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Math 211Math 211

Lecture #32

Higher Order Equations

Harmonic Motion

November 11, 2002

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Higher Order EquationsHigher Order Equations

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Higher Order EquationsHigher Order Equations

• Linear homogenous equation of order n.

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Higher Order EquationsHigher Order Equations

• Linear homogenous equation of order n.

y(n) + a1y(n−1) + · · · + an−1y

′ + any = 0

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Higher Order EquationsHigher Order Equations

• Linear homogenous equation of order n.

y(n) + a1y(n−1) + · · · + an−1y

′ + any = 0

• Linear homogenous equation of second order.

y′′ + py′ + qy = 0

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Higher Order EquationsHigher Order Equations

• Linear homogenous equation of order n.

y(n) + a1y(n−1) + · · · + an−1y

′ + any = 0

• Linear homogenous equation of second order.

y′′ + py′ + qy = 0

• Equivalent system: x′ = Ax, where

x =(

y

y′

)and A =

(0 1−q −p

).

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Linear IndependenceLinear Independence

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Linear IndependenceLinear Independence

• A fundamental set of solutions for the system consists

of two linearly independent solutions.

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Linear IndependenceLinear 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.

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Linear IndependenceLinear 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

y′′ + py′ + qy = 0 ⇔(

u

u′

)&

(v

v′

)are linearly

independent solutions to the equivalent system.

Return LI System

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General SolutionGeneral Solution

Return LI System

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General SolutionGeneral Solution

Theorem: Suppose that y1(t) & y2(t) are linearly

independent solutions to the equation

y′′ + py′ + qy = 0.

Then the general solution is

y(t) = C1y1(t) + C2y2(t).

Return LI System

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General SolutionGeneral Solution

Theorem: Suppose that y1(t) & y2(t) are linearly

independent solutions to the equation

y′′ + py′ + qy = 0.

Then the general solution is

y(t) = C1y1(t) + C2y2(t).

Definition: A set of two linearly independent solutions

is called a fundamental set of solutions.

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Solutions to y′′ + py′ + qy = 0.Solutions to y′′ + py′ + qy = 0.

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Solutions to y′′ + py′ + qy = 0.Solutions to y′′ + py′ + qy = 0.

• The equivalent system has exponential solutions.

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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) = eλt.

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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) = eλt.

• Characteristic equation: λ2 + pλ + q = 0.

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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) = eλt.

• Characteristic equation: λ2 + pλ + q = 0.

� Characteristic polynomial: λ2 + pλ + q.

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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) = eλt.

• Characteristic equation: λ2 + pλ + q = 0.

� Characteristic polynomial: λ2 + pλ + q.

� Same for the 2nd order equation and the system.

Return General solution

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Real RootsReal Roots

Return General solution

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Real RootsReal Roots

• If λ is a root to the characteristic polynomial then

y(t) = eλt is a solution.

Return General solution

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Real RootsReal Roots

• If λ is a root to the characteristic polynomial then

y(t) = eλt is a solution.

� If the characteristic polynomial has two distinct real

roots λ1 and λ2, then y(1t) = eλ1t and y2(t) = eλ2t

are a fundamental set of solutions.

Return General solution

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Real RootsReal Roots

• If λ is a root to the characteristic polynomial then

y(t) = eλt is a solution.

� If the characteristic polynomial has two distinct real

roots λ1 and λ2, then y(1t) = eλ1t and y2(t) = eλ2t

are a fundamental set of solutions.

• If λ is a root to the characteristic polynomial of

multiplicity 2, then y1(t) = eλt and y2(t) = teλt are a

fundamental set of solutions.

Return General solution Two roots

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Complex RootsComplex Roots

Return General solution Two roots

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Complex RootsComplex Roots

• If λ = α + iβ is a complex root of the characteristic

equation, then so is λ = α − iβ.

Return General solution Two roots

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Complex RootsComplex 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) = eλt and z(t) = eλt.

Return General solution Two roots

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Complex RootsComplex 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) = eλt and z(t) = eλt.

• A real valued fundamental set of solutions is

x(t) = eαt cos βt and y(t) = eαt sin βt.

Return General solution Real roots Complex roots

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ExamplesExamples

Return General solution Real roots Complex roots

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ExamplesExamples

• y′′ − 5y′ + 6y = 0

Return General solution Real roots Complex roots

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ExamplesExamples

• y′′ − 5y′ + 6y = 0, with y(0) = 0 and y′(0) = 1.

Return General solution Real roots Complex roots

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ExamplesExamples

• y′′ − 5y′ + 6y = 0, with y(0) = 0 and y′(0) = 1.

• y′′ + 4y′ + 13y = 0

Return General solution Real roots Complex roots

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ExamplesExamples

• y′′ − 5y′ + 6y = 0, with y(0) = 0 and y′(0) = 1.

• y′′ + 4y′ + 13y = 0, with y(0) = −1 and y′(0) = 14.

Return General solution Real roots Complex roots

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ExamplesExamples

• y′′ − 5y′ + 6y = 0, with y(0) = 0 and y′(0) = 1.

• y′′ + 4y′ + 13y = 0, with y(0) = −1 and y′(0) = 14.

• y′′ + 4y′ + 4y = 0

Return General solution Real roots Complex roots

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ExamplesExamples

• y′′ − 5y′ + 6y = 0, with y(0) = 0 and y′(0) = 1.

• y′′ + 4y′ + 13y = 0, with y(0) = −1 and y′(0) = 14.

• y′′ + 4y′ + 4y = 0, with y(0) = 2 and y′(0) = 0.

Return General solution Real roots Complex roots

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ExamplesExamples

• y′′ − 5y′ + 6y = 0, with y(0) = 0 and y′(0) = 1.

• y′′ + 4y′ + 13y = 0, with y(0) = −1 and y′(0) = 14.

• y′′ + 4y′ + 4y = 0, with y(0) = 2 and y′(0) = 0.

• y′′ + 25y = 0

Return General solution Real roots Complex roots

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ExamplesExamples

• y′′ − 5y′ + 6y = 0, with y(0) = 0 and y′(0) = 1.

• y′′ + 4y′ + 13y = 0, with y(0) = −1 and y′(0) = 14.

• y′′ + 4y′ + 4y = 0, with y(0) = 2 and y′(0) = 0.

• y′′ + 25y = 0, with y(0) = 3 and y′(0) = −2.

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The Vibrating SpringThe Vibrating Spring

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The Vibrating SpringThe Vibrating Spring

Newton’s second law: ma = total force.

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The Vibrating SpringThe Vibrating Spring

Newton’s second law: ma = total force.

• Forces acting:

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The Vibrating SpringThe Vibrating Spring

Newton’s second law: ma = total force.

• Forces acting:

� Gravity

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The Vibrating SpringThe Vibrating Spring

Newton’s second law: ma = total force.

• Forces acting:

� Gravity mg.

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The Vibrating SpringThe Vibrating Spring

Newton’s second law: ma = total force.

• Forces acting:

� Gravity mg.

� Restoring force R(x).

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The Vibrating SpringThe Vibrating Spring

Newton’s second law: ma = total force.

• Forces acting:

� Gravity mg.

� Restoring force R(x).

� Damping force D(v).

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The Vibrating SpringThe 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).

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• Including all of the forces, Newton’s law becomes

ma = mg + R(x) + D(v) + F (t)

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• Including all of the forces, Newton’s law becomes

ma = mg + R(x) + D(v) + F (t)

• Hooke’s law: R(x) = −kx.

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• 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.

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• 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

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• 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 x0 = mg/k.

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• 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 x0 = mg/k.

� Set y = x − x0.

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• 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 x0 = mg/k.

� Set y = x − x0. Newton’s law becomes

my′′ = −ky + D(y′) + F (t).

Return Vibrating spring

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• Damping force D(y′)

Return Vibrating spring

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• Damping force D(y′) = −µy′.

Return Vibrating spring

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• Damping force D(y′) = −µy′.

� µ ≥ 0 is the damping constant.

Return Vibrating spring

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• Damping force D(y′) = −µy′.

� µ ≥ 0 is the damping constant.

� Newton’s law becomes

Return Vibrating spring

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• Damping force D(y′) = −µy′.

� µ ≥ 0 is the damping constant.

� Newton’s law becomes

my′′ = −ky − µy′ + F (t)

Return Vibrating spring

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• 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)

Return Vibrating spring

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• 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 =

1m

F (t).

Return Vibrating spring

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• 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 =

1m

F (t).

• This is the equation of the vibrating spring.

Return Vibrating spring equation

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RLC CircuitRLC CircuitL

C

R

E

+

−

I

I

Return Vibrating spring equation

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RLC CircuitRLC CircuitL

C

R

E

+

−

I

I

LI ′′ + RI ′ +1C

I = E′(t)

Return Vibrating spring equation

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RLC CircuitRLC CircuitL

C

R

E

+

−

I

I

LI ′′ + RI ′ +1C

I = E′(t), or

I ′′ +R

LI ′ +

1LC

I =1L

E′(t).

Return Vibrating spring equation

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RLC CircuitRLC CircuitL

C

R

E

+

−

I

I

LI ′′ + RI ′ +1C

I = E′(t), or

I ′′ +R

LI ′ +

1LC

I =1L

E′(t).

• This is the equation of the RLC circuit.

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Harmonic MotionHarmonic Motion

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Harmonic MotionHarmonic Motion

• Spring: y′′ + µmy′ + k

my = 1mF (t).

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Harmonic MotionHarmonic Motion

• Spring: y′′ + µmy′ + k

my = 1mF (t).

• Circuit: I ′′ + RL I ′ + 1

LC I = 1LE′(t).

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Harmonic MotionHarmonic Motion

• Spring: y′′ + µmy′ + k

my = 1mF (t).

• Circuit: I ′′ + RL I ′ + 1

LC I = 1LE′(t).

• Essentially the same equation.

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Harmonic MotionHarmonic Motion

• Spring: y′′ + µmy′ + k

my = 1mF (t).

• Circuit: I ′′ + RL I ′ + 1

LC I = 1LE′(t).

• Essentially the same equation. Use

x′′ + 2cx′ + ω20x = f(t).

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Harmonic MotionHarmonic Motion

• Spring: y′′ + µmy′ + k

my = 1mF (t).

• Circuit: I ′′ + RL I ′ + 1

LC I = 1LE′(t).

• Essentially the same equation. Use

x′′ + 2cx′ + ω20x = f(t).

� We call this the equation for harmonic motion.

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Harmonic MotionHarmonic Motion

• Spring: y′′ + µmy′ + k

my = 1mF (t).

• Circuit: I ′′ + RL I ′ + 1

LC I = 1LE′(t).

• Essentially the same equation. Use

x′′ + 2cx′ + ω20x = f(t).

� We call this the equation for harmonic motion.

� It includes both the vibrating spring and the RLC

circuit.

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The Equation for Harmonic MotionThe Equation for Harmonic Motion

x′′ + 2cx′ + ω20x = f(t).

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The Equation for Harmonic MotionThe Equation for Harmonic Motion

x′′ + 2cx′ + ω20x = f(t).

• ω0 is the natural frequency.

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The Equation for Harmonic MotionThe Equation for Harmonic Motion

x′′ + 2cx′ + ω20x = f(t).

• ω0 is the natural frequency.

� Spring: ω0 =√

k/m.

Return

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The Equation for Harmonic MotionThe Equation for Harmonic Motion

x′′ + 2cx′ + ω20x = f(t).

• ω0 is the natural frequency.

� Spring: ω0 =√

k/m.

� Circuit: ω0 =√

1/LC.

Return

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The Equation for Harmonic MotionThe Equation for Harmonic Motion

x′′ + 2cx′ + ω20x = f(t).

• ω0 is the natural frequency.

� Spring: ω0 =√

k/m.

� Circuit: ω0 =√

1/LC.

• c is the damping constant.

Return

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The Equation for Harmonic MotionThe Equation for Harmonic Motion

x′′ + 2cx′ + ω20x = f(t).

• ω0 is the natural frequency.

� Spring: ω0 =√

k/m.

� Circuit: ω0 =√

1/LC.

• c is the damping constant.

� Spring: 2c = µ/m.

Return

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The Equation for Harmonic MotionThe Equation for Harmonic Motion

x′′ + 2cx′ + ω20x = f(t).

• ω0 is the natural frequency.

� Spring: ω0 =√

k/m.

� Circuit: ω0 =√

1/LC.

• c is the damping constant.

� Spring: 2c = µ/m.

� Circuit: 2c = R/L.

Return

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The Equation for Harmonic MotionThe Equation for Harmonic Motion

x′′ + 2cx′ + ω20x = f(t).

• ω0 is the natural frequency.

� Spring: ω0 =√

k/m.

� Circuit: ω0 =√

1/LC.

• c is the damping constant.

� Spring: 2c = µ/m.

� Circuit: 2c = R/L.

• f(t) is the forcing term.

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Simple Harmonic MotionSimple Harmonic Motion

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Simple Harmonic MotionSimple Harmonic Motion

No forcing

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Simple Harmonic MotionSimple Harmonic Motion

No forcing , and no damping.

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Simple Harmonic MotionSimple Harmonic Motion

No forcing , and no damping.

x′′ + ω20x = 0

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Simple Harmonic MotionSimple Harmonic Motion

No forcing , and no damping.

x′′ + ω20x = 0

• p(λ) = λ2 + ω20

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Simple Harmonic MotionSimple Harmonic Motion

No forcing , and no damping.

x′′ + ω20x = 0

• p(λ) = λ2 + ω20 , λ = ±iω0.

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Simple Harmonic MotionSimple Harmonic Motion

No forcing , and no damping.

x′′ + ω20x = 0

• p(λ) = λ2 + ω20 , λ = ±iω0.

• Fundamental set of solutions:

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Simple Harmonic MotionSimple Harmonic Motion

No forcing , and no damping.

x′′ + ω20x = 0

• p(λ) = λ2 + ω20 , λ = ±iω0.

• Fundamental set of solutions: x1(t) = cos ω0t &

x2(t) = sinω0t.

Return

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Simple Harmonic MotionSimple Harmonic Motion

No forcing , and no damping.

x′′ + ω20x = 0

• p(λ) = λ2 + ω20 , λ = ±iω0.

• Fundamental set of solutions: x1(t) = cos ω0t &

x2(t) = sinω0t.

• General solution: x(t) = C1 cos ω0t + C2 sin ω0t.

Return

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Simple Harmonic MotionSimple Harmonic Motion

No forcing , and no damping.

x′′ + ω20x = 0

• p(λ) = λ2 + ω20 , λ = ±iω0.

• Fundamental set of solutions: x1(t) = cos ω0t &

x2(t) = sinω0t.

• General solution: x(t) = C1 cos ω0t + C2 sin ω0t.

• Every solution is periodic with the natural frequency

ω0.

Return

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Simple Harmonic MotionSimple Harmonic Motion

No forcing , and no damping.

x′′ + ω20x = 0

• p(λ) = λ2 + ω20 , λ = ±iω0.

• Fundamental set of solutions: x1(t) = cos ω0t &

x2(t) = sinω0t.

• General solution: x(t) = C1 cos ω0t + C2 sin ω0t.

• Every solution is periodic with the natural frequency

ω0.

� The period is T = 2π/ω0.

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Amplitude and PhaseAmplitude and Phase

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Amplitude and PhaseAmplitude and Phase

• Put C1 and C2 in polar coordinates:

Return

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Amplitude and PhaseAmplitude and Phase

• Put C1 and C2 in polar coordinates:

C1 = A cos φ, & C2 = A sin φ.

Return

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Amplitude and PhaseAmplitude and Phase

• Put C1 and C2 in polar coordinates:

C1 = A cos φ, & C2 = A sin φ.

• Then x(t) = C1 cos ω0t + C2 sin ω0t

Return

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Amplitude and PhaseAmplitude and Phase

• Put C1 and C2 in polar coordinates:

C1 = A cos φ, & C2 = A sin φ.

• Then x(t) = C1 cos ω0t + C2 sin ω0t

= A cos(ω0t − φ).

Return

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Amplitude and PhaseAmplitude and Phase

• Put C1 and C2 in polar coordinates:

C1 = A cos φ, & C2 = A sin φ.

• Then x(t) = C1 cos ω0t + C2 sin ω0t

= A cos(ω0t − φ).

• A is the amplitude

Return

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Amplitude and PhaseAmplitude and Phase

• Put C1 and C2 in polar coordinates:

C1 = A cos φ, & C2 = A sin φ.

• Then x(t) = C1 cos ω0t + C2 sin ω0t

= A cos(ω0t − φ).

• A is the amplitude; A =√

C21 + C2

2 .

Return

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Amplitude and PhaseAmplitude and Phase

• Put C1 and C2 in polar coordinates:

C1 = A cos φ, & C2 = A sin φ.

• Then x(t) = C1 cos ω0t + C2 sin ω0t

= A cos(ω0t − φ).

• A is the amplitude; A =√

C21 + C2

2 .

• φ is the phase

Return

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Amplitude and PhaseAmplitude and Phase

• Put C1 and C2 in polar coordinates:

C1 = A cos φ, & C2 = A sin φ.

• Then x(t) = C1 cos ω0t + C2 sin ω0t

= A cos(ω0t − φ).

• A is the amplitude; A =√

C21 + C2

2 .

• φ is the phase; tanφ = C2/C1.

Return Amplitude & phase

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ExamplesExamples

Return Amplitude & phase

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ExamplesExamples

• C1 = 3, C2 = 4

Return Amplitude & phase

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ExamplesExamples

• C1 = 3, C2 = 4 ⇒ A = 5

Return Amplitude & phase

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ExamplesExamples

• C1 = 3, C2 = 4 ⇒ A = 5, φ = 0.9273.

Return Amplitude & phase

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ExamplesExamples

• C1 = 3, C2 = 4 ⇒ A = 5, φ = 0.9273.

• C1 = −3, C2 = 4

Return Amplitude & phase

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ExamplesExamples

• C1 = 3, C2 = 4 ⇒ A = 5, φ = 0.9273.

• C1 = −3, C2 = 4 ⇒ A = 5

Return Amplitude & phase

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ExamplesExamples

• C1 = 3, C2 = 4 ⇒ A = 5, φ = 0.9273.

• C1 = −3, C2 = 4 ⇒ A = 5, φ = 2.2143.

Return Amplitude & phase

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ExamplesExamples

• C1 = 3, C2 = 4 ⇒ A = 5, φ = 0.9273.

• C1 = −3, C2 = 4 ⇒ A = 5, φ = 2.2143.

• C1 = −3, C2 = −4

Return Amplitude & phase

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ExamplesExamples

• C1 = 3, C2 = 4 ⇒ A = 5, φ = 0.9273.

• C1 = −3, C2 = 4 ⇒ A = 5, φ = 2.2143.

• C1 = −3, C2 = −4 ⇒ A = 5

Return Amplitude & phase

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ExamplesExamples

• C1 = 3, C2 = 4 ⇒ A = 5, φ = 0.9273.

• C1 = −3, C2 = 4 ⇒ A = 5, φ = 2.2143.

• C1 = −3, C2 = −4 ⇒ A = 5, φ = −2.2143.

Return Amplitude & phase

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ExampleExample

x′′ + 16x = 0, x(0) = −2 & x′(0) = 4

Return Amplitude & phase

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ExampleExample

x′′ + 16x = 0, x(0) = −2 & x′(0) = 4

• Natural frequency: ω20 = 16

Return Amplitude & phase

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ExampleExample

x′′ + 16x = 0, x(0) = −2 & x′(0) = 4

• Natural frequency: ω20 = 16 ⇒ ω0 = 4.

Return Amplitude & phase

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ExampleExample

x′′ + 16x = 0, x(0) = −2 & x′(0) = 4

• Natural frequency: ω20 = 16 ⇒ ω0 = 4.

• General solution:

Return Amplitude & phase

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ExampleExample

x′′ + 16x = 0, x(0) = −2 & x′(0) = 4

• Natural frequency: ω20 = 16 ⇒ ω0 = 4.

• General solution: x(t) = C1 cos 4t + C2 sin 4t.

Return Amplitude & phase

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ExampleExample

x′′ + 16x = 0, x(0) = −2 & x′(0) = 4

• Natural frequency: ω20 = 16 ⇒ ω0 = 4.

• General solution: x(t) = C1 cos 4t + C2 sin 4t.

• IC: −2 = x(0) = C1, and 4 = x′(0) = 4C2.

Return Amplitude & phase

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ExampleExample

x′′ + 16x = 0, x(0) = −2 & x′(0) = 4

• Natural frequency: ω20 = 16 ⇒ ω0 = 4.

• General solution: x(t) = C1 cos 4t + C2 sin 4t.

• IC: −2 = x(0) = C1, and 4 = x′(0) = 4C2.

• Solutionx(t) = −2 cos 2t + sin 2t

=√

5 cos(2t − 2.6779).