I've found a number of questions that concern the Doppler effect, but none that seem to address my question.
I have a background in music. People with a musical ear can generally tell the ratio between two frequencies (as a musical interval). For anyone who's not already aware, we perceive a ratio of 2:1 as an octave, 3:2 as a perfect fifth, 4:3 as a perfect fourth, 5:4 as a major third and 6:5 as a minor third.
Therefore, if a vehicle passes at speed, and I perceive that the frequency of the engine noise drops by a fourth as it does so, then I know that the ratio of the frequencies (approaching:departing) is 4:3.
Is this information (just the ratio of the frequencies) enough, coupled with an assumed speed of sound of around 330m/s, to calculate the speed at which the vehicle passed? We'll assume the car passed quite close by, so can be considered to be coming almost directly towards me when approaching, and almost directly away when departing.
At this point, we don't know the actual frequency of the sound – just the relative frequencies.
Some people (alack not myself) are fortunate enough to have perfect pitch, in which case they could even estimate the exact frequencies. let's assume 220Hz and 165Hz. Is this extra information helpful/needed to ascertain the speed of the passing vehicle?
I'm not interested in telling the difference between 35 and 38mph. More like "By the sound of it, that must have been going at least 80mph!"
Best Answer
Let us consider that you are at rest and the car, which emits at frequency $f_0$, approaches you with speed $v$. The frequency you receive increases to $$f_1=f_0\frac{c}{c-v},$$ where $c$ is the speed of sound. When the car get passed you the perceived frequency is reduced to $$f_2=f_0\frac{c}{c+v}.$$ The ratio is $$\frac{f_1}{f_2}=\frac{c+v}{c-v}.$$ Now solve this equation for $v$, $$v=\frac{r-1}{r+1}c,$$ where $r=f_1/f_2$.
Edit
Let us consider some examples. If the ratio corresponds to an octave (2:1), $r=2$, the car speed is $c/3\approx400\, km/h$, and that should be a Bugatti Veyron. If you notice a fifth (3:2), $r=3/2$ and $v\approx 240\, km/h$, which may be a nice sport car. A minor third (6:5), $r=6/5$, corresponds to $v\approx 110\, km/h$ which can even be a bus. For a difference in frequency corresponding to a semitone, $r\approx 1.06$, the speed is about $36\, km/h$ and for a tone, $r\approx 1.12$, the result is $v\approx 70\, km/h$. In all examples the speed of sound was taken as $c\approx 1240\, km/h$.