How Much Louder Are Four Crying Quadruplets Compared to One?

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toesockshoe
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Homework Statement


When four quadruplets cry simultaneously, how many decibels greater is the sound intensity level than when a single one cries?

Homework Equations


B=10*log(I/I_0)

The Attempt at a Solution


[/B]
I_0 is arbitrary (its just a reference).

When 4 babies cry, I would assume that it is the equivalent of 4 sound sources. Thus the amplitude would be 4 times larger than a single sources (or single crying baby)
Because intensity is proportional to amplitude square, I would assume I/I_o is 16... Putting this into the formula: 10*log(16/1) is 12. The solutions say to do 10*log(4/1) which equals 6 implying that the intensity is only 4 times as great.
 
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toesockshoe said:

Homework Statement


When four quadruplets cry simultaneously, how many decibels greater is the sound intensity level than when a single one cries?

Homework Equations


B=10*log(I/I_0)

The Attempt at a Solution


[/B]
I_0 is arbitrary (its just a reference).

When 4 babies cry, I would assume that it is the equivalent of 4 sound sources. Thus the amplitude would be 4 times larger than a single sources (or single crying baby)
Because intensity is proportional to amplitude square, I would assume I/I_o is 16... Putting this into the formula: 10*log(16/1) is 12. The solutions say to do 10*log(4/1) which equals 6 implying that the intensity is only 4 times as great.
You don't need to do the square thing. If you add two signals, you get double the intensity. The first double gives you +3dB,and the second double (to get you to 4 times) gets you to +6dB. :smile:
 
berkeman said:
You don't need to do the square thing. If you add two signals, you get double the intensity. The first double gives you +3dB,and the second double (to get you to 4 times) gets you to +6dB. :smile:
what? if you add 2 signals, i thought you get 4 times the intensity...

If that's the case, why is it that in a double slit, the central max is 4 times as a single light ray? We also do this using phasor diagrams. Where each additional source of light (or this case sound) represents adding a vector and the combined total length is the total amplitude. We square this to get the intensity.
 
toesockshoe said:
what? if you add 2 signals, i thought you get 4 times the intensity...

If that's the case, why is it that in a double slit, the central max is 4 times as a single light ray? We also do this using phasor diagrams. Where each additional source of light (or this case sound) represents adding a vector and the combined total length is the total amplitude. We square this to get the intensity.
The double-slit has the light rays in phase at the maximum, so you get double the peak electric field and 4x the power at that peak location. But the sound intensity in this problem is not coherent, so double the sources gives double the power.
 
berkeman said:
The double-slit has the light rays in phase at the maximum, so you get double the peak electric field and 4x the power at that peak location. But the sound intensity in this problem is not coherent, so double the sources gives double the power.
do you mind explaining by what you mean when you say "the sound intensity in this problem is not coherent"?. thank you.
 
toesockshoe said:
do you mind explaining by what you mean when you say "the sound intensity in this problem is not coherent"?. thank you.
Sure. In the double-slit example, at the bright spot the E fields are in phase, hence twice the E field gives 4 times the intensity.

But with sound or light waves that are not in phase, you get a doubling of intensity when you double the number of sources. The waves add in an RMS fashion because there are multiple frequencies and phases. Think about if you have two flashlights, you get twice the brightness as compared to just one flashlight...
 
berkeman said:
Sure. In the double-slit example, at the bright spot the E fields are in phase, hence twice the E field gives 4 times the intensity.

But with sound or light waves that are not in phase, you get a doubling of intensity when you double the number of sources. The waves add in an RMS fashion because there are multiple frequencies and phases. Think about if you have two flashlights, you get twice the brightness as compared to just one flashlight...
ok, so were basically saying it averages out to twice the intensity right?
 
toesockshoe said:
ok, so were basically saying it averages out to twice the intensity right?
Yes, when you add the RMS powers of two sources, you get twice the RMS power. :smile:
 
berkeman said:
Yes, when you add the RMS powers of two sources, you get twice the RMS power. :smile:
oh ok. thank you.