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The Expert TA 1 Hurman-l × e The Human Ear Can Det-X e My Account | Chegg.com × Secure | https://usel 1 fl.t Common/TakeTutorialAssignment.asp Class Management | Help HW 15 Begin Date: 11/20/2017 11:00:00 AM - Due Date: 12/4/2017 11:00:00 AM End Date: 12/4 2017 11:00:00 AM (10%) Problem 3: Suppose an oncoming ambulance moving at 105 km h emits a steady 780-Hz sound from its siren. Assignment Status Click here for detailed view 50% Part (a) what frequency, in Hz, is received by a person watching the oncoming ambulance? The speed of sound on this day is 345 m's Grade Summary 0% 100% Problem Status cos0 cotan asinacos0 atan0 acotan0sinh0 cosh0 tanhcotanh tan0 a( Completed Attempts remaining: (5% per attempt) detailed view Degrees Radians NO Hint I give up Hints: 1 % deduction per hint. Hints Feedback: deduction per feedback. 10 50% Part (b) What frequency, in Hz, does she observe after the ambulance has passed? O Type here to search 5:26 PM 11/22/2017

Explanation / Answer

W = U + V is the relative speed of the wave with length L when U is the speed of the source and V is the speed of sound. So the period is t = L/W and the apparent frequency is Fa = 1/t = W/L = (U + V)/L; where L = VT = V/Fs and Fs = 780 Hz is the static frequency and U = 105 kph = (105*1000/3600) m/s.

A) So we have Fa = (U + V)/(V/Fs) = Fs (U/V + 1) = 780*((105*1000/3600)/345 + 1) = 845.94 Hz..

B) And after passing w = V - U is the relative speed of the wave with L. So the apparent period is T = L/w and the apparent frequency fa = 1/T = w/L = (V - U)/(V/Fs) = Fs (1 - U/V) = 780*(1 - (105*1000/3600)/345) = 714.06 Hz

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