Why are there so many Doppler Effect formulas?A Doppler Effect problem with a moving mediumThe Doppler effect in a medium like air (sound) versus the electromagnetic Doppler effectApparent frequency and wavelength in the Doppler effectDoes “apparent frequency” mean the Doppler effect is not an actual physical effect?How to calculate the Doppler effect on a wave reflection when both the source and the listener are moving?Doppler Effect and Energy variation
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Why are there so many Doppler Effect formulas?
A Doppler Effect problem with a moving mediumThe Doppler effect in a medium like air (sound) versus the electromagnetic Doppler effectApparent frequency and wavelength in the Doppler effectDoes “apparent frequency” mean the Doppler effect is not an actual physical effect?How to calculate the Doppler effect on a wave reflection when both the source and the listener are moving?Doppler Effect and Energy variation
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margin-bottom:0;
$begingroup$
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:
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:
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:
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:
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:
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$$Relative$ = $vecV$$Car A$ + $vecV$$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:
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!
waves frequency wavelength relative-motion doppler-effect
asked Aug 9 at 14:55
Kevin NKevin N
5982 silver badges12 bronze badges
$endgroup$
add a comment
|
$begingroup$
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:
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:
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:
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:
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:
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$ = $vecV$$Car A$ + $vecV$$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:
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!
waves frequency wavelength relative-motion doppler-effect
asked Aug 9 at 14:55
Kevin NKevin N
5982 silver badges12 bronze badges
$endgroup$
$begingroup$
The Doppler shift of electromagnetic radiation does obey a single formula that depends only on the relative velocity of source and observer.
$endgroup$
– dmckee♦
Aug 9 at 23:34
add a comment
|
$begingroup$
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:
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:
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:
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:
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:
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$$Relative$ = $vecV$$Car A$ + $vecV$$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:
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!
waves frequency wavelength relative-motion doppler-effect
asked Aug 9 at 14:55
Kevin NKevin N
5982 silver badges12 bronze badges
$endgroup$
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:
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:
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:
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:
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:
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$ = $vecV$$Car A$ + $vecV$$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:
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!
waves frequency wavelength relative-motion doppler-effect
waves frequency wavelength relative-motion doppler-effect
asked Aug 9 at 14:55
Kevin NKevin N
5982 silver badges12 bronze badges
asked Aug 9 at 14:55
Kevin NKevin N
5982 silver badges12 bronze badges
edited Aug 10 at 2:43
Kevin N
asked Aug 9 at 14:55
Kevin NKevin N
5982 silver badges12 bronze badges
asked Aug 9 at 14:55
Kevin NKevin N
5982 silver badges12 bronze badges
asked Aug 9 at 14:55
Kevin NKevin N
5982 silver badges12 bronze badges
5982 silver badges12 bronze badges
$begingroup$
The Doppler shift of electromagnetic radiation does obey a single formula that depends only on the relative velocity of source and observer.
$endgroup$
– dmckee♦
Aug 9 at 23:34
add a comment
|
$begingroup$
The Doppler shift of electromagnetic radiation does obey a single formula that depends only on the relative velocity of source and observer.
$endgroup$
– dmckee♦
Aug 9 at 23:34
$begingroup$
The Doppler shift of electromagnetic radiation does obey a single formula that depends only on the relative velocity of source and observer.
$endgroup$
– dmckee♦
Aug 9 at 23:34
$begingroup$
The Doppler shift of electromagnetic radiation does obey a single formula that depends only on the relative velocity of source and observer.
$endgroup$
– dmckee♦
Aug 9 at 23:34
add a comment
|
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered Aug 9 at 15:12
David WhiteDavid White
5,6832 gold badges7 silver badges22 bronze badges
$endgroup$
$begingroup$
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?
$endgroup$
– Kevin N
Aug 10 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 at 15:49
add a comment
|
$begingroup$
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$$textUse the medium as your frame of reference for all velocity measurements!$$ $$textConvert all velocities to this frame, before substituting them into the equation!$$
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.)
answered Aug 9 at 17:41
DJohnMDJohnM
8,8902 gold badges20 silver badges24 bronze badges
$endgroup$
$begingroup$
What about EM waves where there is no medium?
$endgroup$
– user45664
Aug 9 at 19:24
$begingroup$
Now you're talking relativistic Doppler effect...
$endgroup$
– DJohnM
Aug 9 at 21:31
1
$begingroup$
That's quite a game of chicken.
$endgroup$
– badjohn
Aug 10 at 7:46
add a comment
|
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Aug 9 at 15:12
David WhiteDavid White
5,6832 gold badges7 silver badges22 bronze badges
$endgroup$
$begingroup$
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?
$endgroup$
– Kevin N
Aug 10 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 at 15:49
add a comment
|
$begingroup$
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.
answered Aug 9 at 15:12
David WhiteDavid White
5,6832 gold badges7 silver badges22 bronze badges
$endgroup$
$begingroup$
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?
$endgroup$
– Kevin N
Aug 10 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 at 15:49
add a comment
|
$begingroup$
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.
answered Aug 9 at 15:12
David WhiteDavid White
5,6832 gold badges7 silver badges22 bronze badges
$endgroup$
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.
answered Aug 9 at 15:12
David WhiteDavid White
5,6832 gold badges7 silver badges22 bronze badges
answered Aug 9 at 15:12
David WhiteDavid White
5,6832 gold badges7 silver badges22 bronze badges
answered Aug 9 at 15:12
David WhiteDavid White
5,6832 gold badges7 silver badges22 bronze badges
answered Aug 9 at 15:12
David WhiteDavid White
5,6832 gold badges7 silver badges22 bronze badges
5,6832 gold badges7 silver badges22 bronze badges
$begingroup$
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?
$endgroup$
– Kevin N
Aug 10 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 at 15:49
add a comment
|
$begingroup$
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?
$endgroup$
– Kevin N
Aug 10 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 at 15:49
$begingroup$
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?
$endgroup$
– Kevin N
Aug 10 at 7:14
$begingroup$
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?
$endgroup$
– Kevin N
Aug 10 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 at 15:49
$begingroup$
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 at 15:49
add a comment
|
$begingroup$
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$$textUse the medium as your frame of reference for all velocity measurements!$$ $$textConvert all velocities to this frame, before substituting them into the equation!$$
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.)
answered Aug 9 at 17:41
DJohnMDJohnM
8,8902 gold badges20 silver badges24 bronze badges
$endgroup$
$begingroup$
What about EM waves where there is no medium?
$endgroup$
– user45664
Aug 9 at 19:24
$begingroup$
Now you're talking relativistic Doppler effect...
$endgroup$
– DJohnM
Aug 9 at 21:31
1
$begingroup$
That's quite a game of chicken.
$endgroup$
– badjohn
Aug 10 at 7:46
add a comment
|
$begingroup$
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$$textUse the medium as your frame of reference for all velocity measurements!$$ $$textConvert all velocities to this frame, before substituting them into the equation!$$
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.)
answered Aug 9 at 17:41
DJohnMDJohnM
8,8902 gold badges20 silver badges24 bronze badges
$endgroup$
$begingroup$
What about EM waves where there is no medium?
$endgroup$
– user45664
Aug 9 at 19:24
$begingroup$
Now you're talking relativistic Doppler effect...
$endgroup$
– DJohnM
Aug 9 at 21:31
1
$begingroup$
That's quite a game of chicken.
$endgroup$
– badjohn
Aug 10 at 7:46
add a comment
|
$begingroup$
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$$textUse the medium as your frame of reference for all velocity measurements!$$ $$textConvert all velocities to this frame, before substituting them into the equation!$$
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.)
answered Aug 9 at 17:41
DJohnMDJohnM
8,8902 gold badges20 silver badges24 bronze badges
$endgroup$
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$$textUse the medium as your frame of reference for all velocity measurements!$$ $$textConvert all velocities to this frame, before substituting them into the equation!$$
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.)
answered Aug 9 at 17:41
DJohnMDJohnM
8,8902 gold badges20 silver badges24 bronze badges
answered Aug 9 at 17:41
DJohnMDJohnM
8,8902 gold badges20 silver badges24 bronze badges
answered Aug 9 at 17:41
DJohnMDJohnM
8,8902 gold badges20 silver badges24 bronze badges
answered Aug 9 at 17:41
DJohnMDJohnM
8,8902 gold badges20 silver badges24 bronze badges
8,8902 gold badges20 silver badges24 bronze badges
$begingroup$
What about EM waves where there is no medium?
$endgroup$
– user45664
Aug 9 at 19:24
$begingroup$
Now you're talking relativistic Doppler effect...
$endgroup$
– DJohnM
Aug 9 at 21:31
1
$begingroup$
That's quite a game of chicken.
$endgroup$
– badjohn
Aug 10 at 7:46
add a comment
|
$begingroup$
What about EM waves where there is no medium?
$endgroup$
– user45664
Aug 9 at 19:24
$begingroup$
Now you're talking relativistic Doppler effect...
$endgroup$
– DJohnM
Aug 9 at 21:31
1
$begingroup$
That's quite a game of chicken.
$endgroup$
– badjohn
Aug 10 at 7:46
$begingroup$
What about EM waves where there is no medium?
$endgroup$
– user45664
Aug 9 at 19:24
$begingroup$
What about EM waves where there is no medium?
$endgroup$
– user45664
Aug 9 at 19:24
$begingroup$
Now you're talking relativistic Doppler effect...
$endgroup$
– DJohnM
Aug 9 at 21:31
$begingroup$
Now you're talking relativistic Doppler effect...
$endgroup$
– DJohnM
Aug 9 at 21:31
1
1
$begingroup$
That's quite a game of chicken.
$endgroup$
– badjohn
Aug 10 at 7:46
$begingroup$
That's quite a game of chicken.
$endgroup$
– badjohn
Aug 10 at 7:46
add a comment
|
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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♦
Aug 9 at 23:34