Chapter 2
Small Oscillations and
Coupled Oscillators
Small oscillations are a common phenomenon in physics. They apply to
many systems in which there is a stable equilibrium and the system is perturbed slightly from that equilibrium position. Examples include the displacement from equilibrium of a spring, the angle of a pendulum from the
vertical, a particle in a potential minimum, or the variation in the orbit of
a planet around a star.
2.1
Simple Harmonic Motion
Let us define the position of equilibrium of a system by the coordinate x = 0.
At equilibrium the force vanishes, such that:
F (x = 0) = 0
−
dU
dx
=0
(2.1)
x=0
For small displacements x, we can expand the potential in a series:
U (x) = U (0) + x
dU
dx
+
x=0
x2 d2 U
2 dx2
+ ...
(2.2)
x=0
From the condition of equilibrium (Eq. 2.1) the second term on the right2
hand side of the above equation vanishes. Let us define k = ddxU2
and
choose U (0) = 0, then, neglecting higher order terms:
U (x) =
1 2
kx + . . .
2
x=0
(2.3)
The corresponding force is F (x) = −kx, which for k positive is a restoring
force that drives the system back to equilibrium (if k is negative then the
6
equilibrium is unstable).
The force can also be written as (we will assume a constant mass, m):
F =
d
(mẋ) = mẍ = −kx
dt
(2.4)
which gives:
mẍ + kx = 0
k
ẍ + x = 0
m
ẍ + ω02 x = 0
(2.5)
(2.6)
(2.7)
where we defined ω02 = k/m.
2.1.1
General solution
Equation 2.7 is a second order differential equation. The general solution is
given by the superposition of two linearly independent solutions. As a trial
solution we can take:
x = A cos (ω0 t) + B sin (ω0 t)
(2.8)
where A and B are constants, or,
x = C cos (ω0 t + δ)
(2.9)
where C and δ are constants, which should be determined from the initial
conditions of the problem. Note that one can also use complex exponentials
to write the solution:
x = Cℜe{eiω0 t+iδ } = Cℑm{eiω0 t+iδ+iπ/2 }
(2.10)
We will make use of this form later.
The frequency ω0 is the natural frequency of the oscillator, which is independent of the amplitude.
2.2
Damped Harmonic Motion
Often systems have some mechanism of damping, corresponding to a force
proportional to and in the opposite direction to its velocity:
Fdamping = −kd ẋ
7
(2.11)
Newton’s second law (for a system with constant mass m) thus reads:
ṗ = F + Fdamping
(2.12)
mẍ = −kx − kd ẋ
(2.13)
Let us try first a trial solution of the form x = eλt , where λ is a constant
that we can obtain by substitution into Eq. 2.13:
mλ2 + kd λ + k eλt = 0
(2.14)
We thus have a quadratic equation for λ, which can be solved giving:
q
−kd ± kd2 − 4mk
λ=
(2.15)
2m
However, while our trial solution is a special solution it is not necessarily
the general solution to Eq. 2.13. From the theory of second-order linear
differential equations, we know that the general solution is given by the superposition of two linearly independent solutions. That is, if we can find two
independent solutions x1 and x2 , then the general solution is A1 x1 + A2 x2 ,
where A1 and A2 are constants determined by the boundary conditions (in
our case the initial conditions) of the problem.
From Eq. 2.15 we can identify three different regimes of motion:
if
if
if
kd2 − 4mk > 0
kd2
kd2
− 4mk = 0
overdamped
(2.16)
critically damped
(2.17)
underdamped
(2.18)
− 4mk < 0
Let us study these different regimes in turn.
Overdamped Regime. Here Eq. 2.15 gives two real solutions for λ. We
thus have two independent solutions and the general solution is:
x = A1 eλ1 t + A2 eλ2 t .
(2.19)
Note that both λ1 and λ2 are negative, so this solution represents a rapid
decay of the displacement x.
Critically Damped Regime. In this case Eq. 2.15 gives only one solution,
λ0 = −kd /(2m). We must therefore look for a second (independent) solution
to Eq. 2.13 in order to write the general solution. Considering the solution
x = teλt and substituting into Eq. 2.13 we have:
2λm + λ2 mt + kd + kd λt + kt eλt = 0
(2.20)
λt
2
λt
λ m + kd λ + k te + (2λm + kd ) e = 0
(2.21)
8
x
1.0
Critical Damping
.8
Overdamped
.6
Underdamped
.4
.2
t
0
0.4
0.6
0.8 1.0
1.2
1.4
1.6
-.2
-.4
-.6
Figure 2.1: Overdamped, critically damped and underdamped regimes.
The equation must hold at all values of time, so the terms in each set of
brackets must both vanish. The term in the first set of brackets is the same
as in Eq. 2.14, so λ is again given by Eq. 2.15. For the critically damped
regime, we have already stated that λ = λ0 = −kd /(2m). This causes the
term in the second set of brackets in Eq. 2.21 to vanish, such that x = teλ0 t
is indeed a valid solution. Consequently, the general solution in the critically
damped case is:
x = A1 eλ0 t + A2 teλ0 t
(2.22)
Underdamped Regime. In this regime, Eq. 2.15 gives two solutions.
They may be complex, but they still allow us to write the general solution
in the form A1 x1 + A2 x2 . However, since x is real, A1 and A2 may need to
be complex amplitudes. Taking the values of λ from Eq. 2.15 we have:
q
q
− −(4mk−kd2 )t/(2m)
−(4mk−kd2 )t/(2m)
−kd t/(2m)
+ A2 e
(2.23)
x=e
A1 e
q
4mk − kd2 /(2m), which is a positive quantity in the
Let us define ω0 =
underdamped regime. We then have:
x = e−kd t/(2m) A1 eiω0 t + A2 e−iω0 t
(2.24)
= e−kd t/(2m) [(A1 + A2 ) cos(ω0 t) + i (A1 − A2 ) sin(ω0 t)]
= e−kd t/(2m) [c1 cos(ω0 t) + c2 sin(ω0 t)]
9
(2.25)
(2.26)
where we introduced two different constants c1 and c2 , which must be real
for x to be real (recall that A1 and A2 are in general complex). Note that in
practice we will not find c1 and c2 from A1 and A2 , but instead we can find
them directly from the initial conditions of our problem (that is, the initial
displacement and velocity of the mass).
Note that the general solution in the underdamped case can also be written:
c2
−kd t/(2m)
x = c1 e
cos(ω0 t) + sin(ω0 t)
(2.27)
c1
Again we can redefine our constants by introducing tan α = c2 /c1 (and
again we will be able to find α directly from the initial conditions rather
than through relation to c2 and c1 ). Then:
x = c1 e−kd t/(2m) [cos(ω0 t) + tan α sin(ω0 t)]
c1 −kd t/(2m)
e
[cos α cos(ω0 t) + sin α sin(ω0 t)]
=
cos α
= Ae−kd t/(2m) cos (ω0 t − α)
(2.28)
(2.29)
(2.30)
c1
where A = cos
α . So, we could write the solution as the superposition of two
waves as in Eq. 2.26. Or, equivalently, the sum of two waves of the same
frequency is another wave of the same frequency as in Eq. 2.30.
2.3
Input Forces & Periodic Driving
Using Newton’s second law we can also account for a variety of other forces
that may be in the system. Of particular interest is where an oscillator
is driven by a periodic force Fdrive = F0 cos ωt. The equation of motion
(Eq. 2.13) becomes modified as:
mẍ + kx + kd ẋ = F0 cos ωt
(2.31)
This is a second order inhomogeneous differential equation. Let us recall
that the general solution to an inhomogeneous equation is given by the
linear superposition of the homogeneous solution and a particular solution:
x(t) = xH (t) + X(t)
(2.32)
The homogeneous solution xH (t) (sometimes called the complementary solution) is given by the solution in the absence of the driving term, which we
have already calculated in section 2.2 for the different cases of overdamping, critical damping and underdamping. Note that while xH should satisfy
equation 2.13, it does not by itself satisfy the inhomogeneous equation 2.31.
The particular solution X(t) is any function that satisfies the full inhomogeneous equation (Eq. 2.31).
10
Let us consider a particular solution of the form X(t) = B cos (ωt − φ) where
B and φ are constants to be determined. Substitution into Eq. 2.31 gives:
−mω 2 B cos (ωt − φ) + kB cos (ωt − φ) − kd ωB sin (ωt − φ) = F0 cos ωt
(2.33)
Recalling the identities cos(A − B) = cos A cos B + sin A sin B and sin(A −
B) = sin A cos B − cos A sin B we have:
k − mω 2 B [cos (ωt) cos φ + sin (ωt) sin φ]
− kd ωB [sin (ωt) cos φ − cos (ωt) sin φ] = F0 cos ωt
Collecting terms in cos (ωt) and sin (ωt):
k − mω 2 B cos φ + Bkd ω sin φ − F0 cos (ωt)
+ k − mω 2 B sin φ − kd ωB cos φ sin (ωt) = 0
(2.34)
(2.35)
For our trial solution to be successful, the above equation must hold for all
times, which requires the terms in both sets of square brackets to vanish.
The vanishing of the second term in square brackets requires:
k − mω 2 sin φ = kd ω cos φ
(2.36)
kd ω
→
tan φ =
(2.37)
k − mω 2
Note that:
1 − cos2 φ
1
sin2 φ
=
=
−1
(2.38)
tan2 φ =
2
2
cos φ
cos φ
cos2 φ
1
(2.39)
→
cos2 φ =
1 + tan2 φ
tan2 φ
1 + tan2 φ − 1
=
(2.40)
→
sin2 φ = 1 − cos2 φ =
1 + tan2 φ
1 + tan2 φ
Setting the term in the first set of square brackets in Eq. 2.35 to zero we
then have:
(kd ω)2
k − mω 2 + k−mω
k − mω 2 + kd ω tan φ
2
p
= F0
B
= F0
→
B q
2
(kd ω)
1 + tan2 φ
1 + (k−mω
2 )2
2
k − mω 2 + (kd ω)2
F0
→
Bp
= F0
→
B=p
2
2
2
(k − mω ) + (kd ω)
(k − mω 2 )2 + (kd ω)2
(2.41)
The full solution (in the underdamped regime, for example) can then be
written:
x(t) = Ae−kd t/(2m) cos (ω0 t − α) + B cos (ωt − φ)
(2.42)
The first part of the solution determines the transient dynamics of the system. At long times, the system settles into a periodically oscillating state
given fully by the second part of the solution.
11
2.4
Coupled Oscillators
Let us try to generalize from a single oscillator to a set of oscillators. The
oscillators will be coupled such that each can affect the motion of the others.
k1
m1
k1
m2
x1
k2
x2
Figure 2.2: Two coupled harmonic oscillators.
Let us denote the positions of the oscillators as xi (t) where the index i =
1, . . . , n labels the oscillator. The position of the set of oscillators is then
given by the vector x(t):


x1 (t)
 x2 (t) 

x(t) = 
(2.43)
 ... 
xn (t)
As before let us define xi = 0 as the position of stable equilibrium of the
system. The potential of the system is some function of the position of the
set of oscillators, V (x), and can be expanded in a Taylor series about the
equilibrium point:
X ∂V
1 X ∂2V
xi xj + . . .
(2.44)
xi +
V (x) = V (0) +
∂xi x=0
2
∂xi ∂xj x=0
ij
i
For simplicity, we can set V (0) = 0. At the position of equilibrium, the
first derivative terms vanish. Consequently, the second derivate terms are
the dominant terms in the expansion. Let us define the symmetric matrix
of constants:
∂2V
Kij =
(2.45)
∂xi ∂xj x=0
The force acting on the oscillator with index i is then:
X
∂V
=−
Kij xj
Fi = −
∂xi
j
12
(2.46)
Note that this includes the usual restoring force of a single oscillator (when
j = i) as well as a contribution from each other oscillator.
We can now use Newton’s second law to write the equation of motion of the
coupled mode system:
X
Mi ẍi = −
Kij xj
(2.47)
j
where we assumed that the mass Mi of each oscillator is constant.
2.4.1
Normal Modes
We have seen that a single oscillator exhibits oscillations at a specific natural
frequency. Coupled mode systems have special solutions in which all oscillators have the same frequency. Such solutions are known as normal modes.
To prove this we can make use of the complex exponential formalism and
look for solutions of the form:


A1
 A2  iωt

x(t) = 
(2.48)
 ... e
An
The constants Ai represent the amplitudes of the different oscillators. While
all oscillators have the same frequency, they can have different phases and
amplitudes determined by the Ai . The normal modes are thus determined
by specific ratios of the different Ai . However, note that the total normalization is arbitrary, such that multiplying all the Ai by the same constant
will give the same normal mode.
An obvious question is how many normal modes are there? Well, we have n
oscillators, which will each correspond to a separate second order differential
equation, so we can expect 2n independent solutions. This means that we
need to find n solutions of the form written in Eq. 2.48. The implication of
Eq. 2.48 is that we should later take the real or imaginary part, each corresponding to a different possible solution. Consequently, n normal modes
correspond to 2n real solutions.
Once we find the normal modes, which are characterized by their frequency
ω and amplitudes Ai , we can construct the general solution of the problem.
This is given by some linear superposition of normal modes, which we can
determine from the initial condition of the problem.
To proceed let us write the equation of motion of the coupled modes (Eq. 2.47)
in matrix form:
Mẍ = −Kx
(2.49)
13
where

M1 0 . . . 0
0 M2 . . . 0
..
..
..
..
.
.
.
.
0
0
0 Mn

K11
K21
..
.


M=



K=

K12
K22
..
.
Kn1 Kn2





. . . K1n
. . . K2n
..
..
.
.
. . . Knn
(2.50)





(2.51)
Recall that x = Aeiωt for a normal mode solution (as we wrote in Eq. 2.48).
Consequently ẍ = −ω 2 x, allowing us to re-write the matrix equation of
motion as:
Mω 2 x = Kx
→ M−1 Kx = ω 2 x
(2.52)
(2.53)
where M−1 is the inverse of M. The above equation is an eigenvalue equation, where ω 2 are the eigenvalues and A are the eigenvectors. Note that
since we have an n × n matrix we have n eigenvalues and eigenvectors,
corresponding to the n normal mode solutions we were searching for.
2.4.2
Example: Masses and Springs
As an example, let us consider two masses, m1 = m and m2 = 2m, constrained to move along a straight line. Each mass is connected to a wall by
a spring, with force constant k1 = k for the first mass and k2 = 2k for the
second mass. Consequently inpthe absence of any coupling they have the
same natural frequency ω0 = k/m. The two masses are however coupled
with a third spring, with spring constant k′ = 2k. Let us assume that the
equilibrium position of the system has each spring unstretched; x1 and x2
represent the displacements of the masses from equilibrium.
The potential energy of the system is:
1
1
1
V = k1 x21 + k′ (x2 − x1 )2 + k2 x22
2
2
2
(2.54)
The force on mass m1 is:
F1 = −
∂V
= −k1 x1 − k′ (x1 − x2 ) = 2kx2 − 3kx1
∂x1
(2.55)
and the force on mass m2 is:
F2 = −
∂V
= −k2 x2 − k′ (x2 − x1 ) = 2kx1 − 4kx2
∂x2
14
(2.56)
Of course, we probably could have written down the forces immediately
without first worrying about the potential energy in this example. We anyway find from Newton’s second law:
3k −2k
m1 0
ẍ1
x1
=−
(2.57)
ẍ2
−2k 4k
x2
0 m2
Substituting x = Aeiωt :
3k −2k
A1
m 0
A1
2
=ω
−2k 4k
A2
0 2m
A2
2
mω
3 −2
A1
A1
=
−1 2
A2
A2
k
(2.58)
(2.59)
To solve this eigenvalue equation, let us first define λ = mω 2 /k such that:
3 − λ −2
0
A1
=
(2.60)
−1 2 − λ
A2
0
For there to be a solution, the determinant of the 2 × 2 matrix must vanish,
giving us the quadratic equation:
(3 − λ) (2 − λ) − 2 = 0
2
λ − 5λ + 4 = 0
(2.61)
(2.62)
The p
solutions are λ =p1 and λ = 4, giving the normal mode frequencies
ω = k/m and ω = 2 k/m.
The eigenvectors are found by substituting the values of λ back into Eq. 2.60.
This gives A2 = A1 for λ = 1 and A2 = −A1 /2 for λ = 4. Let us recall
that only the ratio of Ai is determined: multiplying the amplitudes by a
constant gives the same normal mode. It is however common to normalize
the eigenvectors such that they have unit modulus (A21 + A22 = 1):
p
1
1
(2.63)
A= √
ω = k/m
2 1
p
1
2
ω = 2 k/m
A= √
(2.64)
5 −1
Let us now consider the meaning of the result. The first normal mode corresponds to the same frequency as when the masses are uncoupled. In this
case the two masses oscillate in phase and with the same amplitude such
that the central spring neither contracts or extends.
The second mode corresponds to when the masses move out of phase with
each other. In this mode the smaller mass m1 has twice the amplitude of
displacement of the larger mass m2 .
15
2.4.3
Beats
For simplicity, let us now consider the case of two coupled oscillators of equal
mass, m1 = m2 = m, which experience the same force constant k1 = k2 = k.
From symmetry we can then expect one normal mode where the masses oscillate in phase with the same amplitude. In this case, the connecting spring
remains unstretched. A second normal mode corresponds to the situation
where the masses oscillate out of phase.
Let us consider the case where the constant of the connecting spring is
k′ = ǫk, where ǫ ≪ 1. The method of the previous subsection can be
repeated, and one finds that the normal modes are given by:
r
1
k
1
ω1 =
A1 = √
(2.65)
m
2 1
r
1
k
1
ω2 = (1 + 2ǫ)
A2 = √
(2.66)
m
2 −1
We can now write the general solution as a superposition of the normal
modes:
x(t) = c1 A1 cos(ω1 t) + c2 A2 cos(ω2 t) + c3 A1 sin(ω1 t) + c4 A2 sin(ω2 t) (2.67)
If the system starts from rest, then ẋ(t) = 0 giving c3 = 0 and c4 = 0. Let
us consider the initial condition where the first oscillator is displaced:
d
x(0) =
(2.68)
0
c1
c2
1
1
=√
+√
(2.69)
2 1
2 −1
√
The solution to the above equation is given by c1 = c2 = d/ 2. The motion
of the system is then:
d
(cos(ω1 t) + cos(ω2 t))
2
d
x2 (t) = (cos(ω1 t) − cos(ω2 t))
2
x1 (t) =
(2.70)
(2.71)
Noting the trigonometric identities:
1
θ+φ
θ−φ
(cos θ + cos φ) = cos
cos
2
2
2
θ+φ
θ−φ
1
(cos θ − cos φ) = − sin
sin
2
2
2
16
(2.72)
(2.73)
we can re-write the motion of the system as:
ω2 − ω1
ω1 + ω2
x1 (t) = d cos
t cos
t
2
2
ω2 − ω1
ω1 + ω2
x2 (t) = d sin
t sin
t
2
2
(2.74)
(2.75)
For small ǫ the frequencies ω1 and ω2 are similar. We can see that the motion
of each oscillator has a fast oscillation at the average frequency (ω1 + ω2 )/2,
modulated by a slow variation at the difference frequency (ω2 − ω1 )/2. One
can also observe the phenomenon of beating (see Fig. 2.3). Starting the first
oscillator in motion, we can see that the first oscillator gradually transfers
its energy to the second. Eventually, the first oscillator completely stops as
all energy has been transferred to the second oscillator. The process then
repeats and the oscillators periodically transfer energy back and forth.
1.0
x1d
0.5
0.0
-0.5
-1.0
0
5
10
Ω1tH2ΠL
15
0
5
10
Ω1tH2ΠL
15
1.0
x2d
0.5
0.0
-0.5
-1.0
Figure 2.3: Displacements x1 and x2 given by Eqs. 2.74 and 2.75. The first
mass is initially displaced by an amount d. Time is plotted in units of the
period of the lower frequency normal mode. ǫ = 0.1.
17
2.5
The Beaded String
c
In this section we’ll consider a string carrying N beads of equal mass M ,
equally spaced a distance a apart. The ends of the string are fixed, with a
distance a between the first end and the first bead as well as between the
second end and the last bead. The string is also characterized by a tension
T . We’ll consider transverse oscillations of the string.
un-1
u2
u1
un+1
un-1
Figure 2.4: A beaded string.
Let the displacement of the nth bead be un . Using Newton’s second law we
can write the equation of motion, where the force on the nth bead can be
deduced from some trigonometry:
M ün = −T (sin ψ + sin φ)
(2.76)
For small displacements, we can make the approximations:
sin ψ ≈
un − un−1
a
,
sin φ ≈
un − un+1
a
(2.77)
The equations of motion become:
ün =
T
(un−1 − 2un + un+1 )
Ma
(2.78)
This equation is supplemented with the boundary conditions that the ends
of the string are fixed:
u0 = 0
2.5.1
and
uN +1 = 0
(2.79)
Normal Modes
To find the normal modes of the beaded string we can again use the complex
exponential formalism and search for solutions that oscillate at the same
frequency ω:
un = An eiωt
(2.80)
18
where An are some set of constant coefficients. Substituting into the equation of motion (Eq. 2.78) gives:
ω 2 An =
T
(−An−1 + 2An − An+1 )
Ma
(2.81)
To continue, let us first try to solve the case of an infinite chain of beads.
This may seem more challenging, but in an infinite system we can make use
of the property of translational invariance, that is, if we move one step to
the left or right the system looks the same. As we will see this will make it
easier to find the normal modes in the limit of an infinite chain.
Once we know the normal modes of an infinite chain we will select particular
combinations of them that satisfy the boundary conditions so that the ends
of the finite chain are fixed.
2.5.2
Infinite Chain
To make use of translational invariance mathematically, let us first imagine
that we somehow find a normal mode described by a set of amplitudes An .
If we shift the system one step to the left, then the system should look the
same. Consequently, if An gave us a normal mode with frequency ω, there
should be another mode with the same ω and amplitudes:
A′n = An+1
(2.82)
Let us recall that the overall amplitude of a mode is arbitrary up to a
common multiplication factor of the amplitudes. A translationally invariant
mode should reproduce itself when shifted, requiring:
A′n = An+1 = hAn
(2.83)
where h is some constant factor that makes the amplitudes of the shifted
mode proportional to the old ones. If we apply the above relation repeatedly
we can write:
An = hn A0
(2.84)
Here A0 is arbitrary and sets the overall scale. Substituting the above relation into Eq. 2.81:
T
−hn−1 A0 + 2hn A0 − hn+1 A0
Ma
1
T
2
2−h−
→
ω =
Ma
h
ω 2 hn A0 =
19
(2.85)
(2.86)
Note that if we obtain a normal mode described by h then there is another
normal mode with the same frequency given by 1/h. In general, the mode
with frequency ω is then some combination of modes An = hn A0 and An =
h−n A0 , that is:
An = αhn + βh−n
(2.87)
where α and β are constants.
Let us introduce a new constant θ, defined by h = eiθ . Equation 2.86 then
becomes:
T
4T
T iθ
−iθ
2 θ
2
=
2−e −e
(2 − 2 cos θ) =
sin
ω =
(2.88)
Ma
Ma
Ma
2
The corresponding displacements of the mode with given θ are:
un = αeinθ + βe−inθ eiωt
2.5.3
(2.89)
Finite Chain
To describe the finite chain we need to incorporate the boundary conditions
(given in Eq. 2.79). These will restrict θ to specific values, which in turn fixes
ω. The first boundary condition u0 = 0 requires that β = −α. This makes
un proportional to sin(nθ) and our second boundary condition becomes:
uN +1 ∝ sin ((N + 1)θ) = 0
(2.90)
Notice that the allowed values of θ are now constrained to:
θ=
mπ
N +1
(2.91)
where m is an integer, which labels the different modes.
The modes are standing wave solutions, composed of a combination of forward and backward propagating waves. The problem is very similar to the
problem of finding the standing waves for a guitar or violin string. The only
difference is that we considered a discrete system.
For a string with six beads, we obtain the normal modes plotted in Fig. 2.5
by setting m = 1, 2, . . . , 6.
Note that Eq. 2.91 admits values of m greater than six as well. These
however give the same modes; as shown in Fig. 2.6 the modes repeat when
m > 6. Hence a beaded string with six degrees of freedom has six normal
modes.
20
0
1
2
3
4
5
6
7
0
1
2
3
4
5
6
7
0
1
2
3
4
5
6
7
0
1
2
3
4
5
6
7
0
1
2
3
4
5
6
7
0
1
2
3
4
5
6
7
Figure 2.5: Normal modes of a beaded string with six beads.
0
1
2
3
4
5
6
7
0
1
2
3
4
5
6
7
0
1
2
3
4
5
6
7
0
1
2
3
4
5
6
7
Figure 2.6: Repetition of normal modes for mode numbers greater than six.
The modes 3, 11, 17, and 25 are shown. All the modes are equivalent up to
an overall multiplication constant (which could be negative).
The normal mode frequencies are given by inserting Eq. 2.91 into Eq. 2.88:
r
r
θ
mπ
T
T
ω=2
sin
sin
=2
(2.92)
Ma
2
Ma
2(N + 1)
This is known as the dispersion relation of the system.
21