Beaded Strings - Rice Universitycox/vigre/stringtalk_6_11_08.pdf · 2008. 6. 13. · Beaded Strings Forward and Inverse Problems Hunter Gilbert, Walter Kelm, and Brian Leake Physics

Beaded StringsForward and Inverse Problems

Hunter Gilbert, Walter Kelm, and Brian Leake

Physics of Strings PFUG

11 June 2008

Frequencies → Masses + LengthsMasses + Lengths → Frequencies

Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 1 / 36

IntroductionWhy Study Strings?

Consists of a simple physical system

Governed by a system of differential equations

Interesting math for special conditions

Experimental results easy to acquire


IntroductionBrief History of the String Lab

Classes of problems we have worked on:

Dirac Damping of String (Sean Hardesty’s Thesis)

Viscous constant damping

Magnetic damping (Last summer)

Network of strings (Jesse Chan)

’http://cnx.org/content/m16177/latest/’

Physical lab for CAAM 335


IntroductionThis Summer: Beaded Strings

Point masses, taut masslessstring, fixed at both ends

Forward Problem

Given: masses, lengths, andinitial input

Find: the string’s motion

Inverse Problem

Given: the string’s motion forany input

Find: masses and lengths thatuniquely describe the system


Variable Meanings

A Beaded String is uniquely defined by masses and lengths:

mk = Mass of a particular bead.

`k = Distance between the k and k+1 beads.

We will use these expressions to get the equations of motion:

T (y ′) = Total Kinetic Energy as a function of time.

V (y) = Total Potential Energy as a function of time.

yk(t) = Vertical Displacment of a particular bead.


Dynamics of a Beaded String

To find frequencies of vibration given masses and positions ofthe beads, we will use a system of differential equations.

Kinetic energy is the focus of one of the equations.

T (y ′) =1

2

n∑k=1

mk(y ′k(t))2

Symbol meanings:

T (y ′) = Kinetic energy as a function of time.mk = Mass of the given bead.y ′k = Velocity of given bead.

Gives only part of the required system.





T (y ′) =1

2

n∑k=1

mk(y ′k(t))2

Symbol meanings:







T (y ′) =1

2

n∑k=1

mk(y ′k(t))2

Symbol meanings:





Potential Energy is the work done by stretching the string

Work = Force X Distance = Tension X String Elongation

Assume constant tension σ, and measure elongation by:

√`2k + (yk+1 − yk)2 − `k = `k

√1 +

(yk+1 − yk)2

`2k− `k

≈ `k

(1 +

1

2

(yk+1 − yk)2

`2k

)− `k

≈ (yk+1 − yk)2

2`2k.

V (y) =σ

2

n∑k=0

(yk+1 − yk)2

`k



Potential Energy is the work done by stretching the string

Work = Force X Distance = Tension X String Elongation

Assume constant tension σ, and measure elongation by:

√`2k + (yk+1 − yk)2 − `k = `k

√1 +

(yk+1 − yk)2

`2k− `k

≈ `k

(1 +

1

2

(yk+1 − yk)2

`2k

)− `k

≈ (yk+1 − yk)2

2`2k.

V (y) =σ

2

n∑k=0

(yk+1 − yk)2

`k


Dyanamics of a Beaded StringConservation of Energy

We can use the Euler-Lagrange Equation to generate a unifiedsystem of differential equations.

d

dt

∂T

∂y ′j+∂V

∂yj= 0, j = 1, ..., n

d

dt

∂T

∂y ′j(t) = mjy

′′j (t)

∂V

∂yj=

(− σ

`j−1

)yj−1 +

(σ

`j−1+σ

`j

)yj +

(−σ`j

)yj+1


Dynamics of a Beaded StringMatrix Form of Equations

Doing the necessary algebra provides n equations:

uk − uk+1

`k+

uk − uk−1

`k−1−mkλ

2uk = 0, k = 1, 2, ..., n

We can write this system in matrix form

My ′′(t) = −Ky(t)

Here, M is a diagonal matrix holding the masses of the beads.

K is a tridiagonal matrix that gives the elongation of the string



Essential Linear AlgebraInverse: M−1M = I

Eigenvalues (λ) and eigenvectors (u)

Defined for Matrix A by:

Au = λu

They are fundamental properties that describe the matrixbehaviorFor a beaded string, they correspond to vibration frequency(eigenvalue) and vibration mode shape (eigenvector)


The Solution

If we can solve that particular differential equation, we will beable to know the frequencies of vibration.

With the inverse, we can say:

y ′′(t) = −M−1Ky(t)

With the eigenvalues of M−1K , we can rewrite the equation as:

y ′′(t) = −V ΛV−1y(t)

V is the matrix where each column is an eigenvector of M−1K ,and Λ is a diagonal matrix of its eigenvalues.


The Solution

If we can solve that particular differential equation, we will beable to know the frequencies of vibration.

With the inverse, we can say:

y ′′(t) = −M−1Ky(t)

With the eigenvalues of M−1K , we can rewrite the equation as:

y ′′(t) = −V ΛV−1y(t)

V is the matrix where each column is an eigenvector of M−1K ,and Λ is a diagonal matrix of its eigenvalues.


The Solution, pt2

Since Λ is a diagonal matrix, the n × n system reduces to nindependent scalar equations.

γ′′j (t) = −ω2j γj(t)

If we presume that the string begins with no initial velocity, wecan state

γj(t) = γj(0)cos(ωjt)

We then have:

y(t) =∑n

j=1 γj(0)cos(ωjt)vj


The Solution, pt2

Since Λ is a diagonal matrix, the n × n system reduces to nindependent scalar equations.

γ′′j (t) = −ω2j γj(t)

If we presume that the string begins with no initial velocity, wecan state

γj(t) = γj(0)cos(ωjt)

We then have:

y(t) =∑n

j=1 γj(0)cos(ωjt)vj


Summary

The masses are displaced as the superposition of nindependent vectors vibrating at distinct frequencies.These frequencies are in turn given by the eigenvaluesof the matrix M−1K


Hearing the Composition of a String

We are able to go from masses and positions to frequencies ofvibration.

The natural question is now to ask if we can hear the positionsand masses of beads on a string from the vibrations it undergoes.

Using the techniques put forward by Gantmacher and Krein, wecan solve this inverse problem.


Hearing the Composition of a String

We are able to go from masses and positions to frequencies ofvibration.

The natural question is now to ask if we can hear the positionsand masses of beads on a string from the vibrations it undergoes.

Using the techniques put forward by Gantmacher and Krein, wecan solve this inverse problem.


Recurrence

We wish to find values of M and K to solve Ku = λMu.

By simple matrix multiplication, we can show:(− σ

`j−1

)uj−1 +

(σ

`j−1+σ

`j

)uj +

(−σ`j

)uj+1 = λmjuj

Having a fixed left & right end implies u0 = 0 and un+1 = 0

Rearranging the above, we get :

uj+1 =

(− `j`j−1

)uj−1 +

(1 +

`j`j−1− λ`jmj

σ

)uj

Since we know un+1 = 0 when the conditions for a fixed-fixedstring are met...

We can use this condition to make sure λ is an eigenvalue.


Recurrence



`j−1

)uj−1 +

(σ

`j−1+σ

`j

)uj +

(−σ`j

)uj+1 = λmjuj



uj+1 =

(− `j`j−1

)uj−1 +

(1 +

`j`j−1− λ`jmj

σ

)uj




Recurrence



`j−1

)uj−1 +

(σ

`j−1+σ

`j

)uj +

(−σ`j

)uj+1 = λmjuj



uj+1 =

(− `j`j−1

)uj−1 +

(1 +

`j`j−1− λ`jmj

σ

)uj




Polynomial Construction

When j = 1, we can say:

u2 =

(−`1`0

)u0 +

(1 +

`1`0− λ`1m1

σ

)u1

We now create a polynomial p of linear degree, and define itsuch that:u2 ≡ p1(λ)u1

Recalling the formula for a general element of the eigenvector,

uj+1 =

(− `j`j−1

)uj−1 +

(1 +

`j`j−1− λ`jmj

σ

)uj

We can reuse the same trick from above to create a polynomialof degree j.uj+1 ≡ pj(λ)u1


Polynomial Construction

When j = 1, we can say:

u2 =

(−`1`0

)u0 +

(1 +

`1`0− λ`1m1

σ

)u1

We now create a polynomial p of linear degree, and define itsuch that:u2 ≡ p1(λ)u1

Recalling the formula for a general element of the eigenvector,

uj+1 =

(− `j`j−1

)uj−1 +

(1 +

`j`j−1− λ`jmj

σ

)uj

We can reuse the same trick from above to create a polynomialof degree j.uj+1 ≡ pj(λ)u1


Insight

We can build pn by following the recurrence, which requiresknowledge of lengths & masses.

Or, if we have knowledge of the roots of the polynomialbeforehand, we can construct a polynomial of degree n multipliedby a real coefficient that will provide the same behavior.

We already have this knowledge.

The string is fixed at both ends, requiring u0 = 0 and un+1 = 0

Therefore, we can say

pn(λ) = γ∏n

j=1 (λ− λj)


Insight

We can build pn by following the recurrence, which requiresknowledge of lengths & masses.

Or, if we have knowledge of the roots of the polynomialbeforehand, we can construct a polynomial of degree n multipliedby a real coefficient that will provide the same behavior.

We already have this knowledge.

The string is fixed at both ends, requiring u0 = 0 and un+1 = 0

Therefore, we can say

pn(λ) = γ∏n

j=1 (λ− λj)


Fixed-Fixed & Fixed-flat

In order to make the solution unique, we need to find moreinformation.We will break our assumption that the string is fixed at bothends.Allow the string to move at one end, but require it to have 0slope at its last node.

There is now another set of eigenvalues, λ′, that represent the

system.The system is no longer underdetermined.

Figure: Fixed-Flat & Fixed-Fixed Strings


Fixed-Fixed & Fixed-flat

In order to make the solution unique, we need to find moreinformation.We will break our assumption that the string is fixed at bothends.Allow the string to move at one end, but require it to have 0slope at its last node.There is now another set of eigenvalues, λ

′, that represent the

system.The system is no longer underdetermined.

Figure: Fixed-Flat & Fixed-Fixed StringsHunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 18 / 36

Even More Polynomials

We can develop recurrence relations for the slope of the string,like was done for the positions of the beads.

Rearranging our equation developed from matrix multiplication:

uj+1 − uj

`j=

uj − uj−1

`j−1−(λmj

σ

)uj

Using the polynomials already developed, we can rewrite theequation as:

uj+1 − uj

`j=

pj(λ)u1 − pj−1(λ)u1

`j


Even More Polynomials

We can develop recurrence relations for the slope of the string,like was done for the positions of the beads.

Rearranging our equation developed from matrix multiplication:

uj+1 − uj

`j=

uj − uj−1

`j−1−(λmj

σ

)uj

Using the polynomials already developed, we can rewrite theequation as:

uj+1 − uj

`j=

pj(λ)u1 − pj−1(λ)u1

`j


Algebra in Anticipation

Using the same semantic trick as earlier, we define a newpolynomial qn that characterizes the system in fixed-flatoperation.

qj(λ) =1

`j(pj(λ)− pj−1(λ))

Equating expressions for the left and right sides of the equationused to define qj gives us:

qj(λ)u1 = qj−1(λ)−(λmj

σ

)pj−1(λ)u1


Algebra in Anticipation

Using the same semantic trick as earlier, we define a newpolynomial qn that characterizes the system in fixed-flatoperation.

qj(λ) =1

`j(pj(λ)− pj−1(λ))

Equating expressions for the left and right sides of the equationused to define qj gives us:

qj(λ)u1 = qj−1(λ)−(λmj

σ

)pj−1(λ)u1


Continued Fractions

With these characteristic polynomials, we can find masses &displacements.

Using recurrence relationships of the polynomials:

pn(λ)

qn(λ)=`nqn(λ) + pn−1(λ)

qn(λ)

pn(λ)

qn(λ)= `n +

1−mn

σλpn−1(λ) + qn−1(λ)

pn−1(λ)

pn(λ)

qn(λ)= `n +

1

−mn

σλ+

1

`n−1 +1

−mn−1

σλ+ ...+

1

`1 +1

−m1

σλ+

1

`0


Continued Fractions



pn(λ)


qn(λ)

pn(λ)

qn(λ)= `n +

1−mn


pn−1(λ)

pn(λ)

qn(λ)= `n +

1

−mn

σλ+

1

`n−1 +1

−mn−1

σλ+ ...+

1

`1 +1

−m1

σλ+

1

`0


Continued Fractions



pn(λ)


qn(λ)

pn(λ)

qn(λ)= `n +

1−mn


pn−1(λ)

pn(λ)

qn(λ)= `n +

1

−mn

σλ+

1

`n−1 +1

−mn−1

σλ+ ...+

1

`1 +1

−m1

σλ+

1

`0


Continued Fractions



pn(λ)


qn(λ)

pn(λ)

qn(λ)= `n +

1−mn


pn−1(λ)

pn(λ)

qn(λ)= `n +

1

−mn

σλ+

1

`n−1 +1

−mn−1

σλ+ ...+

1

`1 +1

−m1

σλ+

1

`0Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 21 / 36

Practical Problems

With the continued fraction, we have values for masses &lengths.

The process works from a theoretical standpoint...

Fixed-flat is hard to implement.

Fortunately, there is a workaround.

It can be shown that if the beads are symmetric about themidpoint of the string, we can find the fixed-flat eigenvalueswithout any extra work.

.


Practical Problems

With the continued fraction, we have values for masses &lengths.

The process works from a theoretical standpoint...

Fixed-flat is hard to implement.

Fortunately, there is a workaround.

It can be shown that if the beads are symmetric about themidpoint of the string, we can find the fixed-flat eigenvalueswithout any extra work.

.


Symmetry

Experimentally measuring eigenvalues on a symmetric beadedstring gives us a new set of eigenvalues, termed Λ.

Λ has some beautiful properties.

It can be shown that:

The odd indexed eigenvalues of Λ are all symmetric about themiddle of the string.The even indexed eigenvalues of Λ are antisymmetric about themidpoint.

This relationship holds for all symmetric strings.


Symmetry

Experimentally measuring eigenvalues on a symmetric beadedstring gives us a new set of eigenvalues, termed Λ.

Λ has some beautiful properties.

It can be shown that:

The odd indexed eigenvalues of Λ are all symmetric about themiddle of the string.The even indexed eigenvalues of Λ are antisymmetric about themidpoint.

This relationship holds for all symmetric strings.


The Point

The N eigenvalues of a symmetric beaded stringfixed at both ends exactly match the N/2 fixed-fixedand N/2 fixed-flat eigenvalues associated with half of

the string.

Figure: Eigenvalues of Symmetric Beaded Strings


The Inverse Algorithm

Record the eigenvalues for the whole string.

Use the eigenvalues to generate characteristic polynomials pn

and qn.

Use the characteristic polynomials to find the set of masses andlengths.

Figure: The Symmetric Beaded String


Non-symmetric ProblemIntroduction

Masses and lengths may vary arbitrarily

Spectra of entrire string no longer sufficient

Clamp string at some interior point between two masses

Leads to three problems with fixed-fixed boundary conditions


Non-symmetric ProblemSetup

For n1 masses on the left and n2 masses on the right:uk − uk+1

`k+

uk − uk−1

`k−1−mkλ

2uk = 0, k = 1, 2, ..., n1

uk − uk+1

˜k

+uk − uk−1

˜k−1

− mkλ2uk = 0, k = 1, 2, ..., n2

Whole String Boundary Conditions

un1+1 = un2+1

u0 = 0, u0 = 0

un1+1 − un1

`n1

+un2+1 − un2

ñ2

= 0

Clamped String Boundary Conditions

un1+1 = 0, un2+1 = 0

u0 = 0, u0 = 0




`k+

uk − uk−1

`k−1−mkλ

2uk = 0, k = 1, 2, ..., n1

uk − uk+1

˜k

+uk − uk−1

˜k−1

− mkλ2uk = 0, k = 1, 2, ..., n2


un1+1 = un2+1

u0 = 0, u0 = 0

un1+1 − un1

`n1

+un2+1 − un2

ñ2

= 0


un1+1 = 0, un2+1 = 0

u0 = 0, u0 = 0




`k+

uk − uk−1

`k−1−mkλ

2uk = 0, k = 1, 2, ..., n1

uk − uk+1

˜k

+uk − uk−1

˜k−1

− mkλ2uk = 0, k = 1, 2, ..., n2


un1+1 = un2+1

u0 = 0, u0 = 0

un1+1 − un1

`n1

+un2+1 − un2

ñ2

= 0


un1+1 = 0, un2+1 = 0

u0 = 0, u0 = 0




`k+

uk − uk−1

`k−1−mkλ

2uk = 0, k = 1, 2, ..., n1

uk − uk+1

˜k

+uk − uk−1

˜k−1

− mkλ2uk = 0, k = 1, 2, ..., n2


un1+1 = un2+1

u0 = 0, u0 = 0

un1+1 − un1

`n1

+un2+1 − un2

ñ2

= 0


un1+1 = 0, un2+1 = 0

u0 = 0, u0 = 0




`k+

uk − uk−1

`k−1−mkλ

2uk = 0, k = 1, 2, ..., n1

uk − uk+1

˜k

+uk − uk−1

˜k−1

− mkλ2uk = 0, k = 1, 2, ..., n2


un1+1 = un2+1

u0 = 0, u0 = 0

un1+1 − un1

`n1

+un2+1 − un2

ñ2

= 0


un1+1 = 0, un2+1 = 0

u0 = 0, u0 = 0




`k+

uk − uk−1

`k−1−mkλ

2uk = 0, k = 1, 2, ..., n1

uk − uk+1

˜k

+uk − uk−1

˜k−1

− mkλ2uk = 0, k = 1, 2, ..., n2


un1+1 = un2+1

u0 = 0, u0 = 0

un1+1 − un1

`n1

+un2+1 − un2

ñ2

= 0


un1+1 = 0, un2+1 = 0

u0 = 0, u0 = 0


Three spectra Inverse ProblemThe Problem

Relation between characteristic equations

pwhole(λ) = pleft(λ)qright(λ) + pright(λ)qleft(λ)

The roots of pleft and pright are the eigenvalues of the fixed-fixedproblems for the left and right segments

The roots of qleft and qright are the eigenvalues of the fixed-flatproblems

Problem: we can construct pleft and pright and pwhole up to ascaling factor from measured roots, but we cannot measure theroots of qleft and qright , preventing us from forming thecontinued fractions for the left and right segments.


Three spectra Inverse ProblemThe Problem

Relation between characteristic equations


The roots of pleft and pright are the eigenvalues of the fixed-fixedproblems for the left and right segments

The roots of qleft and qright are the eigenvalues of the fixed-flatproblems

Problem: we can construct pleft and pright and pwhole up to ascaling factor from measured roots, but we cannot measure theroots of qleft and qright , preventing us from forming thecontinued fractions for the left and right segments.


Three spectra Inverse ProblemDefinitions

Through the clever use of polynomials, Boyko and Pivovarchikshow that we do not need to know the roots of qleft and qright toconstruct them

Let λk , k = 1, 2, ..., (n1 + n2), be the spectra of the whole stringand let νk,`, k = 1, 2, ..., n1, and νk,r , k = 1, 2, ..., n2, be thespectra of the left and right parts, respectively

Let L be the length of the whole string and let L` and Lr be thelengths of the left and right segments of the clamped string












Three spectra Inverse ProblemConstructing Known Polynomials

Construct the polynomials we know, up to a scaling factor:

pwhole(λ) = Ln1+n2∏k=1

(1− λ

λk

)

pleft(λ) = L`

n1∏k=1

(1− λ

νk,`

)

pright(λ) = Lr

n2∏k=1

(1− λ

νk,r

)


Three spectra Inverse ProblemFinding qleft and qright

Recall the relationship of the characteristic equations:


Notice that pleft(νk,`) = 0 and pright(νk,r ) = 0, as these aresimply the roots we used to construct those polynomials.

Plug in νk,` and νk,r into the relation between the characteristicequations to reveal:

qleft(νk,`) =pwhole(νk,`)

pright(νk,`), k = 1, 2, ..., n1

qright(νk,r ) =pwhole(νk,r )

pleft(νk,r ), k = 1, 2, ..., n2








pright(νk,`), k = 1, 2, ..., n1


pleft(νk,r ), k = 1, 2, ..., n2








pright(νk,`), k = 1, 2, ..., n1


pleft(νk,r ), k = 1, 2, ..., n2


Three spectra Inverse ProblemFinding qleft and qright (continued)

Recall the continued fraction equation:

pn(λ)

qn(λ)= `n +

1

−mn

σλ+

1

`n−1 +1

−mn−1

σλ+ ...+

1

`1 +1

−m1

σλ+

1

`0

pn(0)

qn(0)= `n + `n−1 + ... + `1 + `0 = L

These equations must hold for the two clamped string segmentsas well, giving us the last points, qleft(0) = 1 and qright(0) = 1,needed to completely define the polynomials


Three spectra Inverse ProblemConstructing qleft and qright

Construct qleft and qright :

qleft(λ) =n1∑

k=1

λpwhole(νk,`)

νk,`pright(νk,`)

n1∏j=1,j 6=k

(λ− νj ,`)

(νk,` − νj ,`)+

n1∏k=1

νk,` − λνk,`

qright(λ) =n2∑

k=1

λpwhole(νk,r )

νk,rpleft(νk,r )

n2∏j=1,j 6=k

(λ− νj ,r )

(νk,r − νj ,r )+

n2∏k=1

νk,r − λνk,r

Notice:


pright(νk,`), qright(νk,r ) =

pwhole(νk,r )

pleft(νk,r )

qleft(0) = 1, qright(0) = 1


Three spectra Inverse ProblemConstructing qleft and qright

Construct qleft and qright :

qleft(λ) =n1∑

k=1

λpwhole(νk,`)

νk,`pright(νk,`)

n1∏j=1,j 6=k

(λ− νj ,`)

(νk,` − νj ,`)+

n1∏k=1

νk,` − λνk,`

qright(λ) =n2∑

k=1

λpwhole(νk,r )

νk,rpleft(νk,r )

n2∏j=1,j 6=k

(λ− νj ,r )

(νk,r − νj ,r )+

n2∏k=1

νk,r − λνk,r

Notice:


pright(νk,`), qright(νk,r ) =

pwhole(νk,r )

pleft(νk,r )

qleft(0) = 1, qright(0) = 1


Three spectra Inverse ProblemFinding {mk}, {`k}, {mk}, and {˜

k}

Unique lengths and masses are then determined by continuedfraction expansion of ratio of p’s and q’s, i.e.

pleft(λ)

qleft(λ)= `n1+

1

−mn1

σλ+

1

`n1−1 +1

−mn1−1

σλ+ ...+

1

`1 +1

−m1

σλ+

1

`0


Future Work

Three spectra problem with damping at an interior point

No unique solution

Damping at a mass

Damping at the ends


Mentors

Jeffrey HokansonDr. Steve CoxDr. Mark Embree

References

CAAM 335 Lab 9.http://www.caam.rice.edu/ caam335lab/lab9.pdfCAAM 335 Lab 10.http://www.caam.rice.edu/ caam335lab/lab10.pdfBoyko O and Pivovarchik V. ”The inverse three-spectralproblem for a Stieltjes string and the inverse problem withone-dimensional damping.” Inverse Problems 24 (2008) 1-13.Gantmacher F P and Krein M G. Oscillation Matrices andKernels and Small Vibrations of Mechanical Systems. RevisedEdition. Providence, Rhode Island: AMS Chelsea Pub, 2002.


http://www.caam.rice.edu/~caam335lab/lab9.pdf

http://www.caam.rice.edu/~caam335lab/lab10.pdf

Beaded Strings - Rice Universitycox/vigre/stringtalk_6_11_08.pdf · 2008. 6. 13. · Beaded Strings Forward and Inverse Problems Hunter Gilbert, Walter Kelm, and Brian Leake Physics

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