Exercises

9.4 EXERCISES
Estimate the eigenfrequencies of the oscillating rod of section 9.1.
Firstly we recall that
A=
Ml 2
12
6
3
3
2
and
B=
Mlg
12
6
0
0
3
.
Physical intuition suggests that the slower mode will have a configuration approximating
that of a simple pendulum (figure 9.1), in which θ1 = θ2 , and so we use this as a trial
vector. Taking x = (θ θ)T ,
λ(x) =
xT Bx
3Mlgθ2 /4
9g
g
=
=
= 0.643 ,
T
x Ax
7Ml 2 θ2 /6
14l
l
and we conclude from (9.19) that the lower (angular) frequency is ≤ (0.643g/l)1/2 . We
have already seen on p. 319 that the true answer is (0.641g/l)1/2 and so we have come
very close to it.
Next we turn to the higher frequency. Here, a typical pattern of oscillation is not so
obvious but, rather preempting the answer, we try θ2 = −2θ1 ; we then obtain λ = 9g/l
and so conclude that the higher eigenfrequency ≥ (9g/l)1/2 . We have already seen that the
exact answer is (9.359g/l)1/2 and so again we have come close to it. A simplified version of the Rayleigh–Ritz method may be used to estimate the
eigenvalues of a symmetric (or in general Hermitian) matrix B, the eigenvectors
of which will be mutually orthogonal. By repeating the calculations leading to
(9.18), A being replaced by the unit matrix I, it is easily verified that if
λ(x) =
xT Bx
xT x
is evaluated for any vector x then
λ1 ≤ λ(x) ≤ λm ,
where λ1 , λ2 . . . , λm are the eigenvalues of B in order of increasing size. A similar
result holds for Hermitian matrices.
9.4 Exercises
9.1
Three coupled pendulums swing perpendicularly to the horizontal line containing
their points of suspension, and the following equations of motion are satisfied:
−mẍ1 = cmx1 + d(x1 − x2 ),
−Mẍ2 = cMx2 + d(x2 − x1 ) + d(x2 − x3 ),
−mẍ3 = cmx3 + d(x3 − x2 ),
9.2
where x1 , x2 and x3 are measured from the equilibrium points; m, M and m
are the masses of the pendulum bobs; and c and d are positive constants. Find
the normal frequencies of the system and sketch the corresponding patterns of
oscillation. What happens as d → 0 or d → ∞?
A double pendulum, smoothly pivoted at A, consists of two light rigid rods, AB
and BC, each of length l, which are smoothly jointed at B and carry masses m and
αm at B and C respectively. The pendulum makes small oscillations in one plane
329
NORMAL MODES
under gravity. At time t, AB and BC make angles θ(t) and φ(t), respectively, with
the downward vertical. Find quadratic expressions for the kinetic and potential
energies of the system and hence show that the normal modes have angular
frequencies given by
g
ω2 =
1 + α ± α(1 + α) .
l
9.3
For α = 1/3, show that in one of the normal modes the mid-point of BC does
not move during the motion.
Continue the worked example, modelling a linear molecule, discussed at the end
of section 9.1, for the case in which µ = 2.
(a) Show that the eigenvectors derived there have the expected orthogonality
properties with respect to both A and B.
(b) For the situation in which the atoms are released from rest with initial
displacements x1 = 2, x2 = − and x3 = 0, determine their subsequent
motions and maximum displacements.
9.4
Consider the circuit consisting of three equal capacitors and two different inductors shown in the figure. For charges Qi on the capacitors and currents Ii
Q1
Q2
C
C
Q3
C
L1
L2
I2
I1
through the components, write down Kirchhoff’s law for the total voltage change
around each of two complete circuit loops. Note that, to within an unimportant
constant, the conservation of current implies that Q3 = Q1 − Q2 . Express the loop
equations in the form given in (9.7), namely
AQ̈ + BQ = 0.
Use this to show that the normal frequencies of the circuit are given by
ω2 =
9.5
1 L1 + L2 ± (L21 + L22 − L1 L2 )1/2 .
CL1 L2
Obtain the same matrices and result by finding the total energy stored in the
various capacitors (typically Q2 /(2C)) and in the inductors (typically LI 2 /2).
For the special case L1 = L2 = L determine the relevant eigenvectors and so
describe the patterns of current flow in the circuit.
It is shown in physics and engineering textbooks that circuits containing capacitors and inductors can be analysed by replacing a capacitor of capacitance C by a
‘complex impedance’ 1/(iωC) and an inductor of inductance L by an impedance
iωL, where ω is the angular frequency of the currents flowing and i2 = −1.
Use this approach and Kirchhoff’s circuit laws to analyse the circuit shown in
330
9.4 EXERCISES
the figure and obtain three linear equations governing the currents I1 , I2 and I3 .
Show that the only possible frequencies of self-sustaining currents satisfy either
C
I1
P
Q
U
L
S
9.6
I2
C
L
T
C
I3
R
(a) ω 2 LC = 1 or (b) 3ω 2 LC = 1. Find the corresponding current patterns and,
in each case, by identifying parts of the circuit in which no current flows, draw
an equivalent circuit that contains only one capacitor and one inductor.
The simultaneous reduction to diagonal form of two real symmetric quadratic forms.
Consider the two real symmetric quadratic forms uT Au and uT Bu, where uT
stands for the row matrix (x y z), and denote by un those column matrices
that satisfy
Bun = λn Aun ,
(E9.1)
in which n is a label and the λn are real, non-zero and all different.
(a) By multiplying (E9.1) on the left by (um )T , and the transpose of the corresponding equation for um on the right by un , show that (um )T Aun = 0 for
n = m.
(b) By noting that Aun = (λn )−1 Bun , deduce that (um )T Bun = 0 for m = n.
(c) It can be shown that the un are linearly independent; the next step is to
construct a matrix P whose columns are the vectors un .
(d) Make a change of variables u = Pv such that uT Au becomes vT Cv, and uT Bu
becomes vT Dv. Show that C and D are diagonal by showing that cij = 0 if
i = j, and similarly for dij .
Thus u = Pv or v = P−1 u reduces both quadratics to diagonal form.
To summarise, the method is as follows:
(a)
(b)
(c)
(d)
9.7
find the λn that allow (E9.1) a non-zero solution, by solving |B − λA| = 0;
for each λn construct un ;
construct the non-singular matrix P whose columns are the vectors un ;
make the change of variable u = Pv.
(It is recommended that the reader does not attempt this question until exercise 9.6
has been studied.)
If, in the pendulum system studied in section 9.1, the string is replaced by a
second rod identical to the first then the expressions for the kinetic energy T and
the potential energy V become (to second order in the θi )
T ≈ Ml 2 83 θ̇12 + 2θ̇1 θ̇2 + 23 θ̇22 ,
3 2 1 2
V ≈ Mgl 2 θ1 + 2 θ2 .
Determine the normal frequencies of the system and find new variables ξ and η
that will reduce these two expressions to diagonal form, i.e. to
a1 ξ̇ 2 + a2 η̇ 2
and
331
b1 ξ 2 + b2 η 2 .