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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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:
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Amplitude and PhaseAmplitude and Phase
• Put C1 and C2 in polar coordinates:
C1 = A cos φ, & C2 = A sin φ.
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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
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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).
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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
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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).