Wave Phenomena Physics 15c

Lecture 3 Coupled Oscillators

What We Did Last Time Analyzed a damped oscillator Behavior depends on the damping strength

Weakly, strongly, or critically-damped How many oscillation does it do? Q factor

Studied forced oscillation Oscillation becomes large near the resonance frequency Phase changes from 0 /2 as the frequency increases

through the resonance Resonance is taller and narrower for a larger Q value

Ran out of time Continue today

Goals for Today Wrap up the driven oscillator Initial conditions Power consumption and dissipation

Coupled Oscillators How a pair of harmonic oscillators

behave when they are connected with each other.

Will use some linear algebra

Prepare for real waves

Initial Condition The solution we found was

No free parameters! How can we satisfy the initial condition?

Imagine we didnt have the motor

The equation of motion would be

We know the general solution to it from Lecture #2 Assuming weak damping

x(t) =

r02

02 2 + i

ei t

m

d 2x(t)dt 2

+ fdx(t)

dt+ kx(t) = 0

x(t) = exp f2m

ikm

f 2

4m2

t

exp

2t

exp 0t( )

Initial Condition Add the forced and the damped solutions

The factor x0 on the damped solution has two real parameters

The sum satisfies

This solution can satisfy any initial condition Second term disappears with time We concentrated on the first term (steady-state solution)

x(t) =

r02

02 2 + i

ei t + x0e

2tei0t

m

d 2x(t)dt 2

+ fdx(t)

dt+ kx(t) = rei t

What Happens to Energy? Energy is not conserved because of the friction The energy consumed per unit time = (friction) x (speed)

Motor supplies the energy The work done per unit time = (force) x (speed) at the motor-spring

connection

force = k(x(t) r cos t)speed = r sin t

f x(t){ }2

friction = f x(t)speed = x(t)

kr sin t(x(t) r cos t)

What Happens to Energy? Energy consumed by the friction must equal to the work done by the motor This is true only after averaging over time for one cycle Because the spring and the mass store/release energy

Next two slides show how to confirm Ef = Em Ef = energy consumed by the friction in one cycle Em = energy supplied by the motor in one cycle

Unfortunately, we must do this in real numbers, with sines and cosines This is a bit of pain. Done in the next two slides

Energy Consumed by Friction

Ef = f x2 dt

0

T

= f Re ix0e

i t( )2 dt0

T

= fix0e

i t ix0e i t

2

2

dt0

T

= f 2x0

2e2i t + 2 2x0x0 2x0

2e2i t

4

dt0

T

=fT 2

2r0

2

02 2 + i

r02

02 2 i

=fTr 2

20

4 2

(02 2)2 + 2 2

x(t) = x0ei t

Re(z) = z + z

2

e2i t dt

0

T

= 0

x0 =

r02

02 2 + i

Work Done by Motor

Em = k(x(t) r cos t)r sin t dt0T

= kr Re x0 r( )ei t( )Im(ei t )0

T

dt

= krx0 r( )ei t + x0 r( )e i t

2ei t e i t

2idt

0

T

= krx0 r( )e2i t x0 r( ) + x0 r( ) x0 r( )e2i t

4idt

0

T

= krT

2Im x0( ) = krT2

r02

(02 2)2 + 2 2

=fTr 2

20

4 2

(02 2)2 + 2 2

=fm

km

=02

Energy Balance Sheet Average rate of energy intake equals consumption

Energy flow is proportional to fr2 No energy is consumed if the friction f = 0

Low frequency limit > 0

EfT

=EmT

=fr 2

20

4 2

02 2( )2 + 2 2

fr 202

20

2

0

fr 202

20

2

0

Energy Balance Sheet

On resonance = 0 The motor does large amount of work

only when is close to the resonance frequency 0

The energy flow at the resonance is proportional to Q

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

=

fr 204

22=

mr 203Q

2

fmQ 0

EfT

=EmT

=fr 2

20

4 2

02 2( )2 + 2 2

Coupled Oscillators Two identical pendulums are connected by a spring Consider small oscillation

Spring tension is

Restoring force from gravity

x1 L1 x2 L2

FS = k(x1 x2)

F1 = mg sin1 =

mgL

x1 F2 =

mgL

x2 FS

mg

F1 F2

md 2x1dt 2

= mgL

x1 k(x1 x2)

md 2x2dt 2

= mgL

x2 k(x2 x1)

Symmetry The two equations are symmetric By adding & subtracting them, you get

We can solve the equations for (x1 + x2) and (x1 x2) separately, and find

md 2(x1 + x2)

dt 2=

mgL

(x1 + x2)

md 2(x1 x2)

dt 2=

mgL

+ 2k

(x1 x2)

Each looks like a harmonic oscillator

md 2x1dt 2

= mgL

x1 k(x1 x2)

md 2x2dt 2

= mgL

x2 k(x2 x1)

x1 + x2 = AeiP t

x1 x2 = BeiSt

P =gL

S =gL+ 2 k

m

Parallel and Symmetric

Let B = 0 x1 x2 = 0 Two pendulums are moving in parallel The spring does nothing P = natural frequency of free pendulum

Let A = 0 x1 + x2 = 0 Two pendulums are moving symmetrically The spring gets expanded/ shrunk by twice

the movement of each pendulum S is determined by both the pendulums and

the spring

x1 + x2 = AeiP t

x1 x2 = BeiSt

P =

gL

S =gL+ 2 k

m

Normal Modes The two oscillating patterns are called the normal modes Both are simple harmonic oscillation

Constant frequency & amplitude

Two normal modes for two coupled oscillators Two pendulums have two initial conditions

each (xi and dxi /dt) Two normal modes have two parameters each

(a cost + b sint)

Initial Conditions General solution is a linear combination of the normal modes

Lets find a solution that satisfies an initial condition:

Take the real parts, and remember some trigonometry

x1(0) = a, x1(0) = 0x2(0) = 0, x2(0) = 0

x1 + x2 = AeiP t

x1 x2 = BeiSt

x1 =A2 e

iP t + B2 eiSt

x2 =A2 e

iP t B2 eiSt

A = B = a

x1 =a2

(cosPt + cosSt) = acosP +S

2t

cos

P S2

t

x2 =a2

(cosPt cosSt) = asinP +S

2t

sin

P S2

t

Specific Solution

S = 1.1Px1 x2

Specific Solution

Plot shows the case where P and S are relatively close

The spring constant is small = weak coupling between the two pendulums

P S

Beats Two oscillations of slightly different frequencies produce beats = modulation of amplitude Coupled oscillators change their amplitudes by beats

Beat frequency = difference of two frequencies This is used in tuning piano, guitar, etc

Will come back when we discuss group velocity

Finding Normal Modes Is there a systematic way of finding normal modes? Symmetry is useful. But it does not always work

What if the two pendulums are different? What if we coupled three pendulums?

We want a recipe that is guaranteed to work

You need linear algebra Relax. This is an easy application of what you already know

Rewriting Equation of Motion Write the equations in a 2x2 matrix form

In normal modes, both x1 and x2 oscillate at one same frequency

Substitute these into the equation of motion

d 2

dt 2x1x2

=

gL

km

km

km

gL

km

x1x2

md 2x1dt 2

= mgL

x1 k(x1 x2)

md 2x2dt 2

= mgL

x2 k(x2 x1)

x1x2

=

a1ei t

a2ei t

=

a1a2

ei t

d 2

dt 2x1x2

= 2

a1a2

ei t

Looking for a Normal Mode

Matrix K times vector a equals constant times a itself a is an eigenvector of K The constant 2 is the eigenvalue

In general, an nn matrix has n eigenvectors Barring unfortunate coincidence

We expect to find two normal modes here

2a1a2

ei t =

gL

km

km

km

gL

km

a1a2

ei t

2a = Ka

a a K

Finding Eigenvalues

Equation Ma = 0 can be satisfied for a non-zero a only if det(M) = 0 The determinant in this case is

This gives the two solutions:

2 gL

km

km

km

2 gL

km

a1a2

= 0

2a = Ka

2a +Ka = 0

2 g

L

km

2

km

2

= 2 gL

2 gL 2 k

m

= 0

P =

gL

, S =gL+ 2 k

m

Finding Eigenvectors

For = P

For = S

2 gL

km

km

km

2 gL

km

a1a2

= 0

P =

gL

, S =gL+ 2 k

m

km

km

km

km

a1a2

= 0

a1a2

= a

a

km

km

km

km

a1a2

= 0

a1a2

= a

a

Back to Normal Modes We found the eigenvalues and eigenvectors The normal modes are given by

For P

For S

x1x2

=

a1ei t

a2ei t

=

a1a2

ei t

x1x2

= a

a

eiP t

x1x2

= a

a

eiSt

What Did We Learn? Linear algebra gives you a recipe for finding normal modes by solving an eigenvalue problem We saw how it worked for our simple problem

It guarantees that the normal modes exist There will be n normal modes if we couple n pendulums Subtlety: all eigenvalues 2 must be negative

This is true for any system near a stable equilibrium, because if there were a positive eigenvalue, it will give us a motion that expands exponentially with time

We can now extend our problem knowing that the normal-mode, constant frequency solutions exist This is enough information to let us attack the next problem many

many many coupled oscillators

Many Coupled Pendulums

Connect N pendulums with springs Displacement of the n-th pendulum is xn (n = 1, 2, N) Equation of motion:

Assume that L is very long Ignore the gm/L term m

d 2

dt 2xn =

gmL

xn k(xn xn1) k(xn xn+1)

m

d 2

dt 2xn = k(xn xn1) k(xn xn+1)

Going Continuous Now we make N very large, while making the mass and the spring smaller and smaller

It starts to look like a spring with distributed mass Good model for mechanical waves such as sound

Summary Studied coupled oscillators General solution is an linear combination of normal modes

= patterns of oscillation with constant frequencies Surprising pattern shows up Beats Linear algebra guarantees that the normal modes exist

Eigenvalues Normal frequencies Eigenvectors Normal modes

We are ready to extend coupled oscillators into mass-spring transmission line Next: continuous waves