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.

Question

Quizplus · Accepted Answer

The intensity level is given by:$$ \beta = 10 \log_{10} \frac{I}{I_0} $$where I is the intensity and I0 is the threshold of hearing. Solving for I, we get:$$ I = I_0 10^{\beta/10} $$The intensity is given by:$$ I = \frac{P}{A} $$where P is the power and A is the area. Substituting this into the previous equation, we get:$$ \frac{P}{A} = I_0 10^{\beta/10} $$Solving for P, we get:$$ P = AI_0 10^{\beta/10} $$Substituting the given values, we get:$$ P = (2.0 	ext{ m})(1.0 	ext{ m})(1.0 	imes 10^{-12} 	ext{ W/m}^2) 10^{56/10} $$$$ P = 0.035 	ext{ W} $$Therefore, the correct answer is A.

An Enclosed Chamber with Sound Absorbing Walls Has a 2