Suppose that a public address system emits sound uniformly in all directions and that there are no reflections . The intensity at a location 22 m away from the sound source is 3.0×10−4W/m23.0×10 −4 W/m 2 . What is the intensity at a spot that is 78 m away?

Explanation:

The definition of the intensity in terms of power is given by:

Where:

P is the powerA is the area

If the sound emits uniformly in all directions and that there are no reflections, we can assume the geometry of the wave sound is spherical.

Let's recall the area of a sphere is

To the first location we have:

and to the second location we have:

Now, we can divide each intensity to find the second intensity.

I hope it helps you!

     

I = 2.4*10⁻⁵ W

Explanation:

By definition the intensity of a sound source, is the power emitted from the source divided by the area of the surface covered by the sound waves.Assuming that the source is a point source, so it emits sound uniformly in all directions, and there are no reflections, this area  is just the area of a sphere centered in the point source and with radius equal to the distance between the source and the point where the intensity is measured, as follows:

       

If I and r are givens of the problem, we can solve (1) for P, as follows:

       

Since the power is the same at any distance from the source, we can obtain the new value of I, replacing in (1) the value of P and the new value of r, as follows: