) ¨m 2 x 2 = k 2 (x 1 ?x 2 ) 1 ? ? ? ? ? ? Solution Try solution of the form: x 1 (t) = X 1 sin(ωt+φ), x 2 (t) = X 2 sin(ωt+φ) and see if we can ?nd X 1 , X 2 , ω and φ such that the equation is satis?ed (m 1 ω 2 ?k 1 ?k 2 )X 1 + k 2 X 2 = 0 k 2 X 1 + (m 2 ω 2 ?k 2 )X 2 = 0 or, X 1 0 [A] = X 2 0 where, m 1 ω 2 ?k 1 ?k 2 k 2 [A] = k 2 m 2 ω 2 ?k 2 2 Natural Frequencies For non-trivial solutions we need D(ω) = det[A] = 0 D(ω) := m 1 m 2 ω 4 ?(m 1 k 2 + m 2 (k 1 + k 2 ))ω 2 + k 1 k 2 We can solve for ω 2 , m 1 k 2 + m 2 (k 1 + k 2 )± √ Δ ω 2 = ± 2m 1 m 2 where, Δ = (m 1 k 2 + m 2 (k 1 + k 2 )) 2 ?4m 1 m 2 k 1 k 2 We can show Δ > 0? √ Δ < m 1 k 2 + m 2 (k 1 + k 2 )? . . . ω are real ± 3 ? ? ? Normal Modes X 1 and X 2 will satisfy (m 1 ω 2 1 + k 2 X ± = 0 ± ?k 1 ?k 2 )X ± 2 k 2 X 1 + (m 2 ω 2 2 = 0 ± ?k 2 )X ± . . . the same equation!! X 2 ± m 1 ω 2 ± ?k 1 ?k 2 γ ±δ= =? X 1 k 2 ? ? 2 m 1 m 1 k 1 + k 2 γ = 2m 2 ? k 1 + k 2 , δ = 2m 2 ? 2k 2 + m 1 k 2 2k 2 m 2 4 2 General Solution Unforced Problem 1 sin(ω ? t + φ + )+ X + x 1 (t) = X ? 1 sin(ω + t + φ + ) 2 sin(ω ? t + φ + )+ X + x 2 (t) = X ? 2 sin(ω + t + φ + ) Constants X ± , φ to be determined as a 1 ± function of initial conditions x 1 (0), x 2 (0), x˙ 1 (0) and ˙x 2 (0) Note that X ± 12 are determined once X ± are known 5 Forced Oscillation Spring-Mass System (Undamped) F 1 = ?k 1 x 1 ?k 2 (x 1 ?x 2 )+ F(t) F 2 = k 2 (x 1 ?x 2 ) Newton’s Second Law ¨ m 1 x 1 = ?k 1 x 1 ?k 2 (x 1 ?x 2 )+ F 0 sin ωt ¨m 2 x 2 = k 2 (x 1 ?x 2 ) 6 Solution p c p x 1 (t) = x 1 c (t)+ x 1 (t), x 2 (t) = x 2 (t)+ x 2 (t) c x 1 (t) and x 2 c (t) are solutions of the unforced problem and we already know p p We try a particular solution, x 1 (t) and x 2 (t) of the form p p 1 sin ωt, x 2 (t) = X p x 1 (t) = X p 2 sin ωt (m 1 ω 2 ?k 1 ?k 2 )X p 1 + k 2 X p = F 0 2 1 + (m 2 ω 2 ?k 2 )X p = 0k 2 X p 2 1 and X p We can solve for X p 2 if D(ω) =negationslash 0 7 negationslash negationslash ± Solution (cont’d) X p = (?m 2 ω 2 + k 2 )F 0 , X p = k 2 F 0 1 D(ω) 2 D(ω) Recall ? )(ω 2 ?ω 2 D(ω) = m 1 m 2 (ω 2 ?ω 2 + ) D(ω) = 0 implies that ω = ω Response for m 1 = m 2 , k 1 = k 2 5 4 3 2 1 0 1 2 3 4 5 0 0.5 1 1.5 2 2.5 p p 8 Courtesy of Prof. Martinez-Sanchez. Used with permission. Courtesy of Prof. Martinez-Sanchez. Used with permission. Courtesy of Prof. Martinez-Sanchez. Used with permission. Courtesy of Prof. Martinez-Sanchez. Used with permission. Courtesy of Prof. Martinez-Sanchez. Used with permission. Courtesy of Prof. Martinez-Sanchez. Used with permission. Courtesy of Prof. Martinez-Sanchez. Used with permission.