Intensity, Radius (Calculations)

In summary, the conversation discusses a point source of sound and a person listening from 8m away. When the intensity of the source is halved, the person wishes to know how far they should be to maintain the same loudness. The solution involves understanding the inverse proportionality between intensity and radius, and concludes that the person should be at a distance of r/(2^(1/2)) from the source. The algebraic explanation further confirms this solution.
  • #1
Gunman
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Homework Statement


A point source of sound emits energy equally in all directions at a constant rate and a person 8m from the source listens. After a while, the intensity of the source is halved. If the person wishes the sound to be seem as loud as before, how far should he be now?


Homework Equations


intensity is inversely proportional to radius(distance from the source)


The Attempt at a Solution


I = (K)/(r^2)
I*(r^2)=K
So i presumed that since K has to be constant, and I has been halved. The radius would be (2)^(1/2). So why isn't the answer (2^(1/2))r ? I can understand that the answer is not correct because it is not possible for the intensity to be the same when the radius has increased. So can someone point out to me which part I went wrong? Thank you.
 
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  • #2
Since they are inversely proportional the person needs to get closer by some factor. If it were simply

[tex]I \alpha \frac{K}{r}[/tex]

and the intensity dropped by a factor of two then that is the equivalent of saying that the person went twice as far away, so in the case of the intensity of source being independent then to keep the same sound the person would half to cancel it out by going r/2.

Now in the case of it being inversely proportional to the square of the radius, if the intensity decreases by a half, it could have happened because someone traveled 2^(1/2) away, so if it were actually the source that decreased by a half then the person can cancel that effect just as if they could have caused it by doing the opposite, and traveling 2^(1/2) closer (i.e. r/(2^1/2)).

If it helps to see the algebra in addition to the qualitative observations here is how it would go for the two:
[tex] I \alpha \frac{1}{2} \frac{K}{r/2} = \frac{k}{r}[/tex]

[tex] I \alpha \frac{1}{2} \frac{K}{(r/\sqrt{2})^2} = \frac{k}{r^2}[/tex]
 
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  • #3



Your attempt at a solution is on the right track, but there are a few errors in your calculations. Firstly, the formula for intensity should be written as I = K/r^2, not I = K*r^2. This is because intensity is inversely proportional to the square of the distance, meaning that as the distance increases, the intensity decreases. Secondly, when you halve the intensity, you need to multiply the original intensity by 1/2, not divide it by 2. This is because halving something means dividing it into two equal parts, so the original intensity would need to be divided by 2 to get the halved intensity. Finally, to find the new distance, you would need to take the square root of the original distance multiplied by the halved intensity, which would give you the new distance (r). So the correct equation would be r = (r_initial)*(sqrt(1/2)). Therefore, the person should move to a distance of approximately 5.66m to perceive the sound as loud as before.
 

What is intensity?

Intensity is a measure of the amount of energy per unit area. In the context of radius calculations, it refers to the amount of energy per unit area at a specific distance from the source.

How is intensity related to radius?

Intensity is inversely proportional to the square of the distance from the source. This means that as the distance from the source increases, the intensity decreases.

What is the formula for calculating intensity?

The formula for calculating intensity is I = P/A, where I is intensity, P is power, and A is the area over which the energy is distributed. In terms of radius, the formula can be written as I = P/(4πr^2), where r is the distance from the source.

How is intensity measured?

Intensity is typically measured in watts per square meter (W/m^2).

What factors can affect intensity at a specific radius?

Intensity can be affected by various factors such as the power of the source, the distance from the source, and any intervening objects that may absorb or scatter the energy. Changes in these factors can result in a change in intensity at a specific radius.

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