Question

At a rock concert, a dB meter registered 135 dB when placed 2.9 m in front of a loudspeaker on the stage. The intensity of the reference level required to determine the sound level is 1.0×10−12W/m2 How far away would the sound level be 84 dB ?

Answer 1

Suppose that dB are sound-intensity dB (dB-SIL), then the reference level is I0 = 1.0 x 10^-12 W/m²

so as per the given condition -

135 = 10*log(I/I0) = 10*[log(I) + 12]

=> 13.5 = log(I) + 12

=> log(I) = 1.5,

=> I = 31.6 W/m²

Now, this is covering an area of 4*π*(2.9²), so the total power is 31.6*4*π*(2.9²) W = 3339 W

again for finding the distance, we have the following relations -

84 dB = 10*[log(I) + 12]

=> log(I) = -3.6

=> I = 10^-3.6 W/m²

So, the distance is such that the area covered by 3339 W yields 10^-3.6 W/m²:

3339/(4*π*r²) = 10^-3.6  

=> r = √[3339*10^3.6/(4*π)] = 1028 m

So, the requisite distance is 1028 m.