PHYS20401 Lagrangian Dynamics Terry Wyatt Summary of Coupled and Normal Modes using Lagrangian Methods (Lectures 15–18)

1 Introduction

In the lectures we have considered a system of two identical pendulums ( m and length l ) coupled by a spring with spring constant k. The horizontal displacements of the two from their equilibrium positions were taken to be small, x1/l  1, x2/l  1. This is a very simple example, but it has allowed us to develop most of the essential results and techniques we need to solve more complicated systems.

In terms of position coordinates x1, x2 In terms of normal mode coordinates q1, q2 1 1 q1 = √ (x1 + x2) , q2 = √ (x1 − x2) 2 " 2 # m   k mg   m mg  m 1  mg L = x˙2 + x˙2 − (x − x )2 − x2 + x2 L = q˙2 − q2 + q˙2 − 2k + q2 2 1 2 2 1 2 2l 1 2 2 1 2l 1 2 2 2 l 2

2 2 2 2 L does not decouple into separate x1, x2 terms Lagrangian decouples: L = L1(q1, q˙1) + L2(q2, q˙2) :   mg g 2 g mx¨1 = −k (x1 − x2) − x1 q¨1 = − q1 SHO with ω1 = l l ! " l # mg 2k g 2k g mx¨ = k (x − x ) − x q¨ = − + q SHO with ω2 = + 2 1 2 l 2 2 m l 2 2 m l Formal methods to solve the equations of motion With very simple systems it may be possible to “guess” the form of the normal mode oscillations and their angular . With more complicated systems it may be necessary to use a more formal, systematic approach to solving the equations of motion. When a system oscillates in a particular normal mode, n, all the masses, j, oscillate with the same angular , ωn. Therefore, we can write

iωnt 2 iωnt 2 x j = C j e andx ¨ j = −ωnC j e = −ωn x j. (1)

Substituting from Equation 1 into the equations of motion for x j andx ¨ j, we obtain a set of equations, which 2 2 can be solved to obtain the allowed values of ωn. Substituting for each allowed value of ωn, one can obtain the relationships between the C j corresponding to each normal mode motion of the system.

2 Form

Another general method for solving systems is to write the equations of motion in matrix form

MX¨ = −KX = −ω2 MX or M−1KX = ω2X, (2) where, for our example system, we can write

   mg     1     k + −k     0   m 0     x1  −1 adj M   M =   , K =  l  , X =   , and M = =  m  .    mg    det M  1   0 m   −k k +   x   0  l 2 m Equation 2 is an eigenvalue equation for the matrix    k g k   + −  M−1K =  m l m  .  k k g   − +  m m l

We can associate the eigenvalues and eigenvectors of the matrix M−1K         2 g 2 2k g 1  1  1  1  ω = , ω = + Q1 = √   , Q2 = √   1 l 2 m l     2  1  2  −1  with the normal mode frequencies and coordinates of the system. The Lagrangian can be written in matrix form 1 1 L = X˙ T MX˙ − XT KX. 2 2 Please see the handout on matrices for the mathematical details of matricies that are needed for this section of the course. Notes on eigenvectors/normal mode coordinates

1. The eigenvectors Q , Q are orthogonal. As a matter of convenience, the factors √1 are chosen such that 1 2 2 T they are also normalized. Thus the eigenvectors Qn are orthonormal: Q j Qk = δ jk.

2. For each normal mode (n) the Qn specify the relative amplitude and phase of the oscillations of each degree of freedom ( j). To specify the actual motion we have to give, in addition, the absolute amplitude (An) and phase (φn) for each normal mode. To get the of mass j due to the normal mode n we have (n) i(ωnt+φn) x j = Qn( j) Ane . To get the total displacement of mass j we then have to sum over all the normal modes X   i(ωnt+φn) x j = Qn( j) Ane . n

The initial conditions determine the values of An and φn. (2N conditions needed for N degrees of freedom). 3. For more complicated systems in which two or more of the normal modes are degenerate, i.e., they have the same angular frequency, the choice of precisely which normal mode coordinates to chose is arbitrary. However, it is always possible to choose eigenvectors that are orthonormal and it is usually convenient to do so. 3 Transforming Matrix Form to the Normal Mode Basis

We can form a matrix, B, whose columns are the eigenvectors of the matrix M−1K. For our example system this is given by   1  1 1  B = √   .   2  1 −1  We shall also need the inverse matrix     − adj B 1  1 1  B 1 = = √   , det B   2  1 −1  which in this particular case is identical to B.

We can use the matrix B to transform our matrix treatment from the x1, x2 basis into the q1, q2 basis        1         √ [x1 + x2]    −1 1  1 1   x1   2   q1  B X = √     =   =   = Q. 2      1    1 −1 x2  √ [x1 − x2]  q2 2 In this basis, all of the matrices and matrix equations have a very simple diagonal form

 g    h i0 h i  0  M−1K = B−1 M−1K B =  l  , i.e., a diagonal matrix of the eigenvalues,  2k g   0 +  m l    mg     0  0 −1  m 0  0 −1   M = B MB = M =   , K = B KB =  l  .    mg   0 m   0 2k +  l Muliplying equation 2 from the left by B−1 and inserting BB−1 = I (the unit matrix), the equations of motion become h i h i0 B−1 M−1K BB−1X = ω2B−1X, or M−1K Q = ω2Q, or  g       0         q1  2  q1   l    = ω   .  2k g       0 +   q   q  m l 2 2

The Lagrangian can be rewritten in this basis in the following way

1 . . 1 1 . . 1 1 . . 1 L = X T M X − XT KX = X T BB−1 MBB−1 X − XT BB−1KBB−1X = Q T M Q − QT K0Q, 2 2 2 2 2 2 or        mg         0    1   m 0   q˙1  1   l   q1  L = q˙1 q˙2     − q1 q2     . 2     2  mg     0 m   q˙   0 2k +   q  2 l 2 4 Further Reading

Kibble, T.W.B. and Berkshire, F.H. “Classical Mechanics”: Chapter 11.

Goldstein, H., Poole, C. and Safko, J. “Classical Mechanics”: Chapter 6.

Landau, L.D. and Lifshitz, E.M. “Mechanics”: Sections 21–24.

Matrices: More details on matrices can be found in many mathematics textbooks, e.g., D.W. Jordan and P. Smith, “Mathematical Techniques”, Chapters 7, 8 and 13.