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A bird is flying directly toward a stationary bird-watcher and emits a frequency of 1180 Hz. The bird-watcher, however, hears a frequency of 1260 Hz. What is the speed of the bird, expressed as a percentage of the speed of sound?

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gilcarusolautaro

Answer:

The speed of the bird expressed as a percentage of the speed of sound are %V= 6.39%.

Explanation:

V= 343.2 m/s

Vo= 0

Vf= ?

f'= 1260 Hz

f= 1180 Hz

%V= ?

f'= f* ( (V/ V-Vf))

Clearing Vf:

Vf= 21.94 m/s

%V= (Vf/V)*100

%V= 6.39%

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