541_Physics ProblemsTechnical Physics

541_Physics ProblemsTechnical Physics - Chapter 18 543...

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Chapter 18 543 Additional Problems P18.57 f = 87 0 . H z speed of sound in air: v a = 340 m s (a) λ b = A vf b == 87 0 0 400 1 .. s m ej a f v = 34 8 . ms (b) a aa L = = U V W 4 L v f a = 4 340 4870 0977 1 ms s m . . FIG. P18.57 *P18.58 (a) Use the Doppler formula ′= ± ff vv s 0 bg . With f 1 frequency of the speaker in front of student and f 2 frequency of the speaker behind the student. + = + = f f 1 2 456 343 1 50 343 0 458 456 343 1 50 343 0 454 Hz Hz Hz Hz a f af . . Therefore, fff b =′ −′= 12 399 . H z . (b) The waves broadcast by both speakers have = v f 343 456 0752 s m . . The standing wave between them has d AA 2 0376 . m. The student walks from one maximum to the next in time t 150 0251 . . . m s, so the frequency at which she hears maxima is f T 1 H z . P18.59 Moving away from station, frequency is depressed:
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