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What is the power output of the speaker, and how far away should it be?

At a rock concert, a dB meter registered 130 dB when placed 2.8 m in front of a loudspeaker on the stage. (a) What was the power output of the speaker, assuming uniform spherical spreading of the sound and neglecting absorption in the air. (b) How far away would the sound level be a somewhat reasonable 90 dB?

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1. 130dB = 63.2 Pa Assuming it was bass.. about 30Hz Speed of sound at 343 m/s = Vs half length = 343 m*s^-1 / 60^-1 = Lh pressure = int(63.2Pa*sin(x/Lh*pi),0,Lh)=34.73 Pa Volume in that sphere slice = 4/3 * pi ( (d+Lh/2)^3 - (d-Lh/2)^3) = 758.84 m^3 Time to fill it Lh / Vs = 1/60 s so filling that volume to 34.73Pa in 1/60 s with the density of air 1.2041 kg/m3 And I just got bored...
2. If the dB are sound-intensity dB (dB-SIL), then the reference level is I0 = 10^-12 W/m² 130 = 10*log(I/I0) = 10*[log(I) + 12] 13 = log(I) + 12 log(I) = 1, I = 10 W/m² This is covering an area of 4*π*(2.8²), so the total power is 10*4*π*(2.8²) W. 985 W 90 dB = 10*[log(I) + 12] log(I) = -3 I = 10^-3 W/m² The distance is such that the area covered by 985 W yields 10^-3 W/m²: 985/(4*π*r²) = 10^-3 r = √[985*10^3/(4*π)] r = 280 m Check: The ratio of 130 dB to 90 dB = 40 dB The intensity ratio is 10^-4, so the distance ratio should be 10^2, or 280 m