A fire engine is travelling at a constant velocity towards a stationary observer. Its siren emits a note of constant frequency. As the engine passes close to the observer, the frequency of the note perceived by the observer decreases. Explain this decrease in terms of the wavefronts of the note emitted by the siren.

The frequency of the note emitted by the siren is 400 Hz. After the fire engine has passed, the frequency of the note detected by the observer is 360 Hz. Calculate the speed of the fire engine. (Take the speed of sound in air to be 340 ms^–1.)

diagram showing (non concentric) wavefronts closer together in front/further apart behind source;
frequency is higher as source approaches (because more wavefronts are received per unit of time);
frequency is lower as source recedes (because fewer waverfronts are received per unit of time);

\(360 = 400\left( {\frac{{340}}{{340 + {u_{\rm{s}}}}}} \right)\);
u_s=38ms^–1;
Award [2] for a bald correct answer.