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, 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.

Question

Quizplus · Accepted Answer

The acoustic power entering through the window can be calculated using the formula for sound intensity level $ \beta = 10 \log_{10} \left( \frac{I}{I_0} \right) $, where $ \beta $ is the sound intensity level in decibels (dB), $ I $ is the sound intensity in watts per square meter (W/m$^2$), and $ I_0 $ is the reference sound intensity ($1.0 \times 10^{-12} $ W/m$^2$). Given that $ \beta = 79 $ dB, we can solve for $ I $ and then calculate the power $ P $ entering through the window by multiplying $ I $ by the area of the window.First, solve for $ I $:$$ 79 = 10 \log_{10} \left( \frac{I}{1.0 \times 10^{-12}} \right) $$$$ 7.9 = \log_{10} \left( \frac{I}{1.0 \times 10^{-12}} \right) $$$$ 10^{7.9} = \frac{I}{1.0 \times 10^{-12}} $$$$ I = 10^{7.9} \times 1.0 \times 10^{-12} $$Then, calculate $ P $ by multiplying $ I $ by the area of the window ($ 2.0 \, \text{m} \times 1.0 \, \text{m} = 2.0 \, \text{m}^2 $):$$ P = I \times 2.0 \, \text{m}^2 $$Given the intensity level and the area of the window, the calculation will show that the power entering through the window is closest to 160 µW, making choice A correct. This approach simplifies the process by directly relating the intensity level to the power through the area of the window without needing to calculate the exact intensity value.

Sound Level: an Enclosed Chamber with Sound Absorbing Walls Has