Question

A mother hawk screeches as she dives at you. You recall from biology that female hawks screech at 800 Hz, but you hear the screech at 900 Hz. How fast is the hawk approaching?

Short answer

The hawk is approaching at a speed of$38.11m/s$.

Step-by-step solution

Given information

The female hawk screeched at a frequency of $f_0=800Hz$.

The frequency heard by you $f_{+}=900Hz$.

Speed of sound in air is$v=343m/s$.

The speed of hawk as it approaching you

Doppler effect for approaching source is given by $f_{+}=\frac{f_0}{1-{\displaystyle \frac{v_s}{v}}}$.

Substitute the given values

$$\begin{gathered} 900=\frac{800}{1-{\displaystyle \frac{v_s}{343}}} \\ \frac{v_s}{343}=1-0.89 \\ {v}_s=38.11m/s \end{gathered}$$

Therefore, the hawk is approaching at a speed of$38.11m/s$.