P18_029

# P18_029 - displacement amplitude is the sum of the...

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29. (a) When the right side of the instrument is pulled out a distance d the path length for sound waves increases by 2 d . Since the interference pattern changes from a minimum to the next maximum, this distance must be half a wavelength of the sound. So 2 d = λ/ 2, where λ is the wavelength. Thus λ =4 d and, if v is the speed of sound, the frequency is f = v/λ = v/ 4 d = (343 m / s) / 4(0 . 0165 m) = 5 . 2 × 10 3 Hz. (b) The displacement amplitude is proportional to the square root of the intensity (see Eq. 18–27). Write I = Cs m ,where I is the intensity, s m is the displacement amplitude, and C is a constant of proportionality. At the minimum, interference is destructive and the displacement amplitude is the diFerence in the amplitudes of the individual waves: s m = s SAD s SBD , where the subscripts indicate the paths of the waves. At the maximum, the waves interfere constructively and the
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Unformatted text preview: displacement amplitude is the sum of the amplitudes of the individual waves: s m = s SAD + s SBD . Solve 100 = C ( s SAD s SBD ) and 900 = C ( s SAD + s SBD ) for s SAD and s SBD . Add the equations to obtain s SAD = ( 100 + 900) / 2 C = 20 /C , then subtract them to obtain s SBD = ( 900 100) / 2 C = 10 /C . The ratio of the amplitudes is s SAD /s SBD = 2. (c) Any energy losses, such as might be caused by frictional forces of the walls on the air in the tubes, result in a decrease in the displacement amplitude. Those losses are greater on path B since it is longer than path A....
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