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Working with sound?
An announcer is using a bull horn with a full opening angle of 12ยบ to speak to a crowd. Someone in the crowd measures the volume at 104 dB. How loud would the announcer be without their bull horn and at twice the distance on the same decibel scale?
i wish i knew the stupid distances. god i hate physics with a fiery passion.
perhaps it has something to do with the following: Amplification Factor= 2
/(sin theta)^2 ?
when comparing sounds with and without the horn.
saucerful...fissiks ain't so bad...just hard to spell
Here is the idea, sound intensity decreases as 1/r^2 if you double the distance the intensity decreases by 4, and increases with power.
I = P/Area
if we have the power distributed over a full sphere then area is area of sphere = 4Pir^2, but here we have only 12 deg, so that is only a part of a sphere so Area = 12/360 Area
I1 = 360P/(12 x 4 Pi r^2)
Now we know I2 is twice the distance, but P is the same (I assume the "bull horn" is simply a cone shaped object), so we assume the mouth emits at say 90 deg so
I2 = 360 P/(90 4 Pi (2r)^2
Now divide the two
I1/I2 = 90 x 4/ 12 = 30
now find the dB difference
B = 10 dB log(30) = 14.8 dB
so we would find the volume 104-14.8 dB =89.2dB
Note that I had to make some assumptions about the bull horn and exactly what full opening angle the mouth is...but it is logical...
Also, you may have used B = 10dBlog(I/Io) where Io = 1E-12, I have taken a shortcut and simply replaced B with B2-B1 and Io with I2
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