Question

(a)

What frequency (in Hz) is received by a person watching an oncoming ambulance moving at 106 km/h and emitting a steady 850 Hz sound from its siren? The speed of sound on this day is 345 m/s.

Hz

(b)

What frequency (in Hz) does she receive after the ambulance has passed?

Hz

Answer 1

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 = 850 c/s is the static frequency and U = 106 kph = (106*1000/3600) m/s. =29.444m/s

So we have Fa = (U + V)/(V/Fs) = Fs (U/V + 1) = 850*(29.44/345 + 1) = 922.533 c/s. ANS. a.

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) = 850*(1 - (29.44)/345) = 777.466 cps ANS. b.