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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!"

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    $\begingroup$ Hi M_M, and welcome to Physics Stack Exchange! Interesting question, but here we expect people to have made a serious attempt to solve questions on their own before asking. Have you done any research or calculations to try to work this out yourself? If so, it would help if you edit the question to reflect that. $\endgroup$
    – David Z
    Jul 16, 2017 at 0:21
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    $\begingroup$ Some people (alack not myself) are fortunate enough to have perfect pitch "Perfect pitch," if it means absolute (rather than relative) pitch is not really a good thing to have. People who have it tend to find that it's a nuisance. $\endgroup$
    – user4552
    Jul 16, 2017 at 3:07
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    $\begingroup$ @BenCrowell It's terrible. If I go see a live concert, and they have to transpose the song down to fit into the vocalist's aging vocal range, it's brutal! $\endgroup$
    – Cort Ammon
    Jul 16, 2017 at 19:31
  • $\begingroup$ Logically yes: a former colleague was an on-call firefighter, witha two-tone siren for a ringtone. He ran fast enough that my unmusical ear could hear the Doppler shift between approaching and receding speeds. Based on Diracology's answer even a semitone is plausible from a sprinter (see my comment there too). Apparently the difference in frequency we can hear is ~1% varying with absolute pitch $\endgroup$
    – Chris H
    Jul 17, 2017 at 8:37
  • $\begingroup$ Hi @DavidZ, thanks for the welcome, and for criticising my first post in such a gentle, friendly manner. I have made attempts to work this out for myself since I thought of it about half a year ago. Without a background in physics (or in fact much mathematics), I don't find equations (are they differential, I'm not sure) scary so much as difficult to get to grips with. I've tried visualising the waves in my head in a manner similar to the animations on the Wikipedia page, but I (falsely, I'm glad to find) came to the conclusion that I would need to know the actual pitches, not just the ratios. $\endgroup$
    – M_M
    Jul 18, 2017 at 8:59

3 Answers 3

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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 $c\approx 1240\, km/h$.

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    $\begingroup$ I think your answer and @CortAmmon 's would benefit from working an example to show what velocity is needed to doppler the sound up/down a note. $\endgroup$
    – Dan Is Fiddling By Firelight
    Jul 16, 2017 at 23:18
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    $\begingroup$ Shouldn't that $u$ in the first sentence be a $v$? $\endgroup$
    – user96070
    Jul 17, 2017 at 9:48
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    $\begingroup$ @ChrisH I am not sure I understood you correctly but my I answered was supposed to deal with approaching and receding speeds so that you do not need absolute pitch, only pitch ratios. $\endgroup$
    – Diracology
    Jul 17, 2017 at 11:43
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    $\begingroup$ @Diracology I think my comment was based on a misreading of an earlier version of your post. My own comment makes no sense to me in the light of the current version (no "this post has changed" message on the mobile site might explain it) $\endgroup$
    – Chris H
    Jul 17, 2017 at 15:45
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    $\begingroup$ I'm going to mark this as my accepted answer because of the worked examples (thanks @Dan), which make the answer more accessible to me. That's not to say the other answers aren't valid and interesting. As a personal challenge/learning exercise, I may have to go and waste a lot of paper and pencil nib trying to work out speeds for a tritone, fourth and major third... $\endgroup$
    – M_M
    Jul 18, 2017 at 9:33
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You can do such estimations! It even turns out that you don't need perfect pitch. As a sanity check, picture a train going past you, blowing its horn. Its horn consists of three notes forming a chord (incidentally, the chord was chosen because it was annoying). Now let the train go by you. It's still the same chord, just with a lower root. If the doppler equations you are looking for depended on perfect pitch, it would mean the effect of the changing velocity would affect different pitches differently. Because you observe the same chord as it passes, just shifting downward as a whole, that tells you that the equations you are looking for are frequency independent. It's just the interval that matters!

The equation for doppler shift is

$$f=f_0\frac{c+v_r}{c+v_s}$$

Where $f_0$ is the emitted frequency, $c$ is the speed of the wave (aka speed of sound), $v_r$ is the velocity of the receiver and $v_s$ is the velocity of the source.

Now if you're moving with respect to the air, you'd need to know your speed. Perhaps you could determine this by listening to an object on the ground (like listening to the ding of the railway crossing bells as you go past them). But for simplicity, let's assume we're holding still. $v_r=0$. We can rearrange a bit to get:

$$v_s = c\left(\frac{f_0}{f}-1\right)$$

Now we can also look at the pitch in the opposite direction, which will be $f+\Delta$, where $\Delta$ is the difference in pitch between when its coming to you and away from you. $$ -v_s = c\left(\frac{f_0}{f+\Delta}-1\right)$$

We can combine these equations to get:

$$ \frac{f_0}{f}-1 = -\left(\frac{f_0}{f+\Delta}-1\right)$$ $$ \frac{f_0}{f} = -\frac{f_0}{f+\Delta}+2$$ $$ f_0f+f_0\Delta=-f_0f+2f(f+\Delta)$$ $$ f_0 = f\left(1+\frac{1}{2\frac{f}{\Delta}+1}\right)$$

Substituting in to the earlier equation:

$$ v_s=c\left(1+\frac{1}{2\frac{f}{\Delta}+1}\right)$$

Now the neat thing about this is, if I did my math correctly, the equations for velocity only depend on $\frac{f}{\Delta}$, which is information you'd pick up from the interval alone. No perfect pitch required!

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    $\begingroup$ I think your conclusion is correct, but the interval we tend to perceive is a ratio - a logarithmic not a linear difference. So, you hear $r$ where $1+(v_s/c)=f_0/f$ and $1-(v_s/c)=f_0/(r\,f)$. So I get $r=(1+v_s/c)/(1-v_s/c)$ and $v_s=c\,(r-1)/(r+1)$. Trains that go near my house make almost always just under a full tone shift ($r=9/8$) in passing when I'm waiting at the crossing to walk over the tracks, which I reckon to imply a speed of $30{\rm m\,s^{-1}}$, which is pretty spot on according to a train driver I know. So I've done this question in my head many times for the last 20 years ... $\endgroup$
    – Selene Routley
    Jul 16, 2017 at 0:54
  • $\begingroup$ ... I'd be very surprised if I've slipped up anywhere. Of course, one can calculate $r$ from $f/\Delta$ too, so I've just realized that your answer is, strictly speaking, correct so I've just given it a +1! $\endgroup$
    – Selene Routley
    Jul 16, 2017 at 0:54
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While not explicitly an answer, there is a nice historical connection to your question. In effect, the first public experiment that decisively illustrated the Doppler effect was almost exactly what you're describing.

In 1845, Christrophe Ballot placed one group of trumpet players on a moving train and another group at a station. Having tuned everyone up beforehand, he had both groups play and hold the same note as the train passed the station and observed the effects. Nothing like the classiness of using a group of musicians in a scientific experiment!

Further reading:

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