Lecture 3 Coupled Oscillators - Harvard University

Wave Phe

Physic

Lectu
Coupled O

enomena

cs 15c

ure 3
Oscillators

What We Did La

Analyzed a damped oscillator

  Behavior depends on the dampin
  Weakly, strongly, or critically-d

  How many oscillation does it do?

Studied forced oscillation

  Oscillation becomes large near th
  Phase changes from 0  −π/2 

through the resonance
  Resonance is taller and narrower

Ran out of time  Continue to

ast Time

ng strength
damped
?  Q factor

he resonance frequency
 −π as the frequency increases

r for a larger Q value

oday

Goals for Today

Wrap up the driven oscillator

  Initial conditions
  Power consumption and dissipati

Coupled Oscillators

  How a pair of harmonic oscillator
behave when they are connected
each other.

  Will use some linear algebra

Prepare for real waves

ion

rs
d with

Initial Condition

The solution we found was

x(t ) = r ω 2 eiωt
0

ω 2 − ω2 + iΓω
0

  No free parameters! How can we

Imagine we didn’t have the mo

  The equation of motion would be

  We know the general solution to
  Assuming weak damping

x(t ) = exp ⎛ ⎡ f ± i kf
⎜⎝⎜ ⎢− 2m m − 4m
⎣⎢

e satisfy the initial condition?

otor

e m d 2x(t) + f dx(t ) + kx(t ) = 0
dt 2 dt

it from Lecture #2

⎤⎞
2 ⎥ t ⎠⎟⎟ ≈ exp ⎛ − Γ t ⎞ exp ω0t
⎦⎥ ⎝⎜ 2 ⎠⎟
( )m2

Initial Condition

Add the “forced” and the “dam

x(t ) = r ω 2
0

ω 2 − ω2 + iΓω
0

  The factor x0 on the damped solu

  The sum satisfies m d 2x(t) + f d
dt 2

This solution can satisfy any in

  Second term disappears with tim
  We concentrated on the first term

mped” solutions

eiωt + x e e− Γ t iω0t
2

ω 0

ution has two real parameters

dx(t ) + kx(t ) = re iω t
dt

nitial condition

me
m (“steady-state solution”)

What Happens to

Energy is not conserved becau

  The energy consumed per unit tim

⎪⎧ friction = fx(t) f{
⎨
⎪⎩ speed = x(t)

Motor supplies the energy

  The work done per unit time = (fo
connection

⎪⎧⎨ force = −k(x(t) − r cosωt)
⎪⎩ speed = −rω sinωt

o Energy?

use of the friction

me = (friction) x (speed)

{x(t }) 2

orce) x (speed) at the motor-spring

krω sinωt(x(t) − r cosωt)

What Happens to

Energy consumed by the frictio
done by the motor

  This is true only after averaging o
  Because the spring and the mass

Next two slides show how to c

  Ef = energy consumed by the fric
  Em = energy supplied by the moto

Unfortunately, we must do this
and cosines

  This is a bit of pain. Done in the n

o Energy?

on must equal to the work

over time for one cycle
s store/release energy

confirm Ef = Em

ction in one cycle
or in one cycle

s in real numbers, with sines

next two slides

Energy Consume

∫Ef =T f x 2 dt

0

T Re 2 dt

0
∫ ( )= f
iω x0eiωt

∫= f T ⎛ iω x0eiωt − iω x0∗e−iωt ⎞2
0 ⎜⎝ 2 ⎟⎠

∫= f T ⎛ −ω 2x02e2iωt + 2ω 2x0 x0∗ −
0 ⎝⎜ 4

fTω 2 r ω 2 rω
2 0 ω
=
ω 2 − ω2 + iωΓ ω 2 −
0 0

= fTr 2 (ω 2 ω04ω 2 Γ2ω 2
2 0 − ω 2)2 +

ed by Friction

x(t) = x0eiωt

Re(z) = z + z∗
2
dt

− ω x e2 ∗2 −2iωt ⎞ dt ∫T e±2iωt dt = 0
0 ⎟⎠ 0

ω 2 x0 = r ω 2
0 0

2 − iωΓ ω 2 −ω2 + iΓω
0

Work Done by M

∫Em =T k(x(t) − r cosωt)rω sinω

0

(( ) )∫= krω T Re
0
x0 − r eiωt Im(e

( ) ( )∫= krω
T x0 − r eiωt + x0∗ − r
0 2

( ) (= krω T x0 − r e2iωt − x0 − r

∫0

krωT krωT
2 2
( )=− Im x0 = (ω 2
0

= fTr 2 (ω 2 ω04ω 2 Γ2ω 2
2 0 − ω 2)2 +

Motor

t dt

eiωt )dt

) e−iωt eiωt − e−iωt dt
2i

) ( ) ( )+ x0∗ − r−x0∗ − r e −2 iω t
dt
4i

r ω 02Γω
− ω 2 )2 + Γ2ω 2

f k = ω 2
Γ=m m 0

Energy Balance

Average rate of energy intake

(Ef=Em = fr 2 2
T 2 0
T ω

Energy flow is proportional to f

  No energy is consumed if the fric

Low frequency limit ≈ fr 2ω 2
0
  ω << ω0
2

High frequency limit ≈ fr 2ω 2
0
  ω >> ω0
2

Sheet

equals consumption

ω04ω 2

2− ω 2 2 + Γ2ω 2

)0

fr2

ction f = 0

⎛ ω ⎞ 2
⎜⎝ ω0 ⎠⎟
→ 0

⎛ ω0 ⎞ 2
⎜⎝ ω ⎠⎟
→ 0

Energy Balance

(Ef = Em = fr 2 2
T 2 0
T ω

On resonance = fr 2ω 4 = mr
0
  ω = ω0
2Γ 2

  The motor does large amount o
only when ω is close to the reso
frequency ω0

  The energy flow at the resonan
proportional to Q

  The full-width-at-half-maximum
(FWHM) of the resonance peak
is proportional to 1/Q

Sheet

ω04ω 2 ⎪⎧ Γ ≡ f
m
⎩⎨⎪Q
2 − ω 2 2 + Γ2ω 2 ≡ ω0
Γ
)0

r 2ω03Q
2

of work
onance

nce is

m
k

Coupled Oscillat

Two identical pendulums are c

  Consider small oscillation

x1 ≈ Lθ1 x2 ≈ Lθ2

  Spring tension is

FS = −k(x1 − x2)

  Restoring force from gravity

F1 = mg sinθ1 = mg x1 F2 = m
L L

⎪⎧⎪m d 2x1 = − mg x1 − k(x1 − x
dt 2 L
⎨
⎪⎪⎩m d 2x2 = − mg x2 − k(x2 − x
dt 2 L

tors

connected by a spring

mg x2 FS
L F1

x2 ) mg F2

x1)

Symmetry

The two equations are symme
By adding & subtracting them,

⎪⎪⎧m d 2(x1 + x2 ) = − mg (x1 + x2 )
dt 2 L
⎨
⎪⎩⎪m d 2(x1 − x2 ) = − ⎛ mg + 2k ⎞ (
dt 2 ⎝⎜ L ⎟⎠

  We can solve the equations for (x

and find

⎧⎪ x1 + x2 = Ae
⎨ x1 − x2 = Bei
⎩⎪

⎪⎪⎧m d 2x1 = − mg x1 − k(x1 − x2 )
dt 2 L
⎨
etric ⎪⎪⎩m d 2x2 = − mg x2 − k(x2 − x1)
dt 2 L

, you get

) Each looks like a
(x1 − x2) harmonic oscillator

x1 + x2) and (x1 – x2) separately,

iωP t ⎧g
⎪⎪ωP = L
⎨
iωS t ⎪⎩⎪ωS = g + 2 k
L m

Parallel and Sym

⎪⎧ x1 + x2 = AeiωP t ωP = g
⎨ x1 − x2 = Be iωS t L
⎪⎩

Let B = 0  x1 – x2 = 0

  Two pendulums are moving in pa

  The spring does nothing

  ωP = natural frequency of free pe

Let A = 0  x1 + x2 = 0

  Two pendulums are moving symm

  The spring gets expanded/ shrun
the movement of each pendulum

  ωS is determined by both the pen
the spring

mmetric

ωS = g + 2 k
L m

arallel
endulum

metrically

nk by twice
m

ndulums and

Normal Modes

The two oscillating patterns ar
normal modes

  Both are simple harmonic oscillat
  Constant frequency & amplitud

Two normal modes for two cou
oscillators

  Two pendulums have two initial c
each (xi and dxi /dt)

  Two normal modes have two par
(a cosωt + b sinωt)

re called the

tion
de

upled

conditions

rameters each

Initial Conditions

General solution is a linear com

modes ⎪⎧ x1 + x2 = AeiωP t
⎨ x1 − x2 = Be iωS t
⎩⎪

  Let’s find a solution that satisfies

x1(0) = a, x1(0) = 0

x2(0) = 0, x2(0) = 0

  Take the real parts, and rememb

⎧ x1 = a (cos ω P t + cos ω St ) = a co
⎪ 2
⎪
⎨
⎪ a
⎩⎪ x2 = 2 (cos ω P t − cos ω St ) = −a

s

mbination of the normal

⎧⎪ x1 = eA iωP t + eB iωSt
⎨ x2 = −
⎩⎪ 2 2

eA iωP t eB iωSt

2 2

an initial condition:

A=B=a

ber some trigonometry

os ⎛ ω P + ωS t ⎞ cos ⎛ ωP − ωS t ⎞
⎝⎜ 2 ⎠⎟ ⎜⎝ 2 ⎠⎟

sin ⎛ ωP + ωS t ⎞ sin ⎛ ωP − ωS t ⎞
⎜⎝ 2 ⎠⎟ ⎜⎝ 2 ⎠⎟

Specific Solution

x1 x2

n

2 ωS = 1.1× ωP

Specific Solution

⎧ = a cos ⎛ ωP + ωS t ⎞⎠⎟ cos ⎛ ωP − ω
⎪ x1 2 ⎜⎝ 2 ⎜⎝ 2
⎪
⎨
⎪ a ⎛ ωP + ωS ⎞ ⎛ ωP − ωS
⎪⎩ x2 = 2 sin ⎜⎝ 2 t ⎟⎠ sin ⎜⎝ 2

Plot shows the case where ωP
relatively close

ωP − ωS << ωP + ωS

  The spring constant is small = we
between the two pendulums

n

ωS t⎞⎠⎟

S t ⎞
⎠⎟

P and ωS are ⎧ g
⎪⎪ωP = L
⎨
S ⎪⎪⎩ωS = g + 2 k
L m
eak coupling

small

Beats

Two oscillations of slightly diffe
beats = modulation of amplitud

  Coupled oscillators change their

Beat frequency = difference of

  This is used in tuning piano, guita

Will come back when we discu

erent frequencies produce
de

amplitudes by beats

f two frequencies

ar, etc

uss group velocity

Finding Normal M

Is there a systematic way of fin

  Symmetry is useful. But it does n
  What if the two pendulums are
  What if we coupled three pend

  We want a recipe that is guarante

You need linear algebra

  Relax. This is an easy application

Modes

nding normal modes?

not always work
e different?
dulums?
eed to work

n of what you already know

Rewriting Equati

Write the equations in a 2x2 m

⎪⎧⎪m d 2x1 = − mg x1 − k(x1 − x2 ) d2
dt 2 L dt 2
⎨
⎪⎩⎪m d 2x2 = − mg x2 − k(x2 − x1)
dt 2 L

  In normal modes, both x1 and x2 o

⎛ x1 ⎞⎛ a1e iω t ⎞⎛ a1 ⎞
⎜ x2 ⎟ =⎜ a2e iω t ⎟ =⎜ a2 ⎟ eiωt
⎝⎜ ⎟⎠ ⎝⎜ ⎟⎠ ⎜⎝ ⎟⎠

  Substitute these into the equation

ion of Motion

matrix form

⎛ gk k⎞
⎛ x1 ⎞⎜ −L−m ⎟⎛ x1 ⎞
x2 ⎟ =⎜ k m ⎟⎜ x2 ⎟
2 ⎜ ⎠⎟ ⎜ m gk ⎟ ⎜⎝ ⎠⎟
⎜⎝ ⎝⎜ −L−m ⎟⎠

oscillate at one same frequency ω


d2 ⎛ x1 ⎞⎛ a1 ⎞
⎜ x2 ⎟ = −ω 2 ⎜ a2 ⎟ eiωt
dt 2 ⎜⎝ ⎟⎠ ⎝⎜ ⎟⎠

n of motion

Looking for a No

⎛g
⎛ a1 ⎞⎜ −L−m
−ω 2 ⎜ a2 ⎟ eiωt = ⎜ k
⎟⎠ ⎜ m
⎝⎜ ⎝⎜

a

Matrix K times vector a equals

  a is an eigenvector of K
  The constant –ω2 is the eigenvalu

In general, an n×n matrix has

  Barring unfortunate coincidence…

We expect to find two normal m

ormal Mode

k k⎞ ⎞
⎟⎛ a1 ⎟ eiωt
m m ⎟⎜ a2 ⎟⎠
gk
−L−m ⎟ ⎜⎝
⎠⎟
a
K

s constant times a itself

ue −ω 2a = Ka

n eigenvectors

…

modes here

Finding Eigenval

−ω 2a = Ka ⎛
ω 2a + Ka = 0 ⎜ ω2 −
⎜
⎜
⎝⎜

Equation Ma = 0 can be satisf

det(M) = 0

  The determinant in this case is

⎛ g k⎞ 2 ⎛ k ⎞2 ⎛⎝⎜ ω
⎝⎜ L m ⎠⎟ ⎝⎜ m ⎠⎟
ω 2 − − − =

  This gives the two solutions: ωP

lues

gk k⎞ ⎞
−L−m ⎟⎛ a1 ⎟ =0
m ⎟⎜ a2 ⎠⎟
k
m ω2 − g − k ⎟ ⎝⎜
L m ⎠⎟

fied for a non-zero a only if

ω2 − g⎞ ⎛ ω 2 − g − 2 k⎞ = 0
L ⎟⎠ ⎝⎜ L m ⎠⎟

= g , ωS = g + 2 k
L L m

Finding Eigenvec

⎛ ω 2 − g − k k⎞
⎜ L m ⎟⎛ a1 ⎞
⎜ m ⎟⎜ a2 ⎟=
⎠⎟
⎜k ω2 − g − k ⎟ ⎜⎝
⎝⎜ m L m ⎠⎟

⎛k k⎞
⎜ −m ⎟⎛ a1
For ω = ωP ⎜ m ⎟⎜ a2
For ω = ωS
⎜k − k ⎟ ⎜⎝
⎜⎝ m m ⎠⎟

⎛k k⎞
⎜ ⎟⎛ a1 ⎞
⎜ m m ⎟⎜ a2 ⎟
k ⎟⎠
⎜k m ⎟ ⎝⎜
⎝⎜ m ⎟⎠

ctors

=0 ωP = g , ωS = g + 2 k
L L m

⎞ ⎛ a1 ⎞⎛ a ⎞
1 ⎟ =0 ⎜ a2 ⎟ = a ⎠⎟
2 ⎠⎟ ⎜⎝ ⎟⎠ ⎝⎜
=0
⎛ a1 ⎞⎛ a ⎞
⎜ a2 ⎟ = −a ⎠⎟
⎜⎝ ⎟⎠ ⎝⎜

Back to Normal M

We found the eigenvalues and
The normal modes are given b

⎛ x1 ⎞⎛ a1e iω t ⎞⎛ a1 ⎞
⎜ x2 ⎟ =⎜ a2e iω t ⎟ =⎜ a2 ⎟
⎝⎜ ⎠⎟ ⎝⎜ ⎠⎟ ⎝⎜ ⎠⎟

  For ωP ⎛ x1 ⎞⎛ a ⎞ eiωP t
⎜ x2 ⎟ = a ⎟⎠
⎜⎝ ⎠⎟ ⎝⎜

  For ωS ⎛ x1 ⎞⎛ a ⎞ eiωS
⎜ x2 ⎟ = −a ⎟⎠
⎝⎜ ⎠⎟ ⎝⎜

Modes

d eigenvectors
by
⎞
⎟ eiωt
⎟⎠

St