Practical Guide to Solving Coupled Oscillator Problems November 8, 2013 1 Two and Three Coupled Oscillators 1.1 Formal Preliminaries Consider two particles that are constrained to move in one dimension (for example on an air track). Their positions are specified by their respective distances x 1 and x 2 from their equilibrium positions. These two positions can be represented as a two-component vector x = x 1 x 2 (1) that can be represented as a point in a two-dimensional plane as shown in Fig. (2). An alternative but equivalent representation of x(t) relies on the introduction of vectors e 1 and e 2 aligned along the x 1 and x 2 directions: e 1 = 1 0 ≡ e 1,1 e 1,2 ; e 2 = 0 1 ≡ e 2,1 e 2,2 , (2) where e 1,1 = 1, e 1,2 = 0, e 2,1 = 0 and e 2,2 = 1 or, equivalently, e i,j = δ ij , where δ ij is the Kroncker delta that is equal to 1 if i = j and zero otherwise. These vectors form an orthonormal pair , i.e., they are of unit length and they are perpendicular to each other. They satisfy e i · e j ≡ X k e i,k e j,k = δ ij i, j =1, 2 (3) and X i e i,j ,e i,k = δ jk . (4) By construction any vector x can be uniquely expressed as x = x 1 e 1 + x 2 e 2 ≡ x i e i , (5) where the summation convention over i is understood. Any set of vectors (e 0 1 , e 0 2 ) that provide a unique representation x = x 0 1 e 0 1 + x 0 2 e 0 2 ≡ x 0 i e 0 i , (6) of any vector x is is called a basis. If in addition the vectors satisfy the orthogonality relation e 0 i · e 0 j = δ ij the basis is orthonormal. The vectors (e 1 , e 2 ) form a special basis called the standard basis. The j th component x 0 j , j =1, 2 of x with respect to the (e 0 1 , e 0 2 ) basis is simply x 0 j = e 0 j · x = e 0 j · (x 0 i e 0 i )= x 0 i δ ij = e 0 j,i x 0 i . (7) 1
Transcript
Practical Guide to Solving Coupled Oscillator Problems
November 8, 2013
1 Two and Three Coupled Oscillators
1.1 Formal Preliminaries
Consider two particles that are constrained to move in one dimension (for example on an air track).Their positions are specified by their respective distances x1 and x2 from their equilibrium positions.These two positions can be represented as a two-component vector
x =
(x1x2
)(1)
that can be represented as a point in a two-dimensional plane as shown in Fig. (2). An alternativebut equivalent representation of x(t) relies on the introduction of vectors e1 and e2 aligned alongthe x1 and x2 directions:
e1 =
(10
)≡(e1,1e1,2
); e2 =
(01
)≡(e2,1e2,2
), (2)
where e1,1 = 1, e1,2 = 0, e2,1 = 0 and e2,2 = 1 or, equivalently, ei,j = δij , where δij is the Kronckerdelta that is equal to 1 if i = j and zero otherwise. These vectors form an orthonormal pair , i.e.,they are of unit length and they are perpendicular to each other. They satisfy
ei · ej ≡∑k
ei,kej,k = δij i, j = 1, 2 (3)
and ∑i
ei,j , ei,k = δjk. (4)
By construction any vector x can be uniquely expressed as
x = x1e1 + x2e2 ≡ xiei, (5)
where the summation convention over i is understood. Any set of vectors (e′1, e′2) that provide a
unique representationx = x′1e
′1 + x′2e
′2 ≡ x′ie′i, (6)
of any vector x is is called a basis. If in addition the vectors satisfy the orthogonality relatione′i · e′j = δij the basis is orthonormal. The vectors (e1, e2) form a special basis called the standardbasis. The jth component x′j , j = 1, 2 of x with respect to the (e′1, e
In addition to satisfying the orthogonality relation of Eq. (3), the vectors of an orthonormal basissatisfy the completeness relation, ∑
i
e′i,k, e′i,j = δkj . (8)
Note that the sum is over the first index, indicating the particular vector, rather than over the secondindex, indicating the component of the vector, as in the orthogonality relation. The completenessrelation follows from the fact that (e′1, e
′2) form a basis and any vector can be uniquely represented
in the from of Eq. (6). The j-component of x in the prime basis is
x′j = x′ie′j,i = (x · e′i)e′ij = x′ke
′i,ke′i,j . (9)
The two sides can be equal only if the completeness relation, Eq. (8) is satisfied. (Remember thatthe summation convention is used on repeated indices.)
If the particles are coupled with three springs with respective spring constants ka, kb, kc asshown in Fig. 1, their equations of motion are
This equation can be written in a compact form with the introduction of the stiffness matrix,
kkk =
(ka + kb −kb−kb kb + kc
)≡(k12 k12k12 k22
)(11)
where kij , i, j = 1, 2 are the components of the symmetric stiffness matrix with kij = kji. Then
x(t) = − 1
mkkkx(t) ≡ −DDDx(t), (12)
where
DDD =
((ka + kb)/m −kb/m−kb/m (kb + kc)/m
)≡(ω2a + ω2
b −ω2b
−ω2b ω2
b + ω2c
)(13)
is the dynamical matrix.We expect that the masses will undergo simple harmonic oscillations with two distinct frequen-
cies, so we assume that x(t) has the form
x(t) = e−iωta = e−iωt(a1a2
), (14)
where a1 and a2 are complex amplitudes with independent magnitudes and phases for a total offour independent variables. The physical solutions are obtained by taking the real part of or x(t).The second derivative with respect to time brings down a factor of −ω2 so that Eq. (13) becomes
ω2a = DDDa or (DDD− ω2III)a = 0, (15)
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a homogeneous equation that can only be satisfied when ω2 is an eigenvalue and a is an associatedeigenvector of the the dynamical matrix DDD. III is the identity matrix with components Iij = δij . Ithas the property that IIIa = a for any vector a.
A fundamental theorem of algebra is that a system of homogenous linear equations has a solutionif an only if the determinant of the matrix of its coefficients is zero. To see how this result comesabout in the simplest form, consider an arbitrary 2× 2 matrix
CCC =
(C11 C12
C21 C22
). (16)
The homogenous equation CCCa = 0 in component form is
C11a1 + C12a2 = 0 (17)
C21a1 + C22a2 = 0. (18)
If C22 > 0, then from Eq. (18),
a2 = −C21
C22a1, (19)
which when used in Eq. (17) yields
C11a1 + C12a2 =
(C11 −
C12C21
C22
)a1 =
1
C22(C11C22 − C12C21)a1 =
a1C22
detCCC = 0, (20)
wheredetCCC = C11C22 − C12C21 (21)
is the determinant of CCC. Thus either a1 = 0 or detCCC = 0. If a1 = 0, then a2 = 0 provided C21 6= 0.This is an uninteresting case in which a1 = a2 = 0. If C21 = 0, then C22a2 = 0 and either C22 = 0or a2 = 0. The latter case is the uninteresting one with a1 = a2 = 0 again. If C22 = 0 and C21 = 0and detCCC = 0. A similar line of argument can be used if C11 6= 0. The end result is that in allcases, either a1 = a2 = 0 or detCCC = 0. Clearly the latter case is the only interesting one.
Thus to find the eigenvalues of the dynamical matrix, we need to solve the equation
det(DDD− ω2III) = 0. (22)
This determinant is a second-order polynomial, P (ω2), in ω2, called the characteristic polynomialof DDD, whose roots, which are the eigenvalues, can be calculated for general DDD from the quadraticformula.
1.2 A Simple Two-particle Example
To keep our discussion simple for the moment, we will first consider simple special cases. Inparticular, we consider the case in which ω2
c = ω2b . The characteristic equation in this case is
P (ω2) = det
(ω2a + ω2
b − ω2 −ω2b
−ω2b ω2
a + ω2b − ω2
)= (ω2
a + ω2b − ω2)2 − ω4
b (23)
Its roots are(ω2 − ω2
a − ω2b ) = ±ω2
b , (24)
3
e1
e2
2e� 1e�
e1
e2
x2e�
1e�
x1
x21x� 2x�
θ
1,1e�2,1e�
1,2e�2,2e�
(a) (b)
Figure 2: (a) The standard basis vectors e1 and e2 and the eigenvectors e1 and e2 for the casein which the rotation angle is 45◦. The components eµ,1 and eµ,2 of eµ, µ = 1, 2 are simply theprojects of these vectors onto the e1 and e2 axes. (b) The standard and rotated basis vectors foran arbitrary rotation angle showing how the components of an arbitrary vector can be representedwith respect to either basis.
orω21 = ω2
a ω22 = ω2
a + 2ω2b . (25)
These are the same frequencies we obtained by guessing the form of the displacements with a fixedfrequency from the air track demonstration in class.
Associated with these eigenfrequencies are eigenvectors a1 = (a1,1, a1,2) and a2 = (a2,1, a2,2),which are obtained by requiring DDDa1 − ω2
1a1 = 0 and DDDa2 − ω22a2 = 0. We start with the equation
for a1:
(ω2a + ω2
b − ω21)a1,1 − ω2
ba1,2 = ω2ba1,1 − ω2
ba1,2 = 0
−ω2ba1,1 + (ω2
a + ω2b − ω2
1)a1,2 = −ω2ba1,1 + ω2
aa1,2 = 0. (26)
Both of these equations yield a1,2 = a2,1. Of course if they did not yield the same result, we wouldbe in trouble, but we are guaranteed that they do because detDDD−ω2
1III = 0 guarantees that the twoequations are linearly proportional to each other. Similarly, keeping only one equation,
(ω2a + ω2
b − ω22)a2,1 − ω2
ba2,2 = −ω2ba2,1 − ω2
ba22 = 0 (27)
and a2,1 = −a2,2. If aµ (µ = 1, 2) is an eigenvector, then any constant times aµ is an eigenvector.We will denote the normalized version of aµ as eµ, where
e1 =1√2
(11
)=
(cos θsin θ
), e2 =
1√2
(−11
)=
(− sin θcos θ
), (28)
where for this case θ = 45◦. For more general values of the elastic constants, θ can take on any valueas depicted in Fig. 2. These two vectors are orthogonal, i.e, e1 · e2 = 0, and they were constructedto have unit length. Thus the form an orthonormal pair with
eµ · eν = δµ,ν , µ, ν = 1, 2 (29)
where are before, δµ,ν is the Kronecker delta that is equal to one if µ = ν and zero otherwise.The components eµ,i of eµ are the coefficients of ei in the representation of eµ in the standard
basis:eµ = eµ,iei = Uµ,iei, (30)
4
where as usual the summation convention on repeated indices is understood, Uµ,i = eµ,i are thecomponents of the matrix
UUU =
(e1,1 e1,2e2,1 e2,1
)=
1√2
(1 1−1 1
)=
(cos θ sin θ− sin θ cos θ
). (31)
The set (e1, e2) is merely a rotated version of the basis (e1, e2) as shown in Fig. 2. In this case, therotation angle is 45◦, but in more general problems, it can be any angle, as we shall see shortly.The matrix U is a rotation matrix.
The inverse AAA−1 of any matrix AAA is defined by
AAA−1AAA = AAAAAA−1 = III (32)
Rotation matrices like UUU with components Uµ,i have the property that their inverse is simply theirtranspose UUUT whose components are UTi,µ = Uµ,i. Thus
UUUT =1√2
(1 −11 1
)=
(cos θ − sin θsin θ cos θ
). (33)
and
UUUTUUU =1
2
(1 −11 1
)(1 1−1 1
)=
(1 00 1
)= III. (34)
ThusUUUT is the inverse of UUU.It is clear from the geometry in Fig. 2 that any vector x can be represented in terms of its
components (x1, x2) with respect to the the standard basis or in terms of its components (x1, x2)relative the e1 and e2. In other words, the set (e1, e2) form a basis, and
x = x1e1 + x2e2 ≡ xiei = x1e1 + x2e2 ≡ xµeµ. (35)
Taking the dot produce of both sides of this equation with eµ yields
xµ = x · eµ = xiei · eµ = xieµ,i = Uµ,ixi. (36)
Defining x to be the column vector with components xµ, this equation can be rewritten as
x = UUUx. (37)
Now what can we do with all of this? First, let’s return to the original dynamical equation,Eq. (12), and express x(t) in the “tilde” basis (e1, e2):
x = ¨xµeµ = −DDDxµeµ = −∑µ
ω2µxµeµ (38)
where the last step follows because eµ is the eigenvector of DDD with eigenvalue ω2µ. Taking the dot
produce of both sides of the equation with eµ, ν = 1, 2 leads to
¨xν = −ω2µxν , µ = 1, 2. (39)
In other words x1 and x2 obey the equations of a simple harmonic oscillators with respectivefrequencies ω1 and ω2. We have reduced the coupled oscillator equations to uncoupled singleoscillator equations. x1 and x2 are the normal modes of the system, often denoted as q1(t) and
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q2(t). The solutions to the uncoupled equations are the familiar solutions to the simple oscillator,i.e,.
xµ(t) ≡ qµ(t) = Aµe−iωµt or (40)
qµ(t) = Bµ cosωµt+ Cµsinωµt
ωµ(41)
= Aµ cos(ωµt− δµ) (42)
We now have the general solution for the coupled oscillator problem:
x(t) =∑µ
qµ(t)eµ =∑µ
(Bµ cosωµt+ Cµ
sinωµt
ωµ
)eµ (43)
x(t) =∑µ
(−ωµBµ sinωµt+ Cµ cosωµt) eµ. (44)
The unknown coefficients Aµ (or Bµ and Cµ) are determined by boundary conditions that are
usually specified at t = 0. Setting x(0) = x0 = (x0,1, x0,2) and ˙x(0) = v0 = (v0,1, v0,2), theboundary conditions can be expressed as
x0 = B1e1 +B2e2; v0 = C1e1 + C2e2. (45)
The orthonormal property of eµ allows us to calculatesd the unknown coefficients by simply takingthe dot product of these two equation with e1 and e1:
B1 = x0 · e1 =1√2
(x0,1 + x0,2) (46)
B2 = x0 · e2 =1√2
(−x0,1 + x0,2) (47)
C1 = v0 · e1 =1√2
(v0,1 + v0,2) (48)
C2 = v0 · e2 =1√2
(−v0,1 + v0,2). (49)
Putting these terms back into Eq. (43), we obtain
x1(t) = q1(t)e1,1 + q2(t)e2,1
=1
2(x0,1 + x0,2) cosω1t+
1
2(x0,1 − x0,2) cosω2t+
1
2(v0,1 + v0,2)
sinω1t
ω1+
1
2(v0,1 − v0,2)
sinω2t
ω1
(50)
x2(t) = q2(t)e1,2 + q2(t)e2,2
=1
2(x0,1 + x0,2) cosω1t−
1
2(x0,1 − x0,2) cosω2t+
1
2(v0,1 + v0,2)
sinω1t
ω1+−1
2(v0,1 − v0,2)
sinω2t
ω1
(51)
This equation can easily be applied to special cases. For, example if x0,1 = x0 and x0,2 = 0,v0,1 = v0,2 = 0, the solutions are
x1(t) =1
2x0(cosω1t+ cosω2t)
x2(t) =1
2x0(cosω1t− cosω2t) (52)
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1.3 A More Complex Two-Particle example
In the example we just studied, the normal modes were simple. In one mode the two masses movedin the same direction the same amount, and in the second, they moved in opposite directionsthe same amount. We now consider a somewhat more complicated realization of the two-mass,two-spring problem in which kb = 2ka and kc = 4ka. From Eq. (13),
DDD = ω2a
(3 −2−2 6
), (53)
where ω2a = ka/m. Clearly the eigenvalues of DDD will be proportional to ω2
a, so we set ω2 = ω2as.
The characteristic polynomial is then
P (ω2) = ω40[(3− s)(6− s)− 4] = ω2
0(s2 − 9s+ 14), (54)
and its roots are s1 = 2 and s2 = 7 or ω21 = 2ω2
a and ω22 = 7ω2
a. The eigenvectors a1 = (a1,1, a1,2)and a2 = (a2,1, a2,2) are determined by
(3ω2a − 2ω2
a)a1,1 − 2ω2aa1,2 = = ω2
a[a1,1 − 2a1,2] = 0 (55)
(3ω2a − 7ω2
a)a2,1 − 2ω2aa2,2 = ω2
a[−4a2,1 − 2a2,2] = 0. (56)
Thus the normalized eigenvectors are
e1 =1√5
(21
)=
(cos θsin θ
), e2 =
1√5
(−12
)=
(− sin θcos θ
), (57)
where θ = tan−1(1/2) = 26.6◦. These vectors clearly satisfy the orthonormality and completenessrelations eµ · eν = δµ,ν and
∑µ eµ,ieµ,j = δij .
Set boundary conditions x0 = x0(1, 1) and x(0) = v0 = v0(1,−1), then
B1 = x0 · e1 = x0
(2√5
+1√5
)=
3√5x0
B2 = x0 · e2 = x0
(− 1√
5+
2√5
)=
1√5x0
C1 = v0 · e1 = v0
(2√5− 1√
5
)=
1√5v0
C2 = v0 · e2 = v0
(− 1√
5− 2√
5
)= − 3√
5v0
(58)
and
x1(t) =1
5x0(6 cosω1t− cosω2t) +
1
5v0
(2
sinω1t
ω1+ 3
sinω2t
ω2
)x2(t) =
1
5x0(3 cosω1t+ 2 cosω2t) +
1
5v0
(sinω1t
ω1− 6
sinω2t
ω2
). (59)
7
m m mka kb kc kd
Figure 3: Three-mass oscillator
1.4 A Three Oscillator Problem
In the previous sections, we set up a formalism for solving the eigenvalues and eigenvectors of thedynamical matrix. We focussed on simple two-oscillator examples, but the formalism is valid foroscillator systems with an arbitrary number of degrees of freedom. Basically, each particle hasone translational degree of freedom per spatial dimension. Thus, if there are N particles in ddimensions, there will be dN degrees of freedom, and the dynamical matrix will a dN dimensionalsquare matrix (i.e., a dN × dN matrix). Here we will study a simple three-oscillator system inone-dimension with three degrees of freedom. The system consists of three masses and four springsas shown in Fig. 3.
The springs from left to right have ka = 4k, kb = 3k, kc = 4k, and kd = 3k. The equations ofmotion are
x1(t) = −4ω20x1 − 3ω2
0(x1 − x2)x2(t) = −3ω2
0(x2 − x1)− 4ω20(x2 − x3)
x3(t) = −4ω20(x3 − x2)− 3ω2
0x3, (60)
where ω20 = k/m. The dynamical matrix is then
DDD = ω20
7 −3 0−3 7 −40 −4 7
, (61)
and setting ω2 = s, the characteristic polynomial, det(DDD− ω20s) is
The eigenvalues (in units of ω20) are s = 7, s = 7±
√25 or
s1 = 2, s2 = 7, s3 = 12. (63)
The associated eigenvectors aµ = (aµ,1, aµ,2, aµ,3), µ = 1, 2, 3, are determined by solving (DDD −sµω
20)aaaµ = 0. We solve these equations in order:
s = 2 ,1
ω20
(DDD−2ω20)a1 =
5 −3 0−3 5 −40 −4 5
a1,1a1,2a1,3
=
5a1,1 − 3a1,2−3a1,1 + 5a1,2 − 4a1,3−4a1,2 + 5a1,3
=
000
(64)
The first and third columns yield a1,1 = (3/5)a1,2 and a1,3 = (4/5)a1,2. The equation from thesecond column is automatically satisfied by these relations. The first normalized eigenvector is,therefore,
e1 =1
5√
2
354
(65)
8
The second eigenvector is next:
s = 7 ,1
ω20
(DDD− 7ω20)a1 =
0 −3 0−3 0 −40 −4 0
a2,1a2,2a2,3
=
−3a2,2−3a2,1 − 4a2,3−4a2,2
=
000
. (66)
Thus a2,2 = 0, and a2,3 = −(3/4)a2,1, and the normalized second eigenvector is
(68)Thus, a3,1 = −(3/5)a3,2, a3,3 = −(4/5)a3,2, and the normalized third eigenvector is
e3 =1
5√
2
−35−4
(69)
It is a straightforward exercise to verify the orthogonality and completeness relations: eµ · eν = δµ,νand
∑3µ=1 eµ,ieµ,j = δij .
The general solution for this set of oscillators is thus
x(t) =3∑
µ=1
(Bµ cosωµt+ Cµ
sinωµt
ωµ
)eµ (70)
If the initial conditions x0 and v0 are given for the displacement and velocities, then
Bµ = x0 · eµ, Cµ = v0 · eµ. (71)
If x0 = (x0, 0, 0) and v0 = (0, 0, 0),
B1 =3
5√
2x0, B2 =
4
5x0, B3 = − 3
5√
2, (72)
C1 = C2 = C3 = 0, andx1(t)x2(t)x3(t)
= x0
150(9 cosω1t+ 32 cosω2t+ 9 cosω3t)
310(cosω1t− cosω3t)
625(cosω1t− 2 cosω2t+ cosω3t)
(73)
1.5 Different Masses
So far, we have only considered systems in which the masses are equal. When they are no equal,the most direct approach leads to dynamical matrices that are not symmetric. This matrix, aswe shall see, still has real eigenvalues as it must, but its right an left eigenvectors are different.This is something of a nuisance, and it is advantageous in general to formulate the problem so that
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the dynamical matrix is symmetric. We first look at the the problem of two coupled oscillatorswith different masses using the direct approach, and then we look at it using the approach witha symmetric dynamical matrix. Consider the system depicted in Fig. (1) but with the right masshaving mass m1, the left mass having mass m2, and kc = ka . The equations of motion are
where the stiffness matrix kkk is the same as Eq. (11) and the mass matrix is
mmm =
(m1 00 m2
). (77)
so that
DDD′ =
(ka+kbm1
− kbm1
− kbm2
ka+kbm2
)=
(ω2a + ω2
b −ω2b
−ηω2b −η(ω2
a + ω2b )
)(78)
where ω2α = kα/m1 for α = a, b, c and η = m1/m2. The eigenvalues, which are the roots of the
characteristic polynomial,
P (ω2) = det(−ω2III +DDD′
)= det
(ω2a + ω2
b − ω2 −ω2b
−ηω2b η(ω2
a + ω2b )− ω2
)(79)
= [ω2a + ω2
b − ω2][η(ω2a + ω2
b )− ω2]− ηω4b , (80)
are
P (ω2) =1
2
[(1 + η)ω2
a + ω2b ±
√[(1− η)2(ω2
a + ω2b )
2 − 4ηω4b
]. (81)
The matrix DDD′ is not symmetric. Though it is not necessary for the dynamical matrix to besymmetric, there are advantages to setting up a formalism in which it is. These include dealingwith a matrix form that is guaranteed to have real eigenvalues and one whose “right” and ’‘left”eigenvectors are transposes of each other. If an n×n square matrixAAA has eigenvalues λµ, µ = 1, . . . n,it has right eigenvectors rµ such that AAArn = λnrn and left eigenvectors lµ such that AAAlµ = λµlµ. Forsymmetric matices, lµ = rTµ . To convert the equation of motion, Eq. (75) to one with a symmetric
dynamical matrix, we introduce the square root, mmm1/2, and the inverse square root, mmm−1/2, of themass matrix:
mmm1/2 =
(m
1/21 0
0 m1/22
)mmm−1/2 =
(m−1/21 0
0 m−1/22
). (82)
It is straightforward to verify that mmm1/2mmm1/2 = mmm and that mmm−1/2mmm1/2 = III. We next introduce thevector
y = mmm1/2x =
(m
1/21 0
0 m1/22
)(x1x2
)=
(√m1x1√m2x2
). (83)
This equation can be inverted to give x in terms of y:
x = mmm−1/2y =
(y1/√m1
y2/√m2
). (84)
10
With these definitions, we can reexpress mmmx = mmm1/2y and kkkx = kkkmmm−1/2mmm1/2x = kkkmmm−1/2y. Thus
mmmx = mmm1/2y = −kkkx = −kkkmmm−1/2y or
y = −DDDy, (85)
where
DDD = mmm−1/2kkkmmm−1/2 =
(k11m1
k12√m1m2
k12√m1m2
k22m2
)=
(ω2a + ω2
b −√ηω2b
−√ηω2b η(ω2
a + ω2b )
)(86)
is the now symmetrized dynamical matrix. The only price we had to pay for this transformation isthe introduction of the vector y. It is clear that characteristic polynomial associated with DDD is thesame as that associated with DDD′.
Our next step is to determine the eigenvectors of D. Though we can do this analytically forgeneral ω2
s , s = a, b, c and η, the expressions are messy, so we consider only a special case in whichω2a = 25ω2
0, ω2b = 7ω2
0, and η = 1/4. Then
DDD = ω20
(32 −7/2−7/2 8
)andDDD− ω2III = ω2
(32− s −7/2−7/2 8− s
)(87)
where ω2 = ω20s and the characteristic polynomial is
This result should be contrasted with that we obtained with equal masses. There cos θ = sin θ =1/√
2 so that
x1(t) =1
2x0(cosω1 + cosω2t) = x0 cosωavt cos ∆ωt (96)
x2(t) =1
2x0(cosω1 − cosω2t) = x0 sinωavt sin ∆ωt, (97)
where ωav = (ω1 + ω2)/2 and ∆ω = (ω2 − ω1)/2. These equations describe a beating patternin which the lower frequency parts sin δωt and cos ∆ωt provide an envelope for the more rapidlyvarying parts oscillating with frequency ωav. When sin ∆ωt = 0 (i.e., t = nπ/∆ω for any integer n,x2(t) = 0 and the envelope of x1(t) is a maximum. Similarly, when the cos ∆ωt = 1, the envelopeof x1(t) is zero that of x2(t) is a maximum. The result as we say in class is that in the beginning,mass 2 appears to be at rest, then both masses move, then mass one appears to be at rest and soon periodically as shown in Fig. 4 (a) and (c). In the unequal mass case, cos θ 6= sin θ, and thebeating is less obvious as can be seen by the following alternative (but equivalent) expressions forx1(t) and x2(t):
x1(t) =1
2x0[cosωavt cos ∆ωt+ cos 2θ(sinωavt sin ∆ωt)] (98)
x2(t) =1
2x0 sin 2θ sinωavt sin ∆ωt (99)
Clearly the envelope of x1(t) never goes to zero as shown in Fig. 4 (b) and (d).
2 N-coupled oscillators
2.1 Fixed Boundary Conditions
Consider N particles of mass m connected by springs and constrained to move in one-dimension asshown for three particles in Fig. 3. For the moment, we consider only the case in which all springshave the same spring constant k. There are two boundary conditions on these oscillators; the leftend of the first spring is clamped at the left wall and the right end of the last spring is clamped atthe right wall. These boundary conditions are referred to as fixed boundary conditions.It is usefulto imagine that the walls are clamped particles, which we label as particle 0 and particle N + 1,respectively. The set-up is now that there are N+2 particles 0, 1, ...N,N+1, where the positions ofthe first and last particles are clamped at the walls. The equations of motion for particle 2, ..., N−1all have the same form because they have identical environments, being connected via springs to a
12
5 10 15 20 25 30
-1.0
-0.5
0.5
1.0
(a)
5 10 15 20 25 30
-1.0
-0.5
0.5
1.0
(b)
5 10 15 20 25 30
-1.0
-0.5
0.5
1.0
(c)
5 10 15 20 25 30
-0.5
0.5
(d)
Figure 4: (a) x1(t) and (c) x2(t) for equal-mass coupled oscillators; (b) x1(t) and (d) x2(t) forunequal mass coupled oscillators. Note in the equal-mass case, x1(t) oscillates near zero when x2(t)oscillates near its max and vice versa. In the unequal mass case, the envelope of ov x1(t) nevergoes to zero.
mobile particle on the right and one on the left. Particles 1 and N , on the other hand are connectedto an immobile wall on one side, and their equations of motion look different:
The usual strategy for determining the eigenvalues and eigenvectors of this system of equations isto “guess” (ansatz ) the form of the eigenvectors and use their free degrees of freedom to ensurethat the first and last equations are satisfied. We, therefore, make the following ansatz:
xq,p(t) = Aqe−iωt sin(qp+ φq) (101)
where q is an index (whose form is to be determined) specifying the particular eigenmode, p = 1, ...Nlabels the position of particle p, and φq is a phase which we will use to set boundary conditions. In
13
vector notation, this ansatz corresponds to an vector
xq(t) = Aqe−iωt
sin(q + φq)sin(2q + φq)
.
.
.sin(Nq + φq)
. (102)
If this ansatz is right, then it must be that (2xp − xp+1 − xp−1) be proportional to xp for every p.Consider, therefore,
= 2 sin(qp+ φq)− (sin(qp+ φq) cos q + cos(qp+ φq) sin q + sin(qp+ φq) cos q − cos(qp+ φq) sin q)
= 2(1− cos q) sin(pq + φq). (103)
Thus, provided we satisfy the boundary equations [p = 1 and p = N ], we have (using xq,p =−ω2xq,p), each term in Eqs. (100) have the same form
ω2xq,p = 2(k/m)(1− cos q)xq,p = 4(k/m) sin2(q/2) (104)
Note that this result is independent of φq. This phase will be at our disposal to match differentboundary conditions. To each index q, there corresponds the eigenvalue
ω2q =
2k
m(1− cos q) =
4k
msin2(q/2). (105)
Its associated eigenvectors is
eq = Cq
sin(q + φq)sin(2q + φq)
.
.
.sin(Nq + φq)
, (106)
where the coefficient Cq is chosen to normalize the eigenvector so that e∗q · eq = 1. We will returnto this normalization and the demonstration that the eigenvectors are orthogonal shortly.
Now we have to deal with the boundary conditions. In the current problem, we require thatthe left and right walls be stationary, i.e., that x0 = 0 and xN+1 = 0. To satisfy the first condition,we merely chose φq = 0. The displacements are then proportional to sin qp, which clearly equalszero when p = 0. To match the second boundary condition, we set
sin q(N + 1) = 0. (107)
This constraint quantizes the values of q:
q =nπ
N + 1; n = 1, 2, ..., N. (108)
Notes that n = 0 and n = N + 1 are not counted because sin qp = 0 for every p for these values.There are nonetheless N independent modes as required.
14
Why don’t we have to consider q’s that are outside the range 0 to π? It is because q’s outsidethis range describe exactly the same situations as those for q’s within this range. Consider first thefrequency of Eq. (104), and let q = π ±∆q, then
sinq
2= sin
(π
2± ∆q
2
)= cos
∆q
2(109)
and sin[(π/2) + (∆q/2)] = sin[(π/2) − (∆q/2)]. For each q in the interval (π, 2π), there is acompanion q in the interval (0, π) that that has the same frequency. Similarly for each q in theinterval (π, 2π), there is a companion q in the interval (0, π) with the same eigenfunction moduloa minus sign as can be seen from the relation,
sin(π ±∆q)p = ± cosπp sin ∆q p = ±(−1)p sin ∆q p, (110)
orsin(π + ∆q)p = − sin(π −∆q)p. (111)
Now that we have the quantized values of q, we can demonstrate that the different eigenvectorsare orthogonal and complete, i.e., we need to show that
e∗q′ · eq =
N∑p=1
eq,pe∗q′,p = δq,q′ (112)
∑q
e∗q,peq,p′ = δp,p′ . (113)
To show that the first of these relations is true, we define
where we used various trigonometric identities and the fact that k = nπ/(N+1) for some integer n,so that cos k(N+1) = cosnπ = (−1)n, sin k(N+1) = sinnπ = 0 and limk→0 sin k(N+1)/ sin(k/2) =2(N + 1) Thus,
Jq,q′ =1
2
(−1
2(1 + (−1)n−n
′) + (N + 1)δq−q′,0 +
1
2(1 + (−1)n+n
′) + (N + 1)δq+q′,0
)=
1
2(N+1)δq,q′
(122)where q = nπ/(N + 1) and q′ = n′π/(N + 1). The choice Cq =
√2/(N + 1) then provides the
orthonormal eigenvectors we need.
eq =
√2
N + 1
sin qsin 2q...
sinNq
. (123)
For N = 5, the five eigenvectors {en} are
e1 =
1
2√3
121√3121
2√3
, e2 =
121201212
, e3 =
1√3
01√3
01√3
, e4 =
12−1
2012−1
2
, e5 =
1
2√3
−12
1√3121
2√3
(124)
You should verify directly that these vectors do indeed form an orthonormal set.
2.2 An example with initial conditions
We now have all of the apparatus necessary to solve initial value problems for N coupled oscillatorswith the geometry of Fig. 3. Suppose we impose initial conditions that at t = 0, all velocities arezero and that particle p is at position xp,0. Then we can write
x(t) =∑q
Aqeq cosωqt (125)
where we can choose the cos solutions because we have set the velocities equal to zero. The positionsat t = 0 are, however, non-zero:
x(t = 0) = x0 =
x1,0x2,0...
xN,0
. (126)
We thus havex0 =
∑q
Aqeq. (127)
Our goal is to determine the values of the amplitudes Aq. Once we know them, we know the timeevolution for all times t > 0. Equation 127 is a set of N coupled equations, and because the eq
16
k
m m m m m mk k k k k k
m m m m m mk k k k k
(a)
(c)
(b)
m m
m
mm
m
Figure 5: Schematic of (a) free, (b) hybrid, and (c) free boundary conditions.
form a linearly independent set, this equation has a solution. It is straightforward to solve usingthe orthogonality relations of the basis vectors eq. Simply take the dot product of both sides of theequation with a particular member of this basis e∗s:
e∗s · x0 =∑q
Aqe∗seq =
∑q
Aqδq,s = As, (128)
and the problem is solved. For all time t > 0,
x(t) =∑q
(x0 · e∗q)eq cosωqt. (129)
2.2.1 Other Boundary Conditions
There are three other boundary conditions that are of physical interest: (1) The free chain with freeboundary conditions in which there are no terminal springs confining the particles. There are thenN particles connected by N − 1 springs; (2) A chain with hybrid boundary conditions with one end(say the left end) clamped to the wall but the final particle with no spring to its right; (3) Periodicboundary conditions in which particle N is constrained to have exactly the same displacement asparticle 0. This condition can be thought of as tying the string of particle in a circular loop andwith particle 0 attached to particle N − 1. (Note there are n particle because counting starts at 0,not at 1.)
The first observation is that the equations for particles not at the boundaries are the same forall of these conditions and for the one with fixed boundaries we just considered, so they will all havethe same dependence of eigenfrequencies ωq on q. The solutions with different boundary conditionswill differ in the values of q that are allowed (quantization condition) and possibly on the value ofthe phase φq, which was zero in the case of fixed boundary conditions.
2.2.2 Free Boundary Conditions
We label the first particle as 0 and the Nth particle as N1. The equations of motion for these twoparticles are
where ω20 = k/m. We try a solution in which xp ∼ cos(qp+φq). This form corresponds to changing
the φp defined earlier to −(π/2) + φq. Inserting this form into the two equations above yields:
ω2q cosφq = ω2
0[cosφq − cos(q + φq)] (132)
ω2q cos[q(N − 1) + φq] = ω2
0[cos[q(N − 1) + φq]− cos[q(N − 2) + φq]. (133)
17
We solve the first equation first. With some trivial trigonometric manipulation and using Eq. (105),we find
(1− 2 cos q) cosφq = − cos(q + φq) = − cos q cosφq + sin q sinφq (134)
tanφq =1− cos q
sin q= tan q/2. (135)
Thus, φq = q/2. We now use this result in the second boundary equation:
0 = −ω2q cos q(N − 1
2) + ω20[cos q(N − 1
2)− cos q(N − 32)] (136)
= (−ω2q + ω2
0) cos(N − 12)q − ω2
0(cos q cos(N − 12)q + sin q sin(N − 1
2)q) (137)
= −ω20[(1− cos q) cos(N − 1
2)q + sin q sin(N − 12)q sin q) (138)
= −ω20[cos(N − 1
2)q − cos(N + 12)q] = 2ω2
0 sin qN sin(q/2) = 0. (139)
The “quantization constraints are, therefore,
q =nπ
N, n = 0, 1, ..., N − 1 (140)
Note that q = 0 is include in the set. It must be there because we know that there must be apure translation mode with zero frequency. When q = 0, ωq = −0 as required, and in addition, theamplitude at article p, cos(q(p+(1/2))→ 1, is the same for every p as it is in a uniform translation.
2.2.3 Mixed boundary conditions
As shown in Fig. (5)(b) the left most particle (particle 1) is attached to a spring that is clamped ina fixed position, but the final particle has no spring to its right. To satisfy the boundary conditionthat the first spring is clamped, we use the same ansatz as we used for the spring with fixedboundary conditions at both ends, i.e., xq,p = Aq,p sin qpe−iωqt. The final mass satisfies
mxn = −k(xN − xN−1), (141)
implying
ω2q sin qN = ω2
0[sin qN − sin q(N − 1)] = ω20[sin qN − sin qN cos q = cosNq sin q] (142)
= ω20(1− cos q) sin qN + ω2
0 cosNq sin q. (143)
Using ω2q = ω2
0(1− cos q), we obtain
0 = (1− cos q) sinNq − cosNq sin q = 2 sin2(q/2) sinNq − 2 sin(q/2) cos(q/2) cosNq (144)
There is another boundary condition that is used extensively that in some sense eliminates bound-aries altogether. This is the periodic boundary condition in which displacements satisfy the peri-odicity condition
xp = xp+N (148)
for some integer N for all positions p. You can think of this boundary condition as arising fromwrapping the line of springs around a ring so that the Nth particle and the zeroth one are actuallyidentical. The periodicity condition is trivially satisfied if
xp = Aq,peipq = xp+N = Aq,pe
iq(p+N) (149)
provided
q =2πn
N(150)
for n = 0, 1, · · · (N − 1) . Note that n = 0, which corresponds to q = 0 and to a zero-frequencymode (ωq=0 = 0). This is not surprising because q = 0 implies that all particles move the sameamount, eiqp = 1 when q = 0, implying that no springs are stretched or compressed.
For all of the other boundary conditions, q is restricted to lie between 0 and π. For periodicboundary conditions, q lies between 0 and 2π. What is going on? It turns out that the eigenmodesunder periodic are traveling waves that can move either to the right or left whereas the eigenmodesunder the other boundary conditions are stationary. To see why this is, we first note that periodiceigenfunction satisfies eiqp = ei(q+2π)p for every p, so rather than restricting q to lie between 0 and2π, we could equally well restrict it to lie between −π and π. Or in equivalently restrict n to liebetween −N/2 and N/2. Or course, we have to distinguish between systems with N odd and Neven. In the former case, n = −1
2(N − 1),−12(N − 3), · · · ,−1, 0, 1, · · · , 12(N − 1), and in the latter
case, n = −12N + 1,−1
2N + 2, · · · ,−1, 0, 1, · · · , 12N − 1, 12N . Thus, q can be either positive ornegative. Now, consider the full time dependence of the displacements for a particular mode q:
where δq is the phase of the complex amplitude Aq. This function is a maximum whenever
qp− ωqt− δq = 0; or p = p0q +ωqqt. (152)
where p0q = δq/q. At time t = 0, the maximum occurs at p = p0. Of course p0q is not in generalan integer, but the function xq,p(t) makes sense even when p and p0q are not integers. At time tlater the maximum that was originally at p = p0 has moved to a new position p0q +
ωqq t. In other
words, the maximum move with a constant “velocity” vq = ωqt. The particles themselves do notmove with this velocity: they oscillate back and forth (or up and down) with frequency ωq. It isonly the peak in the amplitude that moves. When q > 0, the peak moves to the right; but it q < 0,the peak positions is at
p = p0q −ωq|q|t. (153)
Thus the peak that was originally at position p0q not moves to the left rather than to the right.
19
m m m m m mka ka ka kakb kb kb
1 2 2 21 1
Figure 6: A linear ball and spring system in with springs of alternating spring constants ka and kb.There is a repeated unit cell (enclosed in the dashed box) with two balls, each of mass m.
It is instructive to return to the other boundary conditions. Their time-dependent displacementsin mode q are
where αq,± = φq ± δq. The first sin function in the last equation describes a wave traveling to theright, and the second to a wave traveling to the left: the standing waves are a linear combinationof right and left-going waves.
2.3 A periodic chain with alternating spring constants
In our study of vibrations of a small number of coupled oscillators, we considered situations inwhich different springs and particles had different spring constants and masses. As the numberor particles grows, the number or arrangements of different springs and masses also grows, andit is an almost impossible task to obtain general expressions for the eigenfrequencies that appliesto all systems. There are, however, special arrangements of different springs and masses that arephysically important. Solids composed of more that one type of atom, for example, have periodicallyrepeated unit cells composed of two or more different atoms. Here we consider the simplest modelof such a system: a ball-and-spring model connected by springs whose spring constants alternatebetween ka and kb as shown in Fig. (6). This give us a repeated pattern of cells consisting of twoequal-mass balls. We label the cell by an integer p and the two balls within each cell by 1 ad 2.The displacement from equilibrium of ball 1 in cell p is xp,1 and that of ball 2 is xp,2.
The equations of motion for the two balls are
d2xp,1dt2
= −ω2a(xp,1 − xp−1,2)− ω2
b (xp,1 − xp,2) (155)
d2xp,2dt2
= −ω2a(xp,2 − xp+1,1)− ω2
b (xp,2 − xp,1), (156)
where ω2a = ka/m and ω2
b = kb/m. We consider periodic boundary conditions and take
xp,α = Aq,αei(qp−ωt), α = q, 2, (157)
where to satisfy the periodic boundary conditions that xp+Nc,α = xp,α, where Nc is the number ofcells, q = 2πn/Nc with n an integer. With this form, the equations of motion become
ω2Aq,1 = (ω2a + ω2
b )Aq,1 − ω2ae−iqAq,2 − ω2
bAq,2 (158)
ω2Aq,2 = (ω2a + ω2
b )Aq,2 − ω2aeiqAq,1 − ω2
bAq,1. (159)
20
-3 -2 -1 1 2 3
0.5
1.0
1.5
2.0
2.5
3.0
q
2 bω
2 aω
( )qω
Figure 7: The vibration (phonon) spectrum (frequency as a function of q) of a two-particle unitcell one-dimensional ball-and-spring model. Note the bandgap at q = π.
This equation can be rewritten in matrix form as
ω2Aq = DDD(q)Aq, (160)
where
DDD =
(ω2a + ω2
b −(ω2b + ω2
ae−iq)
−(ω2b + ω2
aeiq) ω2
a + ω2b
). (161)
is the q-dependent dynamical matrix. Note that the off-diagonal entries are complex, but the 21component is the complex conjugate of the 21 component. This means that the eigenvalues of DDD(q)are real. To determine the, we proceed as usual and calculate the roots of the polynomial,
det[DDD(q)− ω2III] = (ω2a + ω2
b − ω2)2(ω2b + ω2
ae−iq)(ω2
b + ω2aeiq) (162)
= (ω2a + ω2
b − ω2)2 − (ω4a + ω4
b + ω2aω
2b cos q) (163)
The roots, which are plotted in Fig. (7), are
ω2±(q) = ω2
a + ω2b ±
√(ω2a + ω2
b )2 − 4ω2
aω2b sin2(q/2). (164)
The limits of these two solutions as q → 0 and q → π are informative:
ω2− →
{ω2aω
2b
2(ω2a+ω
2a)q2 as q → 0
2ω2b as q → π;
(165)
ω2+ →
{2(ω2
a + ω2b ) as q → 0
2ω2a as q → π.
(166)
(167)
The solutions assume ωa > ωb; the two frequencies are interchanged is the opposite is true. Thusω(q) tends linearly to zero with q, yielding a zero frequency modes at q = 0 as it most. ω+(q = 0) on
the other hand is nonzero. At q = π the two frequencies are different: ω+(π) =√
2ωa > ω−(π) =√2ωb. There is a gap in the spectrum. When ωb → ωa, the gap vanishes and the spectrum is
continuous. The existence of such band gaps, particularly in electronic as opposed to vibrationalproperties, and being able to control them is essential to the electronics industry.
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