Question

The sound level 8.25 m from a loudspeaker, placed in the open, is 115 dB. What is the acoustic power output (W) of the speaker, assuming it radiates equally in all directions?

Short answer

The acoustic power output of the speaker is \(270.27\;{\rm{W}}\).

Step-by-step solution

Determination of power of sound source

The power of the sound source is determined by the product of sound intensity and surface area covered by sound.

Given information

Given data:

The distance of loudspeaker is \({\rm{r}} = 8.25\;{\rm{m}}\).

The sound level is \(\beta = 115\;{\rm{dB}}\).

The standard value for the threshold intensity is \(1.0 \times {10^{ 12}}\;{{\rm{W}} \mathord{\left/{\vphantom{{\rm{W}}{{{\rm{m}}^{\rm{2}}}}}}\right.}{{{\rm{m}}^{\rm{2}}}}}\).

Evaluation of the acoustic power output of the speaker

The intensity of sound is calculated below:

\(\begin{aligned}{c}\beta &= 10\log \left( {\frac{I}{{{I_0}}}} \right)\\I &= {I_0}{10^{\left( {\frac{\beta }{{10}}} \right)}}\end{aligned}\)

Substitute the values in the above equation.

\(\begin{aligned}{l}I &= \left( {1.0 \times {{10}^{ - 12}}\;{{\rm{W}} \mathord{\left/{\vphantom {{\rm{W}} {{{\rm{m}}^{\rm{2}}}}}} \right.} {{{\rm{m}}^{\rm{2}}}}}} \right){10^{\left( {\frac{{115\;{\rm{dB}}}}{{10}}} \right)}}\\I&=0.316\;{{\rm{W}}\mathord{\left/{\vphantom{{\rm{W}}{{{\rm{m}}^{\rm{2}}}}}}\right.}{{{\rm{m}}^{\rm{2}}}}}\end{aligned}\)

The acoustic power output of the speaker is calculated below:

\(\begin{aligned}{c}P &= IA\\ &= I\left( {4\pi {r^2}} \right)\end{aligned}\)

Substitute the values in the above equation.

\(\begin{aligned}{l} &= \left( {0.316\;{{\rm{W}}\mathord{\left/{\vphantom {{\rm{W}} {{{\rm{m}}^{\rm{2}}}}}} \right.} {{{\rm{m}}^{\rm{2}}}}}} \right)\left( {4\pi {{\left( {8.25} \right)}^2}} \right)\\ &= 270.27\;{\rm{W}}\end{aligned}\)

Hence, the acoustic power output of the speaker is \(270.27\;{\rm{W}}\).