Vibrations: Two Degrees of Freedom Systems - Wilberforce Pendulum and Bode Plots

Transcription

1 .003J/.053J Dynaics and Control, Spring 007 Professor Peacock 5/4/007 Lecture 3 Vibrations: Two Degrees of Freedo Systes - Wilberforce Pendulu and Bode Plots Wilberforce Pendulu Figure : Wilberforce Pendulu. Extension of spring coupled to rotation. Fig ure by MT OCW. : ass, : oent of inertia, γ: torsional constant (rotation), k: extension, : coupling of extension and rotation Cite as: Thoas Peacock and Nicolas Hadjiconstantinou, course aterials for.003j/.053j Dynaics and Control, Spring 007. MT OpenCourseWare ( Massachusetts nstitute of Technology.

2 Focus on Spring Figure : Focus on spring. Figure by MT OCW. Force: F = kx + θ (also opposing force in opposite direction) Torque: τ = γθ + x θ and x represents the coupling for sall displaceents Equations of Motion Can use Lagrange, but we will use oentu principles. Equilibriu Points ẍ = g kx θ () θ = γθ x () kx 0 + θ 0 g = 0 γθ 0 + x 0 = 0 θ 0 = x 0 γ g x 0 = k 4 γ Cite as: Thoas Peacock and Nicolas Hadjiconstantinou, course aterials for.003j/.053j Dynaics and Control, Spring 007. MT OpenCourseWare ( Massachusetts nstitute of Technology.

3 3 Look at perturbation around (x 0, θ 0 ) Let x = x 0 + z, θ = θ 0 + φ Substitute in Equation () and () z + kz + φ + kx 0 + θ 0 = g φ + γφ + z + γθ 0 + x 0 = 0 γθ 0 + x 0 = 0 and kx 0 + θ 0 = g are restateents of the equilibriu conditions, so we can reove those ters to obtain z + kz + φ = 0 φ + γφ + z = 0. [ ]{ } [ ] { } { } 0 z k z 0 + = 0 φ γ φ 0 Mass Matrix or nertia Matrix: Stiffness Matrix: Free Response Solution [ ] 0 0 [ ] k γ Free Response: { } { } z c = cos(ωt ψ) φ c Substituting the free response solution into the syste of equations gives: [ ]{ } { } k ω c 0 γ ω = c 0 }{{} Thus: det=0 (3) (k ω )(γ ω ) ( ) = 0 Cite as: Thoas Peacock and Nicolas Hadjiconstantinou, course aterials for.003j/.053j Dynaics and Control, Spring 007. MT OpenCourseWare ( Massachusetts nstitute of Technology.

4 4 k γ ( ω )( ω ) = ( ) Define: k γ = = ω (ω ω ) = ± ω = ω + ω = ω These are two natural frequencies of oscillation. Mode Shapes ω : (k ω )c + c = 0 k ( ω )c + c = 0 (ω ω )c + c = 0 ( )c + c = 0 c c = Figure 3: f you change x, θ increases by a prescribed aount. Fixed by that ratio. Figure by MT OCW. Cite as: Thoas Peacock and Nicolas Hadjiconstantinou, course aterials for.003j/.053j Dynaics and Control, Spring 007. MT OpenCourseWare ( Massachusetts nstitute of Technology.

5 5 ω : (k ω )c + c = 0 ( )c + c = 0 c = c Figure 4: f you change x, θ increases in the other direction based on the ratio. Figure by MT OCW. General Solution for Free Response { } { } { } z = α φ cos(ω t ψ ) + α cos(ω t ψ ) α and ψ are set by initial conditions. At t = 0 s: { } { } { } { } z A ż 0 = = φ 0 φ 0 Stretch spring by a distance A fro resting point { } { } { } A = α cos(ψ ) + α cos(ψ ) (4) 0 { } { } { } 0 = ω α sin(ψ ) + ω α sin(ψ ) (5) 0 Using Equation (5): ω α sin ψ + ω α sin ψ = 0 Cite as: Thoas Peacock and Nicolas Hadjiconstantinou, course aterials for.003j/.053j Dynaics and Control, Spring 007. MT OpenCourseWare ( Massachusetts nstitute of Technology.

6 6 ω α sin ψ ω α sin ψ = 0 ψ = ψ = 0 Using Equation (4) with ψ = ψ = 0: A α = α = { } { } { } z A = A cos ω t + cos ω t φ α and α are the sae. Means the two odes are weighted equally. The equal weighting iplies that when the spring oscillates there is accopanying rotation. Soeties the spring-ass will be still; soeties there will be spring extension without rotation; soeties there will be rotation (both directions) with no spring extension. Analysis For Weak Spring Extension-Rotation Coupling cos α + cos β = cos( α + β )cos( α β ) cos α cos β = sin( α + β )sin( α β ) A ω + ω ω ω z = cos( t)cos( t) << : Weak Coupling ω = ω + = ω + ω ω = ω + 4ω ω = ω 4ω Therefore: z = A cos(ω t)cos( t) 4ω φ = A sin(ω t)sin( t) 4ω Cite as: Thoas Peacock and Nicolas Hadjiconstantinou, course aterials for.003j/.053j Dynaics and Control, Spring 007. MT OpenCourseWare ( Massachusetts nstitute of Technology.

7 Bode Plots 7 Figure 5: Exaple of beating. The rotation is out of phase with the extension. Figure by MT OCW. Beating Bode Plots Response of a degree of freedo syste Figure 6: Cart with spring and dashpot attached. Figure by MT OCW. x = F 0 /k ω ( ) (ζ ω ) Pole: Value of ω that sets denoinator to 0. ωn ω n Cite as: Thoas Peacock and Nicolas Hadjiconstantinou, course aterials for.003j/.053j Dynaics and Control, Spring 007. MT OpenCourseWare ( Massachusetts nstitute of Technology.

8 Bode Plots 8 Zeros: Values of ω that set to nuerator to 0. (None in this exaple) x Express the agnification M = F 0/k in decibels (db). 0 log 0 M = decibels of M. Sall frequency (ω = 0) 0 log 0 = 0 ( ) High frequency ω ω >> 0 log 0 ( ) = 0 log = 40 log ω -40 db/decade of frequency ω n ω /ωn ω n ω n Figure 7: Bode plot of odeled syste. Figure by MT OCW. With little daping in syste, where is axiu? A little below natural frequency ω n. For ore inforation, see Cite as: Thoas Peacock and Nicolas Hadjiconstantinou, course aterials for.003j/.053j Dynaics and Control, Spring 007. MT OpenCourseWare ( Massachusetts nstitute of Technology.