P18_057 - 343 0 343 − 2(30 5 = 608 Hz(c We again pick a...

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57. We use Eq. 18-47 with f = 500 Hz and v = 343 m/s. We choose signs to produce f 0 >f . (a) The frequency heard in still air is f 0 = 500 µ 343 + 30 . 5 343 30 . 5 = 598 Hz . (b) In a frame of reference where the air seems still, the velocity of the detector is 30 . 5 30 . 5 = 0, and that of the source is 2(30 . 5). Therefore, f 0 = 500 µ
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Unformatted text preview: 343 + 0 343 − 2(30 . 5) ¶ = 608 Hz . (c) We again pick a frame of reference where the air seems still. Now, the velocity of the source is 30 . 5 − 30 . 5 = 0, and that of the detector is 2(30 . 5). Consequently, f = 500 µ 343 + 2(30 . 5) 343 − 0) ¶ = 589 Hz ....
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