Coupled Oscillators Concepts of primary interest: Simple

Coupled Oscillators

Contact: [email protected]

Concepts of primary interest:

Simple Harmonic Motion

Coupled Oscillators

Eigenvalue problem

Developing the equations of Motion:

Intuitive Newton’s Law Method

Lagrangian Mechanics

Hamiltonian Mechanics

Work-Energy for Mechanical Engineers

Sample calculations:

A matched equal mass pair of coupled oscillator

– with variable coupling

An oscillator pair with unequal masses

Initial condition matching for near degenerate modes

Comparison of two choices for generalized coordinates

Applications

Tools of the Trade

Cramer’s Rule

Coupled Differential Equation

Principles Underlying the Coupled Oscillator Procedures

Engineers can use this handout by replacing -2 by . A brief discussion of forced couple

oscillations will be added. dd example with damping

The Plan: A system of coupled linear oscillators is to be studied. To anchor the discussion, the

solution to the single simple harmonic oscillator is reviewed. Next a system of two masses coupled

by a linear spring is analyzed by guessing the solution. The result is a pair of two-particle sinusoidal

oscillation patterns each with its characteristic frequency. More general systems are analyzed by

assuming that sinusoidal solution modes exist and applying a diagonalization approach to identify

them. The method can be made more formal by noting that it is a particular example of the matrix

solution to systems of coupled linear differential equations. At several points, physical

interpretations of results are presented that enhance our confidence in the procedure.

12/13/2013 Physics Handout Series.Tank: Coupled Oscillator’s CO-2

Just the single simple harmonic oscillator:

The motion of a mass about its equilibrium position while subject to a linear restoring force is well-

studied. The model equation is: ma m x k x where x is measured relative to the equilibrium

position. The solutions are known to be:

( ) cos( ) sin( ) cos( ) Re i tx t C t D t A t Ae where 2 km

.

The mass hanging on a vertical spring reduces to the same equation when the displacement y is measured relative to the

equilibrium position for the mass. ma m y k y That is: The Hooke’s Law model for the force due to a spring

xF k x for displacements measured relative to equilibrium can be used for all springs. In this model k is the spring

or Hooke’s constant.

A simple example: An equal mass system of coupled oscillators

The springs are linear Hooke’s law springs; there is no friction; x1 and x2 are measured from the

equilibrium positions of the masses. (The springs exert forces on the masses at the points of contact.)

As the spring are linear, the net force on each mass can be deduced by computing the force when (a)

mass one is displaced while the mass two remains at its equilibrium position and adding (b) the force

deduced when mass one is held fixed at equilibrium and mass two is displaced.

(a) Mass one displaced and mass two fixed:

Contribution to net force on mass one: F1a = - k x1 - x1

Contribution to net force on mass two: F2a = + x1

(b) Mass one fixed and mass two displaced:

Contribution to net force on mass one: F1a = + x2

Contribution to net force on mass two: F2a = - k x2 - x2

Newton’s Law for the motions of the masses in the x direction yields:

k k

m

x1 x2

m


1 1 1 2 1 1 1 2

2 2 1 2 2 2 2 1

( ) ( )

( ) ( )

m a m x k x x m x k x x x

m a m x x k x m x k x x x

The equations are symmetric as is the problem. The situation begs for a symmetric solution. Add the

two equations together.

1 2 1 2( ) ( )m x x k x x

That strategy was so successful that as a next try, the two equations are to be differenced.

2 1 2 1( ) ( 2 ) ( )m x x k x x

The system has two modes. The (x1+x2) symmetric mode oscillates at S = ( k/m)1/2

and the (x2–x1)

anti-symmetric mode that oscillates at S = (

k+2/m)1/2.

As a general rule, a symmetric problem is expected to have symmetric and anti-symmetric solutions. A

debate can arise over which solution is symmetric and which is anti-symmetric. Convention is to define the

lowest frequency mode to be a symmetric mode.

Identifying the oscillation pattern: If the two masses move the same distance in the same

direction, the length of the spring joining them does not change. The forces due to the spring witch

constant drop out as they maintain their values at equilibrium. Effectively each mass works

against the one spring with constant k that it contacts. The frequency of the mode is: S = ( k/m)1/2. If

the two masses move with equal magnitude displacements that are oppositely directed, the coupling

spring undergoes a length change double the magnitude of the length changes in the end springs.

Each mass is subject to a force from one end spring (k) and from the coupling spring () that is being

double stretched. The frequency of the mode is therefore: S = (

k+2/m)1/2.

As the problems become more complex, a formal procedure is needed to cover the cases that our

intuition fails to untangle. A multi-step plan of attack is to be introduced and applied to the

symmetric oscillator.

The Formal Method:

Step One - Find the equations of Motion for the Masses: When mass one is displaced by x1 to the

right while mass two is held fixed, forces of – k x1 and – x1 are applied to it by the two springs in

contact with it. When mass one is held fixed at equilibrium while mass two is displaced by x2 to the


right, the center spring exerts a force of + x2 on mass one. The equation of motion for mass one

is: 1 1 2m x k x x . Following the same procedure, the forces on mass two are identified

when mass two is held fixed and mass one is displaced and then added to those that arise when mass

one is fixed and mass two is displaced.

2 1 2( )m x x k x

Equations of Motion: 1 1 2m x k x x and 2 1 2( )m x x k x

An alternative approach uses the Lagrangian approach to generate the equations of

motion. Execute Step-One using your method of choice.

Step Two – Recast the equations of motion in matrix form and identify the Mass and Spring Constant Matrices

Equation of motion summary: 1

2

x

x

= - 1

2

x

x

or x = - x

.

The signs are chosen to resemble the single oscillator equation. is the mass matrix, and is the

spring constant matrix.

1

2

0

0

xm

xm

= - 1

2

xk

xk

where =

0

0

m

m

and =k

k

.

Step Three – Assume Harmonic Solutions in which the masses execute motion at the same frequency.

A harmonic solution is one that varies sinusoidally in time. These solutions can be represented using

complex exponentials in which case it must be understood that only the real part is physically

meaningful.

1

2

( )

( )i tx t a

ex t b

and 1 2

2

( )

( )i tx t a

ex t b

yielding

2

2

0( )

0( )i tak m

ebk m

For engineers: 2

2

0( )

0( )tak m

ebk m

As an alternative, the real sinusoids are to be used.

1

2

( )cos

( )

x t at

x t b

and 1 2

2

( )cos

( )

x t at

x t b

yielding


2

2

0( )cos

0( )

ak mt

bk m

The important point is that in either case the second time derivative is equivalent to multiplication by

-2. Neither i te nor cos( t +) is identically zero, so the equation for the coefficients must have

a non-trivial solution.

A trivial solution is one in which both a and b are zero.

Step Four – Solve the modified eigenvalue problem.

Cramer’s rule specifies that a homogeneous set of n linear equations for n unknowns can have a non-

trivial solution for a and b (a solution in which both a and b are not zero) only if the determinant of

the matrix of coefficients is zero.

2

2

0( )

0( )

ak m

bk m

2

2

( )0

( )

k m

k m

or 2 2 2 0k m k m

This matrix equation is called the eigenvalue equation, and the values of that result in a zero

determinant are the eigenvalues for the problem.

As in so many cases, a little notation is helpful. The equation is rewritten in terms of the reference

or scale value for the frequency squared: 20

km

and the relative spring constant = /k. Taking the

2 across and taking the square root. 2k m

Dividing by m2: 2 2

0 1 .

Solving for 2: 2 2 2 2 20 0 01 and 1 2 1 2( )S AS k

What rule was used to identify the symmetric (S) and the anti-symmetric (AS) mode?

Step Five – Identify the patterns of motion

To identify the symmetric mode, substitute S into the eigenvalue equation.

First try 2 20S

km

.


2

2

0( )

0( )

ak m

bk m

0

0

aa b

b

As is the usual case, the two row equations redundantly state a single relation which for this case is:

a = b. The pattern of motion is: 01

2

( ) 1

( ) 1i tx t

A ex t

. The prose description of this mode is that

the two masses oscillate in phase (in the same direction at the same time) with equal amplitude.

Note that these prose descriptions of the motion are crucial. You need to describe the pattern of

the motion without reference to your choice of x1 and x2. Express your answers in coordinate

independent form! Complex motions are not the order of the day so the real part of the complex

exponential is taken.

0 11

0 12

cos( )

cos( )S

A tx t

A tx t

This answer is expected as both masses would oscillate at 0 is the center spring were removed. In

the motion pattern found, x1(t) – x2(t) = 0 so the length of the center spring and hence its tension

does not change from its value when the masses are at their equilibrium positions.

Next try 2 20 1 2 2( )AS

km m

.

2

2

0( )

0( )

ak m

bk m

0

0

ab a

b

Both row equations state that b = - a. Hence x2(t) = - x1(t). The prose description of this mode is

that the two masses oscillate out of phase (in the opposite direction at any instant) with equal

amplitudes.

0 11

0 12

cos( )

cos( )AS

B tx t

B tx t

In the future, the mode patterns are to be subscripted with their frequencies.

Step Six – Match Initial Conditions

General Solution:

1 1 2 21

1 1 2 22

cos cos( )

cos cos( )Full

t tx tA B

t tx t


Initial Positions:

1 21

1 22

cos cos(0)

cos cos(0)

xA B

x

Initial Velocities:

0 20 11

0 12 0 2

1 2

1 2

sinsin(0)

sin(0) sin

xA B

x

As there are two masses and the initial position and initial velocity must be matched for each, there

are four initial conditions to be matched.

Sample Calculation: An Asymmetric Oscillator

Consider a system with two masses, one of mass m and the other of mass 2m. The displacements x1

and x2 are measured relative to the equilibrium positions of the masses which are coupled as shown

by springs with Hooke constants k, k and 2k.

Step One - Find the equations of Motion for the Masses: When mass one is displaced by x1 to the

right while mass two is held fixed, forces of – k x1 and – k x1 are applied to it by the two springs in

contact with it. When mass one is held fixed at equilibrium while mass two is displaced by x2 to the

right, the center spring exerts a force of + k x2 on mass one. The equation of motion for mass one

is: 1 1 22m x k x k x . Following the same procedure, the forces on mass two are identified

when mass two is held fixed and mass one is displaced and then added to those that arise when mass

one is fixed and mass two is displaced.

2 1 22 3m x k x k x

Equations of Motion: 1 1 22m x k x k x and 2 1 22 3m x k x k x

Step Two – Recast the equations of motion in matrix form and identify the

Mass and Spring Constant Matrices

k k 2 k

m 2m

x1 x2


Equation of motion summary: 1

2

x

x

= - 1

2

x

x

or x = - x

.



1

2

0

0 2

xm

xm

= - 1

2

2

3

xk k

xk k

where =0

0 2

m

m

and = 2

3

k k

k k

.

Step Three – Assume Harmonic Solutions in which the masses execute motion at the same

frequency.

1

2

( )

( )i tx t a

ex t b

and 1 2

2

( )

( )i tx t a

ex t b

yielding

2

2

02

03 2i tak m k

ebk k m


2

2

02

03 2

ak m k

bk k m

2

2

20

3 2

k m k

k k m

or 2 2 22 3 2 0k m k m k

Find the values of for which the determinant vanishes.


or scale value for the frequency squared: 20

km

.

Dividing by m2: 2 2 2 2 4

0 0 02 3 2 0 or 4 2 2 40 02 7 5 0 .

Solving for 2:

22 2 2 2 2

0 0 0 0

7 7 4 2 5 7 3; 2.5

2 2 2 2



In turn, substitute each eigen-frequency into the eigenvalue equation to find the ratio of a to b.

First try 20

km

.

20

20

02

03 2

ak m k

bk k m

2 0

3 2 0

k k k a

k k k b

As is the usual case, the two row equations redundantly state a single relation which for this case is:

a = b. The pattern of motion is: 01

2

( ) 1

( ) 1i tx t

A ex t

. The prose description of this mode is the

that the two masses oscillate in phase (in the same direction at the same time) with equal amplitude.

Note that these prose descriptions of the motion are crucial. You need to describe the pattern of

the motion without reference to your choice of x1 and x2. Express your answers in coordinate

independent form! Complex motions are not the order of the day so the real part of the complex

exponential is taken.

0

0 11

0 12

cos( )

cos( )

A tx t

A tx t

This answer is expected as both masses would oscillate at 0 if the center spring were removed. In

the motion pattern found, x1(t) – x2(t) = 0 so the length of the center spring and hence its tension

does not change from its value when the masses are at their equilibrium positions.

Next try 20

52 .

12

0

02

k k a

bk k

Both row equations state that 12

b a . Hence x2(t) = - 1/2 x1(t). The prose description of this

mode is the that the two masses oscillate out of phase (in the opposite direction at any instant) with

the amplitude of the smaller mass’s motion being twice that of the 2 m mass.

0

0 21

522 0 2

12

52

52

cos( )

( ) cos

tx t

Bx t t


A physical reasoning check of the results builds confidence and heightens your physical intuition.

The oscillation frequency for this mode can be explained if one substitutes x2(t) = - ½ x1(t) into the

equations of motion.

Mass one: 1 1 22m x k x k x 1 1 1 151

2 22 km x k x k x x

Mass two: 2 1 22 3m x k x k x 2 1 2 2 2 22 3 2 3 5m x k x k x k x k x k x

The frequency of oscillation for mass one is 5

2( ) k

m and for mass two is 52

km .

Both mass oscillate at 2 20

52

for this motion pattern.


General Solution:

2 21 11

1 12 2 21

2

coscos( )

cos( ) cosFull

ttx tA B

tx t t

Initial Positions:

211

12 201

2

coscos(0)

cos(0) cos

xA B

x

Initial Velocities:

0 20 11

0 120 2

52

52

12

sinsin(0)

sin(0) sin

xA B

x

As there are two masses and the initial position and initial velocity must be matched for each, there

are four initial conditions to be matched.

Summary of the Pure Matrix Eignevalue Problem:

EIGENVALUE EQUATION: v

= v

= v

or [ - ] v

= 0

This equation has only trivial solutions unless the determinant vanishes.

DETERMINANT CONDITION: | - | = 0.

An nth order equation results, and its n solutions are the eigenvalues. Substitute each eigenvalue into

the eigenvalue equation to determine the ratios of its components. The normalization condition is

used to complete the specification of each eigenvector. The eigenvectors for distinct eigenvalues are


always orthogonal. Eigenvectors for degenerate eigenvalues can be (and should be) chosen to be

orthogonal.

A general problem with n masses coupled by linear springs.

This situation arises effectively for a collection of interacting masses in stable equilibrium for small

displacements from equilibrium. It is assumed that there is a collection { q1, q2, … , qn } of

generalized coordinates (each of which is zero when the system is in its equilibrium configuration)

that describe the configuration of the system. The potential energy is a positive quadratic form in

these coordinates for small displacements from equilibrium leading to general linear restoring forces.

For the examples above, the generalized coordinates were x1 and x2 in each case.

Step One - Find the equations of Motion for the Masses:

Newton’s Second Law or Lagrange’s equations are standard tools employed to generate the

necessary equations of motion. In any case, the equations of motion are considered to be givens for

the current development. An intermediate mechanics course is the proper source for the details of

these equations. The problem is to be studied for small oscillations about equilibrium leading to a set

of n coupled linear differential equations each with second and zero order derivatives only. (See the

problems in the mechanics text by Taylor for some examples that include first order derivatives as

well.)

Step Two – Recast the equations of motion in matrix form and identify the Mass and Spring

Constant Matrices

Equation of motion summary:

1

2

n

q

q

q

= -

1

2

n

q

q

q

or q

= - q

.



Note that the off-diagonal elements can be chosen such that the matrices are real symmetric without

changing the content of the equations of motion. As a first step, adjust the equations of motion such

that each combination of the form


1,i ii iq M q ni has the same dimensions (energy). Maintaining these units, perform algebraic

operations until the symmetric forms appear. NOTE: The eigenvalue method works without this

procedure. You need not transform the equations to find the symmetric matrices. The Lagrangian

approach generates the symmetric forms naturally.

Step Three – Assume Harmonic Solutions in which the masses execute motion at the same

frequency.

A harmonic solution is one that varies sinusoidally in time. These solution can be represent using complex

exponentials in which case it must be understood that only the real part is physically meaningful.

1 1

2 2

( )

( )

( )

k

i tk

n kn

q t a

q t ae

q t a

and

1 1

2 22

( )

( )

( )

k

i tk

n kn

q t a

q t ae

q t a

yielding

As an alternative, the real sinusoids are to be used.

1 1

2 2

( )

( )cos

( )

k

k

n kn

q t a

q t at

q t a

and

1 1

2 22

( )

( )cos

( )

k

k

n kn

q t a

q t at

q t a

yielding

( - 2 )

1

2 cos

k

k

kn

a

at

a

=

0

0

0

The important point is that in either case the second time derivative is equivalent to multiplication by

-2. Neither i te or cos( t +) is identically zero, so the equation for the coefficients must have a

non-trivial solution.


Cramer’s rule specifies that a homogeneous set of n linear equations for n unknowns can have a non-

trivial solution for a and b (a solution other than a = 0 and b = 0) only if the determinant of the

matrix of coefficients is zero.


( - 2 )

1

2

k

k

kn

a

a

a

=

0

0

0

| - 2 | = 0

This matrix equation is the eigenvalue equation, and the values of that result in a zero determinant

are the eigenvalues for the problem. As the determinant condition is an nth order polynomial

equation for , n roots are expected. Each root is to be substituted in turn into the eigenvalue

equation, and the eigenvector that satisfies the eigenvalue equation for that particular eigenvalue is

to be found. As the eigenvalue equation is linear, any solution can be scaled and remain a solution.

An independent normalization process is required to set the overall scale of each eigenvector.


value for the frequency squared such as: 20

km

or 20

g . The relative value of squared

frequencies or of ‘spring’ constants may be useful as well: 2

20

0

A

B


To identify the mode associated with a frequency k , substitute that frequency into the eigenvalue

equation and solve for the ratios of the various aki. As a convention, set the non-zero component aki

with the lowest index i equal to one. The components may be scaled later as part of a normalization

procedure.

First try 2 2

0Sk

m .

( - m2 )

1

2

m

m

mn

a

a

a

=

0

0

0

As is the usual case, the n row equations are not independent so only the ratios of the ami are

determined, not their absolute values.


1 1

2 2

( )

( )

( )

m

m

n mn

m mi t

q t a

q t a

q t a

e

Note that a prose description of the motion is crucial. You need to describe the pattern of the motion

without reference to your choice of q1 , etc. Express your answers in coordinate independent form!

Complex motions are not the order of the day so the real part of the complex exponential is taken.

1 1

2 2

( )

( )cos

( )

m m

n

m

m

mn

q t a

q t at

q t a

Note the each mode is a combination with every one of the generalized coordinates oscillating

sinusoidally at the mode frequency.


General Solution:

1 1

2 2

1

( )

( )cos

( )

n

m mm

n

m

mm

mn

q t a

q t aA t

q t a

Initial Positions:

1 1

2 2

1

(0)

(0)cos

(0)

n

mm

n

m

mm

mn

q a

q aA

q a

Initial Velocities:

1 1

2 2

1

(0)

(0)( ) sin

(0)

n

m mm

n

m

mm

mn

q a

q aA

q a

As there are n degrees of freedom and the initial position and initial velocity must be matched for

each, there are 2 n conditions to be matched.

Sample Calculation 3: Step Six – Match Initial Conditions

Consider the first example problem with the initial conditions:


1

2

(0)

(0) 0

x C

x

; 1

2

(0) 0

(0) 0

x

x

General Solution:

1 1 2 21

1 1 2 22

cos cos( )

cos cos( )Full

t tx tA B

t tx t

Initial Positions:

1 21

1 22

cos cos(0)

cos cos(0)

xA B

x

Initial Velocities:

0 20 11

0 12 0 2

1 2

1 2

sinsin(0)

sin(0) sin

xA B

x

where 1 = S = 0 and 2 = 1 2 0

cos cos

cos cos0S A

S A

CA B

sin sin0

sin sin0S A

S AS A

A B

check signs !!!

The solution of this set of equations yields A = + B = C/2 and S = . The solution is:

1

22 2

cos cos cos cos( )

cos cos cos cos( )S A S A

S A S AFull

C Ct t t tx t

t t t tx t

A little notation renders a clearly picture of the situation.

12 S A and A S plus trig identities

121

12 2

cos cos( )

( ) sin sinFull

t tx tC

x t t t

Note the problem starts by setting mass one in motion. Initially, it oscillates while mass two is nearly

at rest. The oscillation frequency is perturbed by the coupling to be about the average of the

frequencies for the two modes. The coupling spring transfers energy from the motion of particle one

k k

m

x1 x2

m


into the motion of particle two resulting in amplitudes that oscillate in time at half the difference

frequency. As the motions have maximum amplitudes twice in each oscillation the energy cycles

back and forth between the motions of the two masses at the difference frequency.

This effect is spectacular when the mode frequencies are nearly coincident and the coupling is weak.

The point is that even a small coupling can cause a large energy transfer between modes that are

nearly degenerate – have nearly the same frequency.

Exercise: The difference frequency is defined as A - S rather than as S - A. Give a reason

that might have motivated this choice. What does S represent? By convention, how does one

identify the most symmetric mode?

Exercise: Show the steps that lead from the initial conditions to the identification of the constants as:

A = + B = C/2 and S =

Sample Calculation 4: Choice of generalized coordinates does not matter!

Consider the system of masses and springs below. The goal is to demonstrate that the eigen-

frequencies and the patterns of motion associated with each frequency are independent of the

particular set of generalized coordinates chosen.


In the equilibrium configuration, the upper spring has a

length L01 and the lower spring has a length L02. The first

set of generalized coordinates {x1, x2 } are measured from

the equilibrium positions. In the second set {y1, y2 } the

first is measured from the equilibrium position of the

upper mass and y2 is the increase in the length of the

lower spring.

1 1 2

2 1 2

2m x k x k x

m x k x k x

1 1 2

1 2 2

m y k y k y

m y y k y

Exercise: Justify the equations of motion above in each of the two sets of coordinates. Pay special

attention to the acceleration of the lower mass as expressed in the second system.

Using the first set of coordinates, the eigen-frequencies are:

2 20

3 5 3 5

2 2k m

The mode for the highest frequency is:

12 20 5 1

2 2

1( )3 5 3 5cos[ ]

( )2 2 H H HHk m

x tA t

x t

Using the second set of coordinates, the eigen-frequencies are the same:

2 20

3 5 3 5

2 2k m

The mode for the highest frequency is:

12 20 1 5

2 2

1( )3 5 3 5cos[ ]

( )2 2H H H Hk m

y tA t

y t

m

mL01 + x1

L01 +L02 + x2

L01 + y1

L02 + y2


Clearly the frequencies are the same as they should be. The frequencies of the modes and the

patterns of motion are characteristic of the system, not of the coordinates chosen. The mode patterns

can be compared using the relations between the two sets of coordinates: y1 = x1 ; y2 = x2 – x1.

Checking:

1 5 1 52 2cos[ ] cos[ ] and cos[ ] 1 cos[ ]H H H H H H H H H H H HA t A t A t A t

It all works; the physical mode patterns are identical for the high frequency.

Exercise: Using the first set of coordinates, write the equations of motion in matrix form and solve

the eigenvalue problem to find the eigen-frequencies.

Exercise: Using the second set of coordinates, write the equations of motion in matrix form and

solve the eigenvalue problem to find the eigen-frequencies.

Exercise: For each set of coordinates, find the eigen-modes for the lower of the two eigen-

frequencies. Verify that they correspond to the same pattern of motion.

It has been demonstrated that, for a particular physical system executing small vibrations about its

equilibrium configuration, the eigen-frequencies and the patterns of motion associated with each

frequency are independent of the particular set of generalized coordinates chosen. This result is

expected to be true for all physical systems executing small vibrations about their equilibrium

configurations.

Sample Calculation 5: Using coordinates with different dimensions

In the case that all the generalized coordinates have the same dimension (say an Einstein), all the

elements of have same dimension Joules/Einstein2 and all the elements of have the dimensions

(Joule · s2)/Einstein2 . If the coordinates have different dimensions, the various elements of and will

also have different dimensions.


The coordinate x is the displacement of M horizontally to

the right; the coordinate is the angle from the vertical to

the string supporting the mass m measured CCW from

down. The Cartesian coordinates for the two masses are:

xM = x , yM = 0, xm = x+ sin, ym =- cos

For small displacements, the coordinates and velocities are

approximately: Mx x , mx , mx x ,

22 1my and 0my .

For small angles, the tension in the string is approximately mg so the string force on M has a

horizontal component of mg to the right while the horizontal component of the tension force on m

is mg to the left. (sin ≈ )

For small displacements about equilibrium, the equations of motion are:

2

[ ] ( )

[ ]

M x m x k x M m x m k x

m x mg m x m mg

The matrix form of the equations becomes:

2

0

0

M m m x k x

m m mg

Exercise: Verify that as xm= x+ sin, the acceleration of the suspended mass has an x component

2cos sinm xx . Give the form for this acceleration valid for small angles. That is as (

0; 0; 0 ) keep the answer to first order in small things. As the theory of small

oscillations about equilibrium is being studied, ( 0; 0; 0x x x ) as well. One must extract at

least the first order correction to zero in each case. As approaches zero, sin is to be replaced

by , not by zero. Cosine is to be replaced by 1 - 2/2.

Exercise: Verify that the first equation of motion states that the sum of the masses times their

accelerations in the x direction is equal to the net eternal force on the system in the x direction and

that the second states that the mass times the acceleration of the suspended mass is equal to the force

kM

x1

m


that acts on it in the x direction. Note that small-angle approximations correct to the first non-

vanishing order have been used.

This problem is to be simplified by setting M = 2 m and the length of the string equal to mg/k. In this

case: 2 20 P

gk m . Using these values, the eigen-frequencies and eigen-modes are found

to be:

2 2

3 0

0

m m x k x

m m k

22 2021 ; 2 1

1

; 2 1

1

Note that the component displacements have different dimensions. By convention, the first non-zero

entry has been set to one. (If the second entry were the first non-zero value, it would be set to one

(meter)-1 to preserve the dimensions.) Each mode pattern is to be multiplied by a scale factor with

dimensions of length giving the overall components dimensions of length and length/length (= radians).

Developing the Equations of Motion:

Intuitive Newton’s Law Method

This method has been used for the example problems. One simply applies Newton’s Second Law to

each mass using a set of displacements measured relative to an equilibrium configuration.

An equal mass system of coupled oscillators



As the spring are linear, the net force on each mass can be deduced by computing the force when (a)

mass one is displaced while the mass two remains at its equilibrium position and adding (b) the force

k k

m

x1 x2

m


deduced when mass one is held fixed at equilibrium and mass two is displaced.

(a) Mass one displaced and mass two fixed:

Contribution to net force on mass one: F1a = - k x1 - x1

Contribution to net force on mass two: F2a = + x1

(b) Mass one fixed and mass two displaced:

Contribution to net force on mass one: F1a = + x2

Contribution to net force on mass two: F2a = - k x2 - x2

Newton’s Law for the motions of the masses in the x direction yields:

1 1 1 2 1 1 1 2

2 2 1 2 2 2 2 1

( ) ( )

( ) ( )

m a m x k x x m x k x x x

m a m x x k x m x k x x x

Lagrangian Mechanics

For standard cases1 on expresses the difference between the kinetic energy and the potential energy

in terms of a set of coordinates, the q’s, and their time derivatives, the 'sq , measured relative to an

equilibrium configuration.

The kinetic energies of the two masses are 2 21 1 2 2½ and ½m x m x . The leftmost spring is stretched x1

from its equilibrium length, the center spring is stretched by x2 – x1 and the rightmost spring is

compressed by x2. The Lagrangian is the difference between the kinetic and potential energies.

2 2 2 2 21 2 1 2 1 1 2 2 1 2 1 2( , ; , ) ½ ½ (½ ½ [ ] ½ )L x x x x m x m x kx x x kx

The dynamic equations follow from:

1 1

wherej j

i ki k

i i i k

d L L dq q

dt q q dt q q t

[MechAlt.1]

1 Only the most straight forward examples are presented here. Refer to mechanics texts for more general treatments.

k k

m

x1 x2

m


1 1 1 1 1 1 21 1 1

ord L L d L

m x m x kx x xdt x x dt x

and 2 2 2 1 22

Lm x x x kx

x

Hamiltonian Mechanics

The Hamiltonian approach builds on the Lagrangian with a change of independent variables to be

the coordinates qi as before, but with the conjugate momenta, the pi, being substituted for the

coordinate velocities iq . The momentum conjugate to a coordinate is identified as ii

Lqp . It

follows that the momenta for this problem are: 1 1 1 2 2 2andp m x p m x . For simple cases, the

hamiltonian is the total energy expressed in terms of the coordinates qi and the conjugate momenta

pi.

2 2 21 2 1 2 1 2 1 2

2 21 2

1 2(2 ) (2 )( , ; , ) ½ ½ ( ) ½p pm mH x x p p kx x x kx

The dynamical equations follow from:

Hamiltonian: 1

( , , ) ( , , )j

i i i ik

k kH q p t L q q tp q

. [MechAlt.2]

Equations of Motion: andi ii i

H Hq p

p q

[MechAlt.3]

It follows that 1 1 1 1 2 1 1 1 1 21

1and orp

mx p kx x x m x kx x x .

Work-Energy for Mechanical Engineers

We begin by generalizing the problem by applying an external force F1(t) to the right on mass 1 and

an external force F2(t) to the right on mass 2. We choose the generalized coordinates to be x1 and x2 ,

the displacements of each mass from its equilibrium position. 2 ( )F t

1( )F t 2 ( )F t


The total mechanical energy is expressed in terms of the q’s and 'sq is:

2 2 2 2 21 2 1 2 1 1 2 2 1 2 1 2( , ; , ) ½ ½ ½ ½ [ ] ½TE x x x x m x m x kx x x kx

The change in total energy is the work done of the system by F1 and F2 on the particles m1 and m2

with positions x1 and x2 or 1 1 2 2W F x dt F x dt .

The two external forces transfer mechanical energy to the system at the rate 1 1 2 2F x F x . Setting this

rate equal to the total time derivative of the energy:

( )1 1 1 2 2 2 1 1 2 1 2 1 2 2 1 1 2 2[ ][ ]d TE

dt m x x m x x kx x x x x x kx x F x F x

As 1 2andx x are independent variables, the coefficient of each must vanish separately.

2 2 2 1 2( )

1 2 11 1 1 2 1 1 22[ ]( ]) ([ )d TEdt x x m x x x km x kx x x F xFxx

1 1 1 1 2 1 1: [ ]x m x kx x x F

2 2 2 2 1 2 2: [ ]x m x x x kx F

Somewhat magically, our equations of motion have appeared.

1 1 1 1

2 2 2 2

0

0

m x x Fk

m x x Fk

In particular, the equation set above reduces to those developed previously when the external forces

F1 and F2 vanish. The inner product in small oscillation space: the orthogonality of the eigenmodes

Return to the mass-pendulum example with M = 2 m and the

length of the string equal to mg/k. In this case:

2 20 P

gk m . Using these values, the eigen-

frequencies and eigen-modes are found to be:

2 2

3 0

0

m m x k x

m m k

22 2021 ; 2 1

1

2 1

1

The inner product is must include information from the mass matrix . That is, quantity of motion

depends on the variation of the coordinate and the mass (inertia) that is moving. Consider the motion

kM

x1

m


vectors: 1 21 2

1 2

anda a

b b

. The inner product of the vectors 1|2 uses the mass matrix to

weight the coordinates:

* *2 2 11 12 1

2 121 22 1

a b M M a

M M b

For our example,

2 1 2 1

2 2 1

2 2

2 2 2 1 2

1 ( )1 13

)

( ) ( )( ) 0

2(mm m

m m m

m

which demonstrates that the two modes are orthogonal with respect to the mass matrix metric. Given

the form of the result above, a new inner product can be defined by applying a scaling. Any scaling

will do, but we choose m-1 to yield a dimensionless result.

* * * *2 2 2 211 12 1 1

2 1 221 22 1 1

31a b a bM M a am M M b b

As a final step, a normalization convention can be set for the eigenmodes:

k m kme e

The sad story of the coupled oscillator, diagonalization and inner product

NOTE: The normalization of small oscillation modes is not a standard topic in a junior year mechanics course.

The coupled oscillator problem requires that two real symmetric matrices, and , be diagonalized.

The transformation to simultaneously diagonalize the two matrices is more complicated than a

rotation. It is composed of three steps: a rotation to diagonalize ; a set of scalings to each

coordinate to transform the diagonalized into a multiple of the identity and a final rotation to

diagonalize whatever has become after the first two steps. As is a multiple of the identity after

the first two steps, the final similarity transformation does not disturb it. was initially real

symmetric, and it maintains its real symmetric form after the first two steps so it is guaranteed that a

final rotation can render it diagonal. One of the problems is to prove that a similarity transformation


of a real symmetric matrix that uses an orthogonal matrix preserves the symmetry of the matrix. The

proof that a scaling preserves it as well is not difficult.

The sad part of the story is that the scaling alters the inner product such that the components of the

eigenvectors are not evenly weighted in the inner product.

11 12 13

21 22 23

31 32 33

i i i j

j

j

a b c m m m a

I J m m m b

m m m c

where = 11 12 13

21 22 23

31 32 33

m m m

m m m

m m m

An example is shown for three degrees of freedom. This inner product has the dimensions of kg ·m2..

Any inner product that meets the defining behaviors may be used. An alternative of some interest is:

11 12 13

21 22 23

31 32 33

*2

i i i j

ji j

a b c m m m a

I J m m m b

m m m cj

which has dimensions of Energy. Note that the angular frequencies can be chosen to be positive by

convention as it is their squares that arise as eigenvalues.

The difference between this case and the standard eigenvalue problem is evident in the forms:

iq i iq i iq classic eigenvalue problem

iq i2 iq coupled-oscillator problem

The squares of the frequencies are the eigenvalues and the mass/inertia matrix replaces the identity

and becomes the kernel in the inner product for the oscillator mode vectors.

The transpose of the eigenvalue equation is tiq t = i

2t

iq t tiq = i

2t

iq as both matrices

are real symmetric. Hence tiq jq - t

iq jq = 0 = i2

tiq

jq -j2

tiq

jq = (i2-j

2) tiq jq . The

eigen-mode vectors are orthogonal with respect to for distinct eigen-frequencies. As in earlier

cases, the modes for degenerate frequencies can be chosen to be orthogonal with respect to .

P. K. Aravind, Geometrical interpretation of the simultaneous diagonalization of two quadratic

forms, American Journal of Physics (April 1989) 57, 4, pp. 309-311.


Could it get any worse?!

In the coupled oscillator problem, the kinetic energy and the potential energy are represented by

homogeneous quadratic expressions.

1 12 2i j i j i j i jKE m q q and PE k q q

leading to the defining of two matrices:

=

11 12 1

21 22 2

1

...

...

n

n

n nn

m m m

m m m

m m

and =

11 12 1

21 22 2

1

...

...

n

n

n nn

k k k

k k k

k k

Form the generalized coordinate vectors

1 1 2

2 t;

n

n

q q q q

qq q

q

The transformed coordinate components are computed as:

q / = q and ( q /)t = ( q ) t = q t t

where is the n x n matrix representing the transformation. The velocity components q follow the

same transformation rule for time-independent transformations. Next, recall that PE and KE are

physical scalars.

PE = 1/2 q t q = 1/2 ( q /) t /

q / scalar values are invariant

Use the transformation rules.

1/2 q t q = 1/2 ( q /) t /

q / = 1/2 q t t / q

Hence

= t / (the t -object- form is called a congruent transformation)

The congruent transformations for the coupled-oscillator problem have the form:

= � 2 1 where 1 is a rotation that diagonalizes , is a diagonal scaling ( Sij = si ij

) that

maps the diagonalized = 1 ( 1)t to a multiple of the identity c , and �

2 is a rotation to


diagonalize 1 ( 1)t (which is still real symmetric). In this transformed space, the

eigenvectors have the standard metric, a multiple of . The transformed = � 2 1 ( �

2 1)t

is equal to 1 ( 1)t

as 1 ( 1)t is a multiple of , so, after the inverse transformation,

can be used as the metric in the original basis of generalized coordinates.

Transformation Matrix Example with forcing terms:

To begin, one must solve the homogeneous problem finding the eigen-modes and eigen-frequencies.

DO NOT DEPEND ON THIS SECTION. It is a very rough draft. A simple example: An equal mass system of coupled oscillators

f1(t) f2(t)



The energy approach will be used to develop the differential equations for the problem.

Mechanical energy = 2 2 2 2 21 1 2 1 1 2½ ½ ½ ( ) ½ ½kx mx x x mx kx . The rate of change of the

mechanical energy is the power delivered to the system 1 1 2 2( ) ( )f t x f t x .

1 1 1 1 2 1 2 1 2 2 2 2 1 1 2 2( )( ) ( ) ( )kx x mx x x x x x mx x kx x f t x f t x

Collecting the terms that multiply 1x and those that multiply 2x .

1 1 2 1 1 1 2 2 1 2 2 2( ) and ( )mx k x x x x f t x mx k x x x f t x

The two coordinate velocities can be varied independently


1 1 1

2 2 2

( )0

( )0

x x f tm k

x x f tm k

21 1

22 2

( )

( )

x f tk m

x f tk m

We form the transformation matrix by columns with each column being an eigen-mode pattern.

Eigen-modes: 1

1

and 1

1

= 1 2

1 2

1 1

1 1

a a

b b

Multiplying both sides of the equation by this matrix maintains the equality. Engineers sometimes

use the symbol rather than to represent the transformation matrix.

21 1

22 2

( )

( )tt x f tk m

x f tk m

We can always multiply by the identity matrix = 1 .

2

1 1 1 1

22 2

( ) ½ ½; ( )

( ) ½ ½t t t t tx f tk m

x f tk m

The inverse of a matrix can be computed using the row reduction method or as the matrix of

cofactors divided by the determinant. We define our new generalized mode coordinates.

1 1 1 2

2 2 1 2

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )tq t x t x t x t

q t x t x t x t

and

2 21

2 2

0( )

0 2t tk m k m

k m k m

As these relations hold for all time, they can be used to transform the initial condition.

1 1 1 1

2 2 2 2

(0) (0) (0) (0);

(0) (0) (0) (0)t tq x q x

q x q x

Combining, we have the equation for q1 and q2, the normal mode amplitudes.

21 1 1 2

22 2 1 2

( ) ( ) ( )0

( ) ( ) ( )0 2tq f t f t f tk m

q f t f t f tk m

Note that the -2qk kq as multiplying cos[t + ] by -2 yields the second time derivative. For

engineers, it is +2qk kq as the assumed form is et. We find the decoupled equations with forcing

terms.


1 1 1 2 2 2 1 2( ) ( ) and ( 2 ) ( ) ( )mq kq f t f t mq k q f t f t

The mode equations above can be solved using your weapon of choice. In this simple case, it is

almost expected that the symmetric q1 mode is driven by the sum of the driving forces and the q2

mode by their difference. One recovers the original functions as

1 1 1 21

2 2 1 2

( ) ( ) ½[ ( ) ( )]( )

( ) ( ) ½[ ( ) ( )]tx t q t q t q t

x t q t q t q t

Note that the equations have been written assuming that the eigen-modes have the specific form, 1

1

and 1

1

. Substitute the correct eigen-modes for your problem.

Tools of the Trade:

Cramer’s Rule Summary:

Cramer’s rule provides a scheme to compute the solution set of n values to a set of n independent linear

algebraic equations. It also indicates when the solution to a set of equations can exist or when it is unique.

For application to coupled oscillators, it identifies the requirement for a solution to exist to a set of n

homogeneous algebraic equations. In particular the determinant of the matrix of coefficients must vanish, a

condition that yields the spectrum of eigenvalues.

Linear Equations: We introduced the concept of a matrix and hence the determinant by means of

consideration of a system of linear equations. Let us now return to that consideration as an example

of the use of determinants.

Consider the set of n inhomogeneous linear equations in n unknowns.

y1 = a11x1 + a12x2 + ............ + a1nxn

y2 = a21x1 + a22x2 + ............ + a2nxn

.

.

yn = an1x1 + an2x x2 + .......... + annxn

This set of equations can be expressed in matrix form as


Let be the n x n matrix of the coefficients. This system of n equations in n unknowns always has a

unique solution if the determinant of the coefficient matrix is not zero (| | ≠ 0). If this is the case, we

can solve for any unknown xn by replacing the nth column in with the column of values y1 to yn ,

computing the determinant of this new matrix, and dividing that result by the determinant of the

original matrix of coefficients | |. As an example, the value for x2 will be

and similarly for all other unknowns, xi. This prescription is known as Cramer's rule. In the case

where | | = 0, the set of inhomogeneous equations has no solution except in the case where the

determinant of the numerator is also equal to zero, in which event solutions may exist, but they are

not necessarily unique.

In the event that you have a set of homogeneous equations (that is, all of the yn = 0 ) then if

| | ≠ 0 the only possible solution is the trivial solution where all xn = 0. If, however, | | = 0 , then a

non-trivial solution does exist. The last case is special interest for eigenvalue problems such as

finding the normal frequencies and modes of a system of coupled oscillators.

Mega-Application: Coupled Oscillators or Systems of Linear DEs

A set of coupled linear differential equations of the form:

/11 12 11

/21 22 22

a a uu

a a uu

where 2

/2

.....du du d u

u or or ordx dt dt

y1

y2

..yn

=

a11 a12 ... a1n

a21 a22 ... a2n

... ... ... ...an1 ... ... ann

x1

x2

..xn

x2 = 1A

a11 y1 ... a1n

a21 y2 ... a2n

... ... ... ...an1 yn ... ann


can be solved by assuming a solution of the form: 1

2

s t

s t

u eu

u e

for the

du

dt case. The power

of this trial solution is that the process of differentiation becomes effectively multiplication by s

reducing the problem to solving the algebraic eigenvalue problem: 11 12

21 22

a as

a a

For the 2

2

d u

dt case, it is traditional to choose the trial form 1

2

i t

i t

u eu

u e

leading to the

eigenvalue problem 11 12 2

21 22

a a

a a

. The treatment of these problems is to be deferred

in favor of treating the coupled oscillator problem.

Independent Solutions of a Differential Equation:

How many are there? + A verification of the Euler identity.

If everything is well-behaved, an nth order differential equation has n independent solutions. Recall

that our well-behaved functions can be expanded in a Taylor’s Series.

22

2

1 1( ) ( ) ( ) ( ) ( )

2! !o o o

nn

o o o ont t t

df d f d ff t f t t t t t t t

dt dt n dt

A well-behaved function is known if all its derivatives are known for a single value of its argument.

For a function that obeys an nth order differential equation, one need only specify the values of the

function and its first (n-1) derivatives. All the rest of the function’s derivatives can be computed

using the differential equation. For our example above we find

1 2

1 2 01 2

1n n n

n nn n nn

d x d x d xc c c x

dt c dt dt

and after differentiating the equation once, we find:

1 1

1 2 01 1

1n n n

n nn n nn

d x d x d x d xc c c

dt c dt dt dt

and by repeating the process, we can find the derivative of x(t) of arbitrary order.

Hence we expect there to be n independent solutions corresponding to setting the value of the

function and its first (n-1) derivatives.


EXAMPLE: Consider the oscillator DE: 2

2 0d x xdt

2

2

n n

nnd x d x

dtdt

.

From this we see that alternate even (odd) derivatives are just the negative of one another. We can

specify two solutions with the choices

0 0

1 0 2 01 2( ) 1; 0 and ( ) 1; 0

t t

dx dx

dt dtx t x t

Substituting into the Taylor's Series template, we find:

2 21

1 ( 1)( ) 1 ( ) ( ) cos( )

2! (2 )!

nn

o o ox t t t t t t tn

and

12*1 1 2 1

2

( 1) ( 1)( ) ( ) ( ) ( ) sin( )

(2*1 1)! (2 1)!

nn

o o o ox t t t t t t t t tn

It follows that the solution for any arbitrary set of initial values 0

01( ) ;

t

dx

dtx t A B

is:

x(t) = A cos(t-to) + B sin(t-to).

That is all solutions to the DE can be expressed in terms of the two independent solutions that we

have found.

Euler's Identity, ei t = cos(t ) + i sin(t), can now be established if we first show that e

i t is a

solution of the oscillator DE that satisfies 0

0

( ) 1;t

dxx t i

dt for to=0. The completion of the proof

is left to the student as an exercise.

Example Summary:

Example I:

M =0

0

m

m

and K =k

k

; modes:

1

1

and1

1

2 20 1 2

0 ; kk m

Example II:


M =0

0 2

m

m

and K = 2

3

k k

k k

; modes:1

1

and1

2

1

2 2 20 0; 2.5 2

0k m

Example III:

M =0

0

m

m

and K = 2k k

k k

; modes:5 12

1

and1 5

2

1

2 20

3 5 3 5

2 2k m

Example IV: **** This example is non-standard. Ignore it. The results need to be verified.

M =0m

m m

and K = 0

k k

k

; modes:5 12

1

and1 5

2

1

2 20

3 5 3 5

2 2k m

Example V:

M =2

3m m

m m

and K = 2

0

0

k

k

; modes: 2 1

1

and 2 1

1

22 2021

Example VI:

M =0

0 2

m

m

and K = 2

2

k k

k k

; modes:1 3

2

1

and1 3

2

1

2 20

3 3

2

Example VII: 3 mass ring

M = 2

1 0 0

0 1 0

0 0 1

mR

and K = 2

2 1 1

1 2 1

1 1 2

kR

; modes:

1

1

1

,

1

0

1

,

1

2

1

2 2 20 00, 3 ,3


Example VIII: CO2

M =

4 0 0

0 3 0

0 0 4

m

and K =

1 1 0

1 2 1

0 1 1

k

; modes:

1

1

1

,

1

0

1

, 83

1

1

2 2 20 0

1 114 120, ,

Tools of the Trade

For a two mass system

Exercise: Find the matrix inverse of an arbitrary 2 x 2 matrix. Given = 11 12

21 22

a a

a a

, find the

matrix of cofactors. Find adj = [ AijC ]

t. Finally, find -1 = [ A

ijC ]t/| |. Conclude that:

22 12

21 11

11 22 12 21

1

11 12

21 22

a aa a

a a a a

a a

a a

Show that the proposed inverse is correct by direct calculation.

Alternative Method to Generate and : Lagrangian mechanics provides an alternative approach

to developing the equations of motion that govern a physical system. The procedure begins with the

construction of the Lagrangian which is usually the difference between the kinetic and potential

energy of the system. L = K - U. The kinetic energy of a set of particles is most simply stated in terms

of the Cartesian coordinate velocities of each particle.

1 2 31 1

2 2particles particles

K m x y z m x x x

If the problem is described in terms of generalized coordinates related to the Cartesian coordinates

by time-independent coordinate transformations xi (q1, q2, … , qn), then the kinetic energy can be

rewritten as:


K = 1/2 tq q

=

11 12 1 11 2

21 22

1 2

12

...

...

...

nn

n

n n nn n

M M M qq q q

M M q

M M M q

Where the mass matrix has elements: [ ]ij = 3

1

n

i

n

particles n j

x

q

xm

q

. Any functions that appear

in the elements of the mass matrix are replaced by their values in the equilibrium configuration when

a small oscillation problem is studied.

The coupled oscillator problem is studied for the restricted case of small oscillations about

equilibrium. The potential energy is expanded in a Taylor's series about a reference configuration.

2

121 2 10 20 0 0 0 0

1 , 10 0

( , ,..., ) ( , ,..., )n n

n n j j k k j jj j kj k j

U U

q q qU q q q U q q q q q q q q q

The reference configuration is an equilibrium so all the first derivatives vanish. By convention, all

the coordinates are reset to be measured relative to their equilibrium values so qk - qk0 is replaced by

simply qk.

U = U0 + 1/2 tq

q

= 2

121 2 10 20 0

, 10

( , ,..., ) ( , ,..., )n

n n k jj k k j

U

q qU q q q U q q q q q

The constant term has no impact on the equations of motion, and the spring constant matrix has

components [ ]ij which are 2

0i j

U

q q

. As always, the derivatives are to be evaluated at the point

about which the expansion is made which means in the equilibrium configuration for this problem.

Exercise: Beginning with xi = xi (q1, q2, … , qn) and 1 2 3

12

particles

m x x x

show that the

kinetic energy can be represented as 1/2 tq q

where [ ]ij =

3

1

n

i

n

particles n j

x

q

xm

q

.

Exercise: Show that 1/2 tq

q

= 2

12

, 10

n

k jj k k j

U

q qq q

if [ ]ij is 2

0i j

U

q q

.


Exercise: For the simple harmonic oscillator, the kinetic energy is 212 m x and the potential energy is

212 k x . Choose x as the generalized coordinate for the problem, and find and .

Principles Underlying the Coupled Oscillator Procedures

Notation Alert: In this section the double under tilde identifies a square matrix, a single under

tilde identifies a column vector and a under tilde with superscript t identifies a row vector, the

transpose of a column vector.

2 2 1 20; ;K M x Kx M x M Kx x

This suggests that we find the eigenvalues and eigenvectors of the matrix 1M K

. This procedure

begins by setting the determinant 1 21 0M K

. Equivalently 2 0K M

as

1 2 21M M K K M

and the determinant M

is not zero.

Suppose that we have two eigenvectors.

1 ( ) 2 ( )n nM Kx x

1 ( ) 2 ( )m mM Kx x

Rearranging:

( ) 2 ( )( )

n nnKx Mx

( ) 2 ( )

( )n n

mKx Mx

( ) ( ) 2 ( ) ( ) ( ) ( ) 2 ( ) ( )( ) ( );

t t t tm n m n n m n mn mx Kx x Mx x Kx x Mx

( ) ( ) ( ) ( ) 2 ( ) ( ) 2 ( ) ( )( ) ( )

t t t tm n n m m n n mn mx Kx x Kx x Mx x Mx

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )andt t t tm n n m m n n mx Kx x Kx x Mx x Mx

In each pair above, the expressions are transposes of one another. As each expression is a 1 x 1

matrix, it is a scalar value which means its transpose is also that same scalar value. It follows that:

2 2 ( ) ( )( ) ( )0

tn mn m x Mx


This result suggests that we define the inner predict of mode vectors to be:

( ) ( ) ( ) ( )tn m n m

Mx Mx x x

Mass weighted Inner Product

We see that this inner product of modes must vanish for vectors with distinct eigenvalues.

( 2 2 ( ) ( )( ) ( ) 0 0

tn mn m x Mx

This mass weighted inner product is reasonable (even required) because a large mass moving 1 cm is

more significant mechanically that a small mass moving 1 cm.

Exercise: Compute the mass weighted inner product of the eigenvectors with distinct eigenvalues

for each of the cases in the example summary above.

Problems

1.) In sample claculation3, the equal mass oscillator example is set in to motion with mass one initial

displaced C from equilibrium, mass two at its equilibrium position, and with both masses at rest.

Show the steps that lead from the initial conditions to the identification of the constants as: A = - B = C/2 and S =

2.) Continue problem 1 to find: (with 12 S A and A S )

121

12 2

cos cos( )

( ) sin sin

t tx tC

x t t t

Explain clearly the manner in which this relation describes a transfer of energy back and forth

between the masses at the difference frequency .

3. For the system used in sample calculation 4, fill in all the details of the solution of the eigenvalue

problem. Show that the low frequency modes found using the two sets of generalized coordinates

correspond to the same pattern of motion.


4. Develop the equations of motion for

the system to the right and complete a full

eigen-mode solution to find the

frequencies and patterns of motion. Use

the shorthand 02 =

k/m.

Answers: 2 20 1

2

117 3; ;

14

5. Fill in the steps necessary to find the eigen-frequencies and eigen-modes in sample calculation 5.

6. Very detailed considerations lead to the conclusion that the metric of normal mode patterns

includes a factor of the mass matrix . If one is to demonstrate orthogonality, one must use the inner

product prescription: tiq

jq

=Ni ij. Consider sample calculation 2, the first unequal mass

oscillator.

0

0 2

m

m

110 1

12

( ) 1cos( )

( ) 1

x tA t

x t

0

0 2

21

5222

12

52

1cos

( )

( )B t

x t

x t

so: = 0

0 2

m

m

, 1

1

1q

and 2 1

2

1q

.

Show that 1tq

2q

= 0.

7. For each example in the example summary (located just before the problem section) show that

each pair of modes is orthogonal with respect to the mass matrix for the problem (see the previous

problem). That is that tiq

jq

=0 for i≠j.

8. CO2 molecule. Consider the 1D problem with

M = 3/4 m = 3 m0. Use x1 as the displacement of

the leftmost atom to the right, x2 as the

displacement of the center atom, the carbon, to

the right, … . Find the mass and spring

constant matrices, the eigen-frequencies and the mode patterns. A zero frequency mode is usually

associated with a conserved quantity as zero frequency indicates that a value is not changing in time.

k k 2 k

m 2m

x1 x2

k

M mm

x

k


Examine the zero frequency mode pattern to identify the quantity being conserved. What is the value

of this conserved quantity for each of the finite frequency modes? Reckless speculation: For an

isolated system, expect a zero frequency mode in which momentum is constant and higher frequency

modes each with zero net momentum.

9. The circular tube (radius R) with three equal mass beads separated by

identical springs. There is no friction. Adopt a set of three angles as the

set of generalized coordinates. . Find the mass and spring constant

matrices, the eigen-frequencies and the mode patterns. A zero frequency

mode is usually associated with a conserved quantity as zero frequency

indicates that a value is not changing in time. Examine the zero

frequency mode-pattern to identify the quantity being conserved. What is

the value of this conserved quantity for the other modes? 2 2 2

0 00, 3 ,3

Reckless speculation: For an isolated system, expect a zero frequency mode in which momentum is

constant and higher frequency modes each with zero net momentum.

10. The double pendulum: Consider the system illustrated.

Mass one is suspended by a string of length L that

makes an angle with respect to the vertical and

mass two is suspended below mass one by a string

of length D that makes an angle with respect to

the vertical. Choose the origin at the point where the

top string is fixed to the ceiling. Choose horizontal

to the right as the x direction and up as the y

direction. a.) Generate expressions for x1, y1, x2, and

y2 as functions of and .

b.) Generate expressions for 1 1 2 2, , andx y x y as functions of , and .

c.) Generate expression for K and U as functions of , and .

m

m m

k k

k

R

m1

m2

L

D


d.) Note that particle one moves with a velocity of magnitude L relative to the suspension point

and particle two moves at velocity of magnitude D relative to mass one. Identify the angle between

the velocities as - . The velocity of mass two relative to the suspension point is the vector sum of

these velocities. The square of its magnitude is therefore (L )2 + (D )2+ 2 L Dcos( - ).

e.) Return to the expression for U(,) and replace each trig function by its low order expansion

following the patterns: 2

2sin cos 1and . Recall: Include at least the first non-

vanishing correction to the limiting value when you implement an expansion. Extra terms can be

discarded later! Find an expression for U(,) - U() that is correct to second order in and.

f.) Return to the expression for K( , , and replace each trig function by its low order

expansion following the patterns: 2

2sin cos 1and . Recall: Include at least the first

non-vanishing correction to the limiting value when you implement an expansion. Extra terms can be

discarded later! Find an expression for K( , , that is correct to second order in the small

quantities , , Note that cos appears as (1 - 2/2) which is to second order.

g.) Use the results of the previous two parts to give the forms of and that are correct for small

oscillations.

h.) Compare the results with [ ]ij is identified as 2

0i j

U

q q

and [ ]ij =

3

1

n

i

n

particles n j

x

q

xm

q

evaluated in the equilibrium configuration.

11.) Solve the couple oscillator presented in the previous problem after setting the lengths of both

strings equal to L and the two masses equal to m.

12.) Consider three equal masses m strung in a line of four identical springs each with spring

constant k. (Adopt the notation o2 =

k/m.) Find the eigen-frequencies and the eigen-mode patterns.

k k k

m m

x1 x2

m

k

x3


Suspect Answers: 2 = 2 2 ,

1

2

1

;

1

0

1

2 = 2 2 ,

1

2

1

Note the form of the matrix for this problem. It is characteristic of nearest-neighbor coupling.

13.) Consider a general 2 by 2 matrix a b

c d

. Find its inverse as the matrix of cofactors divided by

the determinant. Find the transpose of the matrix. Find the inverse of the transpose. Comment on the

results. Recall that the determinant of the transpose is equal to the determinant of the original square

matrix. Argue that, for square matrices, the transpose of the inverse is the inverse of the transpose.

Search the web. Is the transpose of the inverse equal to the inverse of the transpose?w

References:

1. The Wolfram web site: mathworld.wolfram.com.

2. K. F. Riley, M. P. Hobson and S. J. Bence, Mathematical Methods for Physics and Engineering,

3nd Ed., Cambridge, Cambridge UK (2006).

3. John R. Taylor, Classical Mechanics, University Science Books, Sausalito CA (2005).

2 2

11 11 12 122 2

21 21 22 22

0

0i taK M K M

ebK M K M

Documents

Coupled Oscillators Concepts of primary interest: Simple