76. The observer of the sound (the bird-watcher) is stationary, while the source REASONING (the bird) is moving toward the observer. Therefore, the Doppler-shifted observed frequency is given by Equation 16.11. This expression can be solved to give the ratio of the bird’s speed to the speed of sound, from which the desired percentage follows directly. According to Equation 16.11, the observed frequency is related to the SOLUTION fo frequency of the source, and the ratio of the speed of the source to the speed of sound fs vs v by Solving for gives vs/v This ratio corresponds to . 78. The dolphin is the source of the clicks, and emits them at a frequency The REASONING fs. marine biologist measures a lower, Doppler-shifted click frequency because the dolphin fo, is swimming directly away. The difference between the frequencies is the source frequency minus the observed frequency: We will use (Equation 16.12), where is the speed fs fo. vs − of the dolphin and is the speed of sound in seawater, to determine the difference between v the frequencies. Solving (Equation 16.12) for the unknown source frequency we obtain SOLUTION fs, Therefore, the difference between the source and observed frequencies is 80. Both the observer (you) and the source (the eagle) are moving toward each REASONING other. According to Equation 16.15, the frequency heard by the observer is related to the fo frequency of the sound emitted by the source by fs where and are, respectively, the speeds of the observer and source. vo vs Substituting in the given data, we find that the frequency heard by the observer SOLUTION is
76. REASONING The observer of the sound (the bird-watcher) is stationary, while the source (the bird) is moving toward the observer. Therefore, the Doppler-shifted observed frequency is given by Equation 16.11. This expression can be solved to give the ratio of the bird’s speed to the speed of sound, from which the desired percentage follows directly. SOLUTION According to Equation 16.11, the observed frequency fo is related to the frequency fs of the source, and the ratio of the speed of the source vs to the speed of sound v by Solving for vs/v gives This ratio corresponds to . 78. REASONING The dolphin is the source of the clicks, and emits them at a frequency fs. The marine biologist measures a lower, Doppler-shifted click frequency fo, because the dolphin is swimming directly away. The difference between the frequencies is the source frequency minus the observed frequency: fs − fo. We will use (Equation 16.12), where vs is the speed of the dolphin and v is the speed of sound in seawater, to determine the difference between the frequencies. SOLUTION Solving (Equation 16.12) for the unknown source frequency fs, we obtain Therefore, the difference between the source and observed frequencies is 80. REASONING Both the observer (you) and the source (the eagle) are moving toward each other. According to Equation 16.15, the frequency fo heard by the observer is related to the frequency fs of the sound emitted by the source by where vo and vs are, respectively, the speeds of the observer and source. SOLUTION Substituting in the given data, we find that the frequency heard by the observer is
ch16_hw_76-78-80_solutions.docx