Answer Calculation of Sound Intensity and Power Output

[Noir]

Homework Statement

What is the sound intensity 2.5m away from a 50W speaker? (Assume sound is spread out in a sphere) The intensity level 12.0m from a loudspeaker, placed out in the open, is 100 dB, what is the acoustic power output (in Watts) of the speaker?

Homework Equations

[tex]I =10^{\frac{B}{10} - 12}[/tex]

[tex]v =\frac{P}{A}[/tex]

The Attempt at a Solution

I = P / A
I = 50 / (4 * pi * 2.5 ^ 2)
I = 0.64 W / m ^ 2

If the inensity level is 100 dB, then the Sound intensity is 10 ^ -2 W / m ^ 2.
So, If I = P / A
I x A = P
P = 10 ^ -2 x (4 * pi * 144)
P = 18.2 Watts

Is this correct? The 18.2 Watts looks a little for a speaker putting out 100 dB at 12 m...

 

[Redbelly98]

Looks good.

Note that the 50W speaker produces between 110 and 120 dB at 2.5 m, so 100 dB for 18 W is not out of line.

 

[nlightner]

Yes, your calculations are correct. It is important to note that the sound intensity is inversely proportional to the square of the distance, so as the distance increases, the sound intensity decreases significantly. This is why the power output seems relatively low even for a loudspeaker producing 100 dB at 12 m away.