Question

Two eagles fly directly toward one another, the first at 14.5 m/s and the second at 22.5 m/s. Both screech, the first one emitting a frequency of 3400 Hz and the second one emitting a frequency of 3950 Hz. What frequencies do they receive if the speed of sound is 330 m/s?

Answer 1

Answer: The frequency received by the first eagle is 4425.28Hz and that received by the second eagle is 3798.73Hz

Explanation: Please see the attachments below

Answer 2

Answer:

The first eagle hear a frequency of 4.43kHz and the second eagle hear a frequency of 3.80kHz

Explanation:

Given that

both eagle are flying towards one another

speed of the first eagle v1 = 14.5m/s

speed of the second eagle v2 = 22.5m/s

frequency emitted by the first eagle f1= 3400Hz

frequency emitted by the second eagle f2 = 3950Hz

speed of sound v = 330m/s

first part

$F_1 = (\frac{v + v_2}{v - v_1} )f_1\\F_1 = (\frac{330 + 22.5}{330 - 14.5} )(3400)\\= 3.80 \times 10^3Hz\\= 3.80kHz$

second part

$F_1 = (\frac{v + v_2}{v - v_1} )f_1\\F_1 = (\frac{330 + 14.5}{330 - 22.5} )(3950)\\= 4.43 \times 10^3Hz\\= 4.43kHz$

The first eagle hear a frequency of 4.43kHz and the second eagle hear a frequency of 3.80kHz