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Formulation of Problem in Oscillations - Conservative Systems

Learn about stable and unstable equilibrium, Lagrangian for systems in equilibrium, Lagrange Equations of Motion, and Eigenvalue Equation for oscillatory systems perturbed from equilibrium. Understand the matrix representation and solutions.

jeromebwilliams
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Formulation of Problem in Oscillations - Conservative Systems

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  1. Chapter 6: OscillationsSect. 6.1: Formulation of Problem • Conservative systems  PE V is a function of position only. • Assume:The transformation eqtns defining generalized coords q1,q2, ,,,qn, do not explicitly involve the time. No time dependent constraints. • If the system is in Equilibrium  All generalized forces Qivanish  Qi = (V/qi)0 = 0 (all i)  V has an extremum at the equilibrium configuration: q1,q2, ,,,qn, = q01,q02, ,,,q0n,

  2. Stable Equilibrium  A small disturbance from equilibrium results in small bounded motion around the equilibrium position.  At the equilibrium position qi, the extremum of V is a minimum.  (2V/qiqj)0 > 0 (See figure) Describing the motion which occurs with such a small disturbance away from stable equilibrium is the subject of this chapter!

  3. Unstable Equilibrium  A small disturbance from equilibrium results in unbounded motion.  At the equilibrium position qi, the extremum of V is a maximum.  (2V/qiqj)0 < 0 (See figure)

  4. We’re interested in the case where thesystem is initially in stable equilibrium & is slightly disturbed away from it.  Small departures from equilibrium  Expand V in a Taylor’s series about equilibrium, keep only lowest order terms. Write:qi = q0i + ηi : Expand V about q0i& keep lowest order terms in ηi(summation convention): V(q1,q2, ,,,qn)  V(q01,q02, ,,,q0n) + (V/qi)0ηi + (½)(2V/qiqj)0ηiηj + ... • Note:(V/qi)0 = 0 by the equilibrium conditions. V(q01,q02, ,,,q0n) = aconstant. Does not affect the motion. By shifting the (arbitrary) zero of the PE, we can set it to zero.

  5. Drop terms higher than quadratic in the displacements from equilibrium ηi V(q1,q2, ,,,qn) = (½)(2V/qiqj)0ηiηj • Define:Vij  (2V/qiqj)0 V= (½)Vijηiηj • Note: Vijare symmetric: Vij = Vji . • It is possible for Vij to vanish Vij = 0  Neutral Equilibrium • In what follows, we assume Vij  0

  6. Now, consider the KE, T: • Generalized coords do not contain time explicitly:  Tis a homogeneous, quadratic function of the generalized velocitiesqi. • This is a theorem proven in Ch. 1!  Can write: T = (½)mijqiqj = (½)mijηiηj • Coefficients mij: Not simply masses! Involve masses & transformation eqtns between Cartesian coords & generalized coords. In general, functions of generalized coords. See Ch. 1! Can expand about equilib in similar fashion to V: mij = mij(q1 ,,,qn)  mij(q01,,,q0n) + [(mij)/qk]0ηk +… Keep only the constant term & define: Tij mij(q01,,,q0n)

  7.  T= (½)Tijηiηj • Note: Tijare symmetric: Tij = Tji • Now, consider Lagrangian for system slightly displaced from stable equilibrium: L = T - V = (½)(Tijηiηj - Vijηiηj) (1) • Lagrange Eqtns of Motion, using (1): (d/dt)[(L/ηi)] - (L/ηi) = 0(i = 1,2,3, … n) (2)  Tijηj + Vijηj= 0 (i = 1,2,3, … n) (3) n coupled (simple harmonic oscillator) equations of motion!

  8. In many common cases,T has no cross terms. That is Tij = Tiδij  T= (½)Ti(ηi )2 • So, in this case, the Lagrangian is: L = T - V = (½)[Ti(ηi)2 - Vijηiηj] • Lagrange Eqtns of Motion become:  Tiηi + Vijηj = 0 (i = 1,2,3, … n) No sum on i.

  9. Sect. 6.2: Eigenvalue Equation & Principal Axis Transformation • We want to solve the equations of motion for a system slightly perturbed from stable equilibrium: Tijηj + Vijηj = 0(i = 1,2,3, … n) (3) • (3): n linear (coupled simple harmonic oscillator) differential equations, n unknowns:Try (complex) oscillatory solutions: ηj Caje-iωt • The real part corresponds to the physical motion • Substituting this into (3)  (Vijaj - ω2Tijaj ) = 0 (i = 1,2,3, … n) (4) (4): n homogeneous linear algebraic equations, n unknowns (aj):  The determinant of the coefficients of the aj vanishes!

  10. That is: V11 - ω2T11 V12 - ω2T12 V13 - ω2T13 ….. V21 - ω2T21 V22 - ω2T22 V23 - ω2T23 ….. = 0 (5) V31 - ω2T31 V32 - ω2T32 V33 - ω2T33 ….. ………….. • (5): An nthdegree algebraic equation for ω2: These roots are the allowed frequencies with which the system can vibrate (or oscillate). Once (5) is solved for the nω2 (ω2)kfor each solution, can go back to (Vijaj - ω2Tijaj ) = 0 (i = 1,2,3, … n) (4) and solve for the amplitudes aj • (4) (or (5)) is a type of eigenvalue problem for ω2.

  11. We can rewrite all of this in matrix form. Treating Vij and Tij as square matrices, aj as column vectors & ω2as eigenvalues (λ ω2): (Vijaj - ω2Tijaj ) = 0 (i = 1,2,3, … n) (4) becomes:Va = λTa (6) • (6): This is a variation of the more common eigenvalue problem:  The effect of matrix V acting on an eigenvector a is to produce a multiple (λ) of the matrix T acting on the eigenvector a.

  12. Va = λTa (6) • Can show (for proofs, see text pages 242-245): 1. The eigenvalues λ =ω2 are all real and positive. (Comes from the real & symmetric properties of V & T) 2. The eigenvectors a are all orthogonal and real. 3. Matrix of eigenvectors  A a. A is orthogonal (A-1 = Ã) b. A diagonalizes bothT and V That is, performing a similarity transformation on T: A-1TA =ÃTA = 1 (unit matrix) That is, performing a similarity transformation on V: A-1VA =ÃVA = Vd(diagonal matrix; eigenvalues on diagonal)  If T is already diagonal in our choice of generalized coordinates, to find the eigenvalues, it is sufficient to diagonalize V!!

  13. Simple example: Mass m with 2 degrees of freedom, generalized coords x1 & x2: • Lagrangian: L= (½)m[(x1)2 + (x2)2] - (½)Vijxixj (summation convention) Vij = constants T is already diagonal. So, to find the eigenvalues, we need to only diagonalize V. In this case, λ = mω2  (V - λ1)a = 0 (1)  Determinant: V11 - λV12 V12 V22 - λ = 0 Eigenvalues (Like the simple “two-level” problem in QM!): λ1 = m(ω1)2 = (½){V11 + V22 + [(V11 + V22)2 + 4V12V12]½} λ2 = m(ω2)2 = (½){V11 + V22 - [(V11 + V22)2 + 4V12V12]½}

  14. Allowed oscillation (vibration) frequencies: λ1 = m(ω1)2 = (½){V11 + V22 + [(V11 + V22)2 + 4V12V12]½} λ2 = m(ω2)2 = (½){V11 + V22 - [(V11 + V22)2 + 4V12V12]½} • To get the eigenvectors (oscillation amplitudes) corresponding to these frequencies, put λiinto the original eigenvalue equation & solve for the matrix elements aij of the eigenvalue matrix (whose columns are the eigenvectors) : (V - λ1)a = 0 or: aij(Vij – λiδij) = 0 with normalization: (ai1)2 +(ai2)2 = 1

  15. Allowed oscillation (vibration) frequencies: λ1 = m(ω1)2 = (½){V11 + V22 + [(V11 + V22)2 + 4V12V12]½} λ2 = m(ω2)2 = (½){V11 + V22 - [(V11 + V22)2 + 4V12V12]½} • General solution for the matrix of eigenvectors: a11 a21 A = a12 a22  a11 = V12[(V11 - λ1)2 + (V12)2]-½ a12 = (V11 - λ1)[(V11 - λ1)2 + (V12)2]-½ a21 = (V22 - λ2)[(V22 - λ2)2 + (V21)2]-½ a22 = V21[(V22 - λ2)2 + (V21)2]-½

  16. We have the exact, general solution. However, its useful & educational to look at 2 limiting cases. • Limiting Case 1: Small off diagonal Vij: V11 > V22 > 0 & V12 = V21  0; V12 << (V11 - V22) Define:δ V12(V11 - V22)-1 << 1. Expand various square roots to lowest order in the small quantity δ:  Eigenvalues:λ1 = m(ω1)2 = V11 + V12δ λ2 = m(ω2)2 = V22 - V12δ Eigenvectors: a11 a21 A = a12 a22  a11 = a22 = 1 – (½)δ2 a21 = - a12 = δ - (½)δ3

  17. Limiting Case 2: Large off diagonal Vij: V12 = V21 > V22 > 0 & (V11 - V22) << V12 Define:ε (⅛)(V11 - V22)(V12)-1 << 1. Expand various square roots to lowest order in ε:  Eigenvalues: λ1 = m(ω1)2 = (½)(V11 + V22) + V12 + (V11 - V22)ε λ2 = m(ω2)2 = (½)(V11 + V22) - V12 - (V11 - V22)ε Eigenvectors: a11 a21 A = a12 a22  a11 = a22 = (2)-½(1 + 2ε) a21 = - a12 = (2)-½ (1 - 2ε)

  18. Fig. a: Eigenvalues λivs. δ V12(V11 - V22)-1 for 0 < δ < 3 Includes both limiting cases in this range: (In QM: “Level repulsion” as the off- diagonal interaction is turned on.)

  19. Fig. b: Eigenvector components aijvs. δ V12(V11 - V22)-1 for 0 < δ < 3 Includes both limiting cases in this range:

  20. All of the preceding analysis: Assumed the eigenvalues λiare non-degenerate (no 2 of are equal!). • If there is degeneracy, then, this procedure must be modified. In this case, we cannot uniquely determine the components of the eigenvectors by this procedure. Actually, the eigenvectors in that case are not unique, but can be appropriately chosen from an infinite possible set. • Follow the so-called “Gram-Schmidt” orthogonalization procedure of linear algebra. Discussed in detail on pages 248-249 of the text.

  21. All of this discussion is identical to the Ch. 5 discussion of degenerate eigenvalues of the inertia tensor, and to the procedure for finding the principal axes of the inertia tensor. • Hence the title of the section: “Principal Axis Transformation”

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