What is the intensity of sound 2?

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SUMMARY

The intensity of Sound 2 is calculated based on its intensity level being 2.5 dB greater than Sound 1, which has an intensity of 39.0 W/m². Using the formula 2.5 dB = 10 log(I2/I1), where I1 is 39.0 W/m², the correct calculation yields I2 = 49.0 W/m². The initial attempt at solving the problem incorrectly applied logarithmic principles, leading to an erroneous result of 9.5 x 10^21 W/m².

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eagles12
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Homework Statement



Sound 1 has an intensity of 39.0 W/m^2. Sound 2 has an intensity level that is 2.5 dB greater than the intensity level of sound 1. What is the intensity of sound 2?

Homework Equations



I=energy/area*time
β=(10dB) logI/Io
Io=10^-12 W/m^2

The Attempt at a Solution



Intenisty 1= 39=energy/area*time
β for the second sound= 339.776
Using this i plugged in
339.776=(10dB) logI/Io
33.977=logI/Io
9.49*10^33=I/Io
and found I=9.5*10^21 but it is saying this is incorrect
 
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eagles12 said:

Homework Statement



Sound 1 has an intensity of 39.0 W/m^2. Sound 2 has an intensity level that is 2.5 dB greater than the intensity level of sound 1. What is the intensity of sound 2?

Homework Equations



I=energy/area*time
β=(10dB) logI/Io
Io=10^-12 W/m^2

The Attempt at a Solution



Intenisty 1= 39=energy/area*time
β for the second sound= 339.776
Using this i plugged in
339.776=(10dB) logI/Io
33.977=logI/Io
9.49*10^33=I/Io
and found I=9.5*10^21 but it is saying this is incorrect
I'm not following your logic on that. :rolleyes:

I think you might be making this problem harder than it is. When the problem statement says, "Sound 2 has an intensity level that is 2.5 dB greater than the intensity level of sound 1" that means,

2.5 \ \mathrm{dB} = 10 \ \log \frac{I_2}{I_1}
where I1 is given as 39.0 W/m2. Solve for I2. :wink:
 

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