Hints and answers

NORMAL MODES
9.8
(It is recommended that the reader does not attempt this question until exercise 9.6
has been studied.)
Find a real linear transformation that simultaneously reduces the quadratic
forms
3x2 + 5y 2 + 5z 2 + 2yz + 6zx − 2xy,
5x2 + 12y 2 + 8yz + 4zx
9.9
9.10
to diagonal form.
Three particles of mass m are attached to a light horizontal string having fixed
ends, the string being thus divided into four equal portions each of length a and
under a tension T . Show that for small transverse vibrations the amplitudes xi
of the normal modes satisfy Bx = (maω 2 /T )x, where B is the matrix


2
−1
0
 −1
2
−1  .
0
−1
2
Estimate the lowest and highest eigenfrequencies using trial vectors (3 4 3)T
√
T
T
√
and (3 − 4 3)T . Use also the exact vectors 1
2 1 and 1 − 2 1
and compare the results.
Use the Rayleigh–Ritz method to estimate the lowest oscillation frequency of a
heavy chain of N links, each of length a (= L/N), which hangs freely from one
end. (Try simple calculable configurations such as all links but one vertical, or
all links collinear, etc.)
9.5 Hints and answers
9.1
9.3
9.5
9.7
9.9
See figure 9.6.
√
√
√
(b) x1 = (cos ωt + cos 2ωt), x2 = − cos 2ωt, x3 = (− cos ωt + cos 2ωt).
At various times the three displacements
will reach √
2, , 2 respectively. For exam√
an oscillation
ple, x1 can be written as
√ 2 cos[( 2−1)ωt/2] cos[( 2+1)ωt/2], i.e. √
2 cos[( 2−1)ωt/2];
of angular frequency ( 2+1)ω/2 and modulated amplitude
√
the amplitude will reach 2 after a time ≈ 4π/[ω( 2 − 1)].
As the circuit loops contain no voltage sources, the equations are homogeneous,
and so for a non-trivial solution the determinant of coefficients must vanish.
(a) I1 = 0, I2 = −I3 ; no current in P Q; equivalent to two separate circuits of
capacitance C and inductance L.
(b) I1 = −2I2 = −2I3 ; no current in T U; capacitance 3C/2 and inductance 2L.
= 1.431ξ − 2.097η.
ω = (2.634g/l)1/2 or (0.3661g/l)1/2 ; θ1 = ξ + η, θ2 √
√
Estimated, 10/17 < Maω 2 /T < 58/17; exact, 2 − 2 ≤ Maω 2 /T ≤ 2 + 2.
332
9.5 HINTS AND ANSWERS
1
m
2
M
3
m
(a) ω 2 = c +
d
m
label2
kM
kM
(c) ω 2 = c +
2km
2d
d
+
M
m
Figure 9.6 The normal modes, as viewed from above, of the coupled pendulums in example 9.1.
333