Sound level: An enclosed chamber with sound absorbing walls has a 2.0 m × 1.0 m opening for an outside window. A loudspeaker is located outdoors, 84 m away and facing the window. The intensity level of the sound entering the window space from the loudspeaker is 56 dB. Assume the acoustic output of the loudspeaker is uniform in all directions and that acoustic energy incident upon the ground is completely absorbed and therefore is not reflected into the window. The threshold of hearing is 1.0 × 10^-12 W/m². The acoustic power output of the loudspeaker is closest to A) 0.035 W. B) 0.070 W. C) 0.18 W. D) 0.35 W. E) 0.70 W.

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Quizplus · Accepted Answer

The intensity level is given by:$$ \beta = 10 \log_{10} \left( \frac{I}{I_0} ight) $$where:* $I$ is the intensity of the sound* $I_0$ is the threshold of hearingWe are given that the intensity level is 56 dB, so:$$ 56 	ext{ dB} = 10 \log_{10} \left( \frac{I}{1.0 	imes 10^{-12} 	ext{ W/m}^2} ight) $$Solving for $I$, we get:$$ I = 2.0 	imes 10^{-6} 	ext{ W/m}^2 $$The acoustic power output of the loudspeaker is given by:$$ P = IA $$where:* $A$ is the area of the windowWe are given that the area of the window is 2.0 m × 1.0 m, so:$$ P = (2.0 	imes 10^{-6} 	ext{ W/m}^2)(2.0 	ext{ m})(1.0 	ext{ m}) = 0.035 	ext{ W} $$Therefore, the acoustic power output of the loudspeaker is closest to 0.035 W.

Sound Level: an Enclosed Chamber with Sound Absorbing Walls Has