An ambulance with a siren emitting a whine at 1,190 Hz overtakes and passes a cy

Question

An ambulance with a siren emitting a whine at 1,190 Hz overtakes and passes a cyclist pedaling a bike at 2.1 m/s. After being passed, the cyclist hears a frequency of 971 Hz. How fast is the ambulance moving? (The frequency 1,190 Hz before the ambulance overtakes the cyclist is the frequency observed by the cyclist.)

Explanation / Answer

The frequency - wavelength - velocity of a wave are related via the equation: (1) f * ? = v And the speed of sound at STP = 343m/s From (1), we can form the ratio: (2) fapp / frec = vapp / vrec, where app --> approaching rec --> receding The velocity of the approaching sound as perceived by the bicyclist is: (3) vapp = 343 + (Va - Vb) and the velocity of the receding sound is: (4) vrec = 343 - (Va - Vb) where Va = velocity of the ambulance Vb = velocity of the bicyclist Combining (3) and (4) with (2): (5) fapp / frec = vapp / vrec = (343 + (Va - Vb)) / (343 - (Va - Vb)) Substituting the values we know: (6) 1190 / 971 = (343 + (Va - 2)) / (343 - (Va - 2)) ? (7) 1.113 * (345 - Va) = 341 + Va ? (8) 384 - 1.113Va = 341 + Va ? (9) (384 - 341) = 2.113Va ? (10) Va = 20.4m/s

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