beads on a string: an inverse eigenvalue problem

JAKOB NILSSON

KARLSTAD UNIVERSITY

FACULTY OF HEALTH, SCIENCE AND TECHNOLOGY

FYGB08 - ANALYTIC MECHANICS 1, AUTUMN 2015

Beads on a String: An InverseEigenvalue Problem

Abstract

In this project an introduction to the concept of an inverse eigenvalueproblem is presented by way of a practical example. We present two al-gorithms, which are shown to enable the determination of the weights andpositions of beads placed on a vibrating string, given vibrational data. Inorder to outline the procedure of finding these algorithms, Newton’s equa-tions of motion of the coupled beaded system are also solved, assumingtime-harmonic motion. As an example of a condition which shortens thecalculations involved in the inverse problem, systems possessing symmetryabout the middle point of the string are also discussed, along with the implic-ations of this property.

SUPERVISOR: Igor BuchbergerKEY WORDS: Inverse eigenvalue problem, Beaded string, Coupled oscillations,

Symmetrical load

Contents

1 Introduction 1

2 The Forward Problem 12.1 Problem Specification . . . . . . . . . . . . . . . . . . . . . . . . 12.2 Solving the Linearized Equations of Motion . . . . . . . . . . . . 3

3 The Inverse Eigenvalue Problem 43.1 Problem Specification . . . . . . . . . . . . . . . . . . . . . . . . 43.2 Inner Product Spaces and Construction of a Product . . . . . . . . 53.3 Determining Weights and Nodes . . . . . . . . . . . . . . . . . . 6

3.3.1 Interpolating Polynomials . . . . . . . . . . . . . . . . . 63.3.2 Solving for qn−1(z) by Neumann Boundary Conditon . . . 8

3.4 Determining the Matrix Elements of An . . . . . . . . . . . . . . 103.5 Finding the Masses and Lengths . . . . . . . . . . . . . . . . . . 123.6 Special Case: Symmetrical Load . . . . . . . . . . . . . . . . . . 15

4 Discussion 17

5 Conclusions 18

Analytic Mechanics 1, FYGB08Autumn 2015

Beads on a StringJakob Nilsson

m1

m2

m3

mn

φ0

φ1

φ2

φ3

φn

~r1

x

y

~r2

~r3

~rn

`0 `1 `2 `n

Figure 1: Beads on displaced string, subjected to Dirichlet boundary conditions.Hollowed out circles indicate equilibrium positions.

1 Introduction

An inverse eigenvalue problem is the problem of reconstructing a matrix based onexperimental spectral data. The data are required to give some insight to the ei-genvectors and eigenvalues associated with a system [1]. Examples where theseproblem statements occur range from determining the shape of a body by analysisof its frequencies of vibration (e.g. a drum membrane, see [2]), to problems of con-structing conductivity profiles, [3] and even to problems of describing molecularstructures from Raman or infra-red spectral data [4].

The current paper concerns itself with giving a brief introduction to the inversespectral problem by investigation of a system of beads on a vibrating string. Theprimary goal is to outline a way of entering measured vibrational data into an al-gorithm which gives estimates to the positions and masses of the beads. In orderto justify the algorithm, the solution to the ”forward problem” must be shown byeffectively solving Newton’s equations of motion for the beads.

2 The Forward Problem

2.1 Problem Specification

The system to be analysed is depicted in Figure 1. The n beads are assumed to bepoint masses placed on a massless string, and the string of total length ` is set intomotion by plucking it.

Application of Newton’s second law to the jth mass gives{∑Fx = m jx j = τ [cos(φ j)− cos(φ j−1)]

∑Fy = m jy j = τ [sin(φ j)− sin(φ j−1)](1)

where τ is the known tension of the string and φ j is the angle made by the dis-placement vector ~r j relative to the positive x-direction. Any contribution to the

Page 1 of 19



vibrational motion due to the gravitational force is assumed to be negligible.Without any assumptions made on the magnitude of the displacements of the

beads, the trigonometric functions in Eq. (1), when expressed in terms of the posi-tion coordinates of the beads, read

cos(φ j) =` j +(x j+1− x j)√

`2j +(x j+1− x j)2 +(y j+1− y j)2

, (2)

sin(φ j) =y j+1− y j√

`2j +(x j+1− x j)2 +(y j+1− y j)2

. (3)

Due to the complexity of the above expressions we shall assume that the xand y components of all displacement vectors {~r j}n

j=1 are small as compared to alllengths {` j}n

j=1.This assumption effectively reduces Eq. (2) into

cos(φ j) = 0 { j ∈ N|1≤ j ≤ n},

and thus the first row in Eq. (1) reduces to

∑Fx = 0.

On the other hand, the same assumption implies that Eq. (2) reduces to

sin(φ j) =y j+1− y j

` j,

and thus the equation of motion in the vertical y direction is reduced into

m jy j = τ

(y j+1− y j

` j−

y j− y j−1

` j−1

), (4)

with Dirichlet boundary conditions y0 = 0 and yn+1 = 0 as implied by Figure 1. Inmatrix notation, Eq. (4) is to be written as

My =−Ky, (5)

where M is the symmetric diagonal mass matrix

M =

m1

m2m3

. . .mn

, (6)

and K is the symmetric tri-diagonal stiffness matrix

K = τ

`−1

0 + `−11 −`−1

1−`−1

1 `−11 + `−1

2. . .

. . . . . . −`−1n−1

−`−1n−1 `−1

n−1 + `−1n

. (7)

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2.2 Solving the Linearized Equations of Motion

We now make another assumption, namely that the motion of the beads is harmonicand undamped. We therefore assume that our system of equations (5) has solutionof the form

y(t) = veiωt . (8)

Here, v is some vector which takes into account the configuration of the systemand ω is the eigenfrequency of oscillation.

Insertion of the ansatz (8) into the equation of motion (5) transforms the prob-lem into the generalized eigenvalue problem

λMv = Kv, (9)

where λ = ω2. Defining the matrix An as

An = M−1/2KM−1/2 =

a1 b1b1 a2

. . .. . . . . . bn−1

bn−1 an

, (10)

where the elements a j and b j are of the form{a j = τ(1/` j−1 +1/` j)/m j

b j =−τ/(` j√mkm j+1)

, (11)

and defining the vector u asu = M1/2v, (12)

Eq. (9) is now to be transformed into the standard eigenvalue problem

Anu = λu. (13)

Since the matrices M and K are both symmetric, the matrix An also exhibitsthis property as well. As a direct consequence of the symmetry of An, we knowthat the eigenvalues {λ j}n

j=1 = {ω2j }n

j=1 are real-valued. We also note that eachspecific value corresponds to a unique ”eigendirection”, determined by the mutu-ally orthonormal eigenvectors {u j}n

j=1 [5]. Since it also can be shown that theseeigenvalues indeed are non-degenerate (see end of section 2 of [6]), we can assumethat the amplitude state vector y can be written as the superposition

y =n

∑j=1

γ j(t)v j. (14)

We also note that the orthogonality of the u j’s ensures that the vectors {v j}nj=1 are

M-orthogonal.

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Insertion of the intermediate result (14) into the eq. of motion (5) now yieldsthe uncoupled differential equations

γ j(t) =−ω2j γ j(t). (15)

Thus, we have solutions for the time-dependent components γ j of the form

γ j(t) = c j cos(ω jt)+ s j sin(ω jt). (16)

For a pluck of the string at time t = 0 it is only natural to assume that thestring begins at rest. We therefore put all s j’s in (16) equal to zero and arrive at aparticular solution

y =n

∑j=1

c j cos(ω jt)v j. (17)

It has now been made plausible that the motion of the beads are superpositionsof oscillations with frequencies equal to the n distinct eigenfrequencies {ω j}n

j=1.

3 The Inverse Eigenvalue Problem

3.1 Problem Specification

Having found the harmonic solution (17) to the equations of motion (5), vibrationaldata can now be analysed. The experiment begins by plucking the string of knowntension τ and total length ` while taking measurements of the amplitude over aset time interval. With the spectral data collected and the eigenvalues calculatedusing e.g. an FFT algorithm,1 the problem of finding the masses and their positionsis now essentially a matter of finding the elements of matrix An (see Eq. (10)).This in turn allows for the elements m j in Eq. (6) and the ` j’s in Eq. (7) to bedetermined. Cox et al. outline two different procedures for finding these matrixelements [6]. The first involves an algorithm based on orthogonal polynomials,and through it, the matrix An is reconstructed. In contrast, the second alternative isbased on continued fractions and it bypasses the reconstruction of An, effectivelygoing straight for the masses and positions (see appendix of [6] for this secondmethod). Delving into both methods would however be superfluous for the purposeof giving an introduction to the problem, especially since the methods provideessentially the same results. Thus, the method of orthogonal polynomials is theone to be put under the loupe.

1See e.g. Ch.12 of [7] for a brief introduction to the Discrete Fourier Transform and the FastFourier transform (FFT) algorithm.

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3.2 Inner Product Spaces and Construction of a Product

Let’s begin by defining an inner product space. If α , β and γ are elements of sucha product space, then the following three axioms must be fulfilled [5]:

• Conjugates symmetry: 〈α,β 〉= 〈β ,α〉

• Linearity: 〈α +β ,γ〉= 〈α,γ〉+ 〈β ,γ〉, and 〈aα,β 〉= a〈α,β 〉

• Positive-definiteness: 〈α,α〉= ||α|| ≥ 0 and 〈α,α〉= ||α||= 0 =⇒ α = 0.

Taking these requirements into consideration, we define our discrete innerproduct as

〈p(z),q(z)〉=η

∑j=1

ω j p(ξ j)q(ξ j), (18)

where α(z) and β (z) are polynomials of a certain degree η−1 and {ξ j}η

j=1 are thechosen nodes in the domain of α and β . The ω j’s are weights.2

For our purposes we need to define the elements of our inner product (18) asconsisting of monic polynomials, i.e. polynomials in which the leading coefficientis 1. From this definition, the lowest order polynomial of this inner product spacemust be

p0(z) = 1. (19)

We now want to construct a sequence of orthogonal polynomials up to a givendegree n (the supposed number of beads), in which each polynomial is orthogonalto all lower degree polynomials. Grahm-Schmidt orthogonalization thus yields theremaining higher-order monic polynomials as

pk+1 = zpk(z)−k

∑j=1

〈z pk, p j〉〈p j, p j〉

p j(z). (20)

The denominator of the sum in (20) can however be simplified, owing to theorthogonality between the polynomials of different degree. Thus, only the termswhere j = k and j = k−1 can possibly be different from 0 in the sum in Eq. (20).The recursion now reduces to

pk+1 =zpk(z)−〈z pk, pk〉〈pk, pk〉

pk(z)−〈z pk, pk−1〉〈pk−1, pk−1〉

pk−1(z)

=zpk(z)−〈z pk, pk〉〈pk, pk〉

pk(z)−||pk||2

||pk−1||2pk−1(z)

= (z−ak+1)pk(z)−b2k pk−1(z) , (21)

2Designating weights by ω is an unfortunate convention since it is the same as the eigenfrequen-cies in the forward problem. Its meaning should however be clear from the context.

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where in going from the first to the second row we’ve used the property 〈zpk, pk−1〉=〈pk,zpk−1〉, and the requirement that pk must be orthogonal to all other polynomi-als of lower degree (recall that any such polynomial can be written as a linearcombination of lower degree polynomials {p j}k−1

j=1).Now comes a crucial step: We recognize that by expanding the determinant

of the matrix zI−An around its last row we effectively get the same recurrencerelation as in Eq. (21):

det(zI−An) = (z−an)det(zI−An−1)−b2n−1det(zI−An−2)

Let qk(z) := det(zI−Ak)→

→ qn(z) = (z−an)qn−1(z)−b2n−1qn−2(z) , (22)

with an and bn−1 being the lower-most elements of An. Upon comparison betweenEqs. (21) and (22), we see that the elements q j and p j indeed obey the same recur-rence relation! As a direct consequence of this we would do well to demand thatthe matrix element an is to obey the equation

an =〈zqn−1,qn−1〉〈qn−1,qn−1〉

, (23)

as in accordance with the second row in (21). Eq. (22) can now be rearranged tosolve for b2

n−1:b2

n−1qn−2(z) = (z−an)qn−1(z)−qn(z). (24)

With this result we could, in theory, use the fact that qn−2(z) is monic and thencompare the powers of zn−2 on both sides of the equation. However, we cannotactually calculate the right-hand side of Eq. (24) as of yet, since neither the weightsω j, the nodes ξ j, the determinant qn−1(z), nor an has been identified. We willtherefore return to Eqs. (23) and (24) at a later stage.

3.3 Determining Weights and Nodes

3.3.1 Interpolating Polynomials

It is commonly known that there always exists a unique polynomial r(z) ∈Pn−1that interpolates qn(z) at the n nodes {ξ j}n

j=1 [8], i.e.,

r(ξ j) = qn(ξ j), j = 1, . . . ,n. (25)

Just like as what was said about pn(z) in Section 3.2, we would like qn(z) to beorthogonal to all lower degree polynomials. Hence, as a consequence of Eq. (25)and this additional requirement we have

0 = 〈qn,r〉=n

∑j=1

ω jqn(ξ j)r(ξ j) =n

∑j=1

ω jqn(ξ j)2. (26)

Page 6 of 19



Since weights must be positive, we find that

qn(ξ j) = 0 j = 1,2, . . . ,n. (27)

Also, from the definition of qn in the second row in (22) we can use the identity [5]

qn(z) = det(zI−An) =n

∏j=1

(z−λ j), (28)

and upon comparison between (27) and (28) we deduce that

ξ j = λ j = ω2j . (29)

With the nodes now having been determined, our focus is now shifted to theweights. To this end we introduce the quantity

δk(z) = ∏Jk

z−ξ j

ξk−ξ j, Jk = { j ∈ N|1≤ j ≤ n, j 6= k}. (30)

We also note that since

dqn(z)dz

∣∣∣∣z=ξk

= q′n(ξk) = limz→ξk

q(z)−��:0qn(ξk)

z−ξk=

= limz→ξk

(z−ξk)∏Jk

(z−ξ j)

z−ξk= ∏

Jk

(ξk−ξ j), (31)

this derivative is exactly equal to the product of denominators in (30). Hence theexpression for δk is to be rewritten as

δk(z) =∏Jk

(z−ξ j)

q′n(ξk)=

qn−1(z)q′n(ξk)

+Rn−2(z), (32)

where Rn−2(z) is some unknown polynomial of degree n− 2. It readily followsfrom the above expression that

〈qn−1,δk〉=〈qn−1,qn−1〉

q′n(ξk)=||qn−1||2

q′n(ξk), (33)

where the orthogonality between qn−1 and Rn−2 has been used. At the same time,the weights may be explicitly found by the same inner product:

〈qn−1,δk〉=n

∑j=1

ω jqn−1(ξ j)δk(ξ j) = ωkqn−1(ξk)

⇔

ωk =〈qn−1,δk〉qn−1(ξk)

, (34)

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where in the last step on the first row we’ve used the fact that

δk(ξ j) =

{0 when j 6= k1 when j = k

. (35)

Inserting (33) into the nominator of (34) the weights ωk are written explicitlyas

ωk =||qn−1||2

q′n(ξk)qn−1(ξk). (36)

This expression can however be simplified, as orthogonality properties of the innerproduct are left unchanged if the weights ωk were to be multiplied by a scalar.Therefore, it is convenient to simplify the above expression into

ωk =1

q′n(ξk)qn−1(ξk)=

1(∏

nj=1,6=k(ξ j−ξk)

)qn−1(ξk)

, (37)

while still preserving the orthogonality conditions. The observant reader wouldnow however object that the denominator in (37) would have singular points if ithappens that qn−1(ξk) = 0 for some k. This is fortunately not the case, since

qn−1(z) = det(zI−An−1) =n−1

∏j=1

(z−µ j), (38)

and the eigenvalues {u j}n−1j=1 can be proven to be distinct from all nodes ξ j = λ j. In

fact, Cauchy’s interlacing theorem gives that the two sets of eigenvalues obey thepattern [6]

λ1 < µ1 < λ2 < .. . < λn−1 < µn−1 < λn. (39)

Thus the values of qn−1(ξk) are all separated from 0 and no singular points exist in(37).

But alas, we are still unable to calculate ωk since qn−1(ξk) is an unknown quant-ity. The following section will aim to rectify this shortcoming.

3.3.2 Solving for qn−1(z) by Neumann Boundary Conditon

In order to begin identifying qn−1(z) we need additional data. Section 6 of Coxet al. [6] suggest an approach which at first glance might appear irrelevant to thecurrent problem, but as it turns out, by using this method the missing polynomialqn−1 falls out quite nicely from calculations.

The trick is to apply a Neumann boundary condition to one of the clamped endsof the tensioned string in Figure 1 on page 1. Under the assumption that it is theright-most end where the Neumann condition is to replace the Dirichlet condition,the resulting beaded system would look like the one depicted in Figure 2. TheNeumann condition is justified under our current set of assumptions because theposition of the tensioned, practically massless string must be solely determined by

Page 8 of 19



m1

m2

m3

mn

φ0

φ1

φ2

φ3~r1

x

y

~r2

~r3

~rn

`0 `1 `2 `n

Figure 2: The rearranged system with a Neumann boundary condition replacingthe Dirichlet condition at the right end of the string. A frictionless ring sliding ona pole effectively impose the desired boundary condition.

the masses. This implies that at the right-most end, the position of the mass mn

completely determines the position of the frictionless ring and thus the position ofthe string end as well, effectively bringing the ring to the same height as the nthmass.

Mathematically, what happens to the system is that only the eq. of motion forthe nth bead changes, the resulting equation being

mnyn =−τ

(yn− yn−1

`n−1

), (40)

as can be readily seen by replacing yn+1 = 0 with yn+1 = yn in (4). The newcorresponding system matrix An is thus to be written as

An = An−τ

`nmneneT

n , (41)

whereeT

k = [0,0, . . . , 1︸︷︷︸kth entry

, . . . ,1]T1×n. (42)

We now define a quantity qn(z), in analogue to the original system

qn(z) := det(

zI− An

)=

n

∏j=1

(z− λ j). (43)

By expanding the determinants qn(z) and qn−1(z) around their last rows andtaking differences, Cox et al. go on to show that

qn(z)−qn(z) = (an− an)qn−1(z). (44)

Judging from Eq. (44) above, it would seems as though we’ve now determinedthe eigenvalues µ j of qn−1(z): With both sets of eigenvalues λ j and λ j known

Page 9 of 19



from spectral data, the eigenvalues {µ j}n−1j=1 are simply the roots of the difference

between the left-hand side polynomials, and hence qn−1(z) is found by the iden-tity (38). In practice though, this approach is not a feasible one, since polynomialroots are highly sensitive to noise; which is always present in the experimental datato some extent. Fortunately, for our purposes of finding the weights ωk, there is aneat way of circumventing this problem: Insert the eigenvalues λk = ξk into (44)and get

qn(ξk) =n

∏j=1

(ξk− λ j) = (an− an)qn−1(ξk), (45)

and hence qn−1(ξk) is now given up to a factor (an− an). Insertion into (37) givesthe weights (up to some scaling factor) as

ωk =1(

∏nj=1,6=k(ξ j−ξk)

)qn(ξk)

=1(

∏nj=1,6=k(ξ j−ξk)

) n∏j=1

(ξk− λ j). (46)

Lastly, an explicit expression for qn−1(z) can found from Eq. (44) by an im-portant observation, namely that by expanding the left-hand side polynomials intoproducts of differences between z and the sets of eigenvalues we get

qn(z)−qn(z) =n

∏j=1

(z− λ j)−n

∏j=1

(z−λ j) =

=

(n

∑j=1

(λ j− λ j)

)zn−1 + . . . (47)

Hence, the termn∑j=1

(λ j− λ j) must satisfy the condition

(n

∑j=1

(λ j− λ j)

)= an− an. (48)

Insertion of the result (48) into (44) and rearranging yields

qn−1(z) =qn(z)−qn(z)

n∑j=1

(λ j− λ j). (49)

3.4 Determining the Matrix Elements of An

With the nodes {ξk}nk=1 and weights {ωk}n

k=1 determined by Eqs. (29) and (46)respectively, we’ve effectively decided upon the ”inner workings” of our innerproduct. With knowledge of qn−1(ξk) by way of Eq. (49), the matrix element an

Page 10 of 19



can now be determined by Eq. (23)

an =〈zqn−1,qn−1〉〈qn−1,qn−1〉

=

n∑j=1

ω jξ jqn−1(ξ j)2

n∑j=1

ω jqn−1(ξ j)2. (50)

As was mentioned at the end of Section 3.2, the squared off-diagonal elementb2

n−1 could now theoretically be found by comparing the left and right-hand sidecoefficients in Eq. (24), the equation restated here for convenience:

b2n−1qn−2(z) = (z−an)qn−1(z)−qn(z). (24)

However, while this is a viable way of finding the value of b2n−1, at this stage

we’ve already found the values of qn−1(ξ j). Thus for the purpose of finding anefficient algorithm we multiply (24) by zqn−1(z) and solve for b2

n−1, arriving at theexpression

b2n−1 =

〈zqn−1,zqn−1〉−an〈qn−1,zqn−1〉−〈qn,zqn−1〉〈qn−2,zqn−1〉

=

=〈zqn−1,zqn−1〉−an〈qn−1,zqn−1〉−〈qn,zqn−1〉

〈qn−1,qn−1〉. (51)

In going to the denominator in the last line, we’ve used the monic property ofqn−2(z). By recalling Eq. (11), bn−1 is now determined as the negative root ofb2

n−1.The very last step in order to finish the algorithm is to find the polynomial

qn−2(z) at z = ξ j. This is done by rearranging (24) into

qn−2(ξ j) =(ξ j−an)qn−1(ξ j)−qn(ξ j)

b2n−1

. (52)

Thus to summarize the algorithm for determining the elements of An:

Page 11 of 19



1. Use ωk =1

(∏nj=1,6=k(ξ j−ξk))

n∏j=1

(ξk−λ j)(Eq. (46)) to determine the weights of

the inner product.

2. Use qn−1(ξ j) =qn(ξ j)−qn(ξ j)

n∑j=1

(λ j−λ j)(Eq. (49)) to find the values of qn−1 at all

the nodes {ξ j}nj=1.

3. Compute an =

n∑j=1

ω jξ jqn−1(ξ j)2

n∑j=1

ω jqn−1(ξ j)2(Eq. (50)) to find the lower-most diagonal

matrix element an.

4. Compute b2n−1 =

〈zqn−1,zqn−1〉−an〈qn−1,zqn−1〉−〈qn,zqn−1〉〈qn−1,qn−1〉 (Eq. (51)) and make

the identification bn−1 =−√

b2n−1.

5. Find the values of the next monic polynomial qn−2 at the nodes byqn−2(ξ j) =

(ξ j−an)qn−1(ξ j)−qn(ξ j)

b2n−1

(Eq. (52)).

6. With qn−2(ξ j) determined, determine an−1 as was done for an in step3, followed by bn−2 as seen in step 4. Then compute qn−3(ξ j) usingthe same procedure as in step 5, and so on and so forth, until all matrixelements in An have been computed.

3.5 Finding the Masses and Lengths

With elements of An determined by the procedure outlined in the previous section,we now turn our attention to finding the masses m j and the lengths ` j, given spe-cific values of τ and total string length `. Cox et al. start off their derivations byintroducing the unit vector

e = [1,1,1, . . . ,1]T1×n

and noting thatKe =

τ

`0e1 +

τ

`nen, (53)

which can readily be seen by noting the ”telescoping” nature of the off-diagonalelements in K (see Eq. (7) on page 2). By introducing the vector d as

dT := M1/2eT = [√

m1,√

m2,√

m3, . . . ,√

mn]T (54)

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the above expression can be reformulated as

Ke = M1/2AnM1/2e =τ

`0e1 +

τ

`nen

⇔

And =τ

`0√

m1e1 +

τ

`n√

mnen. (55)

In order to solve this matrix equation, we the denote vectors x and y as thesolutions to the equations

Anx = e1, Any = en. (56)

By using the fact that An is a linear operator obeying the conditions of linearity, wecan readily recast Eq. (55) unto the form

And = An

(τ

`0√

m1x+

τ

`n√

mny). (57)

Therefore, the explicit expression for d is

d =τ

`0√

m1x+

τ

`n√

mny. (58)

In order to find the fractions making up to the right in (58), we use the factthat [6]

tr(An) = tr(

An

)+

τ

`nmn, (59)

which upon rearrangement and multiplication by√

mn yields

τ

`n√

mn=√

mn

(tr(A)n− tr

(An

))=√

mn

n

∑k=1

(λk− λk). (60)

Thus, Eq. (60) determines the value of the coefficient in front of y . Also, from thenth row in (58) we note that

√mn =

τ

`0√

m1xn +

τ

`n√

mnyn

⇔

τ

`0√

m1=

1xn

(√

mn−τ

`√

mnyn

)=

√mn

xn

(1− yn

n

∑k=1

(λk− λk)

). (61)

Thus, Eq. (61) determines the x coefficient in (58), and the equation for d nowreads as

d =

√mn

xn

(1− yn

n

∑k=1

(λk− λk)

)x+√

mn

n

∑k=1

(λk− λk)y. (62)

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A neat reformulation appears if we introduce the quantity

d :=d√

mn=

1xn

(1− yn

n

∑k=1

(λk− λk)

)x+

n

∑k=1

(λk− λk)y (63)

and note that

K/mn =(

M1/2AnM1/2)/mn = diag(d)Andiag(d), (64)

where diag(d) is the matrix where the elements of d are placed along its maindiagonal.

From Eq. (64) it is readily seen that the following equality exist between the( j, j+1) entries on both sides of Eq. (64):

−τ

` jmn= b jd jd j+1

⇔

` jmn =−τ

b jd jd j+1, j = 1,2, . . . ,n−1, (65)

while the (1,1) and (n,n) entries obey the relations

τ

mn

(1`0

+1`1

)= a1d 2

1

⇔

`0mn =τ

a1d 21 + τ/(`1mn)

, (66)

andτ

mn

(1

`n−1+

1`n

)= and 2

n

⇔

`nmn =τ

and 2n + τ/(`n−1mn)

, (67)

respectively. The observant reader may have noticed by now that all the lengths{` j}n

j=0 are determined up to a scaling factor mn in Eqs. (65), (66) and (67). mn

can be computed by using the identity

mn = mn

n∑j=0

` j

`=

n∑j=0

mn` j

`. (68)

With mn determined, one may then solve for the lengths in Eqs. (65), (66) and(67). Also, the remaining masses {m j}n−1

j=1 can now readily be found by comparisonof vector components in (62).

Thus to summarize, the procedure for finding the masses and lengths is asfollows:

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1. Solve the two matrix equations Anx = e1 and Any = en for x and y re-spectively.

2. Define d := 1xn

(1− yn

n∑

k=1(λk− λk)

)x+

n∑

k=1(λk− λk)y (see Eq. (63)).

3. Compute ` jmn =− τ

b j d j d j+1, j = 1,2, . . . ,n−1, (see Eq. (65)).

4. Compute `0mn =τ

a1d 21 +τ/(`1mn)

(see Eq. (66)).

5. Compute `nmn =τ

and 2n +τ/(`n−1mn)

(see Eq. (67)).

6. Compute the mass mn =

n∑j=0

mn` j

` (see Eq (68)).

7. Solve for the lengths using steps 3-5.

8. Solve for the remaining masses by use of Eq. (62).

3.6 Special Case: Symmetrical Load

Having presented an elaborate method to solve for the masses and lengths in theprevious sections one might ask the question ”Do there exist systems for whichthese calculations are somehow simplified?”.

Luckily, the answer to this question is ”yes,” provided an even amount of Nbeads is placed on the string of length L, as to give rise to a resulting symmet-ric load around its middle point L/2. For this condition set on the system, itshould perhaps come as no big surprise that the eigenvectors exhibit alternatingsymmetric and anti-symmetric properties, with odd eigenvalues (Λ1,Λ3,Λ5) cor-responding to symmetric eigenvectors, and eigenvalues (Λ2,Λ4,Λ6) correspondingto anti-symmetric eigenvectors centred around the point of symmetry [6]. Givingan example when N = 6, a specific system is schematically presented in Figure 3.

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M1 M2 M3 M6M5M4

L0 L1 L2 L3 L4 L5 L6

Figure 3: System with symmetrically placed beads: M1 = M6, M2 = M5, M3 = M4& L1 = L6, L2 = L5, L3 = L4.

Cox et al. give schematical representations of the eigenvectors, as can be seenin Figure 4. It can now be readily noted that for the symmetric eigenvectors the sys-tem can be reduced to just its left-most half while imposing a Neumann boundarycondition at L/2. Thus, the corresponding eigenvalues {Λ1,Λ3,Λ5} are the eigen-values {λ1, λ2, λ3} for the Dirichlet-Neumann system of string length `= L/2 andn = N/2 = 3 number of beads.

Figure 4: Symmetrical eigenvectors to the left, with corresponding odd-numberedeigenvalues, and anti-symmetrical eigenvectors with even-numbered eigenvaulesto the right.

In similar fashion, it is readily seen that the anti-symmetric eigenvectors obeythe original Dirichlet boundary conditions, and thus we make the identification{Λ2,Λ4,Λ6}= {λ1,λ2,λ3}.

Hence we conclude that the experimental setup does not need to change if thesystem obeys the above criteria of symmetry. The symmetric system is reduced intoan equivalent n = N/2 bead system on a string of length `= L/2.

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Figure 5: A comparison between the masses and lengths as recovered by the al-gorithms and as measured by directly investigating the system. Source: [6].

4 Discussion

After deriving the two algorithms required for the determination of the masses andthe lengths, one might ask how well our acquired model agrees with the actualsystem. Cox et al. provide some experimental results from the laboratory. As anexample, for a symmetric four bead system, their results are as given in Figure 5.As can be seen from the figure, the positions seem to diverge by no more than acouple of millimetres on a string of length ` ≈ 110 cm, while recovered massesof typical order of magnitude ∼ 10 grams diverge by no more than a fraction of agram as compared to the measured values.

Cox et al. also go on to show the results of a symmetric system consisting of sixbeads (not shown here). However the results now start to deviate more significantly,with deviations on the order of ∼ 1 gram and ∼ 1 cm. This might have beenpredicted from an intuitive standpoint, considering the increased complexity of thespectral data and the risk of generating additional noise.

Even still, these results can by no means be considered as ”inaccurate” byany reasonable standard. The results are even more impressive considering theabundance of assumptions having been made along the way, namely:

• The masses are assumed to be point-like.

• The gravitational force is said to be negligible.

• The string is assumed to be practically massless.

• The displacements are claimed to be small as compared to the lengths.

• No appreciable dampening is said to occur during measurements.

Beginning with the assumption of point-masses, this assumption is clearly onlyvalid for beads of smaller diameters, as any bead of significant size as comparedto the string effectively shortens the length of the vibrating part of the string. One

Page 17 of 19



would thus take care when knowing that the beads used are large in size, and per-haps complement the original data by also presenting results for slightly shorter”effective” lengths. This would theoretically give some indication as to the stabil-ity of the current system with changes in string length.

As for the assumption of a negligible gravitational pull, this assumption canperhaps be made with a little more confidence, since from everyday experience weknow that the vibrations of a string depends on whether it is plucked in a horizontalor vertical plane only in extreme cases, and can be safely disregarded as long asthe string is tensioned tightly enough. The same goes for the assumption of amassless string, since string weights are commonly only of the order of magnitude∼ 0.1− 0.5 grams [9]. Hence in the present case of using beads with weights onthe order of magnitude ∼ 10 grams, the effect of the finite mass of the string isjustifiably neglected.

As for the assumption of small displacements, this is can be investigated qual-itatively by looking at the amplitude data provided by Cox et al. (see datasets in[10]). In the laboratory setup, photo-detectors are capable of measuring amplitudeson the order of ∼ 1−10 microns, which most certainly is smaller than the lengths` j. Hence the assumption of small displacements appears to be justified, providedthat the experimental detection system is sensitive enough to detect the oscillations,of course.

As for the assumption of no dampening, no conclusions as to how much thisapproximation affects the measured masses and lengths have been drawn. How-ever, it suffices to say that any imaginary (dampening) part of the eigenfrequenciesω j must have some effect on the calculated results. As for whether these effectsare quantifiable using a detector of reasonably high sensitivity, this question mustbe left unanswered for the time being.

Also noteworthy is the inevitable inclusion of numerical errors due to calcula-tions of the algorithms using finite precision arithmetic. One should be aware ofhow well-behaved the system matrices are, and also aware of the fact that conditionnumbers depend on both the complexity of the system and perhaps as well on otherunforeseen numerical artefacts.

5 Conclusions

Solving the inverse eigenvalue problem by setting up a model of a linear, time-harmonic system, has proven to be a reasonable undertaking. Two algorithms forthe determination of the system matrix An, and the determination for the massesand lengths between the beads, have been found respectively. The resulting calcu-lations of the masses and positions of the beads have also been shown to be in goodagreement with measurements done directly to the system. It has furthermore beendemonstrated that for symmetrically loaded systems, as described in Section 3.6,no changes in the experimental setup is required. Lastly, arguments have beenpresented, which claim that the additional assumptions of small displacements,

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point-like masses, that no significant perturbations exist due to gravitational pull,and that no discernible dampening occurs during measurements, all appear to bereasonable assumptions for the current application.

References

[1] M. T. Chu. “Inverse Eigenvalue Problems”. In: Society for Industrial an Ap-plied Mathematics 40 (1 1998). 39. DOI: 10.1137/S0036144596303984.

[2] M. Kac. “Can One Hear the Shape of a Drum?” In: The American Math-ematical Monthly 73.4 (1966), pp. 1–23. ISSN: 00029890, 19300972. URL:http://www.jstor.org/stable/2313748.

[3] R. L. Parker and K. A. Whaler. “Numerical methods for establishing solu-tions to the inverse problem of electromagnetic induction”. In: Journal ofGeophysical Research: Solid Earth 86.B10 (1981), pp. 9574–9584. ISSN:2156-2202. DOI: 10.1029/JB086iB10p09574. URL: http://dx.doi.org/10.1029/JB086iB10p09574.

[4] L. Beilina and G. Kuramshina. Inverse problems of vibrational spectroscopy.Chalmers University of Technology. URL: http://www.math.chalmers.se/~hegarty/Compendium_Kuramshina.pdf (visited on 04/01/2016).

[5] D. A. McQuarrie. Mathematical Methods for Scientists and Engineers. Vol. 1.2003. Chap. 9 and 10.

[6] S. J. Cox, M. Embree and J. M. Hokanson. “One Can Hear the Compositionof a String: Experiments with an Inverse Eigenvalue Problem”. In: SIAMReview 54.1 (2012). Also includes appendix, missing in the original paperprovided, pp. 157–178. DOI: 10.1137/080731037. eprint: http://dx.doi.org/10.1137/080731037. URL: http://www.caam.rice.edu/~beads/beads_cf.pdf.

[7] M. T. Heath. Scientific Computing: An Introductory Survey. Ed. by E. M.Munson. 2nd. McGraw-Hill Higher Education, 1996. ISBN: 0070276846.

[8] Polynomial interpolation. URL: https://en.wikipedia.org/wiki/Polynomial_interpolation (visited on 05/01/2016).

[9] D. Achilles. Tensions of Guitar Strings. Dec. 12. URL: https://courses.physics.illinois.edu/phys406/Student_Projects/Fall00/DAchilles/

Guitar_String_Tension_Experiment.pdf.

[10] S. J. Cox, M. Embree and J. M. Hokanson. Inverse Eigenvalue Experimentsfor Beaded Strings. Department of Computational and Applied Mathemat-ics, Rice University, 28. URL: http://www.caam.rice.edu/~beads/(visited on 13/01/2016).

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http://dx.doi.org/10.1137/S0036144596303984

http://www.jstor.org/stable/2313748

http://dx.doi.org/10.1029/JB086iB10p09574



http://www.math.chalmers.se/~hegarty/Compendium_Kuramshina.pdf

http://www.math.chalmers.se/~hegarty/Compendium_Kuramshina.pdf

http://dx.doi.org/10.1137/080731037

http://dx.doi.org/10.1137/080731037

http://dx.doi.org/10.1137/080731037

http://www.caam.rice.edu/~beads/beads_cf.pdf

http://www.caam.rice.edu/~beads/beads_cf.pdf

https://en.wikipedia.org/wiki/Polynomial_interpolation

https://en.wikipedia.org/wiki/Polynomial_interpolation

https://courses.physics.illinois.edu/phys406/Student_Projects/Fall00/DAchilles/Guitar_String_Tension_Experiment.pdf



http://www.caam.rice.edu/~beads/

beads on a string: an inverse eigenvalue problem

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