Sect. 6.5: Forced Vibrations & Dissipative Effects Free vibrations: Occur when the system is displaced from equilibrium & then allowed to oscillate by itself. Now: Forced vibrations (driven oscillators). The system is acted on by an external driving force which continues past t = 0. In this case, the oscillation frequency is determined by driving force, not by resonant frequencies. However, the normal modes are still needed to determine the amplitudes of the forced vibrations. The problem is simplified by using the normal coordinates obtained from the free oscillation problem.

Fj  Generalized Force corresponding to the generalized displacement ηj. A result from Ch. 1  Qi  ∑jFj(rj/qi) = the generalized force associated with the generalized coordinate qi. In the notation of this chapter, in terms of the elements of the eigenvector matrix discussed last time, this becomes: Qi = ∑jajiFj = the generalized force associated with the normal coordinate ζi. Lagrange’s formalism, with generalized forces, easily gives the equations of motion (driven oscillators!): ζi + (ωi)2ζi = Qi (i = 1,2,3, … n)

ζi + (ωi)2ζi = Qi (i = 1,2,3, … n) (1) (1): A set of n inhomogeneous differential eqtns. In general, we can solve them if we know Qi = Qi(t). Normal coordinates allow n uncoupled equations like (1): n uncoupled driven simple harmonic oscillator equations. In coordinates other than normal coordinates, the problem would be n coupled driven simple harmonic equations.  The normal coordinates still uncouple the problem in this case. Note: The following discussion ASSUMES that you know & understand the details of the solution to the driven simple harmonic oscillator equation with a sinusoidal driving force (from undergraduate mechanics!): Also, assumes that you know (math) that the general solution to (1) = general solution to the homogeneous equation + a particular solution to the inhomogeneous one

A common case: Sinusoidal driving force. If this is not the case, we can Fourier transform the force & the equation of motion in time. Since the equation of motion is linear, we can solve the resulting equation Fourier component by Fourier component. Thus, if Qi is sinusoidal, we can view result as being for one Fourier component of a more general time dependent force.  So, it is of general interest to study the case of a sinusoidal driving force: Qi = Q0icos(ωt + δi) ω  frequency of external force. The amplitude Q0i, & phase δi in general are different for each mode i.  Equations of motion are: ζi + (ωi)2ζi = Q0icos(ωt + δi) (i = 1,2,3, … n) (2)

ζi + (ωi)2ζi = Q0icos(ωt + δi) (i = 1,2,3, … n) (2) General solution to (2) = General solution to the homogeneous equation + a particular solution to the inhomogeneous one. Solution to the homogeneous equation is the free vibration problem we just talked about! In real systems, damping (frictional forces, discussed next!) is present  Free vibrations usually die away (exponentially) after enough time. At steady state (after a long enough time), only the particular solution to the driven oscillator equation (2) is present. This is the case we consider.

ζi + (ωi)2ζi = Q0icos(ωt + δi) (i = 1,2,3, … n) (2) From undergrad mechanics (or from straightforward math) it can be shown that the particular solution to (2) is of the form: ζi(t) = Bicos(ωt + δi) with amplitude Bi = Bi(ω) = Q0i[(ωi)2 - (ω)2]-1 The complete steady state solution, in terms of the original generalized coordinate ηj is (sum on i): ηj(t) = ajiζi(t) ηj(t) = ajiQ0i[(ωi)2 - (ω)2]-1 cos(ωt + δi) (i = 1,2,3, … n) (3) Physics: The vibration of each particle = a linear combination of vibrations of the normal modes. Each mode vibrates at the driving force frequency ω. Each mode amplitude depends (inversely!) on (ωi)2 - (ω)2 with ωi = resonant frequency of that mode.

ηj(t) = ajiQ0i[(ωi)2 - (ω)2]-1 cos(ωt + δi) (3) (i = 1,2,3, … n) PHYSICS: The extent to which each mode is excited depends on: 1. (Obviously!) The amplitude of the generalized driving force Q0i. If the driving force has no “component” in the “direction” of the vibration of mode i (if Q0i = 0), that mode will not be excited by the force.  An external force can excite a normal mode only if it tends to move the particles in the same “direction” as in that mode.

ηj(t) = ajiQ0i[(ωi)2 - (ω)2]-1 cos(ωt + δi) (3) (i = 1,2,3, … n) PHYSICS: The extent to which each mode is excited depends on: 2. The closeness of the driving frequency ω to the resonant frequency ωi of the mode. Since the amplitude  [(ωi)2 – (ω)2]-1, the closer ω is to ωi, the larger the mode amplitude will be. From (3), as ω  ωi the amplitude  . This is Physically Unrealistic, as damping has been neglected. Including damping, the amplitude will be finite (but large) as ω  ωi. (Undergraduate result). This is the phenomenon of RESONANCE. (A contradiction! The theory has assumed small amplitudes. It actually breaks down as ω  ωi)

Now, lets go back & include dissipation (frictional forces) in the problem. Assume simple physical systems in which the frictional forces are  the particle velocities.  In this case, they are derivable from a dissipation function like that discussed in Ch. 1. A quick review of this from Ch. 1!

Sect. 1.5: Frictional Forces Model for Friction (or air resistance): Ffx = -kxvx Can Include such forces in Lagrangian formalism by introducing Rayleigh’s Dissipation Function ₣ ₣  (½)∑i[kx(vix)2 + ky(viy)2 + kz(viz)2] Obtain components of the frictional force by: Ffxi  - (₣/vix), etc. Or, Ff = - v₣ Physical Interpretation of ₣ : Work done by system against friction: dWf = - Ff dr = - Ff v dt = -[kx(vix)2 + ky(viy)2 + kz(viz)2] dt = -2₣ dt  Rate of energy dissipation due to friction: (dWf /dt) = -2₣

Rayleigh’s Dissipation Function ₣ ₣  (½)∑i[kx(vix)2 + ky(viy)2 + kz(viz)2] Frictional force: Ffi = - vi ₣ Corresponding generalized force: Qj  ∑iFfi(ri/qj) = - ∑ivi₣ (ri/qj) Note that: (ri/qj) = (ri/qj) Qj = - ∑ivi₣(ri/qj) = - (₣/qj) Lagrange’s Eqtns, with frictional (dissipative) forces: (d/dt)[(L/qj)] - (L/qj) = Qj Or (d/dt)[(L/qj)] - (L/qj) + (₣/qj) = 0 (j = 1,2,3, ..n)

Sect. 6.5 In the notation of this chapter, we have the dissipation function (a homogeneous, quadratic function of the generalized velocities; summation convention): ₣  ₣ijηiηj (1) Can obtain the coefficients ₣ij from the Ch. 1 formalism (equations on previous pages) by changing from velocities to generalized velocities, similarly to how we changed from Cartesian to generalized coordinates to get the formalism used so far. Will get ₣ij as functions of the coordinates. Can show they are symmetric: ₣ij = ₣ji

L = T - V = (½)[Tijηiηj - Vijηiηj] (2) Of course, we still have: L = T - V = (½)[Tijηiηj - Vijηiηj] (2) Equations of motion with the dissipation function (1) are obtained from the generalized Lagrange equations of Ch. 1. In the notation of this chapter these are (₣  ₣ijηiηj) (d/dt)[(L/ηi)] - (L/ηi) + (₣/ηi) = 0 (3) This gives: Tijηi + ₣ijηi + Vijηj = 0 (i = 1,2,3, … n) (4) (4): Equations for n coupled, damped harmonic oscillators.  Following the matrix formalism, to find the normal coordinates, we need to get the transformation which simultaneously diagonalizes T, V, & ₣ . That is, find the simultaneous eigenvalues & eigenfunctions of all 3 matrices. In general, this is not possible: For arbitrary ₣ we cannot find the normal modes!

Tijηi + ₣ijηi + Vijηj = 0 (i = 1,2,3, … n) (4) Eqtns of motion, including dissipative forces: Tijηi + ₣ijηi + Vijηj = 0 (i = 1,2,3, … n) (4) n coupled, damped oscillators In general, it is not possible to find normal mode eigenvalues & eigenvectors because it is not possible to simultaneously diagonalize T, V & ₣. In some special cases, we can do this. For example, if the frictional force  particle mass as well as the velocity. Then ₣ is similar mathematically to the KE T & the diagonalization of T also diagonalizes ₣. When diagonalization is possible, the equations of motion (4) can be decoupled using normal coordinates to give equations for n uncoupled, damped oscillators: ζi + ₣i ζi +(ωi)2 ζi = 0 (i = 1,2,3, … n) (5) ₣i (>0) = diagonal elements of ₣ in the representation of the ζi‘s

ζi + ₣i ζi +(ωi)2 ζi = 0 (i = 1,2,3, … n) (5) Normal Coordinate Equations of motion, with dissipation: ζi + ₣i ζi +(ωi)2 ζi = 0 (i = 1,2,3, … n) (5) (5): A standard damped harmonic oscillator equation of motion! From undergrad mechanics, math gives the solution (using complex exponentials instead of sines & cosines): ζi = Ciexp[-i(ω´)it] (6) where (ω´)i = a complex “frequency” given by: (ω´)i = [(ωi)2 - (¼)₣i]½ - (½)i₣i (7)

ζi = Ciexp[-i(ω´)it] (6) (ω´)i = [(ωi)2 - (¼)₣i]½ - (½)i₣i (7) Normal Coordinate solution, with dissipation: ζi = Ciexp[-i(ω´)it] (6) (ω´)i = [(ωi)2 - (¼)₣i]½ - (½)i₣i (7) PHYSICS: The motion is clearly NOT a pure oscillation, since (ω´)i is complex. (6) & imaginary part of (7) combined give a normal coordinate ζi from (6) which decays exponentially as  exp[- (½)₣it]  Since ₣i > 0 the solution is ALWAYS an exponentially decaying function of time! This makes physical sense: We expect friction to cause damping! (6) & the real part of (7) combined give an apparent oscillatory factor in the normal coordinate. Friction clearly shifts the normal mode frequency from the free vibration result ωi to Ωi  [(ωi)2 - (¼)₣i]½ (8)

ζi = Ciexp[-iΩit] exp[-(½)₣it] (6´) Ωi  [(ωi)2 - (¼)₣i]½ (8)  Normal Coordinate solution, with dissipation: ζi = Ciexp[-iΩit] exp[-(½)₣it] (6´) Ωi  [(ωi)2 - (¼)₣i]½ (8) From (6´) & (8), can have 3 cases: 1. Underdamped motion: Occurs if (ωi)2 > (¼)₣i  Ωi is real  (6´) is oscillatory with an exponentially decaying amplitude. The most common (usual) case. 2. Overdamped motion: Occurs if (ωi)2 < (¼)₣i  Ωi is imaginary  (6´) is not oscillatory, but is a pure exponentially decaying function. 3. Critically damped motion: Occurs if (ωi)2 = (¼)₣i  Ωi is zero  (6´) is not oscillatory, but is a pure exponentially decaying function given by exp[-(¼)₣it]

 Normal Coordinate solution, with dissipation: ζi = Ciexp[-iΩit] exp[-(½)₣it] (6´) Ωi  [(ωi)2 - (¼)₣i]½ (8) The most common case of Underdamped Motion and also, small damping, so (ωi)2 > (¼)₣i  ζi  Ciexp[-iωit] exp[-(½)₣it] (6´´) All of this assumes that the dissipation function matrix ₣ can be diagonalized along with the KE T & the PE V. If not, then the general solution is more difficult to obtain. But the qualitative physics remains the same. We expect an oscillatory solution multiplied by an exponentially damped function.

Tijηi + ₣ijηi + Vijηj = 0 (4) Try an oscillatory solution like: For the general case, where ₣ cannot be diagonalized, go back to the equations of motion with dissipative forces: Tijηi + ₣ijηi + Vijηj = 0 (4) n coupled, damped oscillators (i = 1,2,3, … n) Try an oscillatory solution like: ηi = Cajeγt (9) where γ is complex & given by (Caution, Goldstein notation is confusing. Mine is slightly different!) : γ  - κ - iω (κ, ω are real) (10) Insert (9) into (4). Get (using tensor/matrix formalism, with a = column vector of the ai): Va + γ₣a + γ2Ta = 0 (11)

Proof that κ = -Re(γ) > 0. See text, p. 262-263. For general case, where ₣ can’t be diagonalized, the equations of motion take matrix/tensor form: Va + γ₣a + γ2Ta = 0 (11) Can be solved for the ai only for certain values of γ Proof that κ = -Re(γ) > 0. See text, p. 262-263. (11): Analogous to the eigenvalue equation of before, but containing one more term.

Tijηi + ₣ijηi + Vijηj = Fj (1) Consider forced (sinusoidal) oscillations in the presence of dissipative forces. The equation of motion (in terms of generalized displacements ηj): Tijηi + ₣ijηi + Vijηj = Fj (1) General solution to (1) = General solution to the homogeneous equation + a particular solution to the inhomogeneous one. Solution to homogeneous equation: Free vibration problem! Damping (friction) is present  Free vibrations die away (exponentially) after enough time. At steady state (long enough time), only the particular solution to the driven oscillator equation (1) is present. We consider this case!

Tijηi + ₣ijηi + Vijηj = Fj (1) Equation of motion: Tijηi + ₣ijηi + Vijηj = Fj (1) Complex exponential notation: Generalized driving force corresponding to generalized displacement ηj: Fj  F0je-iωt (F0j complex) (2) Seek a particular solution of (1). Assume a solution of the form: ηj = Aj e-iωt (3) Combining (1), (2), (3) gives: [Vij - iω₣ij - ω2Tij]Aj - F0j = 0 (4) (4): System of coupled, inhomogeneous, linear algebraic equations for the unknown amplitudes Aj. Solution given by Cramer’s Rule from linear algebra.

[Vij - iω₣ij - ω2Tij]Aj - F0j = 0 (4) Formal solution is of the form: Aj = Dj(ω)/D(ω) (5) where: D(ω)  Determinant of the coefficients of the Aj & Dj(ω)  Determinant resulting when the jth column of D(ω) is replaced by F01, F02 ,,, F0n We’re most interested, of course in the denominator D(ω) & its behavior as ω  a free vibration resonant frequency ωi. See text, p. 264, where it is argued that D(ω) must have the form (G  constant, ωi  resonant frequency for the ith normal mode κi  ith normal mode damping constant from the damped oscillator problem, Πi  product over all i): D(ω)  G Πi (ω - ωi + iκ)(ω + ωi + iκ) (6)

[Vij - iω₣ij - ω2Tij]Aj - F0j = 0 (4)  Aj = Dj(ω)/D(ω) (5) where the denominator D(ω) has the form D(ω)  G Πi (ω - ωi + iκ)(ω + ωi + iκ) (6) For a resonance at ω = ωk, a particular free vibration frequency, that factor (resonance denominator) in D(ω) is equal to (iκ)(2ω + iκ).  For small damping (small κ), D(ω) is small  The amplitude Aj from (5) is large. Because of the complicated dependence of D(ω) on ω, we can show that the peak of the amplitude Aj is not exactly at the free vibration frequency ωk. But, if damping is small, the shift from this is small!