Question

An osprey’s call is a distinct whistle at 2200 Hz. An osprey calls while diving at you, to drive you away from her nest. You hear the call at 2300 Hz. How fast is the osprey approaching?

Answer 1

To solve this problem we will apply the concepts related to the Doppler effect. The expression for the observed frequency that is related to the original frequency is given by

`f = f_0 ((v)/(v-v_s))`

Here

`f_0` = Original Frequency

`f` = Observed Frequency

`v_s` = Speed of the Osprey

`v` = Speed of the sound wave

Rearranging the expression to find the Speed of the Osprey

`v_s = v (1-(f_0)/(f))`

Replacing with our values we have that

`v_s = (343m/s)(1-(2200Hz)/(2300Hz))`

`v_s = 14.9m/s`

Therefore the speed of the Osprey when it approaches the listener is
`14.9m/s`