Building Acoustics

170 Building acoustics

1

0

tan( )

j=j 1^2

2

s

kb

Zdd

kb

ρω ρ ω.

′ −

⎛⎞

⎜⎟

=−⎜

′

⎜⎟

⎝⎠

⎟ (5.35)

d

b

z

x

y

Figure 5.10 A single slit of width b in a plate of thickness d.

Using a series expansion in the angular frequency ω, the first three terms will be

222

0

2 0

12 1 6

j

700 5

s

d db

Z d,

b

μ ρω

ωρ

μ

≈+ + (5.36)

where the constant term in the real part corresponds to the value 8μ/a^2 found above for
cylindrical tubes with radius a. We may also observe, looking at the imaginary part, that
the viscosity again results in an added mass for the air in the slit.

5.4.1.4 The Helmholtz resonator. An example using analogies

A complete model for a Helmholtz resonator embedded in a hard wall, as depicted in
Figure 5.8, must include all types of energy losses. The “natural” losses are represented
by two components: the one caused by the viscous losses and the other due to the
reradiated sound energy. Certainly, without the viscous losses the resonator will not act
as an absorber! In designing for good room acoustics conditions it was noted that
diffusers might be just as important as absorbers. In this connection, Helmholtz
resonators are useful, contributing to the diffuse sound by partly reradiating the sound
energy.
There will also be contributions to the viscous losses from the surfaces around the
resonator neck but we shall neglect these for the moment. On the other hand, we shall
have to find an expression for the acoustic impedance describing the sound radiation
from the tube opening. Here we may make use of the formerly derived radiation
impedance for a circular piston placed in an infinite baffle (see section 3.4.4). We then
have to envisage that the air column in the neck moves like a rigid piston. Using a low
frequency approximation for the radiation impedance expressed by its equivalent
acoustic impedance, we get