1. No category

MECH 482- Noise Control
Tutorial 6
1) A window of area 1 m2 in the side of a building has its centre 2 m above the ground
and is radiating sound into the community. The nearest residence is 750 m away. The
average sound intensity over the outside of the window is 0.1W/m2 in the 500 Hz octave
band. Assuming that only noise in the 500 Hz octave band is of interest, calculate the
sound pressure level at the nearest residence if the ground surface between the window
and the residence is concrete. Assume a community location 1.5 m above the ground.
State any other assumptions that you make.
Solution:
2) A room measure 20m x 13m x 5.5m (LxWxH) Assume the speed of sound is 343m/s
a) Calculate the first 3 modes in order of increasing frequency
b) Calculate the number of modes (Nf) up to 16 Hz
c) Calculate the modal density at 16 Hz
Solution:
a)
2
2
2
c   n x   n y   n z  
f         
2  L   W   H  


2
2
2
343m / s   1   0   0  
f100 
 8.875 Hz


  20   13   5.5  
2


2
2
2
343m / s   0   1   0  
f 010 
 13.2 Hz


  20   13   5.5  
2


2
2
2
343m / s   1   1   0  
f110 
 15.7 Hz


  20   13   5.5  
2


b)
N
4 (16 Hz ) 3 1430m 3  (16 Hz ) 2 833m 2 16 Hz (154m)


=3
8(343)
3(343) 3
4(343) 2
c)
dN 4 (16 Hz ) 2 (1430m 3 )  (16 Hz )833m 2
154m



3
2
df
8(343m / s )
(343m / s )
2((343m / s )
3. An electric motor produces a steady state reverberant sound level of 74dB re 20μPa in
a room 3.05 × 6.10 × 15.24m3. The measured reverberation time of the room is 2
seconds.
(a) What is the acoustic power output of the motor in dB re 10-12 W?
(b) How much additional Sabine absorption (in m2) must be added to the room to lower
the reverberant field by 10dB?
(c) What will be the new reverberation time of the room?
Solution:
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