Why Does Doubling Sound Sources Not Simply Double the Decibel Level?

[somecelxis]

Homework Statement

A source prodeuces sound of intensity level of 30dB .If another source prodeuces sound at the same time , what's the resultant level?

Homework Equations

The Attempt at a Solution

the ans is 10 log (2$I_1$ / ($I_0$)) = 33dB , but why not = 10 log (4$I_1$ / ($I_0$)) .. beacuse when 2 sound waves superposed , the resultant amplitude is A+A= 2A , I is directly proportional to (A^2) ...

 

[olivermsun]

Are the two sounds correlated (e.g., in phase) with each other? The sum of the two will only have amplitude A + A (and therefore +6 dB) if they are perfectly in phase with each other. In general they're assumed uncorrelated (they have nothing to do with each other), and so you add the intensities and get +3 dB.

 

[somecelxis]

Are the two sounds correlated (e.g., in phase) with each other? The sum of the two will only have amplitude A + A (and therefore +6 dB) if they are perfectly in phase with each other. In general they're assumed uncorrelated (they have nothing to do with each other), and so you add the intensities and get +3 dB.

it's not 3db am i right? the intensity now it's 2I ?

 

[olivermsun]

33 dB is +3 dB relative to the original 30 dB, right?

Hint: log₁₀2 = ?

 

[HalfLostAndSearching]

The resultant intensity level is not simply the sum of the two individual intensity levels. When two sound waves superpose, the resulting wave has a different amplitude and therefore a different intensity. The equation for intensity is I = A^2, where A is the amplitude. So when two waves with equal intensity (I) are superposed, the resultant intensity is actually four times the original intensity, not two times. This is why the correct equation is 10 log (4*I1/I0), where I1 is the intensity of the first source and I0 is the reference intensity. This equation takes into account the fact that the resultant intensity is four times the original intensity.