posted by Reed .
Point A is 5 m from a loudspeaker. At point B, the loudspeaker sounds half as loud as at point A. How far is point B from the loudspeaker?
I get 50 m, but the answer is supposed to be 15.8 m. I need to know what I'm doing wrong.
By my (apparently flawed) reasoning:
Point A is 10 dB louder than Point B because +10 db = ~2x louder
+10 dB = *10 Intensity (I), because the dB scale is log 10
Therefore, I(A) = 10*I(B) and I(A)/I(B) = 10
Using the formula I(A)/I(B) = r(B)/r(A), I find that
10 = r(B)/r(A)
Plug in 5 m for the value of r(A), and solve for r(B), which comes out to be 50 m.
So what am I screwing up?
Thanks for your help.
It is not clear to me what they mean by "sounding" half as loud. The sound power per area (intensity) will be down by a factor of 2 if you move sqrt2 times farther away, to a distance of 7.07 meters in this case. The sound level there is 3 dB down from Point A.
At 15.8 m the distance is 3.16 times farther and the sound power is 3.16^2 = 10 times less, which is 10 dB down. Apparently they are saying that 10 dB down "sounds like" half as loud. Other websites I have looked at also make the statement that 10 dB down sounds half as loud, so I guess it is an accepted rule of thumb. i had never heard it before.