Total sound intensity from sources

1. Apr 23, 2015

Yoonique

1. The problem statement, all variables and given/known data
A family ice show is held at an enclosed arena. The skaters perform to music with level 80.0 dB. This is too loud for your baby, who yells at 75.0 dB.

(a) What total sound intensity engulfs you?

2. Relevant equations
I ∝ A2

3. The attempt at a solution
Why did the answer sum the individual intensity of the music and the baby for the total sound intensity? Itotal = Imusic + Ibaby?
I thought the sound that one hear should be the superposition of the waves produced by the music and the baby. Assuming it is constructive interference, the total sound intensity should be proportional to the sum of the amplitude squared. Thus the total sound intensity should be Itotal = (Imusic0.5 + Ibaby0.5)2.

Can someone tell me which is the right way? Because I got another question which did the method I proposed and I feel that it makes more sense.

2. Apr 23, 2015

BvU

Permanent constructive interference would be too much of a coincidence. So no $A^2 = (A_1 + A_2)^2$. Equally unlikely is constant destructive interference with $A^2 = (A_1 - A_2)^2$.

What remains is simple addition of the power density (or energy flux) as if the two sources were completely uncorrelated (which is clearly the case in this exercise ). So $A^2 = A_1^2 + A_2^2$.

3. Apr 23, 2015

Yoonique

Okay so if the question is obviously about permanent interference (sources are in phase) I use the former equation. If the question is general and vague (like this question) I use the latter equation?

4. Apr 23, 2015

BvU

Sounds unlikely, doesn't it. I, for one, don't believe it: that way you could generate 4 W with two sources of 1 W each !.

5. Apr 23, 2015

Yoonique

You mean is unlikely for this scenario stated in this question? But it is possible in other scenario where permanent constructive interference occurs right for eg, two speakers in phase, and the path difference of a point is an integer multiple of λ. (d = nλ, n=0,1,2,3..) so at that point you would get 4W from two 1W sources?

6. Apr 23, 2015

tech99

The two sounds are not coherent, so they add on a power basis. The two intensities in W/m^2 are added. The decibel levels must first be converted to intensities and added, then converted back to dB.

7. Apr 23, 2015

Yoonique

What do you mean by not coherent? Means they are not the same at frequency? Mean they are not in phase so complete interference could not happen?

8. Apr 23, 2015

haruspex

You could only add the amplitudes if the sources are in phase all the time. Likewise, subtract if out of phase all the time. Yet other interesting results with other constant phase differences. Since the sounds in this question will be at different frequencies, the phase differences will constantly vary at all points.
With the same frequency from two sources, you get the usual interference pattern. At each point, the phase difference is constant. Where the amplitudes add, and happen to be equal, you get four times the intensity, but integrating over a sufficiently large region the average comes back to simply the sum of the intensities.

I assume you converted from dB to intensity correctly.

9. Apr 23, 2015

tech99

The two sounds are different in their frequency spectrum and are not in any way connected in phase or frequency, so the sound waves will only occasionally be in phase or in anti-phase. The loudness of the sound depends on the intensity in W/m^2, and when we add these two noise-like sounds together the two noise powers are added. Of course, on an oscilloscope you would see occasional peaks occurring when the two waves combine in phase for a moment, but the ear responds only to intensity and is insensitive to the shape of the combined wave.

10. Apr 23, 2015

BvU

No, I mean it's unlikely that you can use $A^2 = (A_1 + A_2)^2$ if the question is obviously about permanent interference (sources are in phase).

perhaps this picture (borrowed from Wikipedia) makes clear why you use simple addition of the I in all cases . And why you can't get 4 W with two times 1 W.

 this was in response to post #5 but didn't make it through on time.

11. Apr 23, 2015

haruspex

Did you mean that? I would have thought that if you are looking at a location where the two sources are always in phase then that is the one place where A=A1+A2 would be true.

12. Apr 24, 2015

BvU

True, but if folks tell me the sound intensity in my office is 40 dB unless I move one foot and it's 120 dB I would not be happy at all. Conflict between daily use and an extremely specific case. My association with sound intensity is not that it's a wildly varying function of position, but you're absolutely right. (No echo's, though ! )