Solution: Accelerating Detector (Doppler Effect)
The detector is moving away. Relative speed of sound is \(v – v_D\). $$ f’ = f \left( \frac{v – v_D}{v} \right) $$ As \(v_D\) increases, \(f’\) decreases.
When \(v_D\) exceeds the speed of sound \(v\), the detector is moving faster than the waves it is chasing. However, it is continuously overtaking the “old” waves that were emitted by the source earlier and are traveling ahead of it.
The relative speed at which it crosses these old wavefronts is \(v_{rel} = v_D – v\). The magnitude of the frequency detected is: $$ f’ = f \left( \frac{v_D – v}{v} \right) $$ As \(v_D\) increases further, this frequency increases.
Eventually, the detector will travel so far and fast that it outruns the very first wavefront ever emitted by the source. Once it crosses this boundary, it enters the “zone of silence” ahead of the entire wave train.