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,78 m away and facing the window.The intensity level of the sound entering the window space from the loudspeaker is 79 dB.Assume the acoustic output of the loudspeaker is uniform in all directions and that the 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 entering through the window space is closest to A)160 µW. B)79 µW. C)320 µW. D)790 µW. E)1600 µW.

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Quizplus · Accepted Answer

The intensity level (IL) in decibels (dB) is given by the formula: $$IL = 10 \log_{10} \left( \frac{I}{I_0} \right)$$ where $I$ is the intensity in watts per square meter ($W/m^2$) and $I_0$ is the threshold of hearing ($1.0 \times 10^{-12} W/m^2$).Given that the intensity level is 79 dB, we can rearrange the formula to solve for $I$: $$79 = 10 \log_{10} \left( \frac{I}{1.0 \times 10^{-12}} \right)$$ $$\Rightarrow \log_{10} \left( \frac{I}{1.0 \times 10^{-12}} \right) = 7.9$$ $$\Rightarrow \frac{I}{1.0 \times 10^{-12}} = 10^{7.9}$$ $$\Rightarrow I = 10^{7.9} \times 1.0 \times 10^{-12} W/m^2$$The power ($P$) entering through the window can be calculated by multiplying the intensity ($I$) by the area ($A$) of the window: $$P = I \times A$$ The area of the window is $2.0 m \times 1.0 m = 2.0 m^2$. Substituting the value of $I$ from above: $$P = 10^{7.9} \times 1.0 \times 10^{-12} \times 2.0 m^2$$ $$P = 2.0 \times 10^{7.9-12} W$$ $$P = 2.0 \times 10^{-4.1} W$$ $$P \approx 160 \mu W$$Therefore, the acoustic power entering through the window space is closest to 160 µW.

An Enclosed Chamber with Sound Absorbing Walls Has a 2