Question: A bat flies toward a moth at speed \({\bf{7}}{\bf{.8}}\,{\bf{m/s}}\) while the moth is flying toward the bat at speed \({\bf{5}}{\bf{.0}}\,{\bf{m/s}}\).The bat emits a sound wave of\({\bf{51}}{\bf{.35}}\,{\bf{kHz}}\). What is the frequency of the wave detected by the bat after that wave reflects off the moth?

The apparent frequency is given as,

Here,\(f\) is the actual frequency emitted from the source,\(v\) is the speed of sound in air,\({v_s}\) is the speed of the source and \({v_o}\) is the speed of the observer.

The speed of the bat is\({v_{bat}} = 7.8\;{\rm{m/s}}\).

The speed of moth towards the bat is\({v_{moth}} = 5.0\;{\rm{m/s}}\).

The sound emitted by the bat is\({f_{bat}} = 51.35\;{\rm{KHz}}\).

There will be two Doppler shifts in this problem. The first for the emitted sound with the bat as the source and the moth as the observer, the second one is for the bat as the observer and the moth as source. So, you can write, the apparent frequency heard by,

Second case gives the apparent frequency is,

Combine these two as,

\({f''_{bat}} = {f_{{\rm{bat}}}}\frac{{\left( {v + {v_{{\rm{moth}}}}} \right)\left( {v + {v_{{\rm{bat}}}}} \right)}}{{\left( {v - {v_{{\rm{bat}}}}} \right)\left( {v - {v_{{\rm{moth}}}}} \right)}}\)

Substitute the values and get,

\(\begin{array}{c}{{f''}_{bat}} = \left( {51.35\,{\rm{kHz}}} \right)\frac{{\left( {343\;{\rm{m/s}} + 5.0\;{\rm{m/s}}} \right)\left( {343\;{\rm{m/s}} + 7.8\;{\rm{m/s}}} \right)}}{{\left( {343\;{\rm{m/s}} - 7.8\;{\rm{m/s}}} \right)\left( {343\;{\rm{m/s}} - 5.0\;{\rm{m/s}}} \right)}}\\ \approx 55.3\,{\rm{kHz}}\end{array}\)

Hence, the apparent frequency detected by the bat is\(55.3\,{\rm{kHz}}\).