lecture #32 harmonic motion november 10, 2003polking/slides/fall03/lecture32p.pdf · • c is the...

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Math 211Math 211

Lecture #32

Harmonic Motion

November 10, 2003

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The Vibrating SpringThe Vibrating Spring

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Newton’s second law: ma = total force.

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• Forces acting:

� Gravity

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• Forces acting:

� Gravity mg.

� Restoring force R(x).

� Damping force D(v).

� External force F (t).

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• 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)

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• Hooke’s law: R(x) = −kx.

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• Hooke’s law: R(x) = −kx. k > 0 is the spring constant.

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� Spring-mass equilibrium

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� Spring-mass equilibrium x0 = mg/k.

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� Spring-mass equilibrium x0 = mg/k. Set y = x − x0.

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Newton’s law becomes

my′′ = −ky + D(y′) + F (t).

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my′′ = −ky + D(y′) + F (t).

• Damping force D(y′)

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my′′ = −ky + D(y′) + F (t).

• Damping force D(y′) = −µy′.

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my′′ = −ky + D(y′) + F (t).

• Damping force D(y′) = −µy′. µ ≥ 0 is the damping

constant.

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my′′ = −ky + D(y′) + F (t).


constant. Newton’s law becomes

my′′ = −ky − µy′ + F (t)

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my′′ = −ky + D(y′) + F (t).



my′′ = −ky − µy′ + F (t), or

my′′ + µy′ + ky = F (t)

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my′′ = −ky + D(y′) + F (t).



my′′ = −ky − µy′ + F (t), or

my′′ + µy′ + ky = F (t), or

y′′ +µ

my′ +

k

my =

1

mF (t).

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my′′ = −ky + D(y′) + F (t).



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 equation

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RLC CircuitRLC CircuitL

C

R

E

+

−

I

I


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C

R

E

+

−

I

I

LI ′′ + RI ′ +1

CI = E′(t)


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C

R

E

+

−

I

I

LI ′′ + RI ′ +1

CI = E′(t), or

I ′′ +R

LI ′ +

1

LCI =

1

LE′(t).


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C

R

E

+

−

I

I

LI ′′ + RI ′ +1

CI = E′(t), or

I ′′ +R

LI ′ +

1

LCI =

1

LE′(t).

• This is the equation of the RLC circuit.

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Harmonic MotionHarmonic Motion

• Spring: y′′ + µm

y′ + km

y = 1m

F (t).

• Circuit: I ′′ + RL

I ′ + 1LC

I = 1L

E′(t).

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y′ + km

y = 1m

F (t).


I ′ + 1LC

I = 1L

E′(t).

• Essentially the same equation. Use

x′′ + 2cx′ + ω20x = f(t).

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y′ + km

y = 1m

F (t).


I ′ + 1LC

I = 1L

E′(t).


x′′ + 2cx′ + ω20x = f(t).

� We call this the equation for harmonic motion.

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y′ + km

y = 1m

F (t).


I ′ + 1LC

I = 1L

E′(t).


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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x′′ + 2cx′ + ω20x = f(t).

• ω0 is the natural frequency.

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x′′ + 2cx′ + ω20x = f(t).


� Spring: ω0 =√

k/m.

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x′′ + 2cx′ + ω20x = f(t).



k/m.

� Circuit: ω0 =√

1/LC.

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x′′ + 2cx′ + ω20x = f(t).



k/m.


1/LC.

• c is the damping constant.

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x′′ + 2cx′ + ω20x = f(t).



k/m.


1/LC.


� Spring: 2c = µ/m.

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x′′ + 2cx′ + ω20x = f(t).



k/m.


1/LC.



� Circuit: 2c = R/L.

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x′′ + 2cx′ + ω20x = f(t).



k/m.


1/LC.



� Circuit: 2c = R/L.

• f(t) is the forcing term.

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Simple Harmonic MotionSimple Harmonic Motion

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No forcing

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No forcing , and no damping.

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x′′ + ω20x = 0

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x′′ + ω20x = 0

• p(λ) = λ2 + ω20

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x′′ + ω20x = 0

• p(λ) = λ2 + ω20 , λ = ±iω0.

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x′′ + ω20x = 0

• p(λ) = λ2 + ω20 , λ = ±iω0.

• Fundamental set of solutions:

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x′′ + ω20x = 0

• p(λ) = λ2 + ω20 , λ = ±iω0.

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

x2(t) = sin ω0t.

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x′′ + ω20x = 0

• p(λ) = λ2 + ω20 , λ = ±iω0.


x2(t) = sin ω0t.

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

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x′′ + ω20x = 0

• p(λ) = λ2 + ω20 , λ = ±iω0.


x2(t) = sin ω0t.


• Every solution is periodic at the natural frequency ω0.

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x′′ + ω20x = 0

• p(λ) = λ2 + ω20 , λ = ±iω0.


x2(t) = sin ω0t.


• Every solution is periodic at the natural frequency ω0.

� The period is T = 2π/ω0.

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Amplitude and PhaseAmplitude and Phase

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• Put C1 and C2 in polar coordinates:

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C1 = A cos φ, & C2 = A sin φ.

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• Then x(t) = C1 cos ω0t + C2 sin ω0t

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= A cos(ω0t − φ).

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• A is the amplitude

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• A is the amplitude; A =√

C21 + C2

2 .

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C21 + C2

2 .

• φ is the phase

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C21 + C2

2 .

• φ is the phase; tan φ = C2/C1.

Return Amplitude & phase

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ExamplesExamples


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ExamplesExamples

• C1 = 3, C2 = 4


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ExamplesExamples

• C1 = 3, C2 = 4 ⇒ A = 5


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ExamplesExamples

• C1 = 3, C2 = 4 ⇒ A = 5, φ = 0.9273.


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ExamplesExamples

• C1 = 3, C2 = 4 ⇒ A = 5, φ = 0.9273.

• C1 = −3, C2 = 4


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ExamplesExamples

• C1 = 3, C2 = 4 ⇒ A = 5, φ = 0.9273.

• C1 = −3, C2 = 4 ⇒ A = 5


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ExamplesExamples

• C1 = 3, C2 = 4 ⇒ A = 5, φ = 0.9273.

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


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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


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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


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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.


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Example of Simple Harmonic MotionExample of Simple Harmonic Motion

x′′ + 16x = 0, x(0) = −2 & x′(0) = 4


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x′′ + 16x = 0, x(0) = −2 & x′(0) = 4

• Natural frequency: ω20 = 16


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x′′ + 16x = 0, x(0) = −2 & x′(0) = 4

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


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x′′ + 16x = 0, x(0) = −2 & x′(0) = 4


• General solution:


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x′′ + 16x = 0, x(0) = −2 & x′(0) = 4


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


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x′′ + 16x = 0, x(0) = −2 & x′(0) = 4



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


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x′′ + 16x = 0, x(0) = −2 & x′(0) = 4



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

• Solution

x(t) = −2 cos 2t + sin 2t

=√

5 cos(2t − 2.6779).

Return Harmonic motion

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Damped Harmonic MotionDamped Harmonic Motion

x′′ + 2cx′ + ω20x = 0


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x′′ + 2cx′ + ω20x = 0

• p(λ) = λ2 + 2cλ + ω20


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x′′ + 2cx′ + ω20x = 0

• p(λ) = λ2 + 2cλ + ω20 ; roots −c ±

√c2 − ω2

0 .


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x′′ + 2cx′ + ω20x = 0

• p(λ) = λ2 + 2cλ + ω20 ; roots −c ±

√c2 − ω2

0 .

• Three cases


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x′′ + 2cx′ + ω20x = 0

• p(λ) = λ2 + 2cλ + ω20 ; roots −c ±

√c2 − ω2

0 .

• Three cases

� c < ω0 — underdamped case


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x′′ + 2cx′ + ω20x = 0

• p(λ) = λ2 + 2cλ + ω20 ; roots −c ±

√c2 − ω2

0 .

• Three cases


� c > ω0 — overdamped case


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x′′ + 2cx′ + ω20x = 0

• p(λ) = λ2 + 2cλ + ω20 ; roots −c ±

√c2 − ω2

0 .

• Three cases


� c > ω0 — overdamped case

� c = ω0 — critically damped case

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Underdamped CaseUnderdamped Case

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• c < ω0

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• c < ω0

• Two complex roots λ and λ, where λ = −c + iω and

ω =√

ω20 − c2 .

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• c < ω0


ω =√

ω20 − c2 .

• General solution

x(t) = e−ct[C1 cos ωt + C2 sin ωt]

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• c < ω0


ω =√

ω20 − c2 .


x(t) = e−ct[C1 cos ωt + C2 sin ωt]

= Ae−ct cos(ωt − φ).

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Overdamped CaseOverdamped Case

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• c > ω0 ,

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• c > ω0 , so two real roots

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λ1 = −c −√

c2 − ω20

λ2 = −c +√

c2 − ω20 .

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λ1 = −c −√

c2 − ω20

λ2 = −c +√

c2 − ω20 .

• λ1 < λ2 < 0.

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λ1 = −c −√

c2 − ω20

λ2 = −c +√

c2 − ω20 .

• λ1 < λ2 < 0.


x(t) = C1eλ1t + C2e

λ2t.

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Critically Damped CaseCritically Damped Case

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• c = ω0

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• c = ω0

• One negative real root λ = −c with multiplicity 2.

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• c = ω0

• One negative real root λ = −c with multiplicity 2.


x(t) = e−ct[C1 + C2t].

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Roots and SolutionsRoots and Solutions

• 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.

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

equation, then z(t) = eλt and z(t) = eλt are a complex

valued fundamental set of solutions.

� x(t) = eαt cos βt and y(t) = eαt sin βt are a real valued

fundamental set of solutions.

lecture #32 harmonic motion november 10, 2003polking/slides/fall03/lecture32p.pdf · • c is the...

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