1 / 22

Acoustics Worksheet Answer Key

Acoustics Worksheet Answer Key. 1. Calculate the wavelengths at the standard octave band center frequencies for sound moving through air. Distance between similar points on a successive wave C=f λ or λ =C/f C=velocity (fps) f=frequency (hz) λ =wavelength (ft)

korene
Download Presentation

Acoustics Worksheet Answer Key

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Acoustics Worksheet Answer Key

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

  3. 1. Calculate the wavelengths at the standard octave band center frequencies for sound moving through air. See S&R p. 730 Calculated for “center band frequencies” 125 250 500 1000 2000 4000 Pay specific attention to 125 hz and 500 hz

  4. 2. Will a standing wave occur at the 500 Hz frequency in a room with parallel 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 fps λ = 2.26’ 28.25/2.26=12.5 Not an increment, no standing wave.

  5. Standing 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”

  6. 3. For a sound source with a directivity factor (Q) equal to 4, what shape best describes the distribution pattern of the sound? • ¼ sphere S&R pg. 766 fig. 18.8

  7. 4. For the following intensities, determine the intensity level (db) if Io=10-16w/cm2(show calculations, round off to whole numbers) • a. 2.1x10-10 • b. 4.1x10-12 • c. 3.7x10-7 • d. 3.7x10-8 • 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 db • b. 46 db • c. 96 db • d. 86 db a.) IL1=10 log (I/I0) IL=10 log(2.1x10-10/10-16) IL=63.2 or 63 db

  8. 5. From the previous problem, combine the following and calculate the new intensity levels (show calculations and round off to whole numbers). • i) a+a 10 log (2.1x10-10+2.1x10-10)/1x10-16 =66.2 or 66 db • ii) a+a+b= 66 db • iii) c+d = 96 db • iv) d+d+a = 89 db

  9. 6. In light of the fact that a 3 db difference is barely noticeable, what can you conclude about combining sound sources of equal intensity? • The addition is just barely noticeable.

  10. Part 2. Acoustical Analysis • 1. For a room 10’x12’x8’ with a room factor equal to 51 ft3. What is the average absorption coefficient for the space? What is the reverberation time 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

  11. 1. For a room 10’x12’x8’ with a room factor equal to 51 ft3. What is the average absorption coefficient for the space? What is the reverberation time for this space? Total area= S=2(10x12) + 8(2)(10+12)=592 R=Sa/1-a 51=592a/1-a a=.08 Tr=.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)

  12. 2. The original total absorption in a room is 46 sabins. The room was redesigned and the new total absorption in the room in 174 sabins. What is the increase in total noise reduction accomplished by this change? • NR=10log (EAz/EA1) • =10log (174/46)=5.8db or 6db

  13. 3. Room A has a total absorption of 75 sabins. Room B has a total absorption of 180 sabins. These rooms are separated by a 12’X10’ wall with a transmission loss of 23 decibels. Determine the noise reduction in room 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)

  14. 3. Room A has a total absorption of 75 sabins. Room B has a total absorption of 180 sabins. These rooms are separated by a 12’X10’ wall with a transmission loss of 23 decibels. Determine the noise reduction in room 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 25db NRb-a=23- 10Log (120/75)=21db

  15. 4. What noise criteria curve is recommended for a small auditorium (50 people max) with no amplification system? • NC 25-30 • See S&R pg. 821 fig. 19.8

  16. 5. Evaluate the design for this proposed symphony hall... • Perform the reverberation time calculation at 500Hz based on the following: • Floor space dimensions: 52’x140’ • Ceiling Height: 30’ • Seating capacity: 840 • Seat type: upholstered • Aisles: 2@5’ wide x 140’ long • Floor covering: Linoleum on concrete • Aisle covering: Heavy carpet on concrete • Walls: Tongue & Groove Cedar • Ceiling: ½” Gyp. Board nailed to 2x4 • Stage opening: 20’x35’ • Air Temperature/RH: 72F/50% RH

  17. b. Comment on the proposed design (what did you find/discover)? • RT500 is appropriate • 140’ is too long, add balcony or change room dimension • RT125 is too low (should be 25-50% longer than RT500)

  18. c. The client has asked about the suitability of this space for listening to religious music. Based on this current design, is it adequate? • RT is too short, should be 1.7 seconds

  19. d. Based on this current design, how much additional absorption (Sabins, ft2) at 500Hz would be needed to comfortably accommodate a performance where speech only is being presented. (Ignore any possible echo effects and possible loud speaker reinforcement systems). • Needed RT=0.85sec • 10702/0.85=12591 needed for RT 7669 existing 4922 additional Sabins needed

  20. e. What will be the effect on reverberation time at 500Hz if the wall material is changed to ½” Gypsum Board nailed to 2x4 studs and the carpet is replaced with linoleum (calculate the new reverberation time)? 7669 exist -196 remove carpet -1514 remove tongue and groove cedar +46 add linoleum +541 add Gyp. Board 6542 Sabins =new total RT=10702/6542 = 1.64 sec

More Related