A fan at a rock concert is 30 m from the edge of the stage. The sound intensity level at the edge of the stage is 110 dB. The fan's eardrums have a diameter of 8.4mm.
How much energy is transferred to each eardrum in one second? (Hint: It may help to asume that the speaker is on the stage located 1.0 m from the edge.)answer in J.
3 answers
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
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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4 answers