Sound Intensity doubles if number of noise sources double?

[Kashmir]

Hello All.
I looked at a question "A noisy machine in a factory produces sound with a level of 80 dB. How many identical machines could you add to the factory without exceeding the 90-dB limit?"

The solution in link below assumes the intensity will increase n fold if I add n machines.
Why is it true that an n fold increase in number means n fold increase in intensity?

Isn't intensity proportional to Amplitude square, so if two machines are perfectly in phase the intensity will quadruple and not double ?

https://homework.study.com/explanat...actory-without-exceeding-the-90-db-limit.html

 

[phyzguy]

How could two noisy machines in a factory be perfectly in phase? You're overthinking this.

 

[tech99]

For the sound of the machines to add in amplitude will require them to be in-phase. Even if the machines rotate together, they are still located in different positions, so that the distance from the observer to each machine will be different. This means that the sound from the machines will add with random phase. In such a case the intensity, or power flux density, is added rather than the amplitude.
The increase in intensity expressed in decibels will then be 10 log N, where N is the number of machines. Had the machines rotated together (in synchronism) and been equally spaced from the observer, the amplitudes would have added and the increase in intensity (power flux density) would be N squared, or equal in decibels to 20 log N.

 

[Lnewqban]

Hello All.
I looked at a question "A noisy machine in a factory produces sound with a level of 80 dB. How many identical machines could you add to the factory without exceeding the 90-dB limit?"

Just a couple of practical considerations:
The relative locations of the machines and the subject or instrument perceiving the sound should be relevant for such question.

In a factory, the production flow and manipulation of materials, forces machines to be far apart.
Also walls, exposed ceiling insulation, partitions, and the space ocupied by each machine, tend to produce shielding or reflecting effects on the sound propagation.

 

[Kashmir]

For the sound of the machines to add in amplitude will require them to be in-phase. Even if the machines rotate together, they are still located in different positions, so that the distance from the observer to each machine will be different. This means that the sound from the machines will add with random phase. In such a case the intensity, or power flux density, is added rather than the amplitude.
The increase in intensity expressed in decibels will then be 10 log N, where N is the number of machines. Had the machines rotated together (in synchronism) and been equally spaced from the observer, the amplitudes would have added and the increase in intensity (power flux density) would be N squared, or equal in decibels to 20 log N.

Thank you .

But lets say they are in perfect phase doesnt the intensity being N squared violate energy conservation?

If individual intensity is $I$ then they'll be quadrupled $4I$ that means we are getting 4 times the energy produced by each source individually?

 

[Nugatory]

But lets say they are in perfect phase doesnt the intensity being N squared violate energy conservation?

No. It's the same amount of energy, just distributed differently: less goes into energy wasted heating the air through which the sound travels and more is delivered directly directly to the human eardrums and other surfaces on which the sound waves impinge.

 

[PeroK]

Thank you .

But lets say they are in perfect phase doesnt the intensity being N squared violate energy conservation?

If individual intensity is $I$ then they'll be quadrupled $4I$ that means we are getting 4 times the energy produced by each source individually?

You're considering something like maximum energy density, not total energy.

 

[Kashmir]

You're considering something like maximum energy density, not total energy.

Could you please explain this further if possible? Thank you