# Small oscillations with many degrees of freedom

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1 Analytical Dynamics - Graduate Center CUNY - Fall 007 Professor Dmitry Garanin Small oscillations with many degrees of freedom General formalism Consider a dynamical system with N degrees of freedom near a minimum of the potential energy The Lagrangian has a general form L = m ij ẋ i ẋ j k ij x i x j, ij where x i q i q 0 i are deviations from the minimum and the mass and stiffness coefficients are symmetric, m ij = m ji and k ij = k ji The Lagrangian can be written in the matrix form as L = ẊT M Ẋ X T K X, where M is the mass matrix, K is the stiffness matrix, X = x,x,,x N a column and X T its transposition a row The Lagrange equations read m ij ẍ j + k ij x j = 0, i = N, 3 or, in the matrix form, We search for the solution in the form j M Ẍ + K X = 0 4 X = a sin ωt 5 [or the same with cos ωt] and obtain the system of linear algebraic equations ω M + K a = 0 6 This system of equations has a nontrivial solution for a only if ω M + K = 0 7 that defines N eigenfrequencies ω α, α = N One can prove that ω α are positive, if the potential energy is a positively defined bilinear form, as it is the case for the energy minimum Then the vectors a α are real We will assume the simplest case of ω α nondegenerate, ω α ω β for α β Technically, Eq 6 is a generalized eigenvalue problem, K a =λm a 8 and a is eigenvector of K with respect to M Algorithm for solving this problem is implemented in Wolfram Mathematica The problem we are solving is resembling an eigenvalue problem It is more complicated, however, since we have two matrices instead of one Whereas in eigenvalue problems eigenvectors corresponding to different eigenvalues are orthogonal, here a generalized orthogonality relation takes place To obtain it, write Eq 6 with two different eigenvalues ω α and ω β : ω α M a α = K a α ω βm a β = K a β 9

2 Now multiply the first equation by a T β from the left, multiply the second equation by at α from the left, and subtract them from each other: ω α at β M a α ω β at α M a β = a T β K a α a T α K a β 0 Using symmetry of matrices M and K, one can be easily show a T β M a α = a T α M a β and a T β K a α = a T α K a β Thus the rhs vanishes and one obtains ω α ω β a T α M a β = 0 This means that for α β one has a T α M a β = 0 It is convenient also to require a T α M a α = This gives the generalized orthogonality condition Now from the second of equations 9 one obtains a T α M a β = δ αβ ω βa T α M a β = ω βδ αβ = a T α K a β 3 One can compose the N N matrix of vectors A by stacking all a α together In terms of A Eq takes the form A T M A = I, 4 where I is the unit matrix, whereas Eq 3 becomes where Ω =diag {ω α } Let us now introduce the normal-coordinate vector ζ defined by These equations can be resolved for ζ and ζ α : A T K A = Ω, 5 X = A ζ, X T = A ζ T = ζ T A T 6 ζ = A X, ζ α = A αi x i 7 Note that, according to Eq 4, A A T and A αi a αi Inserting Eq 6 into the Lagrangian, Eq, and using Eqs and 5, one obtains or L = ζt AT M A ζ ζ T A T K A ζ = T ζt ζ ζ Ω ζ L = N α= 8 ζ α ω αζ α, 9 where ζ α are components of the normal-coordinate vector ζ or normal coordinates One can see that the Lagrangian is separable in terms of the normal coordinates, so that every normal coordinate ζ α oscillates with its own frequency ω α : ζ α t = C sα sinω α t + C cα cos ω α t 0 From Eq 6 one obtains N N x i t = a iα ζ α t = a iα [C sα sin ω α t + C cα cos ω α t] α= α=

3 that can be compared with the tentative Eq 5 To obtain the coefficients from the initial conditions, one writes and at t = 0 from Eqs 7, 0, and obtains C cα = Example: Double pendulum ζ α t = C sα ω α cos ω α t C cα ω α sin ω α t A x i0, C sα = A ẋ i 0 3 αi αi ω α Consider a double pendulum that consists of the pendulum attached in a standard way and the pendulum attached to the end of pendulum For simplicity we consider the case m = m = m and l = l = l Describing the pendula with the angles θ and θ, one writes The potential energy reads and the kinetic energy is given by T = m = x = l sin θ, x = l sin θ + l sin θ y = l cos θ, y = l cos θ l cos θ ẋ + ẏ + ẋ + ẏ = U = mg y + y = mgl cos θ + cos θ [ θ + θ + cos θ cos θ + sin θ sin θ θ θ ] [ θ + cos θ θ cos θ θ + sin θ θ + sin θ θ ] Near the ground state θ = θ = 0 one leaves only quadratic terms that results in L = T U = Identifying this with Eq yields M = θ + θ + θ θ mgl, K = mgl 0 0 θ + θ The generalized eigenvalue problem, Eq 6, has the form ω + ω 0 ω a ω ω + ω 0 a = 0, where we have introduced ω 0 = g l The secular equation is 0 = ω + ω0 [ ω 4 = ω ω 0 + ω ] [ ω ω 0 ω ] [ = + ω ] [ ω 0 ω ] ω 0 Thus the eigenfrequencies are given by ω ω = 0 = ω ω 0, ω = 0 = + ω 0 + 3

4 Let us find now generalized eigenvectors and normal modes of the system Eq?? becomes ω α ω 0 a α + ω αa α = 0 ω α a α + ω α ω 0 a α = 0 For instance, from the second equation one obtains a α = µ α ω α + ω 0 ω α, where µ α are normalization coefficients In particular, ω a = µ + ω 0 ω = µ ω 0 = µ ω 0 = µ ω 0 + = µ ω 0 a = µ ω + ω 0 ω + Let us check the orthogonality of these eigenvectors, Eq : a T M a = µ µ ω 4 0 = µ µ ω 4 0 = µ µ ω 4 0 = 0, + + as expected Now calculate normalization factors: = a T M a = µ ω 4 0 = µ ω = µ ω 4 0 and = a T M a = µ ω = µ ω = µ ω Thus µ ω 0 = µ ω 0 = + / / = = + / / and, finally, a = a = + 4 / /

5 Now, the angles expressed through the eigenmodes are given by Eqs 6 or : θ = a ζ + a ζ = / ζ / + ζ θ = a ζ + a ζ = / ζ + / + ζ The inverse transformation has the form ζ = det det = θ a θ a = θ a θ a = / a a a a + θ + / + θ = / + θ + θ ζ = det det = a θ a θ = / θ / θ = / θ + θ One can see that in the normal mode both pendula are swinging in-phase and in the mode they are swinging antiphase The frequency of the mode is lower 3 Example: Two coupled oscillators Another model that illustrate general principles and has more interesting physical content is the model of two coupled oscillators with the Lagrangian L = m ẋ ω x + ẋ ω x x x 4 We will show that if ω ω, the oscillators behave nearly independently from each other and oscillate with their own frequencies ω,, whereas in the opposite limit they strongly hybridize and both normal modes involve both coordinates Here the matrices M and K are given by 0 ω M =m, K = m / 0 / ω and the eigenvalue problem, Eq 6, is ω + ω / / ω + ω a a = 0 Note that this is a standard eigenvalue problem, not generalized The secular equation has the form 0 = ω ω ω ω 4 /4 = ω 4 ω ω + ω + ω ω 4 /4 and the eigenfrequencies are ω ± = ω ω + ω ± + ω 4ω ω + 4 = ω ω + ω ± ω

6 One can see that, indeed, for ω ω the eigenfrequencies ω ± coincide with ω, On the other hand, for ω = ω = ω 0 one obtains Let us now find the eigenvectors and normal modes From the equation ω ± + ω a ± + / a ± = 0 one obtains ω ± = ω 0 ± 6 a ± = µ ± / ω ± ω where µ α is a normalization coefficient This form is nonsymmetric and it is convenient to symmetrize it To this end, consider the case ω < ω One can see from Eq 5 that the frequencies are ordered as Now a + can be symmetrized as with another µ + Here and ω < ω < ω < ω + a + = µ + ω + ω ω + ω ω + ω = ω + ω +, 7 ω ω + 4 ω + ω = ω + ω + ω ω + 4 ω = ω + ω ω + 4 ω = ω ω + ω + 4 To the contrary, a can be symmetrized as a = µ ω ω ω ω 8 Here and ω ω = ω ω + ω ω + 4 ω = ω ω ω + ω ω + 4 ω = ω ω ω + 4 ω = ω + ω + ω + 4 6

7 Obviously eigenvectors a + and a are orthogonal Let us now calculate the normalization factors = a T + a + = µ + ω ω + ω + ω + 4 ω ω + ω ω + 4 ω + ω + ω ω + 4 ω ω + ω ω + 4 [ = µ + ω ] ω + ω + ω ω ω ω + ω + 4 The equation for µ is the same Thus one obtains µ ± = ω ω + 4 = µ + ω ω + 4 After that Eqs 7 and 8 become a + = ω ω + 4 ω ω + ω +ω + ω ω + 4 ω ω + 4 = + ω ω ω ω + 4 ω ω ω ω + 4 and a = ω ω + 4 ω +ω + ω ω + 4 ω ω + ω ω + 4 = + ω ω ω ω + 4 ω ω ω ω + 4 Both eigenvectors can be written as a ± = ± ± ω ω ω ω + 4 ω ω ω ω + 4 Actually this result is valid for an arbitrary relation between ω and ω For ω ω one has a + = 0, a = 0 that means that ω + = ω corresponds to the second oscillator and ω = ω corresponds to the first oscillator In the resonance case ω = ω one obtains a ± = ± that means that the modes are strongly mixed The original coordinates expressed through the normal modes are given by x i = α=± a iα ζ α, 7

8 ie, x = a + ζ + + a ζ, x = a + ζ + + a ζ The normal modes behave as independent harmonic oscillators with frequencies ω ± Thus the time dependence of both x and x is a superposition of these two oscillations Now consider, as an illustration, the resonance case ω = ω = ω 0 with a small coupling ω 0 Here x = ζ+ + ζ, x = ζ+ ζ 9 and, inversely, ζ + = x + x, ζ = x x Suppose at t = 0 oscillator was in a general state while oscillator was in its ground state Then the initial condition for the normal modes is ζ + 0 = ζ 0 = x 0 ζ + 0 = ζ 0 = ẋ 0 Thus the time dependence of the normal-mode coordinates is given by ζ + t = ζ + 0cos ω + t + ζ + 0 ω + sinω + t = x 0 cos ω + t + ẋ0 sin ω + t ω + ζ t = ζ 0cos ω t + ζ 0 ω sinω t = x 0 cos ω t + ẋ0 sin ω t ω Inserting this into Eq 9 one obtains x t = [ ζ+ t + ζ t ] = { [ x 0[cos ω + t + cos ω t] + ẋ 0 sin ω + t + ω + x t = [ ζ+ t ζ t ] = { x 0[cos ω + t cos ω t] + ẋ 0 ]} sin ω t ω ]} [ ω + sin ω + t ω sin ω t In the small-coupling case the eigenfrequencies are close to each other, see Eq 6, one can approximate = = ω + ω ω 0 On the other hand, one cannot make this approximation in the arguments of sin and cos since it will breakdown at large times From Eq 6 in the weak-coupling case follows ω ± = ω 0 ± ω 0 Usung this and trigonometric relations, one obtains ω+ + ω ω+ ω x t = x 0cos t cos [ = x 0cos ω 0 t + ẋ0 ] sin ω 0 t cos ω 0 8 t = ω 0 ± + ẋ0 t ω 0 ω+ + ω ω+ ω sin t cos t

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