The Doppler Effect (in general)

[Urmi Roy]

I have a problem in regard to the doppler effect,which may be generalised to all waves--sound,water etc.

Please explain why the observed frequency increases as the object approaches an observer and then decreases as the object passes the observer.Actually,I thought that since doppler effect depends only on the relative velocity between observer and source,the observed frequency should be constant throughout the process,as the relative velocity does not change as the distance between the source and observer changes.

I also found on a website that it said that 'the observed frequency of an approaching object declines monotonically from a value above the emitted frequency, through a value equal to the emitted frequency when the object is closest to the observer, and to values increasingly below the emitted frequency as the object recedes from the observer'---Does this mean the same thing as what I asked at the begginning or is it different?

Also,does the intensity of the sound increase as an object approaches an observer and decrease once it passes and recedes from the observer?

 

[tiny-tim]

Please explain why the observed frequency increases as the object approaches an observer and then decreases as the object passes the observer.Actually,I thought that since doppler effect depends only on the relative velocity between observer and source,the observed frequency should be constant throughout the process,as the relative velocity does not change as the distance between the source and observer changes.

I also found on a website that it said that 'the observed frequency of an approaching object declines monotonically from a value above the emitted frequency, through a value equal to the emitted frequency when the object is closest to the observer, and to values increasingly below the emitted frequency as the object recedes from the observer'---Does this mean the same thing as what I asked at the begginning or is it different?

I think you're assuming that the object is heading straight for the observer.

In practice, it usually misses the observer, and so the relative speed does change continuously.

Also,does the intensity of the sound increase as an object approaches an observer and decrease once it passes and recedes from the observer?

Yes.

 

[Doc Al]

Adding to tiny-tim's comments:

Actually,I thought that since doppler effect depends only on the relative velocity between observer and source,the observed frequency should be constant throughout the process,as the relative velocity does not change as the distance between the source and observer changes.

The relative velocity certainly changes: You start out moving toward the source and end up moving away from it.

 

[DeepSeeded]

I think you're assuming that the object is heading straight for the observer.

In practice, it usually misses the observer, and so the relative speed does change continuously.

Well let's say it was heading stright for the observer, and the observer could see both forward and behind. He is right, relativley there is no difference, the doppler effect collapses...

 

[Doc Al]

Well let's say it was heading stright for the observer, and the observer could see both forward and behind. He is right, relativley there is no difference, the doppler effect collapses...

Not quite. See my response above.

 

[DeepSeeded]

Adding to tiny-tim's comments:

The relative velocity certainly changes: You start out moving toward the source and end up moving away from it.

However, in both cases the light is always heading towards the observer after being emitted from the object.

 

[Doc Al]

However, in both cases the light is always heading towards the observer after being emitted from the object.

So? The light/sound always moves toward the observer (relatively) otherwise you won't see/hear anything.

What matters for the Doppler effect is your velocity relative to the source. In one case you move towards the source; in the other, you move away. Big difference.

 

[DeepSeeded]

O well, it was fun while it lasted :)

 

[vin300]

Adding to tiny-tim's comments

The relative velocity certainly changes: :

Oh ho Doc, say yours is the reference frame. The velocity of the train is as it is after it passes you

 

[HallsofIvy]

I think Doc Al and tiny tim are talking about different situations. tiny tim is talking about a situation where the motion of the source of sound is slightly off to one side of the observer, the velocity is continuously changing so the frequency heard is continuously changing. That is also my interpretation of the original question. If the source is moving directly toward the observer, the frequency is constant (above that of the emitted signal) until the source passes then suddenly drops below the frequency of the emitted signal. That is the situation Doc Al is referring to. Note that in the first case, while the change in frequency is continuous, it is not linear. There will be relatively little change in the frequency when the source is farther off, most change when the source is nearest. That effect increases when the point of "closest approach" to the observer is closer to the observer- that is when the source just misses the observer.

 

[tiny-tim]

The velocity of the train is as it is after it passes you

It is as it is, is it? Or is it after being as it is, that it is?

 

[Doc Al]

Oh ho Doc, say yours is the reference frame. The velocity of the train is as it is after it passes you

In this context, "relative velocity" refers to the radial component of the rate of change of the position vector of the source with respect to you (the observer). As the source approaches, the length of that vector is decreasing; as the source recedes, its length increases.

 

[DeepSeeded]

Wait a minute, with light it doesn't matter what your relative velocity is, light is only one speed c, invariant. So... If you had a light emitting diode traveling to you, then away from you, what's the difference? The light is traveling at speed c to you at all times.

Next time I get pulled over I know what my argument is going to be. "Its only a 4 cylinder" isn't working anymore.

 

[Doc Al]

Wait a minute, with light it doesn't matter what your relative velocity is, light is only one speed c, invariant. So... If you had a light emitting diode traveling to you, then away from you, what's the difference? The light is traveling at speed c to you at all times.

Again, it's the speed of the source that matters, not the speed of the light. Same thing with sound: In still air, the speed of sound with respect to a stationary observer will be the same, regardless of the speed of the source. (In fact, that was the example that started this thread.)

 

[Urmi Roy]

situation where the motion of the source of sound is slightly off to one side of the observer, .

I was not aware of any difference between relative motion when the source of sound is slightly off to one side of the observer and when they are not.

What is the cause for this difference?

the velocity is continuously changing so the frequency heard is continuously changing...the frequency of the emitted signal.

What I originally meant to say is that the observed frequency continuously changes even when the source and observer are moving toward each other,and separately when they are moving away from each other,and also that at the point of closest approach,the observed frequency equals the real frequency.

Doc Al said that the observed frequency changes only when the relative motion changes from 'moving toward' to 'moving away'---that's understandable,but I came across a source which said that the observed frequency changes continuously during the interval that the source and observer move toward each other and also during the interval that they move away from each other.
This is what I don't understand--on these individual intervals,there is no change in relative velocity of 'moving toward' or 'moving away'.

Again at the point of closest approach,there is still relative motion between source and observer,so shouldn't there be a different observed freqeuncy even here?

Note that in the first case, while the change in frequency is continuous, it is not linear. There will be relatively little change in the frequency when the source is farther off, most change when the source is nearest. That effect increases when the point of "closest approach" to the observer is closer to the observer- that is when the source just misses the observer.

This seems to confirm that the observed frequency does indeed change continuously during the 'moving towards each other' part and then separately on the 'moving away from each other' part,and that the observed frequency is in some way related to the 'distance between the source and observer.
I really don't understand it!

 

[Urmi Roy]

Note that in the first case, while the change in frequency is continuous, it is not linear...

Is this the same thing that I found on the wesite that said 'the observed frequency of an approaching object declines monotonically from a value above the emitted frequency, through a value equal to the emitted frequency when the object is closest to the observer, and to values increasingly below the emitted frequency as the object recedes from the observer'??

 

[vin300]

The doppler effect as you think holds in the ideal condition that the relative velocity is the same throughout, but in real cases it becomes a monotonically decreasing curve.
The doppler effect would be as you thought if you stood on linear track of an approaching train with constant velocity and the train went through you, but you stand by the side and the component of velocity vector towards you goes on decreasing and becomes zero when closest to you and agin goes on increasing like the velocity vector of a parabolic curve when a ball is thrown up, if you were to see it facing perpendicular to the Earth in the air at the point of 0 velocity.
The formulae still hold true but there is change in frequency as there is change in velocity.

 

[Doc Al]

What I originally meant to say is that the observed frequency continuously changes even when the source and observer are moving toward each other,and separately when they are moving away from each other,and also that at the point of closest approach,the observed frequency equals the real frequency.

Doc Al said that the observed frequency changes only when the relative motion changes from 'moving toward' to 'moving away'---that's understandable,but I came across a source which said that the observed frequency changes continuously during the interval that the source and observer move toward each other and also during the interval that they move away from each other.
This is what I don't understand--on these individual intervals,there is no change in relative velocity of 'moving toward' or 'moving away'.

If the source is moving directly towards you, then the observed frequency is shifted but steady; similarly, if the source moves directly away from you. In such a case the observed frequency doesn't continually change.

Of course, that situation is unrealistic. Usually, the source doesn't come directly at you, otherwise you'll be hit. It passes by you. If you imagine a line drawn from you to the source, the length of that line--which represents the distance between the source and observer--changes as the source approaches you (at an angle), passes by you, and then recedes from you. The rate of change of that distance is the "relative velocity" that we are concerned with. That rate of change varies continuously as the source moves.

Again at the point of closest approach,there is still relative motion between source and observer,so shouldn't there be a different observed freqeuncy even here?

The only relative motion that counts (for the non-relativistic Doppler effect, at least) is motion toward or away from the observer. At the point of closest approach the source is moving sideways with respect to the observer, thus the radial velocity (the rate of change of the distance between source and observer) is momentarily zero. At that moment the source is neither getting closer or farther from the observer.

 

[Urmi Roy]

Usually, the source doesn't come directly at you, otherwise you'll be hit. It passes by you. If you imagine a line drawn from you to the source, the length of that line--which represents the distance between the source and observer--changes as the source approaches you (at an angle), passes by you, and then recedes from you. The rate of change of that distance is the "relative velocity" that we are concerned with. That rate of change varies continuously as the source moves.

I suppose that when the source and observer are in the process of moving past each other,starting from a certain distance,there must be some component of the relative velocity which is continuously changing,resulting to the continuously changing observed frequency.Which component is that?

At the point of closest approach the source is moving sideways with respect to the observer, thus the radial velocity is momentarily zero.

I'll probably understand this better,if I know exactly how a particular component of the relative velocity changes continuously.

 

[Urmi Roy]

The formulae still hold true but there is change in frequency as there is change in velocity.

I suppose to track the continuously changing observed frequency,one would have to apply the same formulae separately for every instant of the motion,right?

 

[jtbell]

Which component is that?

If you put the observer at the center of a polar-coordinate system, you use the radial component of the source's velocity.

 

[cepheid]

I'll probably understand this better,if I know exactly how a particular component of the relative velocity changes continuously.

Well you can just draw a picture. Let's say you, an observer, are located a distance d away from a track (d is your perpendicular distance from the track).

Let's say a train is coming towards you (along the track, of course), and is distance x away from you (in the along the track direction). This means that your actual distance, r, from the train (i.e. the line of sight distance) is given by the hypotenuse of the right triangle:

r = (x² + d²)^1/2​

Your line of sight to the train makes an angle θ with the track, where we have:

tanθ = d/x

cosθ = x/r

sinθ = d/r

​

Now, if the train is moving with speed v along the track (i.e. in the x-direction), then you can resolve this velocity into a component that is parallel to the line of sight (i.e. radial, i.e. toward the observer) and a component that is perpendicular to the line of sight (i.e. tangential). It is clear from the picture that the line of sight component is given by:

v_|| = vcosθ = vx/r = (vx)/(x² + d²)^1/2​

This makes sense because when x > 0, the line of sight component is toward the observer but is decreasing, and so and the frequency is continuously decreased. I guess it would go roughly linearly with x for |x| << d). When x = 0 (the train is at the point right beside you on the track), there is no component of the velocity toward the the observer, and hence the frequency is unshifted at this instant. When x < 0, the line of sight component of the velocity is now away from the observer, and the Doppler shift switches direction.

So your quote from that website was correct. The frequency will vary continuously from being higher than the emitted frequency to lower than the emitted frequency (obviously passing through the emitted frequency along the way).

 

[Urmi Roy]

Wow! That was an excellent and rather simple explanation cepheid!

We can infer from this that the observed frequency will first start to reduce, being the greatest at the point that the whistle of the train is first perceived and slowly passes the real frequency,followed by a gradual decrease until the train's whistle is no longer heard.

Also, this variation follows a parabolic nature, just like the cosine function varies between
-90 and +90 degrees,right?

May I also ask that in the case that the observer is standing on the train's track (this is just a thought experiment!) and the train passes through the observer,the frequency must change abruptly as it drops from a fixed frequency higher than the real frequency (while moving towards the observer) to a frequency equal to the real frequency when it approaches the observer,and vice versa as it finally moves away??

 

[DaveC426913]

May I also ask that in the case that the observer is standing on the train's track (this is just a thought experiment!) and the train passes through the observer,the frequency must change abruptly as it drops from a fixed frequency higher than the real frequency (while moving towards the observer) to a frequency equal to the real frequency when it approaches the observer,and vice versa as it finally moves away??

Yes. Remember that, while sound is perceived as a continuous input, it is actually "granular" in that it oscillates. It is the oscillation that produces the sound. Because of this you could indeed have an apparently instant drop in frequency - the duration of the drop is smaller than the granularity of the sound itself.

At the finest detail, you would measure the motion of the train on a scale that is smaller than the wavelength of the sound.

At some point, the horn/loudspeaker will reach the peak of its compression phase while the train is still coming toward you, yet the next peak will not arrive at the observer's location until after the train has passed. That means the frequency will have dropped froim high to low in one single cycle.

 

[vin300]

At the finest detail, you would measure the motion of the train on a scale that is smaller than the wavelength of the sound.

Yes. applications

 

[JazzFusion]

A picture is worth a thousand words (and a moving animation, 10,000!):

http://lectureonline.cl.msu.edu/~mmp/applist/doppler/d.htm

Click on a single point in the grey box to see animated sound-waves emanate from a "source". Then, click-and-hold on the dot and pull yuor mouse to one side, to form a small vector(start with length bout 0.3). This will simulate a moving sound source. Now pick any point to the side of the moving dot to imagine what an observer would 'hear'. Note how the perceived wavelength changes as the source passes.

Have fun!

 

[vin300]

Look at sonic boom in wiki vrel>1mach
Compressions pass backwards too

 

[Urmi Roy]

Yes. Remember that, while sound is perceived as a continuous input, it is actually "granular" in that it oscillates... Because of this you could indeed have an apparently instant drop in frequency - the duration of the drop is smaller than the granularity of the sound itself.

I'm sorry if I'm being a litlle silly,but I'm having a problem in understanding 'granularity' of sound.I suppose it must be a perceptible segment of the sound wave,but the idea isn't really clear to me.

At some point, the horn/loudspeaker will reach the peak of its compression phase while the train is still coming toward you, yet the next peak will not arrive at the observer's location until after the train has passed. That means the frequency will have dropped froim high to low in one single cycle.

If this drop happens within one cycle,will the separate events of the train approaching and passing by be perceptible to the listener,since the listener receives only the peaks and troughs of the wave.

 

[DaveC426913]

I'm sorry if I'm being a litlle silly,but I'm having a problem in understanding 'granularity' of sound.I suppose it must be a perceptible segment of the sound wave,but the idea isn't really clear to me.

A granule is the analoag of the digital pixel. Granularity is a colloquial term referring to anything that has a lower limit of meaningful resolution.

In an analogue photo it is meaningless to talk about an object in the photo that is smaller than one grain
In a digital photo it is meaningless to talk about an object in the photo that is smaller than one pixel
In a sound wave it is meaningless to talk about the pitch of a sound whose duration is shorter than one cycle of the wave.

If this drop happens within one cycle,will the separate events of the train approaching and passing by be perceptible to the listener,since the listener receives only the peaks and troughs of the wave.

Peaks will come by at frequency of X (per second), then at some point, the (ideal) loudspeaker membrane passes the (ideal) ear of the listener. The very next peak will come by after a delay longer than X seconds but shorter than Y seconds. Thereafter, each peak will follow at the frequency of Y.

There is no continuous drop; there is:
- freq X
- then a single cycle whose peak-to-peak is >X but <Y
- then freqY.

 

[Urmi Roy]

A picture is worth a thousand words (and a moving animation, 10,000!):

http://lectureonline.cl.msu.edu/~mmp/applist/doppler/d.htm

Sorry JazzFusion, I wasn't able to see your link,though I was very eager to,since I don't have Java installed in my computer :(

I'll try installing it,and then perhaps I'll get to see your link.

 

[Urmi Roy]

I think I finally get this idea of granularity.

This concept is applicable to sound waves because there are multiple air molecules that the oscillation has to pass before one single pulse (unit of the wave) is transmitted.

That means that there are different regions of the air, on its path, that the pulse passes through simultaneously.

If we modify any part of the pulse (a unit of the wave),then we can detect a modified frequency.

In the case where the train moves past the observer (which is what DaveC426913 was reffering to in his post)the modification in the sound pulse takes place before the entire pulse passes through the region in concern,so we can detect a wavelength longer than X but shorter than Y.

Is that right?

 

[DaveC426913]

This concept is applicable to sound waves because there are multiple air molecules that the oscillation has to pass before one single pulse (unit of the wave) is transmitted.

No, it would work in a theoretically continuous medium, it has nothing to do with atoms. The oscillation is the phase from peak to trough to peak of the signal.

The infinitely thin diaphragm making the sound is pushing forward then backward, making the peaks and troughs in the air density. As the infinitely thin eardrum passes it, the time interval before the next peak will be delayed.

 

[Urmi Roy]

If I replace 'multiple air molecules that the oscillation...' and put in 'continuous region that the oscillation ...,' and then apply the rest of my argument,will it be okay?

I'm just trying to say that since there is a continuous region through which the sound pulse has to pass,we can modify any part of it within 'one pulse' and thus,the sound pulse that the eardrum detects is changed in its frequency.

 

[DaveC426913]

If I replace 'multiple air molecules that the oscillation...' and put in 'continuous region that the oscillation ...,' and then apply the rest of my argument,will it be okay?

I'm just trying to say that since there is a continuous region through which the sound pulse has to pass,we can modify any part of it within 'one pulse' and thus,the sound pulse that the eardrum detects is changed in its frequency.

Well, you're not "modifying" the pulse.

Also, note that you can only detect a pitch over multiple cycles. One cycle is not a pitch of sound.

 

[Urmi Roy]

Also, note that you can only detect a pitch over multiple cycles. One cycle is not a pitch of sound.

If the drop in frequency first takes place in one cycle,the change may be detected after may more cycles occur,right?

If I'm having a major problem,please correct me.

Also,by 'modifying' I meant that we're changing the frequency of the wave.

Sorry if I'm making a mess!