Answer to Question #2277 in Mechanics | Relativity for chris
Here, Pac =15.3 W (upper case) is the sound power and A the surface area of a sphere of radius r. Thus the sound intensity decreases with 1/r2 the distance from an acoustic point source, while the sound pressure decreases only with 1/r from the distance from an acoustic point source after the 1/r-distance law.
So we have
Ir=Pac/(4πr2 ) → r2=Pac/(4πIr ) → r= √(Pac/(4πIr )) = √(15.3/(4*π*3*106 )) = 0.637*10-3 m.
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