What frequency is received by a mouse just before being dispatched by a hawk flying at it at \(25.0\;{\rm{m/s}}\)and emitting a screech of frequency\({\rm{3500}}\;{\rm{Hz}}\)? Take the speed of sound to be\(331\;{\rm{m/s}}\).

The apparent frequency of a moving source depends on the relative speed between the source and the listener.

The frequency is\(f = 3500\;{\rm{Hz}}\).

The speed of the ambulance is\({v_s} = 25.0\;{\rm{m/s}}\).

The speed of sound is\(v = 331\;{\rm{m/s}}\).

The speed of the listener is zero.

The Doppler Effect tells the apparent frequency is,

\(f' = f\left( {\frac{{v - {v_o}}}{{v - {v_s}}}} \right)\)

Now, the apparent frequency is,

\(\begin{array}{c}f' = 3500\left( {\frac{{331 - 0}}{{331 - 25}}} \right)\\ = 3500 \times \frac{{331}}{{306}}\\ = 3785.94\;{\rm{Hz}}\end{array}\)

Now, the apparent frequency of the passing away ambulance is,

\(\begin{array}{c}f'' = 3500\left( {\frac{{342 - 0}}{{342 - \left( { - 333.3} \right)}}} \right)\\ = 3500 \times \frac{{1140}}{{2251}}\\ = 1772.5\;{\rm{Hz}}\end{array}\)

Hence, the apparent frequencies are \(3785.94\;{\rm{Hz}}\)and \(1772.5\;{\rm{Hz}}\).