Math 211 Math 211 Lecture #33 Harmonic Motion Inhomogeneous - PowerPoint PPT Presentation

1 Math 211 Math 211 Lecture #33 Harmonic Motion Inhomogeneous Equations April 9, 2001

2 Harmonic Motion Harmonic Motion • Spring: y ′′ + µ m y ′ + k 1 m y = m F ( t ) . • Circuit: I ′′ + R L I ′ + LC I = 1 1 L E ′ ( t ) . • Essentially the same equation. Use x ′′ + 2 cx ′ + ω 2 0 x = f ( t ) . • The equation for harmonic motion. Return

3 x ′′ + 2 cx ′ + ω 2 0 x = f ( t ) . • ω 0 is the natural frequency. � ⋄ Spring: ω 0 = k/m. � ⋄ Circuit: ω 0 = 1 /LC. • c is the damping constant. • f ( t ) is the forcing term. Return

4 Simple Harmonic Motion Simple Harmonic Motion • No forcing , and no damping: x ′′ + ω 2 0 x = 0 • p ( λ ) = λ 2 + ω 2 0 , λ = ± iω 0 . • Fundamental set of solutions x 1 ( t ) = cos ω 0 t x 2 ( t ) = sin ω 0 t. & Return Harmonic motion

5 • General solution x ( t ) = C 1 cos ω 0 t + C 2 sin ω 0 t. ⋄ Every solution is periodic with frequency ω 0 . ⋄ ω 0 is the natural frequency. ⋄ The period is T = 2 π/ω 0 . Return

6 Damped Harmonic Motion Damped Harmonic Motion x ′′ + 2 cx ′ + ω 2 0 x = 0 • p ( λ ) = λ 2 + 2 cλ + ω 2 � c 2 − ω 2 0 ; roots − c ± 0 . • Three cases ⋄ c < ω 0 Underdamped ⋄ c > ω 0 Overdamped ⋄ c = ω 0 Critically damped Return Harmonic motion

7 Underdamped Underdamped • c < ω 0 • Two complex roots λ and λ , where λ = − c + iω 0 − c 2 . � ω 2 and ω = • General solution x ( t ) = e − ct [ C 1 cos ωt + C 2 sin ωt ] . = Ae − ct cos( ωt − φ )

8 Overdamped Overdamped • c > ω 0 , so two real roots � c 2 − ω 2 λ 1 = − c − 0 � c 2 − ω 2 λ 2 = − c + 0 . • λ 1 < λ 2 < 0 . • General solution x ( t ) = C 1 e λ 1 t + C 2 e λ 2 t . Return

9 Critically Damped Critically Damped • c = ω 0 • One negative real root λ = − c with multiplicity 2. • General solution x ( t ) = e − ct [ C 1 + C 2 t ] . Return

10 Inhomogeneous Equations Inhomogeneous Equations y ′′ + py ′ + qy = f ( t ) • Corresponding homogeneous equation y ′′ + py ′ + qy = 0 ⋄ We know how to find a fundamental set of solutions y 1 and y 2 . ⋄ The general solution of the homogeneous equation is y h ( t ) = C 1 y 1 ( t ) + C 2 y 2 ( t ) . Return

11 Assume Theorem: • y p ( t ) is a particular solution to the inhomogeneous equation y ′′ + py ′ + qy = f ( t ); • y 1 ( t ) & y 2 ( t ) is a fundamental set of solutions to the homogeneous equation y ′′ + py ′ + qy = 0 . Then the general solution to the inhomogeneous equation is y ( t ) = y p ( t ) + C 1 y 1 ( t ) + C 2 y 2 ( t ) . Return Homogeneous equation

12 Method of Undetermined Coefficients Method of Undetermined Coefficients y ′′ + py ′ + qy = f ( t ) • If the forcing term f ( t ) has a form which is replicated under differentiation, then look for a particular solution of the same general form as the forcing term. Return

13 Exponential Forcing Term Exponential Forcing Term y ′′ + py ′ + qy = Ce at • Example: y ′′ + 3 y ′ + 2 y = 4 e − 3 t • Try y p ( t ) = ae − 3 t ; a to be determined. p + 2 y p = 2 ae − 3 t y ′′ p + 3 y ′ • Particular solution if 2 a = 4 , or a = 2 . y p ( t ) = 2 e − 3 t . Return

14 • Homogeneous equation: y ′′ + 3 y ′ + 2 y = 0 ODE λ 2 + 3 λ + 2 = 0 Ch. poly. ( λ + 2)( λ + 1) = 0 • Fund. set of sol’ns: e − 2 t & e − t . • General solution to the inhomogeneous equation is y ( t ) = 2 e − 3 t + C 1 e − t + C 2 e − 2 t . Return Particular solution

15 Trigonometric Forcing Term Trigonometric Forcing Term y ′′ + py ′ + qy = A cos ωt + B sin ωt • Example: y ′′ + 4 y ′ + 5 y = 4 cos 2 t − 3 sin 2 t • Try y p ( t ) = a cos 2 t + b sin 2 t y ′′ p + 4 y ′ p + 5 y p = ( a + 8 b ) cos 2 t + ( b − 8 a ) sin 2 t. • Particular solution if a + 8 b = 4 , b − 8 a = − 3 . ⇔ a = 28 / 65 and b = 29 / 65 . Return

16 • y p ( t ) = [28 cos 2 t + 29 sin 2 t ] / 65 . • Homogeneous equation: y ′′ + 4 y ′ + 5 y = 0 ⋄ Characteristic polynomial: λ 2 + 4 λ + 5 = 0 ⋄ Roots: λ = − 2 ± i • Fund. set of sol’ns: e − 2 t cos t & e − 2 t sin t . • General solution to the inhomogeneous equation: y ( t ) = [28 cos 2 t + 29 sin 2 t ] / 65 + e − 2 t [ C 1 cos t + C 2 sin t ] . Return Particular solution

17 Complex Method Complex Method x ′′ + px ′ + qx = A cos ωt or y ′′ + py ′ + qy = A sin ωt. • Solve z ′′ + pz ′ + qz = Ae iωt . • x p ( t ) = Re( z ( t )) y p ( t ) = Im( z ( t )) . and Return

18 • Example: x ′′ + 4 x ′ + 5 x = 4 cos 2 t • Solve z ′′ + 4 z ′ + 5 z = 4 e 2 it . • Try z ( t ) = ae 2 it . z ′′ + 4 z ′ + 5 z = (1 + 8 i ) ae 2 it • Particular solution if (1 + 8 i ) a = 4 or 1 + 8 i = 4(1 − 8 i ) 4 = 4 − 32 i a = . 1 + 64 65 Return

19 • Particular solution z ( t ) = (4 − 32 i ) e 2 it / 65 = (4 − 32 i )[cos 2 t + i sin 2 t ] / 65 = [4 cos 2 t + 32 sin 2 t ] / 65 + i [4 sin 2 t − 32 cos 2 t ] / 65 . x p ( t ) = Re( z ( t )) = [4 cos 2 t + 32 sin 2 t ] / 65 . Return Previous

20 Polynomial Forcing Term Polynomial Forcing Term y ′′ + py ′ + qy = P ( t ) • Example: y ′′ − 3 y ′ + 2 y = 1 − 4 t. • Try y ( t ) = a + bt. y ′′ − 3 y ′ + 2 y = ( a − 3 b ) + 2 bt. • Particular solution if a − 3 b = 1 b = − 2 or 2 b = − 4 a = − 5 Return

21 • Particular solution y ( t ) = − 5 − 2 t. • General solution y ( t ) = − 5 − 2 t + C 1 e t + C 2 e 2 t . Return

22 Exceptional Cases Exceptional Cases • Example: y ′′ − 3 y ′ + 2 y = 3 e t . • Try y ( t ) = ae t y ′′ − 3 y ′ + 2 y = 0 . • The method does not work because e t is a solution to the associated homogeneous equation. Return

23 • Try y ( t ) = ate t y ′′ − 3 y ′ + 2 y = − ae t • Particular solution if a = − 3 . • General solution y ( t ) = − 3 te t + C 1 e t + C 2 e 2 t . • If the suggested solution does not work, multiply it by t and try again. Previous

24 Combination Forcing Term Combination Forcing Term Example y ′′ + 5 y ′ + 6 y = 2 e 2 t − 5 cos t • Solve 1 + 6 y 1 = 2 e 2 t y ′′ 1 + 5 y ′ y ′′ 2 + 5 y ′ 2 + 6 y 2 = − 5 cos t • Set y ( t ) = y 1 ( t ) + y 2 ( t ) . Theorem Previous UDC