contestada

An eagle flying at 28 m/s emits a cry whose frequency is 370 Hz. A blackbird is moving in the same direction as the eagle at 11.35 m/s. (Assume the speed of sound is 343 m/s). What frequency does the blackbird hear (in Hz) as the eagle approaches the blackbird?

Respuesta :

Answer:

388.88 Hz

Explanation:

By the Doppler effect, we have the following equation when the source is moving toward the receiver.

[tex]f_o=\frac{f}{1-\frac{v_r}{v}}[/tex]

Where fo is the perceived frequency, f is the emitted frequency, Vr is the relative speed and v is the speed of sound.

The relative speed can be calculated as the difference between the speed of the eagle and the speed of the blackbird, so

vr = 28 m/s - 11.35 m/s

vr = 16.65 m/s

Then, replacing f = 370 Hz, vr = 16.65 m/s, and v = 343 m/s, we get:

[tex]f_o=\frac{370}{1-\frac{16.65}{343}}=388.88\text{ Hz}[/tex]

Therefore, the frequency heard by the blackbird is 388.88 Hz

Otras preguntas