Building Acoustics

Sound absorbers 167

jj

00

a j

ˆˆ

ˆˆ

kd kd

ir

kd kd

xd ir

p cpe pe

Z

vS S pe p e

ρ −

−

=

⎛⎞ +

⎜⎟==⋅

⎝⎠⋅ − j

. (5.20)

As we assume that the closed end is totally reflecting, the sound pressure amplitudes in
the two partial waves must be equal. This gives us

22

00 1 00 00

a jcotg() j j

Zkccdkd

SS

ρρ

ωω

=− ⋅ ⎯⎯⎯⎯<<→− =− c.

dV

ρ

(5.21)

Making the approximation kd << 1 we see that the result is the same as derived above.
The equivalent acoustical stiffness is again ρ 0 c 02 /V.

d

pi

pr

v=0

S

Figure 5.9 Plane waves in a tube. The tube is terminated by a totally reflecting surface.

5.4.1.2 The acoustic mass in a tube

Using a similar procedure, as when deducing the stiffness, we shall find an expression
for the equivalent mass by calculating the acoustical impedance at a distance d from a
pressure release surface, i.e. the pressure at the surface is zero as opposed to the above
setting the particle velocity equal to zero. The same procedure that gave us Equation
(5.21) will now give

(^) a jtg()^00 kd^1 j^0
c
Zkd
SS
ρ << ωρd
=⋅ ⎯⎯⎯⎯→. (5.22)
The equivalent acoustical mass is therefore ρ 0 d/S. We are now in a position to calculate
the resonance frequency of a resonator having a volume V and a “neck” of length d with
a cross sectional area of S. We get:
2
0a 0 00
0
a0

11

.

22 2 2

kcScS

f

mVd

ωρ

ππ πρ π

== = =

Vd

(5.23)

For a more accurate calculation we must take into account that the effective oscillating
mass is larger than the one contained in the neck. We have to add the so-called end
correction. Furthermore, making use of the resonator requires information on the energy
losses of the system. We shall treat the latter item first.