Chapter 18543Additional ProblemsP18.57f=87 0. Hzspeed of sound in air:va=340 m s(a)λb=Avfb==−87 00 4001..s mejafv=34 8. ms(b)aaaL==UVW4Lvfa=−4340487009771mssm..FIG. P18.57*P18.58(a)Use the Doppler formula′=±ffvvs0bg∓.Withf1frequency of the speaker in front of student andf2frequency of the speaker behind the student.+−=−+=ff124563431 5034304584563431 503430454HzHzHzHzafaf..Therefore, fffb=′−′=12399. Hz.(b)The waves broadcast by both speakers have =vf3434560752sm.. The standing wavebetween them has dAA20376.m. The student walks from one maximum to the next intime ∆t1500251...ms, so the frequency at which she hears maxima is fT1Hz.P18.59Moving away from station, frequency is depressed:
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