intensity decay

[chiddler]

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A 50 dB noise is how much further away from a 100 dB noise? (if they are the same in all other respects)

I need math help! This seems so simple but i'm not really understanding how it's done.

so 100 to 50 on a 10*log scale is 10^5 decreasein intensity. Intensity, I = P / pi*r^2.

 

[milski]

Intensity decreases as the square of the distance. You already figured out correctly that you're looking at 10^5 decrease. For that the distance will have to increase sqrt(10^5)=3*10^2 times. So if you were a meter away, you'll have to be 300 meters away. Or if you were 5 meters away, you'll have to be at 1500 meters.

Note that you cannot give an absolute distance (in meters) without knowing how far you were when the intensity was 100 dB, you can only give a relative distance, as in 300 times the original distance.