Posted by Jimmy on Saturday, November 7, 2009 at 10:48pm.
Covert 110 dB to sound wave energy per area. The ear will be 31 times farther from the speaker than the edge of the stage. This will reduce the intensity of the sound by a factor of 31^2, or about 1000. Then multiply that intensity by 1 second and the area of the eardrum.
I= I0*10^(L(i)/10) so..
I(stage)= 10^-12*10^(110dB/10)=.1
I(2)/I(1)=r(1)^2/r(2)^2
I(2)/.1=1^2/30^2, I(2)=.1/(30^2)=1.04*10^-4 W/m^2
Area of the ear= piD^2/4= 3.14* .0084^2/4=55.4e-9m^2
P=I*A= 1.04e-4*55.4e-9= 6.16e-9
E=P*t = 6.16e-9*1= 6.16e-9
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