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I was just introduced at my class, a phenomenon, known as the doppler effect, where the observed frequency increases as the source/observer approaches each other, but decreases, if they were moving away. My teacher then told me this equation:

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At first glance, it was intuitive for me that this applies to all scenarios, whether the source is moving, the observer, or both. An example scenario would be:

enter image description here

Assume, a car moving towards person $Y$, a stationary observer, with a velocity of $7m/s$, producing a frequency of $500Hz$. If $Vsound=330m/s$, inserting the values onto the equation gives us $510.8Hz$ as the observed frequency. So far so good.

The confusion came when my teacher "switched" the scenario , which in this case, the observer moves, while the source remains stationary. Here's the illustration:

enter image description here

Person $Y$ is running towards the car, the source. Let's say with a velocity of $7m/s$, the exact same velocity the car was moving in the previous case, without any further consideration, I wrote the same answer, $510.8Hz$, as the observed frequency. I was pretty confident that How could this be wrong?

Surprisingly, my answer was incorrect, she said you need another formula for this, which she showed:

enter image description here

I took a minute to look at it, and my mind said, "it's not wrong either...". Using that equation, the observed frequency, $f'$ , would be $510.6Hz$, which is very close to the prior observed frequency ($510.8Hz$), but different is different. I didn't want to waster her time, so I carried on. While she was teaching, and even after the class, I kept thinking why my logic was incorrect.

Basically, my logic is based solely upon Relative Velocity, which you are already familiar with, here's just a quick analogy:

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The red car would be moving $80m/s$ relative to the black car, while the black car moves $50m/s$ , but opposite in direction relative to the red one, which basically means

$V$$Relative$ = $\vec{V}$$Car A$ + $\vec{V}$$Car B$

If I were to use this logic in Doppler Effect, it doesn't matter whether the car is moving pass the Stationary Person $Y$, or Person $Y$ moving pass the Stationary Car, the relative velocity would always be $7m/s$. In other words, if the car and Person $Y$ were moving at $3.5m/s$ towards each other, it would be the same as if the Car was moving pass the Stationary Person $Y$, with $v=7m/s$.

This also applies to any other conditions, whether both the source & observer are moving towards each other, away, or even at the same direction(just at different velocities). And just plug it in onto the first equation, and DONE.

But no, it doesn't work that way! Here are all the 8 formulas for each scenario! Although they're not entirely "different formulas", but still, that'll be a pain in my head to remember. I decided to research more and I found a super formula, one that unites them all:

enter image description here

Which is basically, and apparently my teacher didn't tell me, the Main Formula.

Main Question

Nevertheless, going back to my curiosity, why can't we simply find it's relative velocity, for every scenario, and plug the values into the first equation? Or rather how does approaching the source, or the source approaching you, with the same exact relative velocity, create different $f'$ (Observed Frequency)?

Any help would be greatly appreciated!

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  • $\begingroup$ The Doppler shift of electromagnetic radiation does obey a single formula that depends only on the relative velocity of source and observer. $\endgroup$ – dmckee --- ex-moderator kitten Aug 9 '19 at 23:34
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Relative velocity for sound waves is not a "symmetric" situation. For example, in the extreme case of a fighter jet approaching a stationary observer at Mach 1, the jet will be traveling as fast as its sound waves. The observer will not hear anything until the jet gets to his position.

On the other hand, for the case of a stationary sound source and an observer moving toward the sound source at Mach 1, the observer will obviously hear a doppler shifted sound of a much higher frequency than what the sound source is emitting.

Such a non-symmetric situation requires the last formula listed in the question, where corrections are made for both the velocity of the source and the velocity of the observer.

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  • $\begingroup$ If the jet fighter was to fly at Mach 1 and passed over an observer, would the observed frequency be the same as the source's frequency? $\endgroup$ – user226894 Aug 10 '19 at 7:14
  • $\begingroup$ When the jet is just passing by the observer, the answer is yes, but the observed frequency would be obscured by a sonic boom. Right after that, the observed frequency would be somewhat lower than the actual frequency. $\endgroup$ – David White Aug 10 '19 at 15:49
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There is one factor missing from your consideration: the velocity of the medium!

Take the general case of two cars driving at different speeds towards each other while both blow their horns.

You could take either car as being stationary, and wind up with the same relative velocity (car to car).

But substituting this into the various equations you cite would produce different results for the one correct value for the observed frequency. Only one equation can be correct.

The solution: Use the last, most general equation you cite, the super formula, but include in the rules for using the equation$$\text{Use the medium as your frame of reference for all velocity measurements!}$$ $$\text{Convert all velocities to this frame, before substituting them into the equation!}$$

This even handles the situation where the two cars are driving toward each other on the deck of an aircraft carrier steaming at $30$ knots into a $35$ knot headwind...(Some unit conversion required.)

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  • $\begingroup$ What about EM waves where there is no medium? $\endgroup$ – user45664 Aug 9 '19 at 19:24
  • $\begingroup$ Now you're talking relativistic Doppler effect... $\endgroup$ – DJohnM Aug 9 '19 at 21:31
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    $\begingroup$ That's quite a game of chicken. $\endgroup$ – badjohn Aug 10 '19 at 7:46

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