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Two eagles fly directly toward one another, the larger eagle at a speed of 15.0 m/s and the smaller one at a speed of 20.0 m/s. Both eagles screech, the larger emitting a screech of frequency of 3 200 Hz and the smaller one emitting a screech of frequency 3 800 Hz. The speed of sound in the air around the eagles is 330 m/s. What is the frequency of the screech heard by the larger eagle?

Respuesta :

Answer:

Explanation:

It is a problem relating to Doppler's effect , the expression for apparent frequency is as follows

n = [tex]\frac{n_0(V+v_o)}{V-v_s}[/tex]   ; n₀ is frequency of source , V is velocity of sound , v₀ is velocity of observer and v_s is velocity of source

Putting the values

n = [tex]\frac{3800(330+15)}{330 -20}[/tex]

= 4229 Hz .

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