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dnl65078
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Homework Statement
a)A sound wave with an intensity level of 80.1 dB is incident on an eardrum of area 0.600 10-4 m2. Calculate the energy is absorbed by the eardrum in 4.00 minutes in microJ
b)The sound level 25.0 m from a loudspeaker is 69.0 dB. Calculate the rate at which sound energy is produced by the loudspeaker, assuming it to be an isotropic source.
Homework Equations
The Attempt at a Solution
We know tha the refernece intensity (I0) = 10-12W/m2
We know that sound intensity level is
β = 10dB log (I/Io)
Then 80.1dB = 10dB log (I/Io)
8.01 = log (I/Io)
Now finding I then substitue in the equation
I =P/A
= E/AΔt
Then the area covered by the sound is
A = 0.600 10-4 m2.
Then energy is absorbed by the eardrum in 4.00 minutes ( i.e 4*60s = 240s) in μJ.
E = IAΔt
I got 1.44x10-14 (which ended up being wrong))b)
We know tha the refernece intensity (I0) = 10-12W/m2
We know that sound intensity level is
β = 10dB log (I/Io)
Then 69dB = 10dB log (I/Io)
6.9 = log (I/Io)
Now finding I then substitue in the equation
I =P/A
Then the area covered by the sound is
A = 4π(25.0m)2
= 7850m2
Now the total power of the source is
P = IA in W Or J/s
I got 7.85x10-9 please help. what am I doing wrong?