Train Doppler

Puzzle: You’re traveling on a train and another train passes you in the opposite direction. As it does so, it honks its horn and you hear the drop in pitch of the horn from the Doppler effect. You estimate that the drop in pitch is about a minor-third interval. (A minor third is the difference between the note middle-C and the E-flat above it. You can guess this by singing a scale starting at the lower note until you hit the higher note.) How fast are the trains going? Assume both trains are going at the same speed.

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Hint: Do web searches to get the equations for ratios of notes, the Doppler formula for sound, and the quadratic formula.

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Answer: 32 MPH

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Solution: There are 12 steps in an octave, and going up one octave doubles the frequency. All steps are the same ratio from one another. This makes the ratio between each step be the twelfth root of 2, or about 1.059. (You use the twelfth root of 2 so that when you go 12 steps the frequency will have been multiplied by 2.) A minor third is 3 steps. To go up 3 steps (or $1 \over 4$ of an octave) you apply the ratio three times, to get about 1.19. (This is also the fourth root of two because it’s $1 \over 4$ of an octave.) So the ratio of the sound frequencies before and after the train passed is about 1.19, which is all you need. You don’t need the actual frequencies.

The equation for the Doppler effect for sound is:

$$Fl = Fs {1 + {Vl \over c}\over 1 - {Vs \over c}}$$

where $Fl$ is the frequency that the listener hears, $Fs$ is the one the sender sends, $Vl$ is the velocity of the listener relative to the air (towards the sender), $Vs$ is the same for the sender (towards the listener), and $c$ is the speed of sound (330 meters per second).

We assume that the trains are going the same speed, so replace $Vl$ and $Vs$ with $V$, the speed of the trains. This value is positive when the trains are going towards one another and negative when they’re separating. We also assume that the air isn’t moving (no significant wind).

We have two instances of the above equation: one for before the train passes and one for after. $Fs$ is the same in both, and we don’t know it, but it doesn’t matter because it cancels out with itself. The two $Fl$’s are different by the ratio above (1.19), their actual values don’t matter. $V$ is positive in the first equation (trains going towards one another) and negative in the second (same magnitude).

$$Fs {1 + {V \over c} \over 1 - {V \over c}} = 1.19 Fs {1 - {V \over c} \over 1 + {V \over c}}$$

Solving for $V$ gives you a quadratic equation ($A V^2 + B V + C = 0$), which you can solve using the quadratic formula:

$$V = {-B \pm \sqrt{{B^2} - 4 A C} \over 2 A}$$

The plus/minus ($\pm$) means that you get two values for $V$. If you choose minus you get about 32 MPH. If you choose plus you get 17,053 MPH.

The formula for Doppler for light is simpler because you only need the relative velocities of the objects, not their velocities compared to a fixed medium. The results are close enough for speeds that are small compared to the speed of sound or light. If the trains had been going faster (like the TGV in France, about 200 MPH), it would have made a larger difference.

One thing I find unsettling is this plus/minus thing at the end. I was on the train yesterday morning when this happened, so I know that I wasn’t going 17,053 MPH, but what if this number wasn’t something that I could easily guess the answer of, something more abstract? I might have no idea whether 32 or 17,053 is correct. Weird.

(Note added October 7, 2016: Junjiajia Long points out that 17,053 MPH is faster than the speed of sound, meaning that this speed difference would make the Doppler-shifted frequency negative and the train would leave behind a shock wave.)

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