Math 211 Math 211 Lecture #32 Higher Order Equations Harmonic - PowerPoint PPT Presentation

1 Math 211 Math 211 Lecture #32 Higher Order Equations Harmonic Motion April 6, 2001

2 Solutions to y ′′ + py ′ + qy = 0 . Solutions to y ′′ + py ′ + qy = 0 . • Equivalent system: x ′ = A x , where � � � � y 0 1 x = A = . and y ′ − q − p • Look for exponential solutions y ( t ) = e λt . • Characteristic equation: λ 2 + pλ + q = 0 . • Characteristic polynomial: λ 2 + pλ + q . • Same for the 2 nd order equation and the system. • Need two linearly independent solutions. Return

3 Real Roots Real Roots • If λ is a root to the characteristic polynomial then y ( t ) = e λt is a solution. ⋄ y ′′ + 4 y ′ + 3 y = 0 • If λ is a root to the characteristic polynomial of multiplicity 2, then y 1 ( t ) = e λt and y 2 ( t ) = te λt are linearly independent solutions. ⋄ y ′′ + 4 y ′ + 4 y = 0 Return

4 Complex Roots Complex Roots • If λ = α + iβ is a complex root of the characteristic equation , then so is λ = α − iβ . • A complex valued fundamental set of solutions is z ( t ) = e λt z ( t ) = e λt . and • A real valued fundamental set of solutions is x ( t ) = e αt cos βt y ( t ) = e αt sin βt. and ⋄ y ′′ + 4 y ′ + 8 y = 0 Return

5 Examples Examples • y ′′ − 5 y ′ + 6 y = 0 . • y ′′ + 25 y = 0 . • y ′′ + 4 y ′ + 13 y = 0 . Return Real roots Complex roots

6 The Vibrating Spring The Vibrating Spring Newton’s second law: ma = total force. • Forces acting: ⋄ Gravity mg . ⋄ Restoring force R ( x ) . ⋄ Damping force D ( v ) . ⋄ External force F ( t ) . Return

7 • Newton’s law becomes ma = mg + R ( x ) + D ( v ) + F ( t ) • Hooke’s law: R ( x ) = − kx. k > 0 is the spring constant. • Spring-mass equilibrium x 0 = mg/k. • Set y = x − x 0 . Newton’s law becomes my ′′ = − ky + D ( y ′ ) + F ( t ) . Return

8 • Damping force D ( y ′ ) = − µy ′ . • Newton’s law becomes my ′′ = − ky − µy ′ + F ( t ) , or my ′′ + µy ′ + ky = F ( t ) , or y ′′ + µ my ′ + k my = 1 mF ( t ) . Return Vibrating spring

9 RLC Circuit RLC Circuit L I + E C − I R LI ′′ + RI ′ + 1 C I = E ′ ( t ) , or I ′′ + R LC I = 1 1 L I ′ + LE ′ ( t ) . Return Vibrating spring equation

10 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

11 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

12 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

13 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

14 Amplitude and Phase Amplitude and Phase • Put C 1 and C 2 in polar coordinates: C 1 = A cos φ, & C 2 = A sin φ. • Then x ( t ) = C 1 cos ω 0 t + C 2 sin ω 0 t = A cos( ω 0 t − φ ) . � C 2 1 + C 2 • A is the amplitude ; A = 2 . • φ is the phase ; tan φ = C 2 /C 1 . Return

15 Examples Examples • C 1 = 3 , C 2 = 4 ⇒ A = 5 , φ = 0 . 9273 . • C 1 = − 3 , C 2 = 4 ⇒ A = 5 , φ = 2 . 2143 . • C 1 = − 3 , C 2 = − 4 ⇒ A = 5 , φ = − 2 . 2143 . Return Amplitude & phase

16 Example Example x ′′ + 16 x = 0 , x (0) = − 2 & x ′ (0) = 4 • Natural frequency: ω 2 0 = 16 ⇒ ω 0 = 4 . • General solution: x ( t ) = C 1 cos 4 t + C 2 sin 4 t. • IC: − 2 = x (0) = C 1 , and 4 = x ′ (0) = 4 C 2 . • Solution x ( t ) = − 2 cos 2 t + sin 2 t √ = 5 cos(2 t − 2 . 6779) . Return Amplitude & phase

17 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

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

19 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

20 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