Building Acoustics

Sound transmission 221

where kx and ky are the components of the wave number in the medium around the plate
(air). This expression has then to be a solution of the ordinary wave equation:

2

2

22

0

1

0.

p

p

ct

∂

∇ −=

∂

(6.32)

Figure 6.9 Sketch showing a plane bending wave on an infinitely large plate. The plate lies in the x–z plane and
the pressure is calculated in points (x,y).

Inserting Equation (6.31) into (6.32) we immediately see that the wave number k
for the sound field above the plate must be expressed by

22

0

kkkx y.

c

ω

== + (6.33)

A further condition is that the component vy of the particle velocity, i.e. the component
normal to the plate, must be equal to uB at the surface of the plate (y = 0). Since vy is
given by

j( )

00

1 ˆ

,

j

y tkxkyxy

y

p pk

ve

y

ω

ωρ ρ ω

∂ −−

=− ⋅ = ⋅

∂

(6.34)

we get when setting y = 0,

jB j
0

ˆ

ˆ kx y kxx.

pk

ue e

ρω

⋅=⋅− − (6.35)

Hence,

ˆˆ^0 x B.
y

p u and k k

k

ρω

=⋅ =

The sound pressure may thereby be expressed as

(^22) B
00 j( B) j
2
B
2

ˆ

(, ).

1

cu tkx kky

pxy e e

k

k

=⋅⋅ρ ω− −

−

(6.36)

x

z

y

λB

p(x,y)