Civil Engineering Reference
In-Depth Information
When this velocity is known, we may solve the first of these integrals. The second
integral requires knowledge of the pressure
p
(
S
) on the surface. This pressure distribution
cannot be known in advance and involves a generally complicated calculation.
Fortunately enough, when dealing with plane surfaces surrounded by another (infinitely)
stiff surface, identified as an acoustic baffle, the second integral will be zero. We shall
then be left with solving the integral
j(
ω
tkr
−
)
ρ
ck
uS
n
()e
⋅
∫
v
00
pRt
(,) j
=
d.
S
(3.46)
2
π
r
S
It should be seen that we shall be concerned with the radiation to one side of the surface,
which explains the number two instead of four in the denominator. In many cases we
shall only be interested in the pressure far for the surface, i.e. when the distance
r
is
much larger than the dimensions of the source. In these cases, we may substitute
r
by the
distance
R
(see
Figure 3.6)
and, furthermore, place this distance outside the integral. We
then get the famous and very useful Rayleigh integral
ρ
π
ck
∫
v
j(
ω
tkr
−
)
00
pRt
(,) j
=
u S
()e
⋅
d.
S
(3.47)
n
2
R
S
It should be noted that we still have to keep the distance
r
in the exponential function.
(Why is this?). In Chapter 6, where we treat the problem of sound transmission through
walls and floors, we shall use this integral to compute the sound radiation from panels
and walls vibrating in certain patterns, in particular from plates vibrating in their natural
modes. As an introduction to such application we shall calculate the radiation from a
vibrating circular disk or piston set in an infinitely large baffle.
G
p Rt
(,)
r
G
n
G
R
G
dS
S
z
y
x
Figure 3.6
Calculating sound radiation from a vibrating surface of area
S.
3.4.3
Radiation from a piston having a circular cross section
The circular surface depicted in
Figure 3.7
is assumed to be a flat disk set into an
infinitely large baffle. The disk, normally denoted a piston source, has a velocity of
vibration
uu
=⋅
ˆ exp( j
ω
)
