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A bird is flying directly toward a stationary bird-watcher and emits a frequency of 1420 Hz. The bird-watcher, however, hears a frequency of 1456 Hz. What is the speed of the bird, expressed as a percentage of the speed of sound

Respuesta :

Answer:

2.47 %

Explanation:

We are given;

Frequency emitted by source(bird); f_s = 1420 Hz

Frequency heard by observer; f_o = 1456 Hz

From doppler shift frequency, we know that;

f_o = f_s [c/(c - c_s)]

Where c_s is speed of source which is the bird and c is speed of sound.

Thus;

Rearranging the equation, we have;

f_s/f_o = (c - c_s)/c

f_s/f_o = 1 - (c_s/c)

Plugging in the relevant values to get ;

1420/1456 = 1 - (c_s/c)

0.9753 = 1 - (c_s/c)

1 - 0.9753 = (c_s/c)

(c_s/c) = 0.0247

Since we want it expressed im percentage,

Thus, (c_s/c) % = 0.0247 x 100 = 2.47 %

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