Intensity and Power of isotropic sound

[turandorf]

Homework Statement

An electronic point source emits sound isotropically at a frequency of 3000 Hz and a power of 34 watts. A small microphone has an area of 0.74 cm2 and is located 158 meters from the point source.
a) What is the sound intensity at the microphone ?

b) What is the power intercepted by the microphone?

Homework Equations

I=(power)/(area)
Avg power=(1/2)(mu)(v)(w^2)A^2 (not sure what all these mean)

The Attempt at a Solution

I tried Intensity=(power)/(area) but that was wrong. I had converted the area to m^2 too…any ideas?

 

[LowlyPion]

At 158 meters don't you have the surface of a sphere that has a radius of 158 meters?

Maybe if you know what % of this area that the area of the microphone represents on the sphere, you might get an idea of how much of the original 34 watts that's pushing the air around at the speaker is available for detecting?

http://hyperphysics.phy-astr.gsu.edu/hbase/acoustic/invsqs.html#c1

 

[nlightner]

a) The correct equation for sound intensity is I = P/A, where P is the power and A is the area. In this case, the area of the microphone is 0.74 cm2, which is equivalent to 0.000074 m2. So, the intensity at the microphone is 34 watts/0.000074 m2 = 459459.46 watts/m2.

b) To calculate the power intercepted by the microphone, we can use the equation P = (1/2)ρvAw^2, where ρ is the density of the medium (air), v is the speed of sound, A is the area of the microphone, and w is the angular frequency (2πf). Plugging in the given values, we get P = (1/2)(1.2 kg/m3)(343 m/s)(0.000074 m2)(2π(3000 Hz))^2 = 0.0093 watts. So, the power intercepted by the microphone is 0.0093 watts.