Governing Equation 1.003J/1.053J Dynamics and Control I, Spring 007 Proessor Peacoc 5/7/007 Lecture 1 Vibrations: Second Order Systems - Forced Response Governing Equation Figure 1: Cart attached to spring and dashpot subject to orce, F (t). Figure by MIT OCW. mẍ + cẋ + x = F (t) ζ: Damping Ratio : Natural Frequency ẍ + ζ ẋ + x = F (t) (1) m Forced Response - Particular Solution x p (t) Can use Fourier Series or Laplace Transorms Start with a simple case F (t) = =constant Cite as: Thomas Peacoc and Nicolas Hadjiconstantinou, course materials or.003j/1.053j Dynamics and Control I, Spring 007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute o Technology.
F (t) is constant The complementary solution below requires ζ < 1. x c = Ce ζωnt cos(ω d φ) Subscript c or complementary solution. x p =? Try x = At + B. A = 0, B = ( mω n ) = m = ζ A + (At + B) = m Thereore: x = Ce ζωnt cos(ω d t φ) + x c = Ce ( ζωnt) cos(ω d t φ): unnown constants set by initial conditions x p = : determined by orcing; independent o initial conditions Calculating C and φ x(0) = C cos( φ) + = 0 () ẋ(0) = ζ C cos( φ) + Cω d sin φ = 0 (3) The example initial conditions are x(0) = 0, ẋ(0) = 0 Equation (3) gives tan(φ) = ζωn. ω d 1 ζ ω d + ζ (1 ζ ) + ζ 1 cos φ = 1 + tan (φ) = 1 + ωd = = (1 ζ = ) 1 ζ ω d 1 C = 1 ζ Cite as: Thomas Peacoc and Nicolas Hadjiconstantinou, course materials or.003j/1.053j Dynamics and Control I, Spring 007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute o Technology.
3 Complete Solution As t, x = x p [ e ζωnt ] x = 1 cos(ω d t φ) 1 ζ Figure : Solution to dierential equation. Figure by MIT OCW. What actually happens is set by ζ and. x p can be thought o as the steady state response once the transients die down. So we will now ocus on the steady state response. O particular interest is the requency response (i.e. response amplitude and phase as a unction o orcing requency. F (t) is a periodic unction mẍ + cẋ + x = F 0 cos ωt (4) d (T + V ) = (mẍ + x)ẋ = (F (t) Cẋ)ẋ dt In steady state < F (t) x >=< c x >. x p =? Could choose sine and cosine, but use complex exponentials. Easier to wor with phases. Convenient to write and solve or: { } F = Re F 0 e iωt x p = Re { Xe iωt} Cite as: Thomas Peacoc and Nicolas Hadjiconstantinou, course materials or.003j/1.053j Dynamics and Control I, Spring 007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute o Technology.
4 X is a complex number. Substitute in Equation (4). ( mω + ciω + )Xe (iωt) = F 0 e (iωt) F 0 F 0 / X = = [ ] mω + icω 1 ω + iζ ω ωn X : Amplitude e iφ : In phase or out o phase? X = X e iφ With complex numbers, bring complex part to numerator instead o denomina tor. Multiply by complex conjugate. X = X e iφ = ω ω 1 iζ 1 ω iζ ω ωn ω n ( ) ( ) 1 ω + ζ ω F 0 ω n { } { } x p (t) = Re Xe iωt = Re X e iφ e iωt = X cos(ωt φ) ( ) ( ) 1 ω + ζ F ω 0 ωn X = ( ) ( ) 1 ω + ζ ω ω n F 0 1 X = ( ) ( ) 1 ω + ζ ω ω n Cite as: Thomas Peacoc and Nicolas Hadjiconstantinou, course materials or.003j/1.053j Dynamics and Control I, Spring 007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute o Technology.
5 φ =? Ratio o real and imaginary parts. Figure 3: Determining φ using the real and imaginary parts o the solution. Figure by MIT OCW. This diagram corresponds to e iφ. Analysis For ω 0 tan φ = (Forcing Frequency 0) System acts as i it is at steady state. ζ ω 1 ω ω n X = F 0, φ = 0 or π. φ = π is not physically meaningul. Cite as: Thomas Peacoc and Nicolas Hadjiconstantinou, course materials or.003j/1.053j Dynamics and Control I, Spring 007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute o Technology.
6 Analysis For ω I one orces the system too ast, system cannot respond. X 0, lim ω tan φ = 0 Figure 4: φ = π. Approaching 0 rom a negative number so φ = π. System is completely out o phase. Cart moves in opposite direction rom orcing. Figure by MIT OCW. Analysis For ω = Also true or ζ << 1. F 0 / X = = ζ X static ζ φ π π. We start at φ = 0, then we approach tan φ so φ. mω + = 0 n ( m + ic + )Xe iωt = F 0 e iωt Just phase shit and damping: (ic )Xe iωt = F 0 e iωt The maximum requency response is not necessarily the natural requency response. To ind maximum requency response, dierentiate. d [( ω ) ( ω ) ] 1 + ζ = 0 dω Cite as: Thomas Peacoc and Nicolas Hadjiconstantinou, course materials or.003j/1.053j Dynamics and Control I, Spring 007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute o Technology.
7 Minimum o denominator max X ω max = 1 ζ. 0 < ζ. Notice ω max is less than. Figure 5: Summary graph o X vs. (ω/ ) or orced response. X starts out at 1 when (ω/ ) equals zero, and φ equals 0. Then X goes through a maximum at (ω max / ), which is less than 1. At (ω/ ) equals 1, φ equals π/, X equals F 0 /. X continues to diminish and approaches zero or large (ω/ ) and φ equal to π. The dotted line is the observed behavior when ζ = 0, which corresponds to no damping. Figure by MIT OCW. Cite as: Thomas Peacoc and Nicolas Hadjiconstantinou, course materials or.003j/1.053j Dynamics and Control I, Spring 007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute o Technology.