Math 211 Math 211 Lecture #32 Harmonic Motion November 10, 2003 - PowerPoint PPT Presentation

1 Math 211 Math 211 Lecture #32 Harmonic Motion November 10, 2003

2 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 ) . • Including all of the forces, Newton’s law becomes ma = mg + R ( x ) + D ( v ) + F ( t ) Return

3 • 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 ) . • Damping force D ( y ′ ) = − µy ′ . µ ≥ 0 is the damping constant. Newton’s law becomes my ′′ = − ky − µy ′ + F ( t ) , or my ′′ + µy ′ + ky = F ( t ) , or y ′′ + µ my ′ + k my = 1 mF ( t ) . • This is the equation of the vibrating spring. Return

4 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 ) . • This is the equation of the RLC circuit. Return Vibrating spring equation

5 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 ) . � We call this the equation for harmonic motion. � It includes both the vibrating spring and the RLC circuit. Return

6 The Equation for Harmonic Motion The Equation for Harmonic Motion 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. � Spring: 2 c = µ/m . � Circuit: 2 c = R/L. • f ( t ) is the forcing term. Return

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

8 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

9 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

10 Example of Simple Harmonic Motion Example of Simple Harmonic Motion 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

11 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 case � c > ω 0 — overdamped case � c = ω 0 — critically damped case Return Harmonic motion

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

13 Overdamped Case Overdamped Case • 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

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

15 Roots and Solutions Roots and Solutions • If the characteristic polynomial has two distinct real roots λ 1 and λ 2 , then y ( 1 t ) = e λ 1 t and y 2 ( t ) = e λ 2 t are a fundamental set of solutions. • 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 a fundamental set of solutions. • If λ = α + iβ is a complex root of the characteristic equation, then z ( t ) = e λt and z ( t ) = e λt are a complex valued fundamental set of solutions. � x ( t ) = e αt cos βt and y ( t ) = e αt sin βt are a real valued fundamental set of solutions. Return