P18_055

# P18_055 - v + v D ) = ( v + v S ) and f = f = 500 . 0 Hz....

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55. (a) The expression for the Doppler shifted frequency is f 0 = f v ± v D v v S , where f is the unshifted frequency, v is the speed of sound, v D is the speed of the detector (the uncle), and v S is the speed of the source (the locomotive). All speeds are relative to the air. The uncle is at rest with respect to the air, so v D = 0. The speed of the source is v S =10m / s. Since the locomotive is moving away from the uncle the frequency decreases and we use the plus sign in the denominator. Thus f 0 = f v v + v S = (500 . 0Hz) µ 343 m / s 343 m / s+10 . 00 m / s = 485 . 8Hz . (b) The girl is now the detector. Relative to the air she is moving with speed v D =10 . 00 m / stoward the source. This tends to increase the frequency and we use the plus sign in the numerator. The source is moving at v S =10 . 00 m / s away from the girl. This tends to decrease the frequency and we use the plus sign in the denominator. Thus (
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Unformatted text preview: v + v D ) = ( v + v S ) and f = f = 500 . 0 Hz. (c) Relative to the air the locomotive is moving at v S = 20 . 00 m / s away from the uncle. Use the plus sign in the denominator. Relative to the air the uncle is moving at v D = 10 . 00 m / s toward the locomotive. Use the plus sign in the numerator. Thus f = f v + v D v + v S = (500 . 0 Hz) 343 m / s + 10 . 00 m / s 343 m / s + 20 . 00 m / s = 486 . 2 Hz . (d) Relative to the air the locomotive is moving at v S = 20 . 00 m / s away from the girl and the girl is moving at v D = 20 . 00 m / s toward the locomotive. Use the plus signs in both the numerator and the denominator. Thus ( v + v D ) = ( v + v S ) and f = f = 500 . 0 Hz....
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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