515_Physics ProblemsTechnical Physics

515_Physics ProblemsTechnical Physics - Chapter 17 P17.57...

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Chapter 17 517 P17.57 (a) ′= ff v vv diver bg so 1 −= v v f f diver ⇒= F H G I K J f f diver 1 with v = 343 m s , f 1 800 Hz and f = 2150 Hz we find v diver ms =− F H G I K J = 343 1 1800 2150 55 8 .. (b) If the waves are reflected, and the skydiver is moving into them, we have ′′ = + ′′ = L N M M O Q P P + v v v diver diver diver so ′′ = + = f 343 55 8 343 55 8 2500 . . a f af Hz . P17.58 (a) f fv vu −− f fv a f ′ − ′′ = + F H G I K J f v vu vu 11 f fv v u v u uvf v uv f = +−+ = = ej 22 2 2 1 2 1 (b) 130 36 1 km h m s = . ∴= = f 2361 400 340 1 36 1 340 85 9 2 2 . . . a f Hz P17.59 When observer is moving in front of and in the same direction as the source, O S where v O and v S are measured relative to the medium in which the sound is propagated. In this case the
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