Civil Engineering Reference
In-Depth Information
acoustical equations derived previously. We shall use both approaches, starting with a
resonator where the air volume are compressed by a small piston in the neck being driven
by a alternating force
F
as sketched in
Figure 5.8
b).
Incident wave
F
d
Area
S
Piston, area
S
Volume
V
Volume
V
a) b)
Figure 5.8
A simple Helmholtz resonator driven by a) a sound wave. b) a mechanical force.
By differentiating the adiabatic equation of state, we get
d
PV
PV
d
+
γ
=
0.
(5.17)
In this equation we may consider the differential d
P
as the equivalent sound pressure
p
=
F/S
and the pressure
P
as the ambient pressure
P
0
. If the force is giving the piston a
displacement Δ
x
we get
F
γ
Δ ⋅
xS
=−
,
(5.18)
SP
⋅
V
0
and the mechanical stiffness
k
mec
will then be
2
2
2
−
== =
F
γ
PS
ρ
c S
γ
ρ
P
0
0
0
0
k
where
c
=
.
(5.19)
mec
0
Δ
x
V
V
0
We have introduced the sound (phase) speed
c
0
in the last equation. Considering the
relationship in Equations
(5.16)
, the corresponding acoustical stiffness will be equal to
ρ
0
c
0
2
/V
. We may now show by calculating the acoustical impedance in a tube closed at
one end, as depicted in
Figure 5.9,
that we get the same result in the low frequency limit,
i.e. when making the dimensions smaller than the wavelength. We shall use the equations
in Chapter 3 (section 3.5.1), which gives us the acoustic impedance at a distance
d
from
the reflecting closed end as
