Exploring the Doppler Effect: Frequency & Distance

[JeremyG]

Hi all, I was just thinking about the Doppler effect today and I was wondering why distance between the source and listener does not affect the frequency experienced by the listener.

Consider the Doppler formula: fL = fS*(v+vL)/(v+vS), with the positive direction taken from listener to source

and the following situation: Police car with siren driving at a speed of 20m/s behind a criminal getaway car moving at a speed of 15m/s (Police car is behind the criminal's car at this point in time, and frequency of siren emitted by police car is 400Hz). As the police car catches up to the criminal, the frequency heard by the criminal will be shifted above 400Hz, from the Doppler formula.

The opposite effect occurs after the police car catches up to the criminal and is now in front of the criminal (Assume that both cars still travel at their constant velocities). That is, the frequency heard by the criminal will now be less than 400Hz.

What is confusing me is when this transition from >400Hz to <400Hz occurs if both cars are always traveling with different velocities. As the police car catches up to the criminal, from the formula it seems that up till the point that the police car overtakes the criminal, the criminal hears a constant >400Hz frequency siren.

What happens when the police car is side by side with the criminal car? What frequency does the criminal hear then? 400Hz?

Then when the police car overtakes the criminal car, all of a sudden the criminal hears a <400Hz frequency siren? Which then remains constant as the police car travels further and further from the criminal car? That just seems non-intuitive to me. Could it be a logic flaw, or a wrong application of the Doppler formula?

I would appreciate any help in aiding my understanding for this topic! Thanks!

 

[andrewkirk]

The ratio of observed to emitted frequency is $\frac{c+v_r}{c+v_s}$ where $c$ is the speed of sound, $v_r$ is the speed of the receiver (hearer) through the air and $v_s$ is the speed of the source (emitter/siren) through the air. Location and separation do not appear in this formula, so they have no effect.

What happens when the police car is side by side with the criminal car? What frequency does the criminal hear then? 400Hz?

Yes, when the police car pulls alongside the pursued car. A more accurate description of the above formula is that $v_r$ and $v_s$ are the components of the velocities of receiver and source in the direction of the vector connecting the two. In that case, relative location does matter, but only from the point of view of relative direction, not relative distance.

When the police car is alongside, the velocities have zero component in the direction of the vector between the two cars, so there is no Doppler effect.

We all observe this last feature when an ambulance rushes by. The observed frequency ratio smoothly but rapidly changes from the above formula to its reciprocal as the ambulance passes us. The change appears to happen in a second but actually it is an asymptotic transition at both ends.

 

[DrClaude]

What you have to consider is the the relative velocity is itself discontinuous. As the police car catches the getaway car, the relative velocity ($v_L - v_S = v$) goes through zero to $-v$. Only for an infinitesimal instant is it zero, for which indeed there would be no Doppler effect.

Note also that this is strictly true only in 1D. In reality, as the police car gets closer, the relative velocity will vary smoothly as the police car swerves not to collide with the getaway car.

 

[JeremyG]

the velocities have zero component in the direction of the vector between the two cars, so there is no Doppler effect.

What you have to consider is the the relative velocity is itself discontinuous.

Indeed, I have failed to consider that at the point where both cars are at the same position, there is no direction vector to speak of in a 1D situation. Thank you to both gentlemen! :D