Math 211 Math 211 Lecture #35 Forced Harmonic Motion November 18, - PDF document

1 Math 211 Math 211 Lecture #35 Forced Harmonic Motion November 18, 2002 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, where ω � = ω 0 • The solution with initial conditions x (0) = x ′ (0) = 0 : 0 − ω 2 [cos ωt − cos ω 0 t ] = A sin δt A x ( t ) = sin ωt, ω 2 2 ωδ where ω = ω 0 + ω δ = ω 0 − ω and . 2 2 • This is a fast oscillation at frequency ω , with amplitude oscillating slowly with frequency δ . � This phenomenon is called “beats.” Return 1 John C. Polking

4 Forced Undamped Motion (cont.) Forced Undamped Motion (cont.) y ′′ + ω 2 0 y = A cos ω 0 t, where ω = ω 0 • An exceptional case. • Solution with initial conditions x (0) = x ′ (0) = 0 : A x p ( t ) = t sin ω 0 t. 2 ω 0 • The ouput is an oscillation with increasing amplitude. • First example of resonance. � Forcing at the natural frequency can cause oscillations that grow out of control. Return ω � = ω 0 5 Forced, Damped Harmonic Motion Forced, Damped Harmonic Motion x ′′ + 2 cx ′ + ω 2 0 x = A cos ωt Use the complex method. • Solve z ′′ + 2 cz ′ + ω 2 0 z = Ae iωt . • We try z ( t ) = ae iωt and get z ′′ + 2 cz ′ + ω 2 0 z = [( iω ) 2 + 2 c ( iω ) + ω 2 0 ] ae iωt , = P ( iω ) z where P ( λ ) = λ 2 + 2 cλ + ω 2 0 is the characteristic polynomial. 1 P ( iω ) Ae iωt . • The complex solution is z ( t ) = • The real solution is x p ( t ) = Re( z ( t )) . Return 6 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 e 3 it • 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 2 John C. Polking

7 The Transfer Function The Transfer Function • The complex solution is 1 P ( iω ) Ae iωt = H ( iω ) Ae iωt , z ( t ) = 1 where H ( iω ) = P ( iω ) is called the transfer function . • We will use complex polar coordinates to write H ( iω ) = G ( ω ) e − iφ ( ω ) , where G ( ω ) = | H ( iω ) | is the called the gain and φ ( ω ) is called the phase shift . Return 8 The Gain and Phase Shift The Gain and Phase Shift • If P ( λ ) = λ 2 + 2 cλ + ω 2 0 is the characteristic polynomial, then P ( iω ) = Re iφ , where � 0 − ω 2 ) 2 + 4 c 2 ω 2 , ( ω 2 R = and � ω 2 0 − ω 2 � φ = arccot . 2 cω • The transfer function is 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 Return P ( iω ) 9 • 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 ) . � The amplitude of x p is G ( ω ) A , and the phase is φ . Return Transfer function Differential equation 3 John C. Polking

10 • The general solution is x ( t ) = x p ( t ) + x h ( t ) = G ( ω ) A cos( ωt − φ ) + x h ( t ) , where x h ( t ) is the general solution of the homogeneous equation. • x h ( t ) → 0 as t increases, so x h is called the transient term . • x p ( t ) = G ( ω ) A cos( ωt − φ ) is called the steady-state solution. Return Particular solution 11 Example Example x ′′ + 5 x ′ + 4 x = 50 cos 3 t 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 12 The Steady-State Solution The Steady-State Solution x p ( t ) = G ( ω ) A cos( ωt − φ ) . • The forcing function is A cos ωt . • Properties of the steady-state response: � It is oscillatory at the driving frequency. � The amplitude is the product of the gain, G ( ω ) , and the amplitude of the forcing function. � It has a phase shift of φ with respect to the forcing function. Return Steady-state solution Transfer 4 John C. Polking

13 The Gain The Gain 1 G ( ω ) = 0 − ω 2 ) 2 + 4 c 2 ω 2 � ( ω 2 Set ω = sω 0 and c = Dω 0 / 2 (or s = ω/ω 0 and D = 2 c/ω 0 ). Then G ( ω ) = 1 1 ω 2 (1 − s 2 ) 2 + D 2 s 2 � 0 Gain & phase 5 John C. Polking