Question

An eagle flying at 35 m/s emits a cry whose frequency is 440 Hz. A blackbird is moving in the same direction as the eagle at 10 m/s. (Assume the speed of sound is 343 m/s.)
(a) What frequency does the blackbird hear (in Hz) as the eagle approaches the blackbird?
Hz
(b) What frequency does the blackbird hear (in Hz) after the eagle passes the blackbird?
Hz

Answer 1

Answer:

a)  $F=475.7Hz$

b)  $F'=410.899Hz$

Explanation:

From the question we are told that:

Velocity of eagle $V_1=35m/s$

Frequency of eagle $F_1=440Hz$

Velocity of Black bird $V_2=10m/s$

Speed of sound $s=343m/s$

a)

Generally the equation for Frequency is mathematically given by

 $F=f_0(\frac{v-v_2}{v-v_1})$

 $F=440(\frac{343-10}{343-35})$

 $F=475.7Hz$

b)

Generally the equation for Frequency is mathematically given by

 $F'=f_0(\frac{v+v_2}{v+v_1})$

 $F'=440(\frac{343+10}{343+35})$

 $F'=410.899Hz$