Beaded Strings. Forward and Inverse Problems. Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings BeadedPFUG) 11 June 2008

Transcription

1 Beaded Strings Forward and Inverse Problems Hunter Gilbert, Walter Kelm, and Brian Leake Physics of Strings PFUG 11 June 2008 Frequencies Masses + Lengths Masses + Lengths Frequencies Strings 11 June / 36

2 Introduction Why 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 Strings 11 June / 36

3 Introduction Brief 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) Physical lab for CAAM 335 Strings 11 June / 36

4 Introduction This Summer: Beaded Strings Point masses, taut massless string, fixed at both ends Forward Problem Given: masses, lengths, and initial input Find: the string s motion Inverse Problem Given: the string s motion for any input Find: masses and lengths that uniquely describe the system Strings 11 June / 36

5 Variable Meanings A Beaded String is uniquely defined by masses and lengths: m k = Mass of a particular bead. l 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. y k (t) = Vertical Displacment of a particular bead. Strings 11 June / 36

6 Dynamics of a Beaded String To find frequencies of vibration given masses and positions of the beads, we will use a system of differential equations. Kinetic energy is the focus of one of the equations. T (y ) = 1 2 n m k (y k(t)) 2 k=1 Strings 11 June / 36

7 Dynamics of a Beaded String To find frequencies of vibration given masses and positions of the beads, we will use a system of differential equations. Kinetic energy is the focus of one of the equations. T (y ) = 1 2 n m k (y k(t)) 2 k=1 Symbol meanings: T (y ) = Kinetic energy as a function of time. m k = Mass of the given bead. y k = Velocity of given bead. Strings 11 June / 36

8 Dynamics of a Beaded String To find frequencies of vibration given masses and positions of the beads, we will use a system of differential equations. Kinetic energy is the focus of one of the equations. T (y ) = 1 2 n m k (y k(t)) 2 k=1 Symbol meanings: T (y ) = Kinetic energy as a function of time. m k = Mass of the given bead. y k = Velocity of given bead. Gives only part of the required system. Strings 11 June / 36

9 Dynamics of a Beaded String 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: l 2 k + (y k+1 y k ) 2 l k = l k 1 + (y k+1 y k ) 2 l k ( (y k+1 y k ) 2. 2l 2 k l 2 k l k ) (y k+1 y k ) 2 l k l 2 k Strings 11 June / 36

10 Dynamics of a Beaded String 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: l 2 k + (y k+1 y k ) 2 l k = l k 1 + (y k+1 y k ) 2 l k ( (y k+1 y k ) 2. 2l 2 k l 2 k l k ) (y k+1 y k ) 2 l k l 2 k V (y) = σ 2 n (y k+1 y k ) 2 l k k=0 Strings 11 June / 36

11 Dyanamics of a Beaded String Conservation of Energy We can use the Euler-Lagrange Equation to generate a unified system of differential equations. d T + V = 0, j = 1,..., n dt y j y j d T dt y j V y j = (t) = m j y j (t) ( σ l j 1 ) y j 1 + ( σ l j 1 + σ l j ) ) y j + ( σlj y j+1 Strings 11 June / 36

12 Dynamics of a Beaded String Matrix Form of Equations Doing the necessary algebra provides n equations: u k u k+1 l k + u k u k 1 l k 1 m k λ 2 u k = 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 Strings 11 June / 36

13 Dynamics of a Beaded String Essential Linear Algebra Inverse: M 1 M = I Eigenvalues (λ) and eigenvectors (u) Defined for Matrix A by: Au = λu They are fundamental properties that describe the matrix behavior For a beaded string, they correspond to vibration frequency (eigenvalue) and vibration mode shape (eigenvector) Strings 11 June / 36

14 The Solution If we can solve that particular differential equation, we will be able to know the frequencies of vibration. With the inverse, we can say: y (t) = M 1 Ky(t) Strings 11 June / 36

15 The Solution If we can solve that particular differential equation, we will be able to know the frequencies of vibration. With the inverse, we can say: y (t) = M 1 Ky(t) With the eigenvalues of M 1 K, we can rewrite the equation as: y (t) = V ΛV 1 y(t) V is the matrix where each column is an eigenvector of M 1 K, and Λ is a diagonal matrix of its eigenvalues. Strings 11 June / 36

16 The Solution, pt2 Since Λ is a diagonal matrix, the n n system reduces to n independent scalar equations. γ j (t) = ω2 j γ j(t) Strings 11 June / 36

17 The Solution, pt2 Since Λ is a diagonal matrix, the n n system reduces to n independent scalar equations. γ j (t) = ω2 j γ j(t) If we presume that the string begins with no initial velocity, we can state We then have: γ j (t) = γ j (0)cos(ω j t) y(t) = n j=1 γ j(0)cos(ω j t)v j Strings 11 June / 36

18 Summary The masses are displaced as the superposition of n independent vectors vibrating at distinct frequencies. These frequencies are in turn given by the eigenvalues of the matrix M 1 K Strings 11 June / 36

19 Hearing the Composition of a String We are able to go from masses and positions to frequencies of vibration. The natural question is now to ask if we can hear the positions and masses of beads on a string from the vibrations it undergoes. Strings 11 June / 36

20 Hearing the Composition of a String We are able to go from masses and positions to frequencies of vibration. The natural question is now to ask if we can hear the positions and masses of beads on a string from the vibrations it undergoes. Using the techniques put forward by Gantmacher and Krein, we can solve this inverse problem. unter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings BeadedPFUG) Strings 11 June / 36

21 Recurrence We wish to find values of M and K to solve Ku = λmu. By ( simple matrix multiplication, we can show: σ ) ( σ u j σ ) ) u j + ( σlj u j+1 = λm j u j l j 1 l j 1 l j Strings 11 June / 36

22 Recurrence We wish to find values of M and K to solve Ku = λmu. By ( simple matrix multiplication, we can show: σ ) ( σ u j σ ) ) u j + ( σlj u j+1 = λm j u j l j 1 l j 1 l j Having a fixed left & right end implies u 0 = 0 and u n+1 = 0 Rearranging ( the above, we get : u j+1 = l ) ( j u j l j ) u j λl jm j l j 1 l j 1 σ Since we know u n+1 = 0 when the conditions for a fixed-fixed string are met... Strings 11 June / 36

23 Recurrence We wish to find values of M and K to solve Ku = λmu. By ( simple matrix multiplication, we can show: σ ) ( σ u j σ ) ) u j + ( σlj u j+1 = λm j u j l j 1 l j 1 l j Having a fixed left & right end implies u 0 = 0 and u n+1 = 0 Rearranging ( the above, we get : u j+1 = l ) ( j u j l j ) u j λl jm j l j 1 l j 1 σ Since we know u n+1 = 0 when the conditions for a fixed-fixed string are met... We can use this condition to make sure λ is an eigenvalue. unter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings BeadedPFUG) Strings 11 June / 36

24 Polynomial Construction When( j = 1, we can say: u 2 = l ) ( 1 u l 1 ) u 1 λl 1m 1 l 0 l 0 σ We now create a polynomial p of linear degree, and define it such that: u 2 p 1 (λ)u 1 Strings 11 June / 36

25 Polynomial Construction When( j = 1, we can say: u 2 = l ) ( 1 u l 1 ) u 1 λl 1m 1 l 0 l 0 σ We now create a polynomial p of linear degree, and define it such that: u 2 p 1 (λ)u 1 Recalling the formula for a general element of the eigenvector, ( l j ) u j 1 + ( 1 + l j ) u j u j+1 = λl jm j l j 1 l j 1 σ We can reuse the same trick from above to create a polynomial of degree j. u j+1 p j (λ)u 1 Strings 11 June / 36

26 Insight We can build p n by following the recurrence, which requires knowledge of lengths & masses. Or, if we have knowledge of the roots of the polynomial beforehand, we can construct a polynomial of degree n multiplied by a real coefficient that will provide the same behavior. Strings 11 June / 36

27 Insight We can build p n by following the recurrence, which requires knowledge of lengths & masses. Or, if we have knowledge of the roots of the polynomial beforehand, we can construct a polynomial of degree n multiplied by a real coefficient that will provide the same behavior. We already have this knowledge. The string is fixed at both ends, requiring u 0 = 0 and u n+1 = 0 Therefore, we can say p n (λ) = γ n j=1 (λ λ j) unter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings BeadedPFUG) Strings 11 June / 36

28 Fixed-Fixed & Fixed-flat In order to make the solution unique, we need to find more information. We will break our assumption that the string is fixed at both ends. Allow the string to move at one end, but require it to have 0 slope at its last node. Strings 11 June / 36

29 Fixed-Fixed & Fixed-flat unter Gilbert, Walter Kelm, andfigure: Brian Leake Fixed-Flat (Physics of Strings Beaded & PFUG) Strings Fixed-Fixed Strings 11 June / 36 In order to make the solution unique, we need to find more information. We will break our assumption that the string is fixed at both ends. Allow the string to move at one end, but require it to have 0 slope at its last node. There is now another set of eigenvalues, λ, that represent the system. The system is no longer underdetermined.

30 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: u j+1 u j = u ( ) j u j 1 λmj u j l j l j 1 σ Strings 11 June / 36

31 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: u j+1 u j = u ( ) j u j 1 λmj u j l j l j 1 σ Using the polynomials already developed, we can rewrite the equation as: u j+1 u j = p j(λ)u 1 p j 1 (λ)u 1 l j l j Strings 11 June / 36

32 Algebra in Anticipation Using the same semantic trick as earlier, we define a new polynomial q n that characterizes the system in fixed-flat operation. q j (λ) = 1 l j (p j (λ) p j 1 (λ)) Strings 11 June / 36

33 Algebra in Anticipation Using the same semantic trick as earlier, we define a new polynomial q n that characterizes the system in fixed-flat operation. q j (λ) = 1 l j (p j (λ) p j 1 (λ)) Equating expressions for the left and right sides of the equation used to define q j gives us: ( ) λmj q j (λ)u 1 = q j 1 (λ) p j 1 (λ)u 1 σ Strings 11 June / 36

34 Continued Fractions With these characteristic polynomials, we can find masses & displacements. Using recurrence relationships of the polynomials: Strings 11 June / 36

35 Continued Fractions With these characteristic polynomials, we can find masses & displacements. Using recurrence relationships of the polynomials: p n (λ) q n (λ) = l nq n (λ) + p n 1 (λ) q n (λ) Strings 11 June / 36

36 Continued Fractions With these characteristic polynomials, we can find masses & displacements. Using recurrence relationships of the polynomials: p n (λ) q n (λ) = l nq n (λ) + p n 1 (λ) q n (λ) p n (λ) q n (λ) = l 1 n + m n σ λp n 1(λ) + q n 1 (λ) p n 1 (λ) Strings 11 June / 36

37 Continued Fractions With these characteristic polynomials, we can find masses & displacements. Using recurrence relationships of the polynomials: p n (λ) q n (λ) = l nq n (λ) + p n 1 (λ) q n (λ) p n (λ) q n (λ) = l 1 n + m n σ λp n 1(λ) + q n 1 (λ) p n 1 (λ) p n (λ) q n (λ) = l n + m n σ λ l n 1 + m n 1 σ λ l m 1 σ λ + 1 l 0 Strings 11 June / 36

38 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. Strings 11 June / 36

39 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 the midpoint of the string, we can find the fixed-flat eigenvalues without any extra work. Strings 11 June / 36

40 Symmetry Experimentally measuring eigenvalues on a symmetric beaded string gives us a new set of eigenvalues, termed Λ. Λ has some beautiful properties. Strings 11 June / 36

41 Symmetry Experimentally measuring eigenvalues on a symmetric beaded string 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 the middle of the string. The even indexed eigenvalues of Λ are antisymmetric about the midpoint. This relationship holds for all symmetric strings. unter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings BeadedPFUG) Strings 11 June / 36

42 The Point The N eigenvalues of a symmetric beaded string fixed at both ends exactly match the N/2 fixed-fixed and N/2 fixed-flat eigenvalues associated with half of the string. Figure: Eigenvalues of Symmetric Beaded Strings Strings 11 June / 36

43 The Inverse Algorithm Record the eigenvalues for the whole string. Use the eigenvalues to generate characteristic polynomials p n and q n. Use the characteristic polynomials to find the set of masses and lengths. Figure: The Symmetric Beaded String Strings 11 June / 36

44 Non-symmetric Problem Introduction 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 Strings 11 June / 36

45 Non-symmetric Problem Setup For n 1 masses on the left and n 2 masses on the right: u k u k+1 + u k u k 1 m k λ 2 u k = 0, k = 1, 2,..., n 1 l k l k 1 ũ k ũ k+1 l k + ũk ũ k 1 l k 1 m k λ 2 ũ k = 0, k = 1, 2,..., n 2 unter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings BeadedPFUG) Strings 11 June / 36

46 Non-symmetric Problem Setup For n 1 masses on the left and n 2 masses on the right: u k u k+1 + u k u k 1 m k λ 2 u k = 0, k = 1, 2,..., n 1 l k l k 1 ũ k ũ k+1 l k + ũk ũ k 1 l k 1 m k λ 2 ũ k = 0, k = 1, 2,..., n 2 Whole String Boundary Conditions u n1+1 = ũ n2+1 Strings 11 June / 36

47 Non-symmetric Problem Setup For n 1 masses on the left and n 2 masses on the right: u k u k+1 + u k u k 1 m k λ 2 u k = 0, k = 1, 2,..., n 1 l k l k 1 ũ k ũ k+1 l k + ũk ũ k 1 l k 1 m k λ 2 ũ k = 0, k = 1, 2,..., n 2 Whole String Boundary Conditions u n1+1 = ũ n2+1 u 0 = 0, ũ 0 = 0 Strings 11 June / 36

48 Non-symmetric Problem Setup For n 1 masses on the left and n 2 masses on the right: u k u k+1 + u k u k 1 m k λ 2 u k = 0, k = 1, 2,..., n 1 l k l k 1 ũ k ũ k+1 l k + ũk ũ k 1 l k 1 m k λ 2 ũ k = 0, k = 1, 2,..., n 2 Whole String Boundary Conditions u n1+1 = ũ n2+1 u 0 = 0, ũ 0 = 0 u n1+1 u n1 l n1 + ũn 2+1 ũ n2 l n2 = 0 Strings 11 June / 36

49 Non-symmetric Problem Setup For n 1 masses on the left and n 2 masses on the right: u k u k+1 + u k u k 1 m k λ 2 u k = 0, k = 1, 2,..., n 1 l k l k 1 ũ k ũ k+1 l k + ũk ũ k 1 l k 1 m k λ 2 ũ k = 0, k = 1, 2,..., n 2 Whole String Boundary Conditions u n1+1 = ũ n2+1 u 0 = 0, ũ 0 = 0 u n1+1 u n1 l n1 + ũn 2+1 ũ n2 l n2 = 0 Clamped String Boundary Conditions u n1+1 = 0, ũ n2+1 = 0 Strings 11 June / 36

50 Non-symmetric Problem Setup For n 1 masses on the left and n 2 masses on the right: u k u k+1 + u k u k 1 m k λ 2 u k = 0, k = 1, 2,..., n 1 l k l k 1 ũ k ũ k+1 l k + ũk ũ k 1 l k 1 m k λ 2 ũ k = 0, k = 1, 2,..., n 2 Whole String Boundary Conditions u n1+1 = ũ n2+1 u 0 = 0, ũ 0 = 0 u n1+1 u n1 l n1 + ũn 2+1 ũ n2 l n2 = 0 Clamped String Boundary Conditions u n1+1 = 0, ũ n2+1 = 0 u 0 = 0, ũ 0 = 0 Strings 11 June / 36

51 Three spectra Inverse Problem The Problem Relation between characteristic equations p whole (λ) = p left (λ)q right (λ) + p right (λ)q left (λ) The roots of p left and p right are the eigenvalues of the fixed-fixed problems for the left and right segments The roots of q left and q right are the eigenvalues of the fixed-flat problems Strings 11 June / 36

52 Three spectra Inverse Problem The Problem Relation between characteristic equations p whole (λ) = p left (λ)q right (λ) + p right (λ)q left (λ) The roots of p left and p right are the eigenvalues of the fixed-fixed problems for the left and right segments The roots of q left and q right are the eigenvalues of the fixed-flat problems Problem: we can construct p left and p right and p whole up to a scaling factor from measured roots, but we cannot measure the roots of q left and q right, preventing us from forming the continued fractions for the left and right segments. Strings 11 June / 36

53 Three spectra Inverse Problem Definitions Through the clever use of polynomials, Boyko and Pivovarchik show that we do not need to know the roots of q left and q right to construct them Strings 11 June / 36

54 Three spectra Inverse Problem Definitions Through the clever use of polynomials, Boyko and Pivovarchik show that we do not need to know the roots of q left and q right to construct them Let λ k, k = 1, 2,..., (n 1 + n 2 ), be the spectra of the whole string and let ν k,l, k = 1, 2,..., n 1, and ν k,r, k = 1, 2,..., n 2, be the spectra of the left and right parts, respectively Strings 11 June / 36

55 Three spectra Inverse Problem Definitions Through the clever use of polynomials, Boyko and Pivovarchik show that we do not need to know the roots of q left and q right to construct them Let λ k, k = 1, 2,..., (n 1 + n 2 ), be the spectra of the whole string and let ν k,l, k = 1, 2,..., n 1, and ν k,r, k = 1, 2,..., n 2, be the spectra of the left and right parts, respectively Let L be the length of the whole string and let L l and L r be the lengths of the left and right segments of the clamped string Strings 11 June / 36

56 Three spectra Inverse Problem Constructing Known Polynomials Construct the polynomials we know, up to a scaling factor: p whole (λ) = L n 1 +n 2 k=1 n 1 p left (λ) = L l k=1 n 2 p right (λ) = L r k=1 (1 λλk ) ( 1 λ ) ν k,l ( 1 λ ) ν k,r Strings 11 June / 36

57 Three spectra Inverse Problem Finding q left and q right Recall the relationship of the characteristic equations: p whole (λ) = p left (λ)q right (λ) + p right (λ)q left (λ) Strings 11 June / 36

58 Three spectra Inverse Problem Finding q left and q right Recall the relationship of the characteristic equations: p whole (λ) = p left (λ)q right (λ) + p right (λ)q left (λ) Notice that p left (ν k,l ) = 0 and p right (ν k,r ) = 0, as these are simply the roots we used to construct those polynomials. Strings 11 June / 36

59 Three spectra Inverse Problem Finding q left and q right Recall the relationship of the characteristic equations: p whole (λ) = p left (λ)q right (λ) + p right (λ)q left (λ) Notice that p left (ν k,l ) = 0 and p right (ν k,r ) = 0, as these are simply the roots we used to construct those polynomials. Plug in ν k,l and ν k,r into the relation between the characteristic equations to reveal: q left (ν k,l ) = p whole(ν k,l ) p right (ν k,l ), k = 1, 2,..., n 1 q right (ν k,r ) = p whole(ν k,r ) p left (ν k,r ), k = 1, 2,..., n 2 Strings 11 June / 36

60 Three spectra Inverse Problem Finding q left and q right (continued) Recall the continued fraction equation: p n (λ) q n (λ) = l n + m n σ λ l n 1 + m n 1 σ λ l m 1 σ λ + 1 l 0 p n (0) q n (0) = l n + l n l 1 + l 0 = L These equations must hold for the two clamped string segments as well, giving us the last points, q left (0) = 1 and q right (0) = 1, needed to completely define the polynomials Strings 11 June / 36

61 Three spectra Inverse Problem Constructing q left and q right Construct q left and q right : q left (λ) = q right (λ) = n 1 k=1 n 2 k=1 λp whole (ν k,l ) ν k,l p right (ν k,l ) λp whole (ν k,r ) ν k,r p left (ν k,r ) n 1 j=1,j k n 2 j=1,j k n1 (λ ν j,l ) (ν k,l ν j,l ) + k=1 n2 (λ ν j,r ) (ν k,r ν j,r ) + k=1 ν k,l λ ν k,l ν k,r λ ν k,r Strings 11 June / 36

62 Three spectra Inverse Problem Constructing q left and q right Construct q left and q right : q left (λ) = q right (λ) = Notice: n 1 k=1 n 2 k=1 λp whole (ν k,l ) ν k,l p right (ν k,l ) λp whole (ν k,r ) ν k,r p left (ν k,r ) n 1 j=1,j k n 2 j=1,j k n1 (λ ν j,l ) (ν k,l ν j,l ) + k=1 n2 (λ ν j,r ) (ν k,r ν j,r ) + k=1 q left (ν k,l ) = p whole(ν k,l ) p right (ν k,l ), q right(ν k,r ) = p whole(ν k,r ) p left (ν k,r ) q left (0) = 1, q right (0) = 1 ν k,l λ ν k,l ν k,r λ ν k,r Strings 11 June / 36

63 Three spectra Inverse Problem Finding {m k }, {l k }, { m k }, and { l k } Unique lengths and masses are then determined by continued fraction expansion of ratio of p s and q s, i.e. p left (λ) q left (λ) = l n 1 + m n 1 σ λ l n1 1 + m n 1 1 σ λ l m 1 σ λ + 1 l 0 Strings 11 June / 36

64 Future Work Three spectra problem with damping at an interior point No unique solution Damping at a mass Damping at the ends Strings 11 June / 36

65 Mentors Jeffrey Hokanson Dr. Steve Cox Dr. Mark Embree References CAAM 335 Lab 9. caam335lab/lab9.pdf CAAM 335 Lab caam335lab/lab10.pdf Boyko O and Pivovarchik V. The inverse three-spectral problem for a Stieltjes string and the inverse problem with one-dimensional damping. Inverse Problems 24 (2008) Gantmacher F P and Krein M G. Oscillation Matrices and Kernels and Small Vibrations of Mechanical Systems. Revised Edition. Providence, Rhode Island: AMS Chelsea Pub, Strings 11 June / 36