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Linear Chain / Normal Modes
Overview and Motivation chain of N oscillators, where: N We extend our discussion of coupled oscillators to a is some arbitrary number. When N is large it will
become clear that the normal modes for this system are essentially standing waves.
Key Mathematics problems. : We gain some more experience with matrices and eigenvalue
I. The Linear Chain of Coupled Oscillators
Because two oscillators are never enough, we now extend the system that we havediscussed in the last two lectures to N coupled oscillators, as illustrated below. For
this problem we assume that all objects have the same mass m and all springs have
the same spring constant k s.
Our first goal is to find the normal modes of this system. At the beginning we
approach this problem in the same manner as for two coupled oscillators: we find the
net force on each oscillator, find each equation of motion, and then assume a normal-
mode type solution for the system. Let's consider some arbitrary object in this chain,say the j th object. The force on this object will depend upon the stretch of the two
springs on either side of it. With a little thought, you should be able to write down
the net force on this object as
F (^) j = − ks^ (^ qj − qj − 1 )^ − ks ( q^ j − qj + 1 ), (1)
or, upon simplifying,
Fj = ks ( q (^) j − 1 − 2 qj + qj + 1 ). (2)
q 1 (^) = 0 q 2 = 0
q 1 q 2
m ks
… qN = 0
qN
You might worry that this equation is not valid for the first ( j = 1 ) and last ( j = N )
objects, but if we assume that the j = 0 and j = N + 1 objects (the walls) have infinite
mass so that q 0 and q N + 1 are identically zero, then Eq. (2) applies to all N objects.
We shall refer to these two conditions, q 0 = 0 and qN + 1 = 0 , as boundary conditions
(bc's) on the chain of oscillators.
With the expression for the net force on each object we can write down the equationof motion (Newton's second law!) for each object as
q &&^ j − ω~^2 ( qj − 1 − 2 qj + qj + 1 ) = 0 , (3)
where 1 ≤ j ≤ N and, as before, ω~ 2 = k s m. Notice that each equation of motion is
coupled: the equation of motion for the j th object depends upon the displacement
of both the ( j − 1 ) and ( j + 1 ) objects.
II. Normal Mode Solutions We now look for normal-mode solutions (where all masses oscillate at the same
frequency) by assuming that 1
q j = q 0 , jei Ω^ t. (4)
If we substitute Eq. (4) into Eq. (3), after a bit of algebra the equations of motion
become
Ω^2 q 0 , j + ω~ 2 ( q^0 , j − 1 − 2 q 0 , j + q 0 , j + 1 )^ = 0 (5)
Now, keep in mind that what we have here are N equations of motion, one for each
value of j from 1 to N. As in the two-oscillator problem, the set of equations can be
expressed in matrix notation
M M M
L
0 , 4
0 , 3
0 , 2
0 , 1 2 0 , 4
0 , 3
0 , 2
0 , 1
2 2
2 2 2
2 2 2
2 2
0 0 ~ 2 ~
0 ~ 2 ~^0 ~
~ 2 ~^0 ~^0
q
q
q
q
q
q
q
q
(^1) We have slightly changed notation here. We now write the amplitudes q (^) 0 , j with a comma between the zero and the mass index so that terms such as q 0 (^) , j + 1 are unambiguous.
A. Eigenvectors
Now forand eigenvectors by hand is not advisable. N any larger than 3 solving the characteristic equation for the eigenvalues So let's turn to a computer mathematics
program, such as Mathcad, and see what insight we can gain into this problem. Given
the matrix in Eq. (6) (with a specific value for ω~ 2 ), Mathcad can calculate the
eigenvalues and eigenvectors of that matrix.plot the eigenvectors corresponding to the three lowest eigenvalues for the In the graphs on the previous page we N = 50
problem.
The key thing to notice is that these eigenvectors look like standing waves (on a string,for example). That is, as a function of position (i.e., mass index) j , the components
q 0 , j of the eigenvector appear to be a sine function [which must equal zero at the ends
of the chain ( j = 0 and j = N + 1 ) because of the bc's].
This observation inspires the following ansatz for the eigenvectors
q (^) 0 , j = A sin ( φ j ), (8)
where A is some arbitrary amplitude for this sine function (it could be complex
because we are dealing with a complex form of the solutions), and φ is some real
number that will be different for each normal mode.^2 Now Eq. (8) obviously satisfies
the q 0 = 0 bc on the lhs of chain, but not necessarily the rhs bc q N + 1 = 0. To satisfy
this bc we must have
q 0 (^) , N + 1 = A sin[ φ ( N + 1 )] = 0 , (9)
which is true only for φ ( N + 1 ) = n π, where n is an integer. That is, we must have
= (^) N + 1
φ n n^ π (10)
where the integer n labels the (normal-mode) solution. Now because any integer n in
Eq. (10) produces a value for φ that satisfies Eq. (9), it looks like φ n can take on an
infinite number of values; this seems to imply an infinite number of normal modes.
Well, this can't be right because we know that there are only two normal modes for
(^2) Recall, for a standing wave on a string the spatial part of the standing wave can be written as sin( (^2) λπ x ), so the
parameterlater). φ is obviously related to the wavelength of the normal modes (in some manner – more detail on this
the two-oscillator problem. In fact, because the N oscillator problem involves an N
dimensional eigenvalue problem, there are exactly N normal modes. The solution to
this conundrum lies in the fact that the sine function [in Eq. (9)] is periodic. It can beshown that there are N + 1 unique solutions, but because the n = 0 solution is trivial,
there are only N unique, nontrivial solutions. In fact, we can specify these N unique
solutions by choosing n such that
1 ≤ n ≤ N. (11)
Combining Eqs. (8), (10), and (11) we can write the N eigenvectors as
q 0 , j = A sin Nn +^ π^1 j , 1 ≤ n ≤ N (12)
B. Eigenvalues
So we have now specified the eigenvectors. What about the eigenvalues? We can
obtain these by inserting Eq. (8) into Eq. (5), which produces
Ω 2 sin ( φ j ) +ω~^2 {sin [φ ( j − 1 )] − 2 sin(φ j ) +sin[ φ ( j + 1 )]} = 0. (13)
Now it looks like this equation depends upon j , but it does not. Using some trig
identities it is not difficult to show that Eq. (13) simplifies and can be solved for Ω 2 as
Ω 2 = 4 ω ~ 2 sin^2 ^ φ 2 . (14)
And remembering that φ only takes on the discrete values given by Eq. (10), we have
the N eigenvalues
( )
Ω 2 = 4 ~^2 sin^22 N + 1
n ω^ n^ π ,^ (15)
where n = 1 , 2 , 3 ,L, N. The normal-mode frequencies are thus given by
( )
Ω (^) ± =± 2 ~sin 2 N + 1
n ω^ n^ π.^ (16)
It is interesting to plot the (positive) frequencies as a function of mode number n.
Such a graph is shown below for several values of N. As with previous graphs we
have set m^ =^1 and k s = 1.
where Ω n =Ω n +, and the constants An and Bn (which have replaced A above) are
arbitrary complex numbers. 3 The general solution can thus be written as a linear
combination of the normal modes as
( ) ( ) ( ) ( )
∑ (^ )
=
Ω −Ω
N + + n ni t n i t Nn
Nn
Nn
Nn
N
Ae n^ Be n q t N
q t
q t
q t
1 1
1
1
1 3
2
1
sin
sin 3
sin 2
sin 1
π
π
π
π
M M
[Equation (18) is the extension of Eq. (3) of the Lecture (4) notes.] As before, the
arbitrary amplitudes An and Bn depend upon the initial conditions of all the oscillating
objects.
To see exactly how the An and Bn are determined, let's consider the N = 3 case. As in
the two-oscillator case, let's make the normal modes explicitly real by setting B n = A * n.
For three oscillators Eq. (18) then becomes
( ) ( ) ( )
∑= (^ Ω^ + −Ω )
1
4
4
4 3
2
1 sin 3
sin 2
sin 1 n n i t n i t n
n
n Ae n^ Ae n q t
q t
q t π
π
π
As in Lecture Notes 4 for the two-oscillator problem, we can rewrite An ei Ω n^ t^ + An * e − i Ω nt
as 2 [ Re ( An ) cos(Ω (^) nt ) −Im( An ) sin( Ω nt )]and apply the initial conditions, which gives us
( ) ( ) ( )
∑= (^ )
1 44
4 3
2
(^1) Re sin 3
sin 2
sin 1 2 0
n n n
n
n A q
q
q π
π
π
and
( ) ( ) ( )
∑= (^ )
(^144)
4 3
2
(^1) Im sin 3
sin 2
sin 1 2 0
n n n
n
n qq n A
q π
π
π &
(^3) problem. (^) Notice that the column vector of the rhs of Eq. (17) is the n th eigenvector of the associated eigenvalue
So we see that the real part of the amplitudes An depend upon the initial positions of
the three objects, while the imaginary part of the amplitudes depend upon their initial
velocities.oscillator problem we applied the normal-mode transformation to the equivalent of So where do we go from here? You may remember that for the two-
Eqs. (20) and (21), which allowed us to find the amplitudes (see p. 5-6 of the Lecture
4 notes). There is an equivalent transformation here that will allow us to find the An 's.
To most easily see what it is, let's explicitly write out Eq. (20) as
( ) ( ) ( )
( ) ( ) ( )
( )
( ) ( ) ( )
( )
( ) ( ) ( )
( )
3 (^94)
(^32)
(^34) 2 (^32)
2 1 (^34) 2
4 3
2
(^1) Re sin
sin
sin Re sin
sin
sin Re sin
sin
sin 2 0
A A A
q
q
q π
π
π π
π π
π
π
and evaluate the sine functions, which gives us
( ) ( ) ( )
( ) ( ) ( )
1 2 3 3
2
(^1) Re 2
Re 2
Re 2
A A A
q
q
q
Now notice what now happens if we multiply this equation by the first eigenvector
( ) ( ) ( )
(^34) 2
4 sin
sin
sin π
π
π when it is written as a row vector (sin ( π 4 ) sin( )π 2 sin( 34 π)) = 21 ( 2 2 2 ). We
obtain
21 (^2 q^1^^ ( )^0 +^2 q^2 ( )^0 +^2 q^3 ( )^0 )^ =^4 Re(^ A^1 ).^ (24)
Notice the very nice result that the terms containing the amplitudes A 2 and A 3
produce zero when multiplied by the first eigenvector (in row form).solve for the real part of We can now
A 1 in terms of the initial positions as
Re ( A 1 (^) ) = 81 ( 2 q 1 ( ) 0 + 2 q 2 ( ) 0 + 2 q 3 ( ) 0 ). (25)
This equation is equivalent to the first row of Eq. (18) [or Eq. (20a)] in the Lecture 4
notes for the two-oscillator problem. To obtain Re ( A 2 ) and Re( A 3 ) it should be
obvious that one needs to multiply Eq. (23) by the respective (row) eigenvectors.
( a ) Starting with Eq. (15) and using the angle-addition formula for the sine function,
show, for example, that Ω^2 N + 2 is the same as Ω (^2) N. [Hint: write N as ( N + 1 ) − 1 and N + 2 as ( N + 1 ) + 1 .]
( b ) Starting with Eq. (12) show also that the eigenvector for n = N + 2 is the same as
the eigenvector for n = N.
* 5.3 Similar to the second figure in the notes, graph the three eigenvectors for three
coupled oscillators.
* 5.4 Four coupled oscillators
(a) What are the eigenvalues Ω^2 n for four coupled oscillators?
(b) Similar to the second figure in the notes, find and then graph the four
eigenvectors for four coupled oscillators.
* 5.5 Similar to Eq. (25), find the imaginary part of A 2 for the N = 3 system of
coupled oscillators.
* 5.6 Apply Eq. (26), the normal-mode transformation, to Eqs. (20) and (21) to obtain
the two equations for three oscillators that are equivalent to Eqs. (18) and (19) of the
Lecture 4 notes for two oscillators.
* 5.7 Show that the square of Eq. (26), the normal-mode transformation, is
proportional to the identity matrix. Thus find the inverse of the normal-mode
transformation.
- 5.8 For the three-oscillator problem find the normal-mode coordinates Q 1 ( ) t , Q 2 ( t ), and Q 3 ( ) t in terms of the displacements q 1 ( t ), q 2 ( t ), and q 3 ( t ).
