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Halliday 2 Cap 18 - P18 099

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    December 2018

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99. (a) The problem asks for the source frequency f . We use Eq. 18-47 with great care (regarding its ± sign conventions).   340 − 16 f = f 340 − 40 Therefore, with f  = 950 Hz, we obtain f = 880 Hz. (b) We now have  f =f so that with f = 880 Hz, we find f  = 824 Hz.  340 + 16 340 + 40