Question

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?

Answer 1

388.88 Hz

Step-by-step explanation:

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

`f_o=(f)/(1-(v_r)/(v))`

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:

`f_o=(370)/(1-(16.65)/(343))=388.88\text{ Hz}`

Therefore, the frequency heard by the blackbird is 388.88 Hz