Civil Engineering Reference
In-Depth Information
reflecting. The surface impedance for these lids is thereby infinite and the particle
velocity is zero. Our task is to find an expression for the pressure
p
in the tube.
As a starting point we assume that there is no sound source inside the tube, which is
wholly analogous to the case of free vibration in a mechanical system treated in Chapter
2. Now we have to solve the homogeneous wave equation, the Helmholtz equation
(3.7),
which in one-dimensional form is given by
2
d
p
2
+
kp
=
0.
(3.88)
2
d
x
The general solution of this equation is
px
( )
=
A
sin(
kx
)
+
B
cos(
kx
),
(3.89)
where
A
and
B
are constants to be determined by the boundary conditions. Having
assumed that the particle velocity is zero at each end of the tube, this implies that the
gradient d
p
/d
x
of the pressure is zero at both tube ends. Setting the coordinate
x
equal to
zero at one end and equal to
L
at the other end, these conditions imply that the constant
A
must be equal to zero and that the wave number
k
may only attain a given set of values.
The only possible solutions, the
eigenfunctions
will be
px B
()
=
cos(
kx
),
(3.90)
n
n
where the wave numbers
k
n
and the associated natural frequencies are given by
n
⋅
π
nc
⋅
0
k
=
and
f
=
.
(3.91)
n
n
L
2
L
The lowest natural frequency of a tube having a length of one metre is then
f
1
≈ 170 Hz
(20°C). (What is the lowest natural frequency of a similar tube having an open end as an
organ pipe? To calculate this one may for simplicity assume that the sound pressure in
the open end is equal to zero.)
How does one calculate the sound pressure caused by a given sound source placed
at certain point inside the tube? We have to solve the wave equation
(3.88)
but now
having a source term on the right hand side. Assuming a monopole source inside the tube
we shall first have to modify Equation
(3.2)
, which is the equation of conservation of
mass in the system. We shall now write
∂
ρ
=− ∇⋅ +
ρ
v
qt
(, ,
r
(3.92)
0
∂
t
where
q
represents the source, a mass flux in an area having coordinates
r
0
. This implies
that the sought after right-hand source term of the wave equation will be a time
derivative of
q
. We shall not go into details on how to obtain a solution of the wave
equation in this case. Suffice to say that one expresses both the pressure and the source
strength by sums of the eigenfunctions
(3.90)
and then adjusting the coefficients in these
sums. Assuming that the source area is very small in comparison with the other
dimensions, the pressure in a position
x
caused by a source placed in position
x
0
may be
written as
