problem16_77

# University Physics with Modern Physics with Mastering Physics (11th Edition)

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16.77: Taking the speed of sound to be s, m 344 the wavelength of the waves emitted by each speaker is m. 00 . 2 a) Point C is two wavelengths from speaker A and one and one-half from speaker B , and so the phase difference is rad. 180 π = ° b) , m W 10 98 . 3 m) (4.00 4 W 10 00 . 8 4 2 6 2 4 2 - - × = × = = π πr P I and the sound intensity level is dB. 0 . 66 ) 10 log(3.98 dB) 10 ( 6 = × Repeating with 57.2dB. and W 10 31 . 5 gives m 00 . 3 and W 10 00 . 6 7 5 = ×
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Unformatted text preview: = × =--β I r P c) With the result of part (a), the amplitudes, either displacement or pressure, must be subtracted. That is, the intensity is found by taking the square roots of the intensities found in part (b), subtracting, and squaring the difference. The result is that dB. 62.1 and W 10 60 . 1 6 = × =-β I...
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• Frequency, sound intensity level, phase difference

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