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P214soln8

# P214soln8 - Solutions(HW#8 Prepared by Seong Chan Park I...

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Solutions (HW #8) Prepared by Seong Chan Park I. 16.40 (a) f 0 A = v - 15m / s v + 0m / s f A = 340 - 15 340 × 392Hz = 375Hz (b) f 0 B = v + 15m / s v - 35m / s f B = 340 + 15 340 - 35 × 392Hz = 371Hz (c) f 0 A - f 0 B = 4Hz II. 16.69 (a)Let v insect be the speed of the insect and f insect the frequency with which the sound waves both strike and are reflected from the insect. The frequency of the wave strike the insect is given by: f insect = f bat v + v insect v - v bat . And the frequency at which the bat receives is shifted as: f refl = f insect v + v bat v - v insect . Combining these two relations, we could solve for v insect , v insect = v f relf ( v - v bat ) - f bat ( v + v bat ) f relf ( v - v bat ) + f bat ( v + v bat ) . (1) (b) v insect = 340m / s × 83 . 5(340 - 3 . 9) - 80 . 7(340+3 . 9) 83 . 5(340 - 3 . 9)+80 . 7(340+3 . 9) = 2 . 0m / s 1

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III. 16.71 (a) f R = f S r c - v c + v = f S p 1 - v/c p 1 + v/c = f S (1 - v/c ) 1 / 2 (1 + v/c ) - 1 / 2 (b) For v/c ¿ 1, (1 v/c ) ± 1 / 2 1 - v/ 2 c ,so f L f S (1 - v/ 2 c ) 2 f S (1 - v/c ) (2) where we also used (1 - x ) 2 1 - 2 x for small x . (c) The speed of an ordinary airplane is much less than the speed of light and the ap- proximation v/c ¿ 1 is valid. f R = 243MHz + 46 . 0 Hz (1 + v/c ) f S = (1 + v/c ) × 243MHz . (3) So v/c = 1 . 89 × 10 - 7 or v = 56 . 8m / s. IV. 16.73 (a)The thunderclouds, moving toward the installation, encounter more wavefronts per time than would a stationary cloud, and so an observer in the frame of the storm would detect a higher frequency.
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