Calculation for finding distance with regard to sound intensity

In summary: So can you now answer the question in post #1?In summary, the sound level from an amplifier when it is connected to a speaker will be the same as when the speaker was connected to the amplifier that has a fixed power output of 300 W. If the speaker is connected to an amplifier that has a variable power output, the sound level from the amplifier will be closer to the speaker when it is connected to the amplifier with the fixed power output.
  • #1
Aleisha
25
0

Homework Statement


Two amplifiers can be used to power a sound system.
Amplifier 1: Has a fixed power output of 300 W.
Amplifier 2: Has a variable power output of between 15 W and 175 W.
You may assume that the power output of the amplifier is equal to the power of the sound waves that are produced.
The speaker is disconnected from Amplifier 1 and is instead connected to Amplifier 2, which is operated at maximum power.
Previous calculated question sound level of amplifier 1= 121.15 dB
Calculate how close to the speaker must you stand for the sound level to be the same as when the speaker was connected to Amplifier 1.

Homework Equations


A=4 x Pi x r^2
I=P/A
dB=10log(I/Io)

The Attempt at a Solution


I want to calculate a distance, so the the sound decibel of amplifier 2 (121.15 dB) will equal amplifier 1.
I don't know an equation with distance that would relate to the equations above.
 
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  • #2
Aleisha said:
I don't know an equation with distance that would relate to the equations above
Yet you managed to quote an equation which is relevant.
When sound has traveled a distance r from its source, how thinly is its power spread?
 
  • #3
haruspex said:
Yet you managed to quote an equation which is relevant.
When sound has traveled a distance r from its source, how thinly is its power spread?
Right ofcourse I didn’t realize I was looking for radius. Don’t i need to know the area to get the radius? Or could I equate the two equations together?
 
  • #4
Aleisha said:
Right ofcourse I didn’t realize I was looking for radius. Don’t i need to know the area to get the radius? Or could I equate the two equations together?
A wavefront from the source forms an expanding sphere. The power is spread out over the surface of the sphere. As the radius doubles, what happens to the power density?
 
  • #5
The power density will increase by a factor of 4, if the radius is doubled?
 
  • #6
Aleisha said:
The power density will increase by a factor of 4, if the radius is doubled?
Power density is power per unit area. The power total is constant. If the radius doubles what happens to the area?
 
  • #7
Ohhh the area is 1/4 of the original amplifier?
 
  • #8
Aleisha said:
Ohhh the area is 1/4 of the original amplifier?
No.
Consider a single pulse of sound emitted from a point source. After some time it has traveled a distance r, so the wavefront forms a spherical shell radius r centred on the source. The total energy in the pulse is spread evenly over the shell, producing a certain energy density, ρ. The same time again later it has expanded to radius 2r. What is the energy density now?
 
  • #9
Twice as large
 
  • #10
Aleisha said:
Twice as large
The total energy is the same. It is spread over a larger area.
Do you understand the concept of density?
 
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Likes Aleisha
  • #11
It must be half then? That’s what I thought I meant because if 1/4 because the radius its r squared not just r... I’m getting too confused haha
 
  • #12
Aleisha said:
It must be half then?
That is moving in the right direction, but as you say it is radius squared, not half but ...?
(I have the feeling that you have, in your mind, already given the right answer but kept writing it down wrongly.)
 
  • #13
haruspex said:
That is moving in the right direction, but as you say it is radius squared, not half but ...?
(I have the feeling that you have, in your mind, already given the right answer but kept writing it down wrongly.)
Ah ha! It’s 4 times bigger!
 
  • #14
Aleisha said:
Ah ha! It’s 4 times bigger!
What is four times bigger, the area or the energy density?
 
  • #15
haruspex said:
What is four times bigger, the area or the energy density?
It must be the area according to the equation A=4 x Pi x r^2
 
  • #16
Aleisha said:
It must be the area according to the equation A=4 x Pi x r^2
Right, so what happens to the energy density?
 
  • #17
I=P/A If the area is four times bigger then the intensity (energy density) then the intensity with decrease by 4 times, correct?
 
  • #18
Aleisha said:
I=P/A If the area is four times bigger then the intensity (energy density) then the intensity with decrease by 4 times, correct?
Right.
So can you now answer the question in post #1?
 
  • #19
Yes yes! Thank you I get the correlation now :)
 

FAQ: Calculation for finding distance with regard to sound intensity

1. How is sound intensity related to distance?

Sound intensity decreases as distance increases. This is because the same amount of sound energy is spread out over a larger area as it travels, resulting in a decrease in intensity.

2. What is the equation for finding distance using sound intensity?

The equation for finding distance using sound intensity is: distance = √(power / (4π * intensity)). This equation can also be written as d = √(P / (4π * I)), where P is the power of the sound source and I is the sound intensity at the given distance.

3. Can sound intensity be measured directly?

No, sound intensity cannot be measured directly. It can only be calculated using the sound source's power and the distance from the source. Sound intensity is typically measured in decibels (dB), which is a logarithmic scale.

4. How does the distance affect the perceived loudness of sound?

As distance increases, the perceived loudness of sound decreases. This is because as sound travels, the intensity decreases, causing the sound to become quieter. This can also be affected by other factors such as the environment and the frequency of the sound.

5. Are there any other factors that can affect the calculation for finding distance with regard to sound intensity?

Yes, there are other factors that can affect the calculation for finding distance with regard to sound intensity. These include the frequency of the sound, the medium through which the sound is traveling, and any obstacles or reflections that may alter the sound's path. These factors can impact the accuracy of the calculation and should be taken into consideration when using the equation.

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