 PRACTICE PROBLEM ONE: From Previous lecture
A 16’ x 20’ x 9’ room has absorptive coefficients as
follows: Ignore doors & windows.
Walls
Floor
Ceiling
.30
.25
.40
A. Find total room absorption
B. Find reverberation time
The finishes then change to the following coefficients:
Walls
.46
Floor
.40
Ceiling
.86
C. Find the noise reduction due to the new finishes
D. Find the difference in reverberation time.
PROBLEM SOLUTION
 Wall area =
648 x .30 =
194.4
 Floor area =
320 x .25 =
80.0
 Ceiling area =
320 x .40 =
128.0
total = 402.4 sabins
T = .05 x (2880 / 402.4) = .36 seconds
NEW
Wall area =
Floor area =
Ceiling area =
648 x .46
320 x .40
320 x .86
=
298.08
=
128.0
=
275.2
total = 701.28
Noise Reduction = 10log (701.28/402.4) = 10 log 1.7427 =
10 x .2412 = 2.41 db. T = .05 x 4.1068 = .205 sec
Time difference = .36 - .205 = .1550 sec, = 43%
PRACTICE PROBLEM TWO:
SOUND PROJECTION
Say a fire alarm horn sounds
at a loudness of 100 decibels at a distance of 12 feet. At
what intervals must horns be placed so that a minimum of
90 decibels can be heard. Find the distance, x, from the
horn that the sound be only 90 decibels. Consider no help
from sound reflection.
Solution: First, find the amount of sound energy in
watts/cm2 , created at the distance of 12 feet. Second,
find the amount of energy required to produce 90 decibels.
Third, find the distance where the 90 db sound energy is
produced.
 First: EQUATION 1: 100 db = 10 log ( I.E.12 / 10-16 )
simplify the equation by dividing both sides by 10; so

10 db = log (I.E.12 / 10-16 ) now, since you cannot
solve for I.E.12 inside the log function, take the anti-log of
each side of the equation to get rid of the log function;
anti-log of 10 = 1010 ;
 anti-log of the right side, log (I.E.12 / 10-16 ) is simply,
I.E.12 / 10-16 , so the resulting equation is
 1010 = I.E.12 / 10-16, rearrange and solve for I.E.12;
I.E.12 = 1010 x 10-16 = 10-6 watts/cm2
Second: EQUATION 1 Find the energy required to
produce 90 decibels. Follow the same procedure as
previous:

90 = 10 log ( I.E.x / 10-16 ) ; divide both sides by 10,
and 9 = log ( I.E.x / 10-16 )

Take anti-log of both sides, and 109 = I.E.x / 10-16 ;
rearrange and I.E.x = 109 x 16-16 ;

 so the amount of energy required to produce 90 decibels at
the x distance;
I.E.x
= 10-7 watts/cm2
 Realize here that 10-7 is a smaller number than 10-6
 Third: EQUATION 2
With the Inverse Square Law
Formula, find the x distance;
I.E.12 / I.E.x = ( x / 12 )2 ; 10-6 / 10-7 = x2 / 144;
rearrange the formula to solve for x2 , then x ;
 divide the left side, so
10-6 x 107 = x2 / 144 ;
 so
10 = x2 / 144 ; and solving for x2 ; x2 = 10 times 144
= 1440, so

x = √1440 , and
x = 38 feet.
 So the Interval of fire horns is two times that distance, or =
2 x 38’ = 76 feet
 FORMULAS FOR SOUND PROJECTION, ABSORPTION, AND
ISOLATION:
 IL = 10 log IE / 10-16 SOUND PROJECTION
 IE1 / IE2 = [ d2 / d1 ]2 SOUND ENERGY & DISTANCE
 NR = 10 log [ a2 / a1 ] SOUND REDUCTION BY
ABSORPTION ON SURFACES
 NR = TL+10 log ( a
R/
S ) SOUND REDUCTION BY
AN ISOLATION BARRIER
SOUND REDUCTION
BY A PHYSICAL BARRIER
SOUND ISOLATION
 Sound isolation is the restriction of sound from areas where
it is not desirable. Examples include:
Privacy within spaces
Security
Annoyance in task areas
Annoyance due to vibration & sound of equipment
 Isolation is accomplished by:
Sound reduction by Isolation Noise Reduction
Sealing of openings and cracks
Rerouting and absorption
Vibration absorption
Masking by other sounds
 Sound isolation by use of a barrier that prevents
transmission of sound is effective in providing privacy and
security of speech.
 All construction assemblies used as sound barriers have
been tested for their integrity in reducing transmission of
sound energy.
 Tested assemblies are given a rating, over the six sound
frequencies, called “Sound Transmission Class.” TL, or
Transmission Loss is the number of decibels loudness the
assembly will prevent from passing through.
 Continuous sounds which manage to travel through an
assembly will reverberate within an adjacent space if the
surfaces are reflective, and reinforce the sound level
within that space.
 NOISE REDUCTION BY A BARRIER is measured by the
formula,

NR = TL + 10 log ( a R / S )
where
NR = noise reduction in decibels
TL = sound transmission loss of the barrier in
decibels, found from the chart
a R = the room absorption of the receiving room
S = surface area of the barrier in square feet

Noise Reduction becomes the DIFFERENCE in the
loudness of the original sound, and the loudness of the
sound in the receiving room.
 Do not confuse Noise Reduction with Intensity Loudness.
WALL DESIGN NUMBER 5 TRANSMISSION LOSS
EQUALS 52 DECIBELS AT FREQUENCY OF 500 HZ
5/8"
gypsum bd
both sides
8
decibels
hat channels
@ 2'-0" oc
3" sound attenuation
fiberglass insulation
Initial sound of
60 decibels
emerges at 8 db
2" x 4" studs
@ 16" oc
 But realize that sound that gets through, or around a
barrier into an adjacent room is available to reverberate
within the receiving room and build up because of
reflection.
 Short duration sounds, such as a door closing do not last
long enough to increase the noise in a receiving room. But
steady sound such as that from the operation of machinery
can build up in a receiving room if the surfaces are
reflective, and the noise level will be louder than when it
first enters the space.
 For that reason, room absorption in the reduction of sound
is important in addition to noise reduction by a barrier.
 Example Problem:
 Say an office is 20’ x 20’ x 10’ ceiling height. One wall
separates the room from a mechanical space where
machinery produces 75 decibels of steady sound. The
barrier wall has a Transmission Loss of 50 decibels.
Find the noise level in the room if the room absorption
equals 300 sabins;
NR = TL+10 log ( a R / S )
TL = 50, a R = 300, S = 200 sq.ft.
Initially, the sound enters the room at a level of 25
decibels, because 50 decibels are stopped by the wall, so
NR = 50 + 10 log ( 300/200) = 50 + 10 log (1.5)
NR = 50 + 10(.1761) = 50 + 1.76 51.76 decibels
Noise in the room is 75 – 51.76 = 23.24 decibels
As an alternate to this problem,
Say that we tripled the absorption of the room by adding
more absorptive surfaces, and ar = 900 sabins
Then Noise Reduction = 50 + 10 log (900/200)
NR = 50 + 10 log (4.5) = 50 + 10 (.6532)
NR = 50 + 6.532 = 56.53 decibels,
So the noise level in the room equals 75 – 56.53 = 18.47
decibels.
which represents a reduction of about 20%.
NOW AS A NOVEL EXAMPLE, CONSIDER A SITUATION THAT
INVOLVES A COMBINATION OF
SOUND PROJECTION
SOUND ABSORPTION
SOUND ISOLATION
EXAMPLE PROBLEM A motorcycle at point B backfires
with a Loudness Intensity of 110 decibels at point A.
Find the Intensity Loudness in Room One.
Ignore distances of sound travel in rooms one and Two.
absorbent coefficients of doors same as walls
32'
16'
Room One
ceiling ht. 8'
absorption
coefficients:
floor .44
ceiling .36
walls .90
roadway
absorption
coefficients:
.22
floor
ceiling .26
walls .18
TL=12
wall
D
40'
280'
B
A
TL=52
Room Two
12' ceiling ht.
24'
sound from motorcycle gets into rooms two and
one only through walls C and D
wall
C
A motorcycle at B
backfires with a sound
level of 110 decibels
at point A
PROCEDURE: Find IE at wall C, then IL at wall C, then IL
in Room Two, then IL in Room One . . .
 IL = 10 log IE / 10-16
 110 = 10log IE40 / 10-16
 11 = log IE40 / 10-16
 1011 = IE40 / 10-16 ; IE40 = 10
40'
280'
B
-5
The intensity energy at point A
A
TL=52
wall
C
A motorcycle at B
backfires with a sound
level of 110 decibels
at point A
 IE1 / IE2 = [ d2 / d1 ]2
Intensity ENERGY at C:
 10 -5 / IEC = [280/40] 2 = 49
 IEC = 1/49 x 10 -5 = 2.04 x 10 -2 x 10 -5 = 2.04 x 10
The amount of Intensity Energy at wall C
-7
Then find Intensity Loudness at point C:
 IL C = 10log [2.04 x 10 -7 / 10-16 ] = 10log [ 2.04 x 10 9 ]
 IL C = 10 [ 9.309 ] = 93 decibels at wall C.
Find absorption of Rm 2 in order
To find intensity loudness in Rm 2
ABSORPTION OF ROOM
TWO:
32’
Floor = 24 x 32 x .22 = 169
absorption
coefficients:
.22
floor
ceiling .26
walls .18
TL=12
wall
D
wall
C
Walls = (48+64)12 x .18 = 242
Ceiling = 24x32x.26 = 200
ROOM ABSORPTION = 611
sabins
NOISE REDUCTION IN RM 2:
TL=52
NR = TL + 10log [ 611 / 12x32]
NR = 52 + 10log 1.59
Room Two
12' ceiling ht.
NR = 52 + 10 ( .20 )
24'
SOUND INTENSITY LEVEL IN
NR = 52 + 2.0 = 54 decibels
ROOM TWO EQUALS
93 – 54 = 39 DECIBELS
TOTAL ROOM ABSORPTION IN
ROOM ONE:
16'
Room One
ceiling ht. 8'
Floor: 16 x 32 x .44 = 225
Walls: (32 + 64) x 8 x .90 = 691
Ceiling: 16 x 32 x .36 = 184
TL=12
wall
D
32'
absorption
coefficients:
.44
floor
ceiling .36
walls .90
ROOM ABSORPTION = 1100
SABINS
NOISE REDUCTION IN ROOM
ONE:
NR = TL + 10log [ 1100 / 8 x 32 )
NR = 12 + 10log 4.30 = 12 + 10 x
.633
NR = 12 + 6.33 = 18.33 decibels
SOUND INTENSITY LEVEL IN
ROOM ONE = 39 – 18.33
Equals 20.67 DECIBELS
Other considerations must be given to
components within a building to assure restriction
of sound. Sound travels through openings as does
water through a leak.
Imagine if you filled a room with water, and it
leaks, then sound will also leak out.
 SOUND ENERGY THROUGH OPENINGS
 Since sound energy emanates much like light energy,
openings through a barrier are similar to a water leak –
sound and light get out through openings in barriers and
through cracks between doors and frames.
 Diffracted sound increases as a percentage as the opening
size is reduced. The smallest opening has the largest
percentage of diffracted sound.
Thru-Wall Outlet
Where two adjacent rooms
have electrical or
telephone outlets, thruwall devices should be
avoided because they
create an opening in the
wall.
Some codes, and hotel
franchise companies will
require back to back
outlets to be separated by
a sound barrier to stop
sound and maintain the
fire integrity of the wall.
Two Separate Outlets
A device called an “automatic door bottom” has a sound
insulating strip that drops down when the door is closed,
and retracts when the door is opened.
 MASKING OF SOUND
 Sounds heard, that have no message of interest can be
blocked out by the brain, as no handle is present to garner
attention. Background sounds such as instrumental music
can mask sounds that have a variation of pitch and loudness
if the loudness is minimal, and there is no message.
 But conversation that has a subject of the listener’s
interest, will be picked up by the brain, and those sounds
will mask the background, simply because of the interest in
the message.
 Mechanical equipment that grows unbalanced in its
operation often causes vibrations of movement and sound,
and can be of great annoyance to some people, while with
others it will not.
 Sounds that can be electronically produced called “pink
noise” and “white noise” are useful in masking certain
irritating sounds for some people.
 Privacy and security of sound are essentially one in the
same, except the circumstances for solution might not be
the same. For instance, masking of sound with other types
of sound can solve privacy concerns, but masking is not
necessarily sufficient for security.
 Annoyance of unwanted sound in task areas may be relative
to the unwanted sound. Disturbance during tasks can be
not only an annoyance for the ear, but also for the brain,
particularly when unwanted sounds contain messages that
changes one’s level of concentration.
 The human mind cannot concentrate on more than one
thing simultaneously. Regardless of how well one thinks he
or she can study in a crowded restaurant with a Chai tea
latte and laptop computer, the audible sounds and the
visual activity act doubly to compete with your level of
concentration on a particular subject. Think about it.

Background sounds with no message of interest can
help to mask distractions if the background sounds hold no
interest - - - such as Tchaikovsky’s third symphony.
 PRACTICE PROBLEM:
A 16’ x 20’ x 9’ supply room has absorptive coefficients as
follows:
Walls
.46
Floor
.40
Ceiling
.86
The 20’ wall side of the room is adjacent to another room,
in which mechanical equipment sounds with a steady hum
of 70 decibels. The separating wall has a Transmission Loss
of 46 decibels.
Find the level of sound in the supply room.
 SOLUTION:
 ROOM ABSORPTION
Walls: 648 x .46 = 298
Floor: 320 x .40 = 128
Ceiling: 320 x .86 = 275
total = 701
NR = TL + 10log (701/180) = 46 + 10 log 3.89 =
46 + 10 ( .59) = 46 + 5.9 = 51.9 decibels
Sound in receiving room = 70 – 51.9 = 18.1 decibels