A bird is flying directly toward a stationary bird-watcher and emits a frequency of 1140 Hz. The bird-watcher, however, hears a frequency of 1190 Hz. What is the speed of the bird, expressed as a percentage of the speed of sound?

Question

Physics

Sierra Cline

30 June, 03:03

A bird is flying directly toward a stationary bird-watcher and emits a frequency of 1140 Hz. The bird-watcher, however, hears a frequency of 1190 Hz. What is the speed of the bird, expressed as a percentage of the speed of sound?

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Answers (1)

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Evelyn Stevens · Answer 1

Vs = 4.20%*V

Explanation:

let fs be the frequecy of the bird flying emits toward the stationary bird, fo be the frequecy that the stationary bird hears, Vs be the speed of the flying bird and Vo be the speed of the stationary bird and V be the speed of sound.

From doppler effect we know that for an observer at rest (stationary bird), when the source (flying bird) is moving towards the observer, then the observer will hear an increasing frequency by:

fo = [ (V + Vo) / (V - Vs) ]*fs

fo = [ (V) / (V - Vs) ]*fs, since Vo = 0m/s.

1190 = [V / (V - Vs) ]*1140

1190/1140 = V / (V - Vs)

(1190/1140) * (V - Vs) = V

V - Vs = (1140/1190) V

1 - Vs/V = 1140/1190

Vs/V = 1 - 1140/1190

Vs/V = 5/119

Vs = (5/119) * V

Therefore, the speed of the bird is Vs = 4.20%*V