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Typical oscillatory systems

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Typical oscillatory systems
9.1 TYPICAL OSCILLATORY SYSTEMS
corresponding to a kinetic energy, is positive definite; that is, whatever non-zero
real values the q̇i take, the quadratic form (9.1) has a value > 0.
Turning now to the potential energy, we may write its value for a configuration
q by means of a Taylor expansion about the origin q = 0,
∂V (0)
1 ∂2 V (0)
qi +
qi qj + · · · .
V (q) = V (0) +
∂qi
2 i j ∂qi ∂qj
i
However, we have chosen V (0) = 0 and, since the origin is an equilibrium point,
there is no force there and ∂V (0)/∂qi = 0. Consequently, to second order in the
qi we also have a quadratic form, but in the coordinates rather than in their time
derivatives:
V =
bij qi qj = qT Bq,
(9.2)
i
j
where B is, or can be made, symmetric. In this case, and in general, the requirement
that the potential is a minimum means that the potential matrix B, like the kinetic
energy matrix A, is real and positive definite.
9.1 Typical oscillatory systems
We now introduce particular examples, although the results of this section are
general, given the above restrictions, and the reader will find it easy to apply the
results to many other instances.
Consider first a uniform rod of mass M and length l, attached by a light string
also of length l to a fixed point P and executing small oscillations in a vertical
plane. We choose as coordinates the angles θ1 and θ2 shown, with exaggerated
magnitude, in figure 9.1. In terms of these coordinates the centre of gravity of the
rod has, to first order in the θi , a velocity component in the x-direction equal to
l θ̇1 + 12 l θ̇2 and in the y-direction equal to zero. Adding in the rotational kinetic
energy of the rod about its centre of gravity we obtain, to second order in the θ̇i ,
1
T ≈ 12 Ml 2 (θ̇12 + 14 θ̇22 + θ̇1 θ̇2 ) + 24
Ml 2 θ̇22
6 3
1
= 16 Ml 2 3θ̇12 + 3θ̇1 θ̇2 + θ̇22 = 12
Ml 2 q̇T
q̇,
3 2
where q̇T = (θ̇1 θ̇2 ). The potential energy is given by
V = Mlg (1 − cos θ1 ) + 12 (1 − cos θ2 )
so that
V ≈ 14 Mlg(2θ12 + θ22 ) =
T
1
12 Mlgq
6 0
0 3
(9.3)
(9.4)
q,
(9.5)
where g is the acceleration due to gravity and q = (θ1 θ2 )T ; (9.5) is valid to
second order in the θi .
317
NORMAL MODES
P
P
P
θ1
θ1
θ1
l
θ2
θ2
θ2
l
(a)
(b)
(c)
Figure 9.1 A uniform rod of length l attached to the fixed point P by a light
string of the same length: (a) the general coordinate system; (b) approximation
to the normal mode with lower frequency; (c) approximation to the mode with
higher frequency.
With these expressions for T and V we now apply the conservation of energy,
d
(T + V ) = 0,
dt
(9.6)
assuming that there are no external forces other than gravity. In matrix form
(9.6) becomes
d T
(q̇ Aq̇ + qT Bq) = q̈T Aq̇ + q̇T Aq̈ + q̇T Bq + qT Bq̇ = 0,
dt
which, using A = AT and B = BT , gives
2q̇T (Aq̈ + Bq) = 0.
We will assume, although it is not clear that this gives the only possible solution,
that the above equation implies that the coefficient of each q̇i is separately zero.
Hence
Aq̈ + Bq = 0.
(9.7)
For a rigorous derivation Lagrange’s equations should be used, as in chapter 22.
Now we search for sets of coordinates q that all oscillate with the same period,
i.e. the total motion repeats itself exactly after a finite interval. Solutions of this
form will satisfy
q = x cos ωt;
(9.8)
the relative values of the elements of x in such a solution will indicate how each
318
9.1 TYPICAL OSCILLATORY SYSTEMS
coordinate is involved in this special motion. In general there will be N values
of ω if the matrices A and B are N × N and these values are known as normal
frequencies or eigenfrequencies.
Putting (9.8) into (9.7) yields
−ω 2 Ax + Bx = (B − ω 2 A)x = 0.
(9.9)
Our work in section 8.18 showed that this can have non-trivial solutions only if
|B − ω 2 A| = 0.
(9.10)
This is a form of characteristic equation for B, except that the unit matrix I has
been replaced by A. It has the more familiar form if a choice of coordinates is
made in which the kinetic energy T is a simple sum of squared terms, i.e. it has
been diagonalised, and the scale of the new coordinates is then chosen to make
each diagonal element unity.
However, even in the present case, (9.10) can be solved to yield ωk2 for k =
1, 2, . . . , N, where N is the order of A and B. The values of ωk can be used
with (9.9) to find the corresponding column vector xk and the initial (stationary)
physical configuration that, on release, will execute motion with period 2π/ωk .
In equation (8.76) we showed that the eigenvectors of a real symmetric matrix
were, except in the case of degeneracy of the eigenvalues, mutually orthogonal.
In the present situation an analogous, but not identical, result holds. It is shown
in section 9.3 that if x1 and x2 are two eigenvectors satisfying (9.9) for different
values of ω 2 then they are orthogonal in the sense that
(x2 )T Ax1 = 0
and
(x2 )T Bx1 = 0.
The direct ‘scalar product’ (x2 )T x1 , formally equal to (x2 )T I x1 , is not, in general,
equal to zero.
Returning to the suspended rod, we find from (9.10)
Mlg
ω 2 Ml 2
6 0
6 3 = 0.
−
12
0 3
3 2 12
Writing ω 2 l/g = λ, this becomes
6 − 6λ −3λ ⇒
λ2 − 10λ + 6 = 0,
−3λ 3 − 2λ = 0
√
which has roots λ = 5 ± 19. Thus we find that the two normal frequencies are
given by ω1 = (0.641g/l)1/2√and ω2 = (9.359g/l)1/2 . Putting the lower of the two
values for ω 2 , namely (5 − 19)g/l, into (9.9) shows that for this mode
√
√
x1 : x2 = 3(5 − 19) : 6( 19 − 4) = 1.923 : 2.153.
This corresponds to the case where the rod and string are almost straight out, i.e.
they almost form a simple pendulum. Similarly it may be shown that the higher
319
NORMAL MODES
frequency corresponds to a solution where the string and rod are moving with
opposite phase and x1 : x2 = 9.359 : −16.718. The two situations are shown in
figure 9.1.
In connection with quadratic forms it was shown in section 8.17 how to make
a change of coordinates such that the matrix for a particular form becomes
diagonal. In exercise 9.6 a method is developed for diagonalising simultaneously
two quadratic forms (though the transformation matrix may not be orthogonal).
If this process is carried out for A and B in a general system undergoing stable
oscillations, the kinetic and potential energies in the new variables ηi take the
forms
µi η̇i2 = η̇T Mη̇, M = diag (µ1 , µ2 , . . . , µN ),
(9.11)
T =
i
V =
νi ηi2 = ηT Nη,
N = diag (ν1 , ν2 . . . , νN ),
(9.12)
i
and the equations of motion are the uncoupled equations
µi η̈i + νi ηi = 0,
i = 1, 2, . . . , N.
(9.13)
Clearly a simple renormalisation of the ηi can be made that reduces all the µi
in (9.11) to unity. When this is done the variables so formed are called normal
coordinates and equations (9.13) the normal equations.
When a system is executing one of these simple harmonic motions it is said to
be in a normal mode, and once started in such a mode it will repeat its motion
exactly after each interval of 2π/ωi . Any arbitrary motion of the system may
be written as a superposition of the normal modes, and each component mode
will execute harmonic motion with the corresponding eigenfrequency; however,
unless by chance the eigenfrequencies are in integer relationship, the system will
never return to its initial configuration after any finite time interval.
As a second example we will consider a number of masses coupled together by
springs. For this type of situation the potential and kinetic energies are automatically quadratic functions of the coordinates and their derivatives, provided the
elastic limits of the springs are not exceeded, and the oscillations do not have to
be vanishingly small for the analysis to be valid.
Find the normal frequencies and modes of oscillation of three particles of masses m, µ m,
m connected in that order in a straight line by two equal light springs of force constant k.
This arrangement could serve as a model for some linear molecules, e.g. CO2 .
The situation is shown in figure 9.2; the coordinates of the particles, x1 , x2 , x3 , are
measured from their equilibrium positions, at which the springs are neither extended nor
compressed.
The kinetic energy of the system is simply
T = 12 m ẋ21 + µ ẋ22 + ẋ23 ,
320
9.1 TYPICAL OSCILLATORY SYSTEMS
m
x1
k
µm
x2
m
k
x3
Figure 9.2 Three masses m, µm and m connected by two equal light springs
of force constant k.
(a)
(b)
(c)
Figure 9.3 The normal modes of the masses and springs of a linear molecule
such as CO2 . (a) ω 2 = 0; (b) ω 2 = k/m; (c) ω 2 = [(µ + 2)/µ](k/m).
whilst the potential energy stored in the springs is
V = 12 k (x2 − x1 )2 + (x3 − x2 )2 .
The kinetic- and potential-energy

1 0
m
A=  0 µ
2
0 0
symmetric matrices are thus


0
1
−1
k
0 ,
2
B =  −1
2
1
0
−1

0
−1  .
1
From (9.10), to find the normal frequencies we have to solve |B − ω 2 A| = 0. Thus, writing
mω 2 /k = λ, we have
1−λ
−1
0 −1
2 − µλ
−1 = 0,
0
−1
1−λ which leads to λ = 0, 1 or 1 + 2/µ. The corresponding eigenvectors are respectively






1
1  1 
1  1 
1
1
2
3

1
0
−2/µ .
x = √
,
x = √
,
x = 3
2
2 + (4/µ2 )
1
−1
1
The physical motions associated with these normal modes are illustrated in figure 9.3.
The first, with λ = ω = 0 and all the xi equal, merely describes bodily translation of the
whole system, with no (i.e. zero-frequency) internal oscillations.
In the second solution the central particle remains stationary, x2 = 0, whilst the other
two oscillate with equal amplitudes in antiphase with each other. This motion, which has
frequency ω = (k/m)1/2 , is illustrated in figure 9.3(b).
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