! 4 4! o! +! h 4 o=0! ±= ± p i And back-substituting into the linear equations gave us the ratios of the amplitudes of oscillation:.»» = A p e i! +t»»

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1 Topic 6: Coupled Oscillators and Normal Modes Reading assignment: Hand and Finch Chapter 9 We are going to be considering the general case of a system with N degrees of freedome close to one of its stable equilibrium points. We are going to show that for small displacements from equilibrium, the system acts like N independent SHOs, usually with N different characteristic frequencies. One or more of these SHOs cn be excited, depending on the initial conditions. If only one oscillation frequency is excited, the N degrees of freedom move synchronously at a common mode frequency. The ratios of the displacements for the different degrees of freedom, known as mode displacement ratios are an intrinsic characteristic of the normal mode that is oscillating. Any motion can be described as a linear combination of modes. We will now develop this formally.. The Double Pendulum Revisited Recall our solution for the double pendulum. The Lagrangian for this system is: L = ml h _ + _ + cos ( ) _ _ i + mgl ( cos + cos ) From this, by considering small displacements (throwing out high order terms and making small angle approximations) we got two coupled differential equations: + =! o + = g! o We looked explicitly for solutions where both masses oscillate with the same frequency: ; (t) = ; e ±i!t Note that we did not nd a general solution to the problem where the oscillations of ; are not necessarily at the same frequency. By substituting our solution into the EOM, we found two linear equations (! o! )! =0! +! o! =0 which can have non-trivial solutions only if the determinant of the coefcients vanishes. So, we got two roots corresponding to two different frequencies:»» (! o! )! =0! (! o! ) The determinant vanishing gives us:! o!! 4 =0

2 ! 4 4! o! +! h 4 o=0! ±= ± p i And back-substituting into the linear equations gave us the ratios of the amplitudes of oscillation:.»» = A p e i! +t»» = A p e i! t We are now going to develop this more formally for arbitrary systems with N generalized co-ordinates.. A Review of Matrix Algebra The formalism we are about to develop requires some matrix algebra, so lets do a quick review. A collection of basic facts: Matrix multiplication is not commutative: if we consider two matrices A and B:! o C = AB 6= BA We will use Einstein summation notation. In this convention, repeated indices are summed over. So, we would write the product of two matrices as: c ij = a ik b kj where k is summed over, and c ij refers to the ij th element ofthematrix C. The transpose of a matrix is dened by interchanging rows and columns. We write it as then The inverse of a matrix is dened such that if so we can easily see that ~A ~x 0 = A~x A ~x 0 = ~x A A = AA = μ where μ is the identity matrix 0 0 : In the particular case where A = A, ~ then the matrix is said to be orthogonal. The adjoint is formed by taking the complex conjugate of A ~ A y = A unitary matrix is dened as one where ~A Λ A y A = and further, if A y = A, the matrix is self-adjoint or hermitian.

3 .3 Small Oscillations Now return to the problem at hand. Consider a conservative system no explicit time dependence to L, and V = V (q). We will be looking at systems with small displacements from equilibrium, so that we can linearie the EOM (as we discussed in the last topic). As we sawfrom expanding the Lagrangian, Taylor's theorem guarantees that most systems are linear if displacements are small enough. Our condition for static equilibrium states that the generalized forces Q i = io This must be true for all dimensions (no saddle points). In general we can write the Lagrangian as: L = X m ij (~q) _q i _q j V (~q) Why does the mass term depend on q? This will be the case if the co-ordinates are not Cartesian. Take the case of the double pendulum. Here we had the term cos ( ) _ _, so we have a coefcient that depends on the q's. Our stability =0 If we change co-ordinates in a singular manner: ~q! 0 =0 implying the static solution is co-ordinate independent. Lets look at small perturbations about the equilibrium position: ~q! ~q o + ~ (t) Expand the potential about the equilibrium position, ~q V (~q) =V(~q o )+ i j + h:o:t: jo The kinetic energy T = m ij _q i _q j = m ij _ i _ j 3

4 Expand the m ij mij (~q) k o k + ::::: Now T is already second order in the _ 's! since we want lowest order, keep only rst term in the m ij. Let m ij be elements of a square, real, symmetric matrix T, and be the same for a matrix V. How do we know V is symmetric? It doesn't matter which order we take the partial derivatives in, so exchange of indeces leaves the value unchanged. So, we can express the Lagrangian as o The equations of motion are: or in matrix notation: L = T ij _ i _ j V ij i j = d~ dt T d ~ V dt T ij j + V ij j =0 T + V =0 alinear ODE We have N equations, each may contain all the co-ordinates. Note: T must be positive denite, because the kinetic energy must be positive: d~ dt T d dt > 0 In matrix language, T being positive denite (having positive eigenvalues) guarantees the kinetic energy will be positive no matter what is (in other words, any matrix that obeys the above equation always is called positive denite). As for our double pendulum example, we will seek solutions for which all the co-ordinates have the same oscillation frequency the normal modes. = a k e i! kt where we should have asmany normal modes as we have degrees of freedom. So, the EOM are V = T =! T V ij a j! T ij a j =0 and we have N linear homogeneous equations for the a's. These have non-trivial solutions only if the determinant of the matrix formed by the coefcents vanishes: V! T V! T ::: V! T V! T ::: ::: ::: ::: =0 4

5 We can write this as a modied eigenvalue equation. Let! = jv Tj =0 which is a polynocial equation for. From the EOM, we get a characteristic (secular) equation for k th eigenvector a k for the k th normal mode (ie. for each root of for each eigenvalue substitute that root into the equations to get the ratio of the a i 's, and these form the components of the eigenvector. We therefore have N eigenvalues/eigenvectors. Our modied eigenvalue equation is written V a k = k T a k (called a modifed eigenvalue eqn. since usually V a = a). We can show that the k must be real. Take the transposed complex conjugate equation: a y l V = Λ l ay l T and dot it into a k a y l V a k Λ l ay l T a k =0=( k Λ l)a y l T a k and for l = k ( k Λ k) a y k T a k =0 let a k = ff k + i k a y T a k k = ~ff k T ff k + ~ k T k + i ~ff k T k ~ k T ff k Now the last term vanishes because T is a symmetric matrix. So this proves a y T a k k is real (= twice the kinetic energy so additionally is >0). For ( k Λ k) a y T a k k = 0, we must have then k Λ k =0=) k real. Now k = ay k V a k a y k T a k The denominator is T > 0, so if V is positive denite, the numerator is > 0 and k are both real and positive, so we get stable oscillations and stable equilibria. Note however that if there exists an a k such that a y k V a k < 0=) <0;! <0 and! is imaginary. The solution is then a k _ e j!jt, and we have exponential growth. So this proves that V is a minimum for stable equilibrium. ie. a y k V a k = V > 0 for >0 In other words - we took our equilibrium to be at V = 0, so for stable equilibrium any other valid state must have larger potential energy. 5

6 Suppose that all the 's are distinct (ie. there are no degenerate roots) now k Λ l 6= 0ifk6=l ( k Λ l ) ~a k T a l =0 =)~a k T a l = 0 for k 6= l We therefore have an orthogonality condition for non-degenerate roots. We can determine the ratio of elements of a k, but the normalization is undetermined. We usually normalize by requiring for each a k! we have N such equations. A = ~a k T a k = 0 and we can write the normalization condition In general, let a a ::: a n j j j j j j j j j j j j ~A T A = μ.4 Example: Two coupled oscillators C A Consider two coupled oscillators with the Lagrangian given by: L = _x + _y! o x + y + ffxy where ffxy is a coupling term. This type of Lagrangian could apply to two pendula of unit mass coupled by a spring, or electrical circuits coupled by mutual inductance. Note that if ff =0,wehave two independent oscillators with angular frequency! o. 6

7 The equations of motion are: @x =0 Welook for the normal mode solutions where x; y~e i!t :! o! x ffy =0! o! y ffx =0 we get nontrival solutions to these linear equations only if so! o! ff ff! o! =0! o! = ±ff! ; =! o ff We nd the ratios of the oscillation amplitudes for the eigenvectors by plugging the eigenfrequencies back into the linear equations:! =! o ff! A x = A y the eigenvectors are therefore:! =! o + ff! A x = A y! = p! o ff x y = p for the symmetric mode (where both pendula oscillate in the same direction), and p x! =! o + ff = p y This is a common char- If we call the eigenvector for the antisymmetric mode, where they oscillate out of phase. acteristic that the antisymmetric mode has higher frequency. solutions ; then x = + p y = p For weak coupling, variations in x and y are the superposition of oscillations with nearly identical frequencies. We get a beat frequency phenomenon. To see this, look at the limit of weak coupling where ff<<! o ;! ß! o ff 7! o

8 so in this case! ß! o + ff x _ e i!ot ff! o! o + e i!ot + = p e i!ot e i! bt + e i! bt ff! o where we let! b = ff!o, so x / e i!ot cos! b t and y _ e i!ot ff! o e i!ot = p e i!ot e i! bt e i! bt + ff! o so y / p e i!ot sin! b t So we can see that the beats are out of phase, and energy is exchanged between the two oscillators..5 Normal Co-ordinates We now illustrate the concept of normal co-ordinates, which are just the a k 's. We can expand our generalized coordinates (the 's) in terms of the eigenvectors (as we did above for x; y (t) =C k a k e i! kt where c k is a complex amplitude factor, and a k is a vector with real components. So the physical coordinates are a superposition of the normal modes. The complex amplitudes are then determined by the initial conditions, but the frequencies and phases of each mode, and the relative amplitudes of the motion of the 's for that mode are characteristic of the mode. 8

9 A transformation into the normal coordinate system diagonalizes T and V, so that we can write for the k th mode k =Re c k a k e i!tλ (note here k isn't summed over), where k is the normal co-ordinate for that mode, and L = m k _ k! k k (summed over k) in our previous example (ignoring the phase factor for now); and x / ( + ) y / ( ) x / Re [(cos! o t + i sin! o t) cos(! b t)] / cos! o t cos! b t y / Re [(cos! o t + i sin! o t)( i sin(! b t))] / sin! o t sin! b t so x = x max at t = 0, and y =0att=0. But, we can shift the phase (time) by multiplying by a complex constant..6 Degeneracy So far we have assumed that all roots of the secular equation are distinct. Nothing guarantees of course that the roots will be distinct, and we can have multiple l = k. We can then get two normal eigenvectors which will not, in general, be orthogonal. In this case we have to construct an orthogonal set. We can do this by taking a linear combination of the degenerate eigenvectors (which are linearly independent but not orthogonal). Lets look at an example for the caseof two degenerate roots corresponding to the normalized eigenvectors a ;a (remember we normalize them so that ~a T a = ). We construct a third vector a 3 = c a + c a We want to make ~a T a 3 =0. also, ~a 3 T a =c +c (~a T a )=0 c c = ~a T a ~a 3 T a 3 ==c +c +c c (~a T a )= (Note: ~a T a = ~a T a since T is symmetric). We have two independent equations that we can solve for c and c. We then use a 3 and a as our orthogonal eigenvectors. They are automatically orthogonal to the other (non-degenerate) vectors. This is called the Gram-Schmidt process. We can follow the same procedure for higher multiplicity roots also. Note that even though the 's are degenerate, the modes describe different motions of the system (which happen to have the same frequency). 9

10 .7 Molecular Vibrations Consider a diatomic molecule a physical system that can be approximated by two masses coupled by a spring: The Euler-Lagrange equations give us and we get the eigenfrequencies from the roots are L = m _x + m _x k (x x ) m ẍ +k(x x )=0 m ẍ +k(x x )=0 m! k k k m! k =0 m! k m! k k =0! =0; k μ where μ = m m m +m. What does the root! =0correspond to? It is associated with a nonoscilliatory motion of the system, ie. translation of the center of mass.! = k=μ corresponds to vibration about the center of mass. ur normal coordinates are for! =0;x =x. x =x clearly corresponds to a translation of the system. We can remove this by considering only motion about the CM, thereby reducing the number of degrees of freedom. This will help with larger numbers of atoms..7. The linear, symmetric, triatomic molecule As seen in the previous example, we would like to rst suppress the center of mass motion. To do this, we make the total momentum = 0. The CM at rest implies m x + m x + m 3 x 3 =0 x = m m (x +x 3 ) 0

11 We can use this to eliminate one or our coordinates, reducing the dimension of the problem by one L = m _x + _x 3 + m _x Λ k (x x ) +(x x 3 ) eliminate x, and do a lot of algebra, and we nd L = m M _Q a + m Q m _ s km Q a k Q s m where Q a = x + x p 3 Q s = x x p 3 and M = m + m is the total mass. These motions correspond to our normal modes which we could have guessed physically: q Our "masses/spring constant" are m =k and m M m = km m k 0 m 0 are so the normal frequencies! 0 =! (symmetric) = k m! (antismmetric) = km m m There is also a transverse mode of oscillation. Again, we reduce the degrees of freedom by removing the CM motion, but in this case also rotation: m (y + y 3 )=m y suppress translation Now we also want to suppress rotation in the plane of the molecule, so use conservation of angular momentum r a = r a0 + u a

12 where u a is the displacement from equilibrium, and the angular momentum L = X a = d dt m a r a v a ß X m a r a0 _u a X m a r a0 u a and since we can choose the origin arbitrarily, X m a r a0 u a =0 For our molecule, this implies: =0 y = y 3 (supress rotation) Now we write the Lagrangian. Let f be the deviation of from ß.» y y f = l + y 3 y l The potential energy V = k0 l f and k = k 0 l L = m We use our two constraints _y + _y 3 + m _y k0 (y + y 3 +y ) y = y 3 ; y m m = y to remove y and y 3 : L = m _y + m m _y 4 k0 y + m m y

13 From the E-L equations: so m + m ÿ m = 4k + m 0 y m! M =k 0 m m In general, if we have N atoms, we have 3N degrees of freedom. 3N-6 correspond to vibration (we subtract 6 3 for translation and 3 for rotation). For co-linear atoms, we have 3N-5 dof, since rotation about the lineof the atoms is not signicant. For a linear molecule, n- are in line, so n-4 bring the atom out of line. Here both have the same frequency ( mutually perpendicular planes). 3