Question

Two eagles fly directly toward one another, the first at 15.0 m/s and the second at 17.2 m/s. Both screech, the first emitting a frequency of 3543 Hz and the other one of 3800 Hz. What frequencies do they receive if the speed of sound is 330 m/s? (Enter your answers to at least the nearest 10 Hz.)

First Eagle = Hz

Second Eagle= Hz

Answer 1

frequency received by the first eagle,

E.1 = f*(v+vr)/(v-vs) [ Doppler equation]

given values for the first eagle,

(f=3800 Hz,v=330 m/s,vr=15 m/s,vs=17.2 m/s)

putting all the values,

E.1= 3800 * (330+15)/(330-17.2)= 4191.18 hz

second eagle receive ,

E.2=f*(v+vr)/(v-vs) [ Doppler equation]

now the given values for second eagle,

(f=3543 Hz,v=330m/s,vr=17.2 m/s,vs=15 m/s)

E.2 = 3543 * (330+17.2)/(330-15)= 3905.17 hz

so, the frequency received by both the eagles,

first eagle = 4191.18 hz

second eagle = 3905.17 hz

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