Answer:
f' = 3665.51 Hz
Explanation:
given,
speed of the hawk = 24.7 m/s
frequency of screech emitted by the hawk = 3400 Hz
speed of sound = 331 m/s
By Doppler's effect
[tex]f' = (\dfrac{v}{v-v_s}})f[/tex]
f' is the frequency received by the mouse
v is the speed of the sound
v_s is the speed of the hawk
now,
[tex]f' = \dfrac{341}{341-24.7}}\times 3400[/tex]
f' = 1.078 x 3400
f' = 3665.51 Hz
The frequency received by the stationary mouse is equal to 3665.51 Hz