Transcript Recitation #11 - AMMON Template

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=fλ
or λ=C/f
C=velocity (fps)
f=frequency (hz)
λ=wavelength (ft)
Lower frequency: longer wavelength
1. Calculate the wavelengths at the
standard octave band center frequencies
for sound moving 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. 730
Calculated for “center band frequencies”
125
250
500
1000
2000
Pay specific attention to 125 hz and 500 hz
4000
2. Will a standing wave occur at the 500 Hz
frequency in a room with parallel walls that
are 28’-3” apart?
Sound travels at different
►
C=fλ or λ=C/f
speeds through various media.
Media Speed (C)
►
C=velocity (fps)
Air:
1,130 fps
Water: 4,625 fps
►
f=frequency (hz)
Wood: 10,825 fps
Steel: 16,000 fps
►
λ=wavelength (ft)
λ = 2.26’
28.25/2.26=12.5
Not an increment, no standing wave.
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”
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
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
► a.
63 db
► b. 46 db
► c. 96 db
► d. 86 db
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.) IL1=10 log (I/I0)
IL=10 log(2.1x10-10/10-16)
IL=63.2 or 63 db
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
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.
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
Absorption coefficient
A=Sα
α=Iα/Ii
A=total absorption (sabins)
S=surface area (ft2 or m2)
α=absorption coefficient
sabins
(m2)=
10.76 sabins (sf)
(w/cm2)
(w/cm2)
α=absorption coefficient
Iα=sound power intensity absorbed
Ii=sound power impinging on material
1.0 is total absorption
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
Average
Absorption
51=592a/1-a
αavg=ΣA/S
a=.08
A=Sα
Tr=.05V/Sa = 1.02
Period of time required for a 60 db drop after sound
source stops
TR= K x V/ΣA
(metric)
TR: reverberation time (seconds)
K: 0.05 (English) (0.049 in RR-7) or 0.16
V: volume (ft3 or m3)
ΣA: total room absorption, sabins (ft2 or m2)
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
►=10log
(EAz/EA1)
(174/46)=5.8db or 6db
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)
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
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
5. Evaluate the design for this
proposed symphony hall...
A.
Perform the reverberation time calculation at 500Hz based on the following:
A.
Floor space dimensions:
B.
Ceiling Height:
30’
C.
Seating capacity:
840
D. Seat type:
52’x140’
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 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)
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
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
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