# Homework Help: Sound power in room

1. Jun 12, 2012

### Ry122

effective sound power from cube

if there was a sound source that radiates equally in all directions in a cube enclosure, and the cube was completely absorbative, how would you tell how much sound is coming out of a small gap in one of the walls and find the effective sound power of the cube.
Attached is a diagram of the room and opening.

I think to do this problem i need to find out how much sound is incident upon the gap in the wall.
And because the sound radiates away from the source in a spherical pattern, you need to determine what area of the sphere will come into contact with the gap in the wall. Am I on the right track?

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• ###### Screen Shot 2012-06-12 at 5.41.23 PM.jpg
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Last edited: Jun 12, 2012
2. Jun 12, 2012

### cepheid

Staff Emeritus
Re: effective sound power from cube

Yeah.

3. Jun 13, 2012

### Ry122

okay, I'm just not exactly sure of how to apply the math here.

If i simplify this to a 2D problem and draw a line from the edges of the point source to the opening and measure the angle made, and divide that by 360 degrees that will be the percentage of the power that's being let out (because if it were 360 degrees that would mean all the power is being let out).

I'm not sure how to do this in 3d though?

4. Jun 13, 2012

### cepheid

Staff Emeritus
What is the surface area of a sphere of radius r, where r is the distance from the centre of the cube to the centre of one of its faces?

5. Jun 14, 2012

### Ry122

That piece of information won't be particularly insightful in this situation, because what I need to do is find the surface area within the limits defined by the opening in the cube. Probably would require integration, but not sure how to do it.

6. Jun 14, 2012

### cepheid

Staff Emeritus
Actually what I said was highly relevant, because if you know the power of the source, then you can figure out the power arriving per unit area (in W/m2) at a radial distance r from the source. Then a pretty good approximation to the power radiating out of the opening is just the area of the opening multiplied by this power per unit area (or intensity, I). The the smaller the opening is, the better this approximation.

The problem with an opening of non-negligible size, I guess, is that not all parts of the opening are perpendicular to the source, and the intensity is the power arriving per unit of perpendicular area. Or, to put it another way, the radial distance to the source, r, is different at different points on the surface of the square opening. But what I said above about the sphere is still very important, because it gives you intensity as a function of the radial distance: I(r). To compute the total power across the opening, instead of simply assuming that I(r) is constant across the opening and multiplying it by the area of the opening, instead you'd integrate the function I(r) over the opening (it would be a 2D integral). You need to figure out how the radial distance to the source, r, varies across the square. I.e. if the square has a coordinate system (x,y) on its surface with the point (0,0) being at the centre of the square, what is r(x,y)?

A hint for setting up this integral is as follows: at any point (x,y) on the square, you have a right triangle that is formed by:

1. a line going from this point (x,y) directly to the source (has length r = some value)

2. a line going from the centre of the square (0,0) to the source (has length r = L/2 where L is the cube length)

3. a line along the surface of the square connecting these two points (x,y) and (0,0).

Since you know the length of side 2, and you can figure out the angle between side 2 and side 1 from the geometry, you therefore know the length of side 1, which is r(x,y).

Here is a 1D drawing of this:

Code (Text):

S = "source"

*    | y = y
*           |
r                  *                  |
*                         |
*                               |
*                                     |
*                                           |
S  -------------------------------------------|  y = 0    <----  square opening
L/2                    |
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7. Jun 14, 2012

### dimension10

Re: effective sound power from cube

Flux?

8. Jun 14, 2012

### Ry122

I have an easier method that I think should work.

You know the overall power of the source, and you know that any amount of the power radiating away from point source will be nullified when it strikes the wall.

The percentage of the sphere not striking the wall is also the percentage of the power escaping.

So you just have to calculate the part of the sphere which is able to escape and divide that by the surface area of the whole sphere, then multiply that by the total power of the source.

Or is there something wrong with this approach?

9. Jun 14, 2012

### cepheid

Staff Emeritus
What you describe is exactly the same as what I said here:

Think about it: dividing the power of the source by the surface area of the sphere gives you the power per unit area. Then multiplying that by the area of the square gives you the power incident on that opening.

As I pointed out, this is just an approximation (for reasons that I explained in my previous post). But it may well be good enough for your purposes, depending on what is expected.