Math 211 Math 211 Lecture #35 Forced Harmonic Motion November 19, - PowerPoint PPT Presentation

1 Math 211 Math 211 Lecture #35 Forced Harmonic Motion November 19, 2001

2 Forced Harmonic Motion Forced Harmonic Motion Assume an oscillatory forcing term: y ′′ + 2 cy ′ + ω 2 0 y = A cos ωt • A is the forcing amplitude • ω is the forcing frequency • ω 0 is the natural frequency. • c is the damping constant. Return

3 Forced Undamped Motion Forced Undamped Motion y ′′ + ω 2 0 y = A cos ωt • Homogeneous equation: y ′′ + ω 2 0 y = 0 � General solution y ( t ) = C 1 cos ω 0 t + C 2 sin ω 0 t. • ω � = ω 0 : Particular solution A x p ( t ) = 0 − ω 2 cos ωt. ω 2 Return

4 • ω � = ω 0 � Initial conditions x (0) = x ′ (0) = 0 ⇒ A x ( t ) = 0 − ω 2 [cos ωt − cos ω 0 t ] . ω 2 � Set ω = ω 0 + ω δ = ω 0 − ω . and 2 2 x ( t ) = A sin δt sin ωt. 2 ωδ � Fast oscillation with frequency ω with amplitude oscillating slowly with frequency δ . ◮ Beats. Return

5 • ω = ω 0 y ′′ + ω 2 0 y = A cos ω 0 t. � An exceptional case. Particular solution A x p ( t ) = t sin ω 0 t. 2 ω 0 � Oscillation with increasing amplitude. � First example of resonance. ◮ Forcing at the natural frequency can cause oscillations that grow out of control. Return

6 Forced, Damped Harmonic Motion Forced, Damped Harmonic Motion x ′′ + 2 cx ′ + ω 2 0 x = A cos ωt • Homo. equation: x ′′ + 2 cx ′ + ω 2 0 x = 0 • Ch. polynomial: P ( λ ) = λ 2 + 2 cλ + ω 2 0 • Assume the underdamped case, where c < ω 0 . c 2 − ω 2 � • Roots λ = − c ± 0 = − c ± iη where � ω 2 0 − c 2 . η = • Fundamental set of solutions x 1 ( t ) = e − ct cos ηt and x 2 ( t ) = e − ct sin ηt Return

7 Inhomogeneous equation Inhomogeneous equation x ′′ + 2 cx ′ + ω 2 0 x = A cos ωt • Use the complex method. Solve z ′′ + 2 cz ′ + ω 2 0 z = Ae iωt . � Try z ( t ) = ae iωt . z ′′ + 2 cz ′ + ω 2 0 z = [( iω ) 2 + 2 c ( iω ) + ω 2 0 ] ae iωt . = P ( iω ) z � P ( iω ) = ( iω ) 2 + 2 c ( iω ) + ω 2 0 = [ ω 2 0 − ω 2 ] + 2 icω. 1 P ( iω ) Ae iωt . • Complex solution: z ( t ) = • Real solution: x p ( t ) = Re( z ( t )) . Return Previous

8 Example Example x ′′ + 5 x ′ + 4 x = 50 cos 3 t • P ( λ ) = λ 2 + 5 λ + 4 . � P ( iω ) = P (3 i ) = − 5 + 15 i 1 P ( iω ) · 50 cos 3 t • z ( t ) = = − [(cos 3 t − 3 sin 3 t ) + i (sin 3 t + 3 cos 3 t )] • x p ( t ) = Re( z ( t )) = 3 sin 3 t − cos 3 t. Return Particular solution

9 Transfer Function Transfer Function • Complex solution: 1 P ( iω ) Ae iωt = H ( iω ) Ae iωt . z ( t ) = 1 • H ( iω ) = P ( iω ) is called the transfer function . � We will write H ( iω ) = G ( ω ) e − iφ ( ω ) . ◮ G ( ω ) = | H ( iω ) | is the gain and φ is the phase . Return

10 • The characteristic polynomial P ( iω ) = Re iφ 0 − ω 2 ) 2 + 4 c 2 ω 2 � R = � ( ω 2 � ω 2 0 − ω 2 � � φ = arccot . 2 cω • Transfer Function P ( iω ) = 1 1 Re − iφ = G ( ω ) e − iφ . H ( iω ) = � The gain G ( ω ) = 1 1 R = 0 − ω 2 ) 2 + 4 c 2 ω 2 . � ( ω 2 � ω 2 0 − ω 2 � � The phase shift φ = arccot . 2 cω P ( iω ) Return

11 • The complex particular solution is z ( t ) = H ( iω ) Ae iωt = G ( ω ) e − iφ · Ae iωt = G ( ω ) Ae i ( ωt − φ ) . • The real particular solution is x p ( t ) = Re( z ( t )) = G ( ω ) A cos( ωt − φ ) . Return Transfer function Differential equation

12 • General Solution x ( t ) = x p ( t ) + x h ( t ) = G ( ω ) A cos( ωt − φ ) + e − ct [ C 1 cos ηt + C 2 sin ηt ] . • Transient term . � x h ( t ) = e − ct [ C 1 cos ηt + C 2 sin ηt ] . • Steady-state solution. � x p ( t ) = G ( ω ) A cos( ωt − φ ) . Return Homogeneous equation Particular solution

13 • Example: x ′′ + 5 x ′ + 4 x = A cos ωt 1 • G ( ω ) = and (4 − ω 2 ) 2 + 25 ω 2 � � 4 − ω 2 � φ = arccot . 5 ω � With ω = 3 , 1 √ G (3) = 10 ≈ 0 . 0632 5 φ = arccot( − 3 / 5) ≈ 2 . 1112 . � SS solution x p ( t ) = G (3) A cos(3 t − φ ) . Return Gain & phase

14 Steady-State Solution Steady-State Solution x p ( t ) = G ( ω ) A cos( ωt − φ ) . • The forcing function is A cos ωt . • The steady-state response is oscillatory. � The amplitude is G ( ω ) times the amplitude of the forcing term. � At the driving frequency. � With a phase shift of φ/ω . Return Steady-state solution Transfer

15 Gain Gain 1 G ( ω ) = 0 − ω 2 ) 2 + 4 c 2 ω 2 � ( ω 2 • Set s = ω ω = sω 0 or ω 0 c = Dω 0 D = 2 c . or 2 ω 0 Then G ( ω ) = 1 1 ω 2 (1 − s 2 ) 2 + D 2 s 2 � 0 Gain & phase