2.  PRACTICE PROBLEM ONE: From Previous lecture A 16’ x 20’ x 9’ room has absorptive coefficients as follows: Ignore doors & windows. Walls.30 Floor.25 Ceiling.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.

3. 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 total = 402.4 sabins T =.05 x (2880 / 402.4) =.36 seconds NEW Wall area =648 x.46=298.08 Floor area =320 x.40=128.0 Ceiling area = 320 x.86=275.2 total = 701.28 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%

4. 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/cm 2, 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.

5.  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 = 10 10 ;  anti-log of the right side, log (I.E. 12 / 10 -16 ) is simply, I.E. 12 / 10 -16, so the resulting equation is  10 10 = I.E. 12 / 10 -16, rearrange and solve for I.E. 12 ; I.E. 12 = 10 10 x 10 -16 = 10 -6 watts/cm2

6.  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 10 9 = I.E. x / 10 -16 ; rearrange and I.E. x = 10 9 x 16 -16 ;  so the amount of energy required to produce 90 decibels at the x distance; I.E. x = 10 -7 watts/cm 2  Realize here that 10 -7 is a smaller number than 10 -6

7.  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 = x 2 / 144; rearrange the formula to solve for x 2, then x ;  divide the left side, so 10 -6 x 10 7 = x 2 / 144 ;  so 10 = x 2 / 144 ; and solving for x 2 ; x 2 = 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

8.  FORMULAS FOR SOUND PROJECTION, ABSORPTION, AND ISOLATION:  IL = 10 log IE / 10 -16 SOUND PROJECTION  IE 1 / IE 2 = [ d 2 / d 1 ] 2 SOUND ENERGY & DISTANCE  NR = 10 log [ a 2 / a 1 ] SOUND REDUCTION BY ABSORPTION ON SURFACES  NR = TL+10 log ( a R / S ) SOUND REDUCTION BY AN ISOLATION BARRIER

9. SOUND REDUCTION BY A PHYSICAL BARRIER

10. 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

11.  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.

12.  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.

14. WALL DESIGN NUMBER 5 TRANSMISSION LOSS EQUALS 52 DECIBELS AT FREQUENCY OF 500 HZ

15.  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.

16.  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 ) 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

17. As an alternate to this problem, Say that we tripled the absorption of the room by adding more absorptive surfaces, and a r = 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%.

18. NOW AS A NOVEL EXAMPLE, CONSIDER A SITUATION THAT INVOLVES A COMBINATION OF SOUND PROJECTION SOUND ABSORPTION SOUND ISOLATION

19. 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. PROCEDURE: Find IE at wall C, then IL at wall C, then IL in Room Two, then IL in Room One... roadway

20.  IL = 10 log IE / 10 -16  110 = 10log IE 40 / 10 -16  11 = log IE 40 / 10 -16  10 11 = IE 40 / 10 -16 ; IE 40 = 10 -5 The intensity energy at point A  IE 1 / IE 2 = [ d 2 / d 1 ] 2 Intensity ENERGY at C:  10 -5 / IE C = [280/40] 2 = 49  IE C = 1/49 x 10 -5 = 2.04 x 10 -2 x 10 -5 = 2.04 x 10 -7 The amount of Intensity Energy at wall C 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.

21. ABSORPTION OF ROOM TWO: Floor = 24 x 32 x.22 = 169 Walls = (48+64)12 x.18 = 242 Ceiling = 24x32x.26 = 200 ROOM ABSORPTION = 611 sabins NOISE REDUCTION IN RM 2: NR = TL + 10log [ 611 / 12x32] NR = 52 + 10log 1.59 NR = 52 + 10 (.20 ) NR = 52 + 2.0 = 54 decibels SOUND INTENSITY LEVEL IN ROOM TWO EQUALS 93 – 54 = 39 DECIBELS 32’ Find absorption of Rm 2 in order To find intensity loudness in Rm 2

22. TOTAL ROOM ABSORPTION IN ROOM ONE: Floor: 16 x 32 x.44 = 225 Walls: (32 + 64) x 8 x.90 = 691 Ceiling: 16 x 32 x.36 = 184 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

23. 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.

24.  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.

25. Thru-Wall Outlet Two Separate Outlets Where two adjacent rooms have electrical or telephone outlets, thru- wall 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.

26. 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.

27.  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.

28.  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.

29.  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.

30.  SOLUTION:  ROOM ABSORPTION Walls: 648 x.46 = 298 Floor: 320 x.40 = 128 Ceiling: 320 x.86 = 275total = 701 NR = TL + 10log (701/180) = 46 + 10 log 3.89 = 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