Math 211 Math 211 Lecture #32 Higher Order Equations Harmonic - PDF document

1 Math 211 Math 211 Lecture #32 Higher Order Equations Harmonic Motion November 11, 2002 2 Higher Order Equations Higher Order Equations • Linear homogenous equation of order n . y ( n ) + a 1 y ( n − 1) + · · · + a n − 1 y ′ + a n y = 0 • Linear homogenous equation of second order. y ′′ + py ′ + qy = 0 • Equivalent system: x ′ = A x , where � y � 0 � 1 � x = and A = . y ′ − q − p Return 3 Linear Independence Linear Independence • A fundamental set of solutions for the system consists of two linearly independent solutions. Definition: Two functions u ( t ) and v ( t ) are linearly independent if neither is a constant multiple of the other. • u ( t ) and v ( t ) are linearly independent solutions to � u � v � � y ′′ + py ′ + qy = 0 ⇔ & are linearly u ′ v ′ independent solutions to the equivalent system. Return 1 John C. Polking

4 General Solution General Solution Suppose that y 1 ( t ) & y 2 ( t ) are linearly Theorem: independent solutions to the equation y ′′ + py ′ + qy = 0 . Then the general solution is y ( t ) = C 1 y 1 ( t ) + C 2 y 2 ( t ) . Definition: A set of two linearly independent solutions is called a fundamental set of solutions. Return LI System 5 Solutions to y ′′ + py ′ + qy = 0 . Solutions to y ′′ + py ′ + qy = 0 . • The equivalent system has exponential solutions. • Look for exponential solutions to the 2nd order equation of the form 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. Return 6 Real Roots Real Roots • If λ is a root to the characteristic polynomial then y ( t ) = e λt is a solution. � 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. Return General solution 2 John C. Polking

7 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 Return General solution Two roots 8 Examples Examples • y ′′ − 5 y ′ + 6 y = 0 , with y (0) = 0 and y ′ (0) = 1 . • y ′′ + 4 y ′ + 13 y = 0 , with y (0) = − 1 and y ′ (0) = 14 . • y ′′ + 4 y ′ + 4 y = 0 , with y (0) = 2 and y ′ (0) = 0 . • y ′′ + 25 y = 0 , with y (0) = 3 and y ′ (0) = − 2 . Return General solution Real roots Complex roots 9 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 3 John C. Polking

10 • Including all of the forces, 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 11 • 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 Vibrating spring 12 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 4 John C. Polking

13 Harmonic Motion Harmonic Motion • Spring: y ′′ + µ m y ′ + k m y = 1 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 14 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 15 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 with the natural frequency ω 0 . � The period is T = 2 π/ω 0 . Return 5 John C. Polking

16 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 ) . � • A is the amplitude ; A = C 2 1 + C 2 2 . • φ is the phase ; tan φ = C 2 /C 1 . Return 17 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 18 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 6 John C. Polking