Vibrations: Second Order Systems with One Degree of Freedom, Free Response

Single Degree of Freedom System 1 2.003J/1.053J Dynamics and Control I, Spring 2007 Professor Thomas Peacock 5/2/2007 Lecture 20 Vibrations: Second Order Systems with One Degree of Freedom, Free Response Single Degree of Freedom System Figure 1: Cart attached to spring and dashpot. Figure by MIT OCW. mẍ + cẋ + kx = F (t) System response? What is x(t)? Use 18.03 Background. x(t) = Free Response � �� � Complementary Solution, when F (t)=0 This lecture will cover the Free Response. + Response Due to Forcing � �� � Particular Solution Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. Free Response 2 Free Response Look at k → 0 Figure 2: Cart with dashpot only. Figure by MIT OCW. mẍ + cẋ = 0 Assume conditions x(0) = x0 and ẋ(0) = v0 . mẍ + cẋ = mv̇ + cv = 0 v = v0 e(−ct/m) already used ẋ(0) = v0 Integrate v(t) once. Using x(0) = x0 , we obtain: x = x0 + c � mv0 � 1 − e− m t c Figure 3: Solution to differential equation. Solution attenuates to a steady state value. Figure by MIT OCW. Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. Free Response 3 Figure 4: Velocity profile of solution. Velocity attenuates to zero. Figure by MIT OCW. No oscillations. Because k = 0, there was no restoring term. Look at m → 0 cẋ + kx = 0 or k ẋ = − x c x(0) = x0 Therefore: k x(t) = x0 e− c t Figure 5: Solution to differential equation. Position decays to zero. Figure by MIT OCW. Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. Free Response 4 Figure 6: Velocity profile of solution. Value attenuates to steady state value. Figure by MIT OCW. ẋ = − kx0 − k t e c c No oscillations in this system. Dashpot force balances the spring force as x → 0, spring force → 0. Vibrations require a restoring force (e.g. spring) and inertia (e.g. mass). Full Free Response Problem So let us consider the full problem: mẍ + cẋ + kx = 0 (1) Note that cẋ(c > 0) is a damping term and is responsible for decay of oscillations. Examination of Energy � � 1 d d 1 (T + V ) = mẋ2 + kx2 = mẋẍ + kxẋ = ẋ(mẍ + kx) = ẋ(−cẋ) = −cẋ2 dt dt 2 2 For c > 0: d (T + V ) < 0 dt Damping. Mechanical energy is dissipated. For c < 0: Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. Free Response 5 d (T + V ) > 0 dt Energy input (Control system providing energy) Solution of the Equation with Engineering Quantities Rewrite mẍ + cẋ + kx = 0 as: ẍ + 2ζωn ẋ + ωn2 x = 0 (2) k m c ζ= 2mωn ωn2 = ωn : Natural Frequency ζ: Damping Ratio To solve, we assume a solution of the form x = Ae(λt) Substitute in Equation (2): λ2 + 2ζωn λ + ωn2 = 0 λ = −ζωn ± ωn 2 2 � ζ2 − 1 (3) When ζ > 1 and ζ < 1, the behavior is different. Assume c ≥ 0. (ζ ≥ 0) We have the following cases. Case 1: Overdamped ζ > 1 ⇒ λ1 , λ2 = Real Negative Numbers x = A± e “ ” √ −ζωn ± ζ 2 −1 → 0 as t → ∞ Case 2: Critically Damped ζ = 1 ⇒ λ1 , λ2 = −ωn x = (A1 + A2 t)e−ωn t → 0 as t → ∞ (4) Equation (4) is the fastest approach to the set point. That is why it is named critically damped. Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. Free Response 6 Case 3: Underdamped 0≤ζ<1 λ1 , λ2 = −ζωn ± iωd � ωd = ωn 1 − ζ 2 Underdamped (Not enough damping to prevent oscillations). When ζ → 0, ωd → ωn (Natural frequency). � � x = A1 eiωd t + A2 e−iωd t e−ζωn t Must have that A1 and A2 are complex conjugates because x is real. x =[A1 (cos ωd t + i sin ωd t) + A2 (cos ωd t − i sin ωd t)]e−ζωn t =[(A1 + A2 ) cos ωd t + i(A1 − A2 ) sin ωd t]e−ζωn t � �� � � �� � A3 A4 A1 + A2 = A3 i(A1 − A2 ) = A4 � � A4 x = A3 cos ωd t + sin ωd t e−ζωn t A3 � � x = A3 cos ωd t + tan φ sin ωd t e−ζωn t x= � � A3 cos ωd t cos φ + sin ωd t sin φ e−ζωn t cos φ Note the trigonometric identity. x(t) = Ce−ζωn t cos(ωd t − φ) (5) e−ζωn t : Decaying in time cos(ωd t − φ): Oscillatory Behavior C and φ can be found from initial conditions. C= A3 cos φ φ = arctan A4 A3 (6) (7) Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. Free Response 7 Equations (6) and (7) relate C and φ to A3 and A4 . But 1 cos2 φ = 1 + tan2 φ. 1 A2 = 1 + 42 2 cos φ A3 1 = cos φ � � A23 + A24 ⇒ C = A23 + A24 A3 If 0 ≤ ζ < 1, the solution will show decaying oscillations. How do we determine (C and φ) or (A3 and A4 )? Often easier to relate A3 and A4 to initial conditions. Initial Conditions: x(0) = x0 , ẋ(0) = v0 x = [A3 cos ωd t + A4 sin ωd t]e−ζωn t At t = 0, x0 = A3 (using x(0) = x0 ) ẋ =[−A3 ωd sin ωd t + A4 ωd cos ωd t]e−ζωn t − ζωn [A3 cos ωd t + A4 sin ωd t]e−ζωn t At t = 0: v0 = A4 ωd − ζωn A3 = A4 ωd − ζωn x0 A4 = C= � x20 v0 + ζωn x0 ωd + tan φ = � v0 + ζωn x0 ωd �2 v0 + ζωn x0 ωd x0 (8) (9) Examine solution. x(t) = Ce−ζωn t cos(ωd t − φ) Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. Free Response 8 e−ζωn t : Decay cos(ωd t − φ): Oscillating Figure 7: Solution both decays and oscillates given the presence of exponential solution and sinusoidal solution. Figure by MIT OCW. Calculate Amplitude. e−ζωn t x(t) = [−ζω (t+nτ )] = eζωn nτd n d x(t + nτd ) e � ln � 2π x(t) ωn 2π ωn 2π = nζ � = nζωn τd = nζ = nζ � x(t + nτd ) ωd ωn 1 − ζ 2 1 − ζ2 (10) For ζ << 1: ln � � x(t) = 2πnζ x(t + nτd ) (11) Need ωn , ζ to define system. Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. Free Response 9 Example Experiment: Flexible Rod. Figure 8: Flexible rod. Figure by MIT OCW. Measure frequency of oscillation: ωd . x(t) . This ratio is related Measure amplitude over several periods to obtain x(t+nτ d) to the damping ratio ζ by the equations (10) or (11) if ζ << 1. With ωd and ζ, one can calculate the natural frequency ωn . Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].