"PLAY IT LOUD" For an outdoor concert we want the sound intensity to be 1 W/m^2 at a distance of 20 m from the speaker array. If the sound intensity is uniform in all directions, what is the required average acoustic power output of the array?

Question

"PLAY IT LOUD"

For an outdoor concert we want the sound intensity to be 1 W/[latex]m^2[/latex] at a distance of 20 m from the speaker array. If the sound intensity is uniform in all directions, what is the required average acoustic power output of the array?

Accepted Answer

IDENTIFY, SET UP, and EXECUTE:

This example uses the definition of sound intensity as power per unit area. The total power is the target variable; the area in question is a hemisphere centered on the speaker array. We assume that the speakers are on the ground and that none of the acoustic power is directed into the ground, so the acoustic power is uniform over a hemisphere 20 m in radius. The surface area of this hemisphere is[latex]\left(\frac{1}{2}\right)(4 \pi)(20 \mathrm{~m})^{2}[/latex], or about 2500 [latex]m^2[/latex] . The required power is the product of this area and the intensity:

[latex]\left(1 \mathrm{~W} / \mathrm{m}^{2}\right)\left(2500 \mathrm{~m}^{2}\right)=2500 \mathrm{~W}=2.5 \mathrm{~kW}[/latex].

EVALUATE: The electrical power input to the speaker would need to be considerably greater than 2.5 kW, because speaker efficiency is not very high (typically a few percent for ordinary speakers, and up to 25% for horn-type speakers).

Question 16.7: "PLAY IT LOUD" For an outdoor concert we want the sound inte...

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