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?

Question

Physics

Conrad Reilly

12 November, 11:02

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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Answers (1)

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Mya Branch · Answer 1

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%