Sound absorbers - Building Acoustics - page 167

− j kd j kd

ˆ ˆ

⎛ p ⎞ ρ c pe + pe

00 i r

Z = = ⋅

. (5.20)

⎜ ⎟ a

− j kd j kd

vS ⋅ S ˆ ˆ

⎝ ⎠ pe − p e

xd = i r

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

2 2

ρ c ρ

ω

c ρ c

kd << 1

00 00 00

Z =− j ⋅ cotg( kd ) ⎯⎯⎯ →− j =− j .

(5.21)

a

S S d ω V

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 0 2 / V .

d

v =0

p i

S

p r

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

ρ c ωρ d

00 kd << 1 0

Z = j ⋅ tg( kd ) → j . (5.22)

a

S S

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

ω 1 k 1 ρ c S c S

0 a 0 0 0

f == = = .

(5.23)

0

22 ππ m 2 πρ π V d 2 Vd

a 0

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.