Acoustics Worksheet Acoustics Worksheet Answer KeyAnswer Key

1. Calculate the wavelengths at the 1. Calculate the wavelengths at the standard octave band center standard octave band center frequencies for sound moving frequencies for sound moving

through air.through air.Distance between similar points on a successive wave

C=fλ or λ=C/f

C=velocity (fps)f=frequency (hz)λ=wavelength (ft)

Lower frequency: longer wavelength

1. Calculate the wavelengths at the 1. Calculate the wavelengths at the standard octave band center standard octave band center frequencies for sound moving frequencies for sound moving

through air.through air.Hz 62.5 125 250 500 1000 2000 4000 8000

Ft 18.1 9.04 4.25 2.26 1.13 0.57 0.28 0.14

See S&R p. 730Calculated for “center band frequencies”

125 250 500 1000 2000 4000

Pay specific attention to 125 hz and 500 hz

2. Will a standing wave occur at the 500 2. Will a standing wave occur at the 500 Hz frequency in a room with parallel Hz frequency in a room with parallel

walls that are 28’-3” apart?walls that are 28’-3” apart?► C=fλ or λ=C/f ► C=velocity (fps)► f=frequency (hz)► λ=wavelength (ft)

Sound travels at different speeds through

various media.Media Speed (C)Air: 1,130 fps

Water: 4,625 fps

Wood: 10,825 fps

Steel: 16,000 fpsλ = 2.26’ 28.25/2.26=12.5

Not an increment, no standing wave.

Standing WavesStanding Waves

If a room dimension is an integer multiple of a wave length, a standing wave forms for that wave length

Sound energy cancels itself

500 hz 2’-3” … 4x = 9’-0” 5x = 11’-3”

8x = 18’-0” 10x = 22’-6”

3. For a sound source with a directivity 3. For a sound source with a directivity factor (Q) equal to 4, what shape best factor (Q) equal to 4, what shape best

describes the distribution pattern of the describes the distribution pattern of the sound?sound?

►¼ sphere¼ sphere

S&R pg. 766 fig. 18.8

4. For the following intensities, 4. For the following intensities, determine the intensity level (db) if determine the intensity level (db) if

IIoo=10=10-16-16w/cmw/cm22

(show calculations, round off to whole numbers)(show calculations, round off to whole numbers)►a. 2.1x10a. 2.1x10-10-10

►b. 4.1x10b. 4.1x10-12-12

►c. 3.7x10c. 3.7x10-7-7 ►d. 3.7x10d. 3.7x10-8 -8

Extreme range dictates the use of logarithms

IL=10 log (I/I0)

IL: intensity level (dB)I: intensity (W/cm2)I0: base intensity (10-16 W/cm2, hearing threshold)Log: logarithm base 10►a. 63 dba. 63 db

►b. 46 dbb. 46 db►c. 96 db c. 96 db ►d. 86 dbd. 86 db

a.) IL1=10 log (I/I0)

IL=10 log(2.1x102.1x10-10-10/10-

16) IL=63.2 or 63 db

5. From the previous problem, combine the 5. From the previous problem, combine the following and calculate the new intensity following and calculate the new intensity levels (show calculations and round off to levels (show calculations and round off to

whole numbers).whole numbers).► i) a+a 10 log i) a+a 10 log

(2.1x10(2.1x10-10-10+2.1x10+2.1x10-10-10)/1x10)/1x10-16-16

=66.2 or 66 db=66.2 or 66 db

ii) a+a+b= 66 dbii) a+a+b= 66 db

iii) c+d = 96 dbiii) c+d = 96 db

iv) d+d+a = 89 dbiv) d+d+a = 89 db

6. In light of the fact that a 3 db difference is 6. In light of the fact that a 3 db difference is barely noticeable, what can you conclude barely noticeable, what can you conclude about combining sound sources of equal about combining sound sources of equal

intensity?intensity?►The addition is just barely noticeable.The addition is just barely noticeable.

Part 2. Acoustical AnalysisPart 2. Acoustical Analysis

►1. For a room 10’x12’x8’ with a room 1. For a room 10’x12’x8’ with a room factor equal to 51 ftfactor equal to 51 ft33. What is the . What is the average absorption coefficient for the average absorption coefficient for the space? What is the reverberation time space? What is the reverberation time for this space?for this space?Absorption

A=Sα

A=total absorption (sabins) S=surface area (ft2 or m2)α=absorption coefficient

sabins (m2)= 10.76 sabins (sf)

Absorption coefficient

α=Iα/Ii

α=absorption coefficient Iα=sound power intensity

absorbed (w/cm2)Ii=sound power impinging on

material (w/cm2)

1.0 is total absorption

1. For a room 10’x12’x8’ with a room factor 1. For a room 10’x12’x8’ with a room factor equal to 51 ftequal to 51 ft33. What is the average absorption . What is the average absorption

coefficient for the space? What is the coefficient for the space? What is the reverberation time for this space?reverberation time for this space?

Total area= S=2(10x12) + 8(2)Total area= S=2(10x12) + 8(2)(10+12)=592(10+12)=592

R=Sa/1-aR=Sa/1-a

51=592a/1-a51=592a/1-a

a=.08a=.08

TTrr=.05V/Sa = 1.02=.05V/Sa = 1.02

Average Absorption

αavg=ΣA/S

A=Sα

Period of time required for a 60 db drop after sound source stops

TR= K x V/ΣA

TR: reverberation time (seconds)K: 0.05 (English) (0.049 in RR-7) or 0.16

(metric)V: volume (ft3 or m3)ΣA: total room absorption, sabins (ft2 or

m2)

2. The original total absorption in a room is 46 sabins. 2. The original total absorption in a room is 46 sabins. The room was redesigned and the new total The room was redesigned and the new total

absorption in the room in 174 sabins. What is the absorption in the room in 174 sabins. What is the increase in total noise reduction accomplished by this increase in total noise reduction accomplished by this

change?change?►NR=10log (EANR=10log (EAzz/EA/EA11))

►=10log (174/46)=5.8db or 6db=10log (174/46)=5.8db or 6db

3. Room A has a total absorption of 75 sabins. Room B 3. Room A has a total absorption of 75 sabins. Room B has a total absorption of 180 sabins. These rooms are has a total absorption of 180 sabins. These rooms are separated by a 12’X10’ wall with a transmission loss separated by a 12’X10’ wall with a transmission loss

of 23 decibels. Determine the noise reduction in room of 23 decibels. Determine the noise reduction in room B for sounds originating in room A.B for sounds originating in room A.

Combined effect of TL and absorption

NR=TL-10 Log (S/AR)

NR: noise reduction (db)TL: transmission loss (db)S: area of barrier wall (ft2)AR: total absorption of receiving

room (sabins, ft2)

3. Room A has a total absorption of 75 sabins. Room B 3. Room A has a total absorption of 75 sabins. Room B has a total absorption of 180 sabins. These rooms are has a total absorption of 180 sabins. These rooms are separated by a 12’X10’ wall with a transmission loss separated by a 12’X10’ wall with a transmission loss

of 23 decibels. Determine the noise reduction in room of 23 decibels. Determine the noise reduction in room B for sounds originating in room A.B for sounds originating in room A.

Combined effect of TL and absorption

NR=TL-10 Log (S/AR)NR: noise reduction (db)

TL: transmission loss (db)S: area of barrier wall (ft2)

AR: total absorption of receiving room

(sabins, ft2)

NR=TL-10 Log (S/AR)NRa-b=23- 10Log (120/180)=24.8 or 25dbNRb-a=23- 10Log (120/75)=21db

4. What noise criteria curve is 4. What noise criteria curve is recommended for a small auditorium recommended for a small auditorium (50 people max) with no amplification (50 people max) with no amplification

system?system?►NC 25-30NC 25-30

See S&R pg. 821 fig. 19.8See S&R pg. 821 fig. 19.8

5. Evaluate the design for this 5. Evaluate the design for this proposed symphony hall...proposed symphony hall...

A. Perform the reverberation time calculation at 500Hz based on the following:

A. Floor space dimensions: 52’x140’

B. Ceiling Height: 30’

C. Seating capacity: 840

D. Seat type: upholstered

E. Aisles: 2@5’ wide x 140’ long

F. Floor covering: Linoleum on concrete

G. Aisle covering: Heavy carpet on concrete

H. Walls: Tongue & Groove Cedar

I. Ceiling: ½” Gyp. Board nailed to 2x4

J. Stage opening: 20’x35’

K. Air Temperature/RH: 72F/50% RH

b. Comment on the proposed b. Comment on the proposed design (what did you design (what did you

find/discover)?find/discover)?►RRT500T500 is appropriate is appropriate

►140’ is too long, add balcony or 140’ is too long, add balcony or change room dimensionchange room dimension

►RRT125T125 is too low (should be 25-50% is too low (should be 25-50% longer than Rlonger than RT500T500))

c. The client has asked about the c. The client has asked about the suitability of this space for listening to suitability of this space for listening to religious music. Based on this current religious music. Based on this current

design, is it adequate?design, is it adequate?►RRTT is too short, should be 1.7 seconds is too short, should be 1.7 seconds

d. Based on this current design, how much additional d. Based on this current design, how much additional absorption (Sabins, ft2) at 500Hz would be needed to absorption (Sabins, ft2) at 500Hz would be needed to

comfortably accommodate a performance where comfortably accommodate a performance where speech only is being presented. (Ignore any possible speech only is being presented. (Ignore any possible echo effects and possible loud speaker reinforcement echo effects and possible loud speaker reinforcement

systems).systems).

►Needed RNeeded RTT=0.85sec=0.85sec

►10702/0.85=12591 needed for R10702/0.85=12591 needed for RTT

7669 existing7669 existing

4922 additional Sabins 4922 additional Sabins neededneeded

e. What will be the effect on reverberation time e. What will be the effect on reverberation time at 500Hz if the wall material is changed to ½” at 500Hz if the wall material is changed to ½” Gypsum Board nailed to 2x4 studs and the Gypsum Board nailed to 2x4 studs and the carpet is replaced with linoleum (calculate the carpet is replaced with linoleum (calculate the new reverberation time)? new reverberation time)?

7669 exist7669 exist

-196 remove carpet-196 remove carpet

-1514 remove tongue and groove cedar-1514 remove tongue and groove cedar

+46 add linoleum+46 add linoleum

+541 add Gyp. Board+541 add Gyp. Board

6542 Sabins =new total6542 Sabins =new total

RRTT=10702/6542 = 1.64 sec=10702/6542 = 1.64 sec