Sound insulation and sound radiation
Wolfgang Kropp
11.1
Sound radiation and sound insulation
We actually do not need this chapter  You have learned everything you need to
know:
When waves propagating from one medium to the next it is the change of impedance
which determines what happen. In our case this will be the radiation impedance. Even
here Snell’s law will be valid
Along the plate there is a pattern given by the bending wavelength λB. This pattern has to
be identical with the projection of the
wavelength of the radiated sound. Only if the
wavelength on the plate is bigger then the
wavelength in air, sound radiation is possible
(this is only true for infinite plates as we see
later).
In the case where the wavelength on the
plate is shorter than the wavelength in air
you cannot find an angle where the projection
fits.
This a relatively
simple derivation
of what is
following in
detail in later in
the text.
Radiation from
infinite plates
Three areas:
! B < ! air
No radiation
Only a nearfield
! B = ! air
Sound radiation
parallel to the
plate surface
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! B > ! air
Sound radiation
under a certain angle
Sound insulation and sound radiation
Wolfgang Kropp
11.2
This would be a good qualitative description. The effect that there is no sound radiation
for ! B < ! air could be explained by the “short-circuiting” between positive and negative
displacement on the plate. Instead of getting the air in a wave motion, it is just moved
parallel to the surface.
When ! B = ! air a sound wave propagates parallel to the plate surface. For frequencies
with ! B > ! air sound is radiated under a certain angle.
Radiation efficiency
To characterise the efficiency of a radiator the radiated power is compared with the
radiation of a plane wave with the same surface and the same averaged velocity. The
radiation efficiency
!=
Wradiated
"cS v
where S is the radiating surface and v
2
2
is the averaged effective value of the velocity
of the radiator (i.e. average over the surface). The radiation efficiency defined in this way
can become bigger than unity, since it is not relating for instance mechanical power to
radiated power, but compares the radiation
from a structure with what would be radiated
from a plate, vibrating in phase and with the
same amplitude as the structure has in
average.
In our case the radiation efficiency has to
look as this =>
For this analysis we can summarise that there is one frequency where c B , the speed of
sound on the plate, is identical with c , speed of sound in air. This frequency is called
critical frequency and can be calculated as c air = c B or c air =
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4
B
m
! 2 =>
Sound insulation and sound radiation
Wolfgang Kropp
fc =
2
c air
m
2!
B
=
2
c air
(
"P h 1 # $ 2
2!
3
hE
) =c
11.3
2
air
1
2! h
12
(
"P 1 # $ 2
E
) = 0.55 c
2
air
hc L
=
1
h
Kc
12
c L is the speed of sound for longitudinal waves in structures. The factor K c depends on
the material only. Typical values are
material
concrete
stone
light concrete
gypsum
steel
aluminium
glass
K c [m/s]
18
22
36
32
12
12
14
Many of the values are in the same order. This means that it is mainly the thickness of the
material, which will lead to differences in the critical frequency.
Radiation from finite plates
From our daily experience we know that all structures radiate independently where the
critical frequency is situated. The results before are only valid for infinite plates. The
main difference between infinite and finite plates become obvious when looking at a plate
(e.g. a window) built into a bigger and more
rigid structure (e.g. a wall).
It is obvious that this so-called
hydrodynamic short-circuiting does not
work perfectly at the edges of the plate.
There is some air, which cannot move from
areas of high pressure to areas of low
pressure. Indeed, one can show by theory
that the edges of the plate are responsible for the radiated sound below the critical
frequency.
$
( For a slightly damped plate the radiation efficiency
1
Uc
f
&
for f << fc & can be calculated as
2
& # Sfc fc
& where U is the circumference of the plate and S is its
&
& surface.
&
&
Uc
! " % 0.45
for f " fc )
f
&
&
c
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&
&
'
*
Sound insulation and sound radiation
Wolfgang Kropp
11.4
An increase of the radiated sound power
does although occur when discontinuities
are present such as stiffeners, material
changes or close to the forces acting on the
structure. Examples for the radiation
efficiency are shown here
For a vibrating plate built into a rigid wall: with decreasing size, the radiation efficiency is
increasing. It approaches the radiation efficiency of a point source in front of a rigid wall.
From radiation to excitation
The radiation efficiency of walls, floors,
ceiling etc. is and important factor for
radiation of vibrations transported
through the structure of the building.
so-called flanking transmission is
strongly depended on the radiation
efficiency of the structures.
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the
The
Sound insulation and sound radiation
Wolfgang Kropp
11.5
The source in the sending room excites a part of the structure. The structure is vibrating.
These vibrations are transported to the next room and radiated into the room. Both,
excitation and radiation depend on the radiation efficiency of the structure, this means on
the relation between the wavelength on the structure (bending waves) and the wavelength
in air.
What can we do to keep the radiation efficiency low?
One idea would b to chose the critical frequency in such a way that it will be very high.
How to do this?
Looking at the equation for the
critical frequency fc =
2
c air
m
2!
B
It is clear that we need something
heavy with low bending stiffness.
Some materials might deliver this, eg.
Heavy rubber plates.
Alternative would be to make the plate softer by
cutting small grooves into it. This will not affect the
mass but making the structure “softer”. It is also
clear that stiffening a structure by rips or beams
might have a negative effect since it increases
the bending stiffness strongly but will increase
the mass only slightly.
Is it not possible to modify the radiating wall
or ceiling, one might cover the it with a second
wall freely suspended. This second wall should
then have a low radiation efficiency
TASK: After all what you have learned draw
your conclusions concerning sound insulation
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Sound insulation and sound radiation
Wolfgang Kropp
11.6
Mathematical derivation of what we have learned
(just for fun )
Sound radiation from bending waves on infinite plates
In the Figure below a situation is shown, where a plate is vibrating with the wave length
()
! jk y
! B , and the velocity v P y = v B e B from the plate a wave might be radiated under a
certain angel. Only one dimension of the plate is considered to simplify the problem. The
radiated wave can be described as
(
)
p rad x, y, ! = p A e " jk x x e
" jk y y
At the surface of the plate the velocity in the
normal direction to the surface of the plate
has to be identical with the particle velocity
in air. This means that the wave number of
the plate k B and the wave number in the y
direction k y have to be identical (see Snell's
law). The velocity in the field directly above the plate is
(
)
kx
)
k cos "
v rad ,x x = 0, y, ! =
(
v rad ,x x = 0, y, ! =
"!
p A e # jk x x e # jkB y = v B e # jkB y which is equivalent with
( )p
#!
A
e
( ) e $ jk y = v e $ jk
B
$ jk cos " x
By
B
This velocity has to be identical with the velocity of the plate. This allos us to calculate
()
the unknown amplitude p A ! =
"!
kx
vB .
The wave number in the x direction however, is not independent from the wave number in
y-direction. For both the relation k 2 = k 2x + k 2y has to be valid as we learned when
discussing the oblique incidence of waves on a surface. Replacing k y by k B we obtain
k 2x = k 2 ! k B2 or k x = k 2 ! k B2 where k is the wave number in air. The radiated pressure
can finally be written as
(
)
p rad x, y, ! = v B
"!
k 2 # k B2
e
# j k 2 #kB2 x
e # jkB y
Together with the velocity on the surface of the plate the intensity in the normal
direction of the plate surface can be calculated
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Sound insulation and sound radiation
Wolfgang Kropp
11.7
Ix =
1
2
{(
) ( )}
Re p x = 0, y v P* y
=
#% !" '%
2
v B Re$ (
%& k x %)
2
1
Three different cases can be distinguished.
Case 1 : k air > k B or ! air < ! B
In this case the wavelength in air is smaller than the wavelength on the plate. The
root k 2 ! k B2 is real and consequently also the wave number k x . The radiated power per
unit area is given as I =
1
2
vB
2
!"
kx
=
1
2
vB
!"
2
()
k cos #
=
1
2
vB
2
!c
()
cos #
This can be interpreted as the effective value of the velocity times the radiation
impedance for a wave radiated under the angle ! . The angle ! is determined by Snell's
2
2
()
2
law k cos ! = k " k
2
B
2
()
which gives cos ! = 1 "
k B2
k2
#
&
#
&
2
c2 (
%
%
k
(
or for the angle ! ! = cos "1 % 1 " B ( = cos "1 % 1 "
which we already have obtained
2
2 (
%
(
k
cB (
%
$
'
$
'
for Snell's law.
If the speed of sound in air is much smaller than the speed of sound for the bending waves
on the plate, the angle ! will become close to 0 degrees, i.e. the wave is radiated as a
plane wave in normal direction to the plate. The radiation efficiency is
1
!=
W radiated
"cS v
2
=
2
vB
"cS
2
"#
kx
1
2
S
vB
2
=
1
()
cos $
which becomes in this case close to unity.
Case 2 : k = k B , ! = ! B
In this case the wavelength on the plate equals the wavelength in air. This also means that
the speed of sound on the plate is identical with the speed of sound in air.
This case gives in some way strange results. First of all the amplitude of the pressure
wave as well as the radiated power and the radiation efficiency becomes infinity, since the
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Sound insulation and sound radiation
Wolfgang Kropp
11.8
root
k !k
2
2
B
becomes zero. The angle ! is in this case 90 degrees, which means that the
wave is radiated in parallel to the plate. The somewhat strange results concerning pressure
amplitude and radiated power is a consequence of that we do not include radiation loading.
Creating a high pressure in front of the plate certainly means, that the plate will
experience a force, which will act against the vibration. In most of the cases the radiation
loading is neglected. However, in the case of very light structures or plates in a heavy
medium like water, radiation load has to be considered.
Case 3 k < k B , ! < ! B
In this case the wavelength on the plate is shorter than the wavelength in air. The speed
of sound on the plate is smaller than the speed of sound in air. The root
k 2 ! k B2
becomes imaginary and consequently also the wave number in x-direction. Instead of a
radiated wave only a nearfield (i.e. an in x-direction decreasing exponential function) is
created. Since the wave number in x-direction is imaginary the intensity yields zero. This
means there is no power radiated from the surface of the plate.
The radiation efficiency consequently equals zero. The air particles in front of the surface
move on circles as can be seen when considering both the velocity in x-direction and the
velocity in y direction. Since the velocity amplitude in x-direction becomes imaginary, while
the velocity amplitude in y-direction is still real, there is a phase difference of 90 degrees
between both.
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Sound insulation and sound radiation
Wolfgang Kropp
11.9
Sound insulation
In order to have a quiet room one needs possibilities to hinder the propagation of sound
power W in into this room. The sound insulation of the walls (or whatever is around the
room) determines how much sound power Wtr is coming into the room.
Definitions:
The transmission factor
of sound power is ! =
! is a measure for the capability to isolate against the transport
Wtr
W in
.
The reduction index R is 10log
1
!
= 10log
W in
Wtr
.
The reduction index can be measured for rooms where the sound field has statistical
properties by measuring
•
the averaged sound pressure level LS in the sending room
•
the averaged sound pressure level LR in the receiving room
•
the equivalent absorption area AR in the receiving room
the surface area S through which the sound power is expected to be transported.
•
The reduction index is R = LS ! LR ! 10log
AR
S
.
The reduction index weighted (i.e. compared with the reference curve) is called R w . A
reduction index measured in field is indicated by a prime ( R ! ).
If a construction consists of several parts with different reduction index the resulting
reduction index is
R res = 10log
S 1 + S 2 + ...S n
S 1 10
!
R
10
+ ...S n 10
!
Rn
= 10log
10
"S
"S #
i
i
where S1, S2= the area of the different parts andR1, R2 = the reduction index of the
different parts.
TASK: IS THERE A RELATION BETWEEN SOUND INSULATION AND SOUND
RADIATION?
Have a look on the notes from the last lecture on sound radiation. We found for a
very big (infinite) plate:
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Sound insulation and sound radiation
Wolfgang Kropp
11.10
•
•
•
no sound radiation below the critical frequency
fc
very high radiation parallel to the surface of
the plate at the critical frequency
radiation at frequencies above the critical
frequency
HAS THIS ANY CONSEQUENCE FOR THE SOUND
INSULATON?
Derivation of the theory for sound insulation of plates
Airborne sound incident at a panel, wall or some other type of partition mainly excites
bending waves in the construction. The sound insulation of plates can be derived in several
ways. One way follows the approach we used for Snell's law (see notes).
One assumes an incident wave, a reflected wave and a transmitted wave. All three waves
together have to fulfil the boundary conditions at the plate:
continuity of particle
velocity
the
differe
nce
betwee
n the
pressure on both sides of the plate excites the plate to vibrations.
Since we assume the same medium on both sides of the plate, the wavenumbers are the
same as well as the angle of incidence and the angle under which the wave is radiated into
the second medium.
( ) e " jk sin(# )y
+ jk cos(" )x # jk sin(" )y
!e
e
( ) ()
p (x, y, ! ) = rp ( )
p (x, y, ! ) = tp (! )e
p i x, y, ! = p A, i ! e
r
t
A, i
A, i
" jk cos # x
( ) e " jk sin(# )y
" jk cos # x
The sin- and cosine functions come from the projection of the waves in the different
directions. Each of the individual exponential functions gives the changes of the phase due
to a change in the x- or y- co-ordinate.
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Sound insulation and sound radiation
Wolfgang Kropp
11.11
The particle velocities of the waves (in the directions of the waves) are obtained from the
pressure using the relation p/u = !c . The components normal to the plate are a factor cos
ϑ smaller.
( ) !e
( ) #c ( )
p cos(# )
u (x, y, ! ) = "r
! )e
(
$c
p cos(" )
u (x, y, ! ) = t
! )e
(
#c
u i x, y , ! =
p A, i cos "
( ) e $ jk sin(" )y
$ jk cos " x
A, i
( ) e " jk sin(# )y
+ jk cos # x
r
A, i
( ) e $ jk sin(" )y
$ jk cos " x
t
When discussing Snell's law we had continuity of the pressure. However, this is not the
case when a plate is separating both sides. There the pressure is different on both sides.
Otherwise the plate would not reduce sound; there would be complete transmission
through the plate.
The vibration of the plate is described by the bending wave equation
B
( ) + m ! "(x,t ) = p
! 4" x , t
2
4
!x
!t
2
1
(x,t ) # p (x,t ) Assuming harmonic vibrations the
2
displacement on the plate has the form
( )
()
( ) ()
! j"#(x, " )
B
$ " mj"#(x, " ) = j" [p (x, " ) $ p (x, " )]
!x
! x, " = !B " e # jkB y or u p x, ! = u B ! e " jkB y becomes after multiplication with j! :
4
2
1
4
2
j!" is just the velocity u p of the plate.
B
( ) # " mu
! 4u P x, "
!x
4
2
P
(x, " ) = j" [p (x, " ) # p (x, " )]
which is
1
2
[ ( ) ( )]
Bk B4u B e ! jkB y ! " 2mu B e ! jkB y = j" p 1 x, " ! p 2 x, "
Now we can replace the pressure by p 1 = p i + p r and p 2 = pt .
Since the velocity on the plate has to be identical with the velocity of the sound field on
both sides for all y. This means that
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Sound insulation and sound radiation
Wolfgang Kropp
e
( )
! jk sin " y
11.12
()
= e ! jkB y or k B = k sin ! has to be valid.
______________________________________________________
Comment the impedance is Z =
plate when exposed to a wave:
Z =
p
up
p
[
=
1
! p2
up
] = Bk
4
B
p
u
, from the equation above we get the impedance of a
! " 2m
j"
Using all equation for the pressures and velocities the two unknowns (reflection
coefficient and transmission coefficient) can be determined.
Task: try this by your own
1
Transmission coefficient: t =
1+
( ) $" m + Bk
cos !
[
2 j"#c
2
4
sin 4 !
]
The transmission factor is the squared magnitude of the transmission coefficient:
!=
1
()
% cos "
'
1+'
'& 2#c$
(2
*
2
4
4
* +$ m + Bk sin "
*)
[
]
2
Physic behind the equation:
The relation shows that the transmission through the plate depends on the angle of
incidence.
We see that there is a possibility for total transmission when the expression in the second
parenthesis in the nominator becomes zero.
2
4
4
4
4
Total transmission occurs for: !" m + Bk sin # = 0 or when k sin ! =
()
Or for cos ! = 0 .
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" 2m
B
Sound insulation and sound radiation
Wolfgang Kropp
11.13
From this expression we can calculate the frequency at which total transmission occurs.
This frequency is called the coincidence frequency and the phenomenon the coincidence
phenomenon. The forced bending wave velocity coincides with the sound velocity in air. Or
expressed in wave number, the projection of the wave number of the exciting wave in air
onto the surface of the plate coincides with the wave number of the free bending wave in
the plate, compare with what we
have learned about critical
frequency
k sin ! =
4
" 2m
B
= kB .
For each angle of incidence we will
find one coincidence frequency.
The lowest coincidence frequency
is obtained for a wave propagating
parallel to the plate. This lowest
frequency is the critical frequency)
Reduction index as function of
frequency and angel of incidence
fcoincidence =
c2
m
2! sin 2 "
B
.
The coincidence phenomenon is a special
type of resonance. We are used to link the
resonance concept to the case where a driving force or pressure has a frequency, which
equals a natural frequency (eigenfrequency) of the driven system. Here, instead, it is a
driving field the spatial distribution of which equals a (natural) bending wave field of the
driven system.
As we recall from physics (see Physics, V2), it is the losses that limit the response of a
system driven at resonance. So, at coincidence, the losses in the system get important.
The reduction index does not drop to zero because there are always losses (damping) in
real systems, which then determine the reduction index.
We see that for 90 degrees incidence (gracing incidence, parallel to the plate) the
reduction is zero over the whole frequency range since cos ϑ = 0 and therefore τ = 1.
However, this result is somewhat artificial since for a parallel wave there is no normal
component to the plate and therefore there is no sound power transmitted through the
plate.
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Sound insulation and sound radiation
Wolfgang Kropp
11.14
4
4
2
When we are well below the coincidence frequency, Bk sin ! << " m . The wall or panel
has mass character. In this case the transmission factor becomes ! =
1
()
%
cos #
'
1 + '"m
2$c
'&
(2
*
*
*)
The sound insulation of the plate follows a mass law, which says that it increases with 6 dB
when doubling the mass per unit area.
This does not mean that the plate as the whole is vibrating like one rigid mass, but
each element vibrates more or less independently from its neighbour following the
vibration pattern imposed by the sound field.
Summary:
For the sound reduction index we get quite different expressions below and above the
critical frequency. We can divide the frequency range into three ranges; f <fc, f ≈ fc, and
f > fc. We get
f < fc
The construction has mass character. The reduction index follows a mass law.
We call this Non resonant transmission.
f ≈ fc The construction gets sound transparent for gracing incidence if the losses can be
neglected. Incident sound is totally transmitted. Resonant transmission.
f > fc For a certain angle of incidence the sound transmission is total if the losses are
zero. Also for constructions with losses the transmission becomes big. Resonant
transmission.
Diffuse sound field
(Sound comes from all directions with the same probability).
For f > fc we always have sound waves fulfilling the coincidence condition if the sound field
is diffuse. This cannot happen for f < fc.
We consider diffuse sound incidence at a wall, i.e. all
angles of incidence ϑ are equally probable. We have to
take the average over the transmitted sound power and
the incident sound power for all angels
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sin!
!
d!
Sound insulation and sound radiation
Wolfgang Kropp
11.15
# /2
!d =
$0
W in (" )! (" ) cos " sin "d"
# /2
$
=2
# /2
$
! (" ) cos " sin "d"
0
W in (" ) cos " sin "d"
0
________________________________________________________
How to do :
The total solid angle is 4π. The fraction of the sound field having an angle of incidence ϑ is
then
2! sin "d"
=
4!
sin "d"
2
Projected area against this incident sound is S cos ϑ. We then have to take the average
of the transmission factor for diffuse sound incidence, τd as
# /2
!d =
$0
! (" ) cos " sin "d"
# /2
$
=2
# /2
$
! (" ) cos " sin "d"
0
cos " sin "d"
0
Using this expression, τd may be calculated for different cases. For f ≥ fc we have to
introducee the losses of the plate or wall through a complex Youngs modulus E = E0 (1 + jη)
where η is the loss factor. This expression for τ (ϑ) is then inserted in the integral above.
It is a more tricky derivation. However, in this case it is found that the total transmission
is dominated by that part of the incident sound field that fulfils the coincidence condition.
The integration only needs to be performed over this resonance peak. The derivation is
found in Cremer’s paper from 1942.
The results are: ! d = ! (0)2, 3log
1
! (0)
or Rd ≈ R(0) – 10 log (0,23 R(0))
With τ(0) obtained from the formula for the transmission factor setting ϑ = 0 we obtain
for the three frequency areas of interest:
For f < fc
Rd mass law ≈ 20 log m + 20 log f – 49 dB
For f ≈ fc
R d = R d , masslaw + 10log ! + 8dB
η is the loss factor of the wall material.
Finally for f > fc
f
R ! R (0) + 10log( " 1) + 10log # " 2dB
fc
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Sound insulation and sound radiation
Wolfgang Kropp
11.16
Comments:
R(0) increases with 6 dB/octave. We can then see that above the critical frequency Rd
increases with 9 dB/octave if we can assume the loss factor η to be frequency
independent.
For η = 0,01 the depth in the coincidence valley is approximately 12 dB.
The influence of the coincidence phenomenon may be observed in the reduction index
already around fc/2. The pure mass law is therefore valid to approximately 0,5fc.
Thick brick- and concrete walls have such a low critical frequency that the expression
describes the reduction index in most of the frequency range from 100 to a couple of
1000 Hz.
shows in summary the schematical reduction index for a single wall of limited size.
Observe that the frequency scale is normalised with regard to the critical frequency. For
thin constructions such as gypsum boards, thin steel plates and thin glass plates the sound
reduction index is determined by the mass law.
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Sound insulation and sound radiation
Wolfgang Kropp
Problems to section 11
11.17
a) A vibrating machine is mounted on a quadratic steel plate of thickness 10 mm and
surface area of 10m2. The machine is vibrating with a frequency of 500 Hz and the rms
value of the velocity has been measured to 10-3 m/s.
a) What is the critical frequency for the plate?
b) What is the radiation factor at this frequency?
c) How large is the radiated sound power from the plate?
b) The steel plate in 11.1 is changed to a 16 cm thick quadratic concrete plate with the
same area as in 11.1. Assume that the vibration velocity in the concrete plate is the
same as for the steel plate. The concrete has the density 2300 kg/m3 and a Young’s
modulus of 26 GPa. The Poisson’s number can be set to 0,3.
a. What is the critical frequency for the concrete plate?
b. What is the radiation factor at this frequency?
d) How large is the radiated sound power from the concrete plate in this case?
c) An infinite plate vibrates with a frequency that is higher than the critical frequency.
a)
Why are sound waves radiating from a certain angle from the plate?
c. Why does the infinite plate not radiate anything at all for frequencies below the
critical frequency?
d) Plane sound waves with a frequency of 250 Hz hits a 12 cm thick concrete plate
( E ! 2, 5"101 0 Pa, ρ=2300 kg/m3, ! " 0.3 ) with an angle of 45°.
a) What is the wave length of the forced bending waves in the plate?
b) What is the wave length of the free bending waves in the plate at 250 Hz?
c) At what angle of incidence will the wave length of the forced bending waves be equal to
the free bending waves in the plate?
e) A large plate vibrates in phase over the entire surface. The radiation factor is 1 then.
The rms-value for the vibrational velocity is then veff = 3,5 !10"5 m/s and the plate area
is 20m2.
How large is the radiated power?
"8
f) The mean level was measured to 58 dB (re 5 !10 m/s) for a wall of area 10m2. Note the
reference value for the velocity level. The wall is the only surface radiating to a room
with the reverberation time 1s and dimensions 4 x 5 x 2,5 m3. The resulting sound
"5
pressure level in the room was measured to 55 dB ( 2 !10 Pa).
Calculate the radiation factor for the wall.
CHALMERS UNIVERSITY OF TECHNOLOGY
Division of Applied Acoustics
E-mail: [email protected]
Sound insulation and sound radiation
Wolfgang Kropp
11.18
g) A machine is mostly vibrating below 700 Hz. The machine is built-in a sealed box in
order to prevent radiation of sound from the machine to the surrounding.
You shall now choose the material, which gives lowest amount of radiated sound. The
largest available thickness is 80 mm due to limited available space.
The available material is steel plate and aluminium plate with the following data:
Steel:
ρ=7880 kg/m3, E=210 GPa, ν=0.3, thickness=44 mm.
Aluminium: ρ=2700 kg/m3, E=70 GPa, ν=0.3, thickness=15 mm.
2
V,T
S
v
2
The floor of a car (see figure above) works as a source to the compartment when the
car is driven on a road.
Calculate the total sound level inside the compartment caused by the floor vibrations!
Calculate the Schröder frequency.
The following information is given for the indicated octave bands.
100
200
400
800
1600
fcenter [Hz]
radiation-factor, ! 0,02
0,15
T6 0 [s]
2
3,29x10-6
!v " [(m/s)2]
Floor surface, S=3,52 m2
Compartment volume, V=5 m3
CHALMERS UNIVERSITY OF TECHNOLOGY
Division of Applied Acoustics
E-mail: [email protected]
0,01
0,10
0,30
0,08
0,80
0,09
5,22x10-6
9,81x10-8
3,21x10-7
1,00
0,07
1,98x10-6