Row 1 is 3.3 feet from the speakers and each row is 3.3 feet behind the row in f
ID: 2832682 • Letter: R
Question
Row 1 is 3.3 feet from the speakers and each row is 3.3 feet behind the row in front.
This rock group typically puts on a 2 hour concert.
Note: The equation above would not be valid for indoor concerts as the sound would reverberate from the walls so it would depend on acoustics and size of room. The rate of change with distance would be much less.
Find the sound level for rows 1, 10 and 20.
Find the rate of change of the decibel level as a function of the distance x from the speakers.
Investigate the rate of change of the volume as you move from the front row back to a place where you will not damage your hearing. Make a statement about what you find including the row on which you will purchase tickets.
THis link will give a decibel chart: http://www.dangerousdecibels.org/education/information-center/decibel-exposure-time-guidelines/
Explanation / Answer
Given,
D = 120 + 20log(3.3/x), where x is the distance from the source
(i) For first row, x = 3.3, hence
D = 120 + 20*log(3.3/3.3)
= 120 + 20*log(1)
= 120 DB
(ii) For 10th row, x = 3.3*10 = 33, hence
D = 120 + 20*log(3.3/33)
= 120 + 20*log(1/10)
= 120 + 20*(-1)
= 100 DB
(iii) For 20th row, x = 3.3*20 = 66, hence
D = 120 + 20*log(3.3/66)
= 120 + 20*log(1/20)
= 120 + 20*(-1.301)
= 93.98 DB
Rate of change of Decible level as a function of x, is
dD/dx = d(120+20*log(3.3/x))/dx
= 20*d(log(1/x))/dx
= 20*d(-log(x))/dx
= -20/x
For the safe distance where we don't hurt ears while listening for 2 hours is 91 DB, hence
91 = 120 + 20*log(3.3/x)
-29 = 20*log(3.3/x)
10^(-29/20) = 3.3/x
x = 3.3/10^(-29/20)
= 3.3*10^(1.45)
= 93.006 or 93/3.3 = 29th row
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