Civil Engineering Reference
In-Depth Information
∞
g(q
r
)
=
g(r)
exp
(iωq
r
)
d
r
(10.118)
−∞
The axes
X
and
Y
can be chosen with the origin on
.Let
τ
be the angle between
Y
and
. In Equation (10.116)
q
r
r
can be replaced by
q
x
x
+
q
y
y
,where
q
x
=
q
r
cos
τ
and
q
y
=
q
r
sin
τ
. Equation (10.116) becomes
∞
∞
1
4
π
2
h(ω)g(q
r
)
exp
(jω(t
τ
zz
(r,t)
=
−
q
x
x
−
q
y
y))ω
d
q
r
d
ω
(10.119)
−∞
−∞
The vertical displacement for a given direction
of the source is a superposition
of
elementary
displacements
related
to
excitations
with
a
space
time
dependence
exp
(jω(t
−
q
x
x
−
q
y
y))
.Foragiven
ω
let
q
R
be the the slowness at the Rayleigh pole.
When
q
r
→
q
R
the elementary displacement tends to
∞
. Iterative methods can be used
to obtain the related slowness component in the plane
XY
. Many poles exist, and the
iterations must start close to the true Rayleigh pole. For the usual porous frames of
sound absorbing materials which are much heavier than air, the Rayleigh pole is close
to the Rayleigh pole for the frame in vacuum. As an illustration the slowness
q
R
is
predicted for the three geometries of Figure 10.8. For these geometries
q
R
for the frame
in vacuum is a root of a polynomial in
q
−
R
.
In Figure 10.8(a) the line source L is on the horizontal surface perpendicular to the axis
of symmetry
Z
and the Rayleigh wave propagates on this surface. In Figure 10.8(b) the
Rayleigh wave propagates in a meridian plane in a direction perpendicular to
Z
,andin
Figure 10.8(c) the Rayleigh wave propagates in a meridian plane in a direction parallel to
Z
. The slowness component
q
R
on the free surface where the Rayleigh wave propagates
is a root of a polynomial (see Equation 5.61 and Figure 5.1 for the definition of the free
surface in Royer and Dieulesaint 1996) which can be rewritten with the present notations:
Case (a)
G
(
2
G
q
−
R
ρ
1
q
−
R
ρ
1
2
C
G
−
q
−
R
ρ
1
+
A)
−
−
(
2
G
q
−
R
ρ
1
2
F
2
C
−
+
A)
−
=
0
(10.120)
Case (b)
G
(
2
G
q
−
R
ρ
1
q
−
R
ρ
1
2
A)
G
q
−
R
ρ
1
+
A)
−
−
(
2
G
+
−
(
2
G
q
−
R
ρ
1
2
A
2
2
G
+
A)
−
A
−
=
0
(10.121)
+
Case (c)
G
C
q
−
R
ρ
1
q
−
R
ρ
1
2
A)
G
−
q
−
R
ρ
1
−
−
(
2
G
+
C
q
−
R
ρ
1
2
F
2
2
G
−
A
−
=
0
(10.122)
+
