An outside loudspeaker (considered a small source) emits sound waves with a power output of 100 W.
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An outside loudspeaker (considered a small source) emits sound waves with a power output of 100 W.

[From: ] [author: ] [Date: 11-12-21] [Hit: ]
comparing to the reference level,100 W / 1 W/m² = 100 m², ie,r = 2.......
(a) Find the intensity 9.0 m from the source.

(b) Find the intensity level in decibels at that distance.

(c) At what distance would you experience the sound at the threshold of pain, 120 dB?

I found (A) to be correct at 0.098 W/m2
And I found (C) to be correct at 2.82 m....
I need help with (B)...I cant seem to get the correct answer!

-
Thanks for telling me it is chapter 14, that is very important.

The area of a sphere 9 meter in radius is 4πr² = 1018 m²
intensity = 100W /1018 m² = 0.098 W/m²

b) to convert to dB, you need to state a reference.
I'll use dB SIL (Sound Intensity Level), reference to 10^-12 W/m²
0.098 W/m² / 1e-12 W/m² = 0.098e12
dB = 10 log 0.098e12 = 109.8 dB

c) same as b in the reverse
ratio = 10^(120/10) = 1e12
comparing to the reference level, that is
ratio = 1e12 x 1e-12 = 1 W/m²
so at what distance is the sound 1 W/m²
100 W / 1 W/m² = 100 m², ie, where the area of the sphere is that
A = 4πr² = 100
r = 2.82 meters
1
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