Calculating Sound Intensity 24 m from a Loudspeaker

[Jtappan]

Homework Statement

The sound level 24 m from a loudspeaker is 66 dB. What is the rate at which sound energy is produced by the loudspeaker, assuming it to be an isotropic source?

____W

Homework Equations

?

Something to do with Intensity?

The Attempt at a Solution

I don't know where to begin on this problem. My book doesn't describe any problems that are related to distance nor does it have any equations that are related to distance.

 

[mezarashi]

This tests your understanding of "sound intensity", nature of sound propagation, and a bit of math.

What is sound intensity? What are its units? (hint: W/area).

At 24 meters, what is the surface area if sound radiates isotropically (i.e. like an expanding sphere).

You'll have to convert from the dB to linear scale to get power in Watts.

 

[ogb p]

I would first start by defining some terms and concepts that are relevant to this problem. Sound intensity is a measure of the amount of sound energy passing through a unit area per unit time. It is usually measured in watts per square meter (W/m^2). In this case, the sound level of 66 dB is a measure of the sound pressure level, which is a logarithmic scale used to describe the intensity of sound.

To solve this problem, we can use the equation for sound intensity:

I = P/A

where I is the sound intensity, P is the sound power, and A is the area over which the sound is spread. In this case, we are given the distance from the loudspeaker (24 m) but not the area. To find the area, we can use the formula for the surface area of a sphere:

A = 4πr^2

where r is the radius of the sphere. In this case, the loudspeaker can be approximated as an isotropic source, meaning it radiates sound equally in all directions. Therefore, the radius of the sphere would be 24 m.

Now, we can substitute these values into the equation for sound intensity:

I = P/4πr^2

To solve for P, we can rearrange the equation to:

P = I*4πr^2

Substituting in the given value for sound intensity (66 dB or 10^-6 W/m^2) and the distance (24 m), we get:

P = (10^-6 W/m^2)*4π*(24 m)^2

P = 9.07 W

Therefore, the rate at which sound energy is produced by the loudspeaker is approximately 9.07 watts. Keep in mind that this is an approximation as we are assuming the loudspeaker is an isotropic source and not accounting for any other factors that may affect the sound intensity at a distance of 24 m.