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Physics Practice Test 3: Acoustics Questions and Answers - Prof. Scott H. Hawley, Assignments of Physics

Practice test questions and answers from chapter 7, 9, and 13 of a physics textbook, focusing on acoustics. Topics include sound diffusion, absorption, reverberation time, and measuring absorption coefficients. Students can use this document to review and prepare for exams, quizzes, or assignments.

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Pre 2010

Uploaded on 12/12/2009

absentvirtue
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PHYSICS 2010 “Practice” Test 3 (Ev Ch 7, 9, & 13)
1. Sound at “normal” incidence is scattered with nearly perfect 180 degree diffusion from a
single monocylindrical diffuser. So why doesn’t everybody just make an entire wall to be one
giant monocylinder? (Answer in terms of the diffusion properties, not so much the
construction or aesthetic aspects.)
a. Because in coming incidence is not limited to nominal incidence, and the diffusion
properties of a monocylinder are much worse at other angles. Also a very large cylinder
would have a large radius of curvature, making it fairly flat to high frequency sounds,
thus not offering much diffusion at these frequencies.
2. In an example in the book, the absorption due to a coca cola bottle acting as a Helmholtz
absorber was measured. It was found to have an absorption of 5.9 Sabins -- equivalent to a
whole adult person – at resonant frequency was 185 Hz, and its Q Value was 276. Why was
the coke bottle judged to not have a significant effect on the acoustics of a room?
a. Because the bandwidth of the resonator is
b. ∆f=f0/Q=185/276=.67 Hz
c. This is such a narrow peak that it wouldn’t have a noticeable effect on the frequency
response of the room.
3. What assumptions are implied when we speak of a “reverberation time” for a room?
a. High modal density, which implies uniformity of distribution of sound energy and
randomization of direction of propagation.
4. What are two reasons why it is easier to measure reverb times at high frequencies than at low
frequencies?
a. Mode decay variations are less prominent at higher modal density
b. Low frequency sounds can be difficult to generate and thus it is more difficult to obtain
good signal to noise ratio at low frequencies.
5. You’re noticing a problem with flutter echo in your pimped out wii playing basement room.
Your dad won’t let you splay the walls and you’re already maxed out on absorption from the
shag carpet and your mom’s beanie babies on the shelves. What else can you add to the room
in order to control the flutter without degrading the acoustical quality of the room
a. Add Diffusion
6. Hey intern! Build me a bass trap to attenuate this 67Hz room mode! Then get me some Coffee!
Explain how you would do this including dimensions, materials used, and their placements.
a. Build it out of fairly dense wood and make it a quarter wavelength deep,
b. which at 67 Hz is
i. 1140 / 67 / 4 = 4.25 Ft
c. Have coming out from the wall and put porous absorber like glass fiber across the face
and have some panels of the absorber arranged inside the trap as well.
7. Your best bud says, “hey Hombre, we can use some of this egg create foam stuff to make
absorbers and outfit our studio! Describe two different ways you could measure the
absorption coefficient of the egg-crate foam stuff.
a. The reverberation chamber method: Measure RT60 of a room with and without the
foam, then use the Sabine equation and the area of the foam to algebraically solve the
absorption coefficient.
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1. Sound at “normal” incidence is scattered with nearly perfect 180 degree diffusion from a single monocylindrical diffuser. So why doesn’t everybody just make an entire wall to be one giant monocylinder? (Answer in terms of the diffusion properties, not so much the construction or aesthetic aspects.) a. Because in coming incidence is not limited to nominal incidence, and the diffusion properties of a monocylinder are much worse at other angles. Also a very large cylinder would have a large radius of curvature, making it fairly flat to high frequency sounds, thus not offering much diffusion at these frequencies. 2. In an example in the book, the absorption due to a coca cola bottle acting as a Helmholtz absorber was measured. It was found to have an absorption of 5.9 Sabins -- equivalent to a whole adult person – at resonant frequency was 185 Hz, and its Q Value was 276. Why was the coke bottle judged to not have a significant effect on the acoustics of a room? a. Because the bandwidth of the resonator is b. ∆f=f 0 /Q=185/276=.67 Hz c. This is such a narrow peak that it wouldn’t have a noticeable effect on the frequency response of the room. 3. What assumptions are implied when we speak of a “reverberation time” for a room? a. High modal density, which implies uniformity of distribution of sound energy and randomization of direction of propagation. 4. What are two reasons why it is easier to measure reverb times at high frequencies than at low frequencies? a. Mode decay variations are less prominent at higher modal density b. Low frequency sounds can be difficult to generate and thus it is more difficult to obtain good signal to noise ratio at low frequencies. 5. You’re noticing a problem with flutter echo in your pimped out wii playing basement room. Your dad won’t let you splay the walls and you’re already maxed out on absorption from the shag carpet and your mom’s beanie babies on the shelves. What else can you add to the room in order to control the flutter without degrading the acoustical quality of the room a. Add Diffusion 6. Hey intern! Build me a bass trap to attenuate this 67Hz room mode! Then get me some Coffee! Explain how you would do this including dimensions, materials used, and their placements. a. Build it out of fairly dense wood and make it a quarter wavelength deep, b. which at 67 Hz is i. 1140 / 67 / 4 = 4.25 Ft c. Have coming out from the wall and put porous absorber like glass fiber across the face and have some panels of the absorber arranged inside the trap as well. 7. Your best bud says, “hey Hombre, we can use some of this egg create foam stuff to make absorbers and outfit our studio! Describe two different ways you could measure the absorption coefficient of the egg-crate foam stuff. a. The reverberation chamber method: Measure RT60 of a room with and without the foam, then use the Sabine equation and the area of the foam to algebraically solve the absorption coefficient.

b. The tone burst method: Measure the intensity of direct sound and the intensity of sound reflected off the foam. The ratio of these two intensities is alpha, the energy absorption coefficient. c. Impedance tube method (described in text)

8. You’ve got some plywood with a surface density of 2 lb/ft^2 , and you want to make a piece o wood panel absorber to absorb frequencies around 130 Hz. How deep should you make the airspace? a. F 0 =170/√(md) b. So md = (170/f 0 )^2 c. d=1/m(170/f 0 )^2 d. = ½(170/130)^2 e. =.86 inches 9. Stephen has some pegboard with 5% perforation and thickness of 1” (from stacking 8 sheets of 1/8” pegboard). The hole diameter is .25”. He wants to mount it near a wall to make a Helmholtz absorber with a resonant frequency of 96 Hz. How deep should he make the air gap? a. F 0 =200√p/dt b. Where t= thickness + .8 (hole diameter) = 1” + .8(.25) c. D=(p/t)(200/f 0 )^2 = 5/1.2(200/96)^2 = (4.17)(4.34) = 18. d. 18. 10. Name 4 characteristics of a perfectly diffuse sounds field. a. The frequency and spatial irregularities obtained via steady state measurements must be negligible b. Beats in decay should be negligible c. Decays should be perfectly exponential d. The decays should be spatially homogenous, i.e. reverb times should be the same everywhere e. The decays should be isotropic, i.e. dependent of the mic direction f. The character of the decay will be essentially the same for all frequencies. 11. You mount some porous absorber away from the wall with an air gap of 4”. Name the three lowest frequencies at which absorption will be boosted the most due to this air gap. a. f=v/λ b. 4” = .33333 ft, so λ = 1.333333 ft c. ….there is more to this problem.. 12. Sound is recorded in one room, and then played back in a room with a longer reverberation time. How does the reverberation time of the reproduced (played back) sound in the new room compare with the reverb time of the two rooms. a. It will be longer than the TR of either room, but closer to that of the room with the larger TR. 13. Write a paragraph describing how you would measure the reverb time in Neely dining hall. a. Open