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svHyLyte · 15-03-2013, 05:48

Quote:

Originally Posted by Richard_W

Interesting, and simple explanation even without the formulas ... the frequency shifts of the source and the receiver created by moving medium cancel each other.

Right ...

Yup, exactly right--its the relative speeds of transmitter and receiver that determine Doppler effect frequency shifts. No formulas needed. Only attenuation--the decay of the sound--will be effected by wind speed/direction, faster up-wind than down, which is why one can hear sounds from further off when down wind of the transmitter--like waves crashing on a breakwater, eh?

FWIW...

jongleur · 15-03-2013, 09:25

Consider an observer (not the listener) watching the emitter and receiver travelling from right to left. Let's say the observer can see the sound waves.

It would appear to the observer that the waves from the emitter are squeezed together on the left of the emitter and the waves would be stretched out on the right of the emitter.

In absolute terms, the frequency on the left of the emitter is 'higher', but the listener is also moving to the left, 'catching' the waves a bit of time later each wave, so the 'higher' frequency would be perceived by the listener as lower than they are in absolute terms.

Now consider a vacuum. No sound waves. Is the medium absolutely important or absolutely unimportant?

jongleur · 15-03-2013, 09:47

I hit post too quickly. I meant to add:

Now, if both the emitter and listener are stationary and the medium (air) is moving from left to right, is that not the same as the emitter and listener moving right to left through the medium?

**Paul Elliott** · 15-03-2013, 09:57

Quote:

Originally Posted by jongleur

Consider an observer (not the listener) watching the emitter and receiver travelling from right to left. Let's say the observer can see the sound waves.

It would appear to the observer that the waves from the emitter are squeezed together on the left of the emitter and the waves would be stretched out on the right of the emitter.

In absolute terms, the frequency on the left of the emitter is 'higher', but the listener is also moving to the left, 'catching' the waves a bit of time later each wave, so the 'higher' frequency would be perceived by the listener as lower than they are in absolute terms.

But you aren't describing "absolute terms". You are describing a system with a third reference point, one which is moving relative to the emitter.

Quote:

Now consider a vacuum. No sound waves. Is the medium absolutely important or absolutely unimportant?

It is absolutely important that there be a medium. With a medium, any medium, the density of that medium will have no effect on frequency when the emitter and detector are fixed (relative to each other).

About the density of the medium, here's a thought experiment: Say I have a ham transmitter, and you have a receiver. I transmit on 14.3MHz. The signal leaves my radio, runs through 20 ft of coaxial cable, through my antenna, into the air, bounces off the ionosphere and down to your antenna, through your coax, and and finally into your receiver. The propagation velocity of most coaxial cable is between 0.66c and 0.88c (c = speed of light in vacuum). The Vprop of radio signals in air is about 0.999c. As the signal moves through these different media, the speed of propagation changes, as does the physical wavelength, (the vProp determines the wavelength). At no point would we say that the frequency changes, because it really doesn't. The physical wavelength changes, and the phase changes (relative to some uniform reference).

You can probably treat the moving air (wind) in a similar manner to a more, or less dense, medium. It will affect the Vprop, but not the frequency.

**Paul Elliott** · 15-03-2013, 09:59

Quote:

Originally Posted by jongleur

I hit post too quickly. I meant to add:

Now, if both the emitter and listener are stationary and the medium (air) is moving from left to right, is that not the same as the emitter and listener moving right to left through the medium?

Yes. For the purposes of this discussion they are exactly the same.

I think that most of us are in violent agreement.

And this has little to do with sailing. So what? It's a fun discussion.

**David M** · 15-03-2013, 10:33

Since the medium that the sound is traveling on is also itself traveling towards the listener, then the number of cycles that pass the listeners ear per second would be greater....more cycles per amount of time passing a particular point would be a higher frequency.

Imagine two ticker tapes with equally spaced dots on them. You look at one ticker tape that is stopped and the second ticker tape whose dots are moving towards you. The second tape is going to have more dots per second passing your eyes.

Sound is not limited to the grid of space like electromagnetic waves are. Sounds medium is not limited to standing still with respect to two fixed points. If sound and EM waves were limited like this then what Paul is saying would be true. No offense Paul....just trying to set up the argument.

**Paul Elliott** · 15-03-2013, 11:13

Quote:

Originally Posted by David M

Since the medium that the sound is traveling on is also itself traveling towards the listener, then the number of cycles that pass the listeners ear per second would be greater....more cycles per amount of time passing a particular point would be a higher frequency.

Imagine two ticker tapes with equally spaced dots on them. You look at one ticker tape that is stopped and the second ticker tape whose dots are moving towards you. The second tape is going to have more dots per second passing your eyes.

Sound is not limited to the grid of space like electromagnetic waves are. Sounds medium is not limited to standing still with respect to two fixed points. If sound and EM waves were limited like this then what Paul is saying would be true. No offense Paul....just trying to set up the argument.

But -- the medium is moving past the sound source as well as the listener. With sound, the effect of the moving medium is to change the propagation velocity, not the perceived frequency.

I'm going to try an experiment with my shopvac (as a blower), some hoses, and a tiny speaker inside the hose. I'll bet that the frequency of the sound I hear when the air leaves the hose will be unchanged. In fact, I will put a constriction at the hose exit, so the air velocity will be different at my ear than it was at the speaker. I predict that even then the pitch will be unchanged.

David, if I understand you correctly, you are saying that for either of these cases the frequency I hear will be different than what I put into the speaker.

At least we're not arguing about relativistic effects. That gets (to me, anyway) non-intuitive very quickly. Example: If you're in a spaceship moving away from Earth at almost the speed of light, and you shine your flashlight forwards. How long does it take for the light to reach the nose of the ship? Point the flashlight back at Earth. How long until your signal is seen back on Earth?

Let's stick with sound, OK?

**Paul Elliott** · 15-03-2013, 11:21

Quote:

Originally Posted by David M

Imagine two ticker tapes with equally spaced dots on them. You look at one ticker tape that is stopped and the second ticker tape whose dots are moving towards you. The second tape is going to have more dots per second passing your eyes.

But in this analogy, we have a blank ticker tape, at one end a source (punching machine), and at the other end a detector (your eyes). Regardless of what the tape speed may be (as long as it is constant), if the source punches one hole per second, you will detect one hole per second. Depending on the tape speed, these holes will be further apart or closer together (wavelength), but they will still be passing your eyeball at one hole per second (frequency).

**Paul Elliott** · 15-03-2013, 11:26

I really enjoy a good-natured argument, especially when I'm right! Even when I'm wrong (hypothetically speaking), I get to learn something.

svHyLyte · 15-03-2013, 12:01

Quote:

Originally Posted by Paul Elliott

But in this analogy, we have a blank ticker tape, at one end a source (punching machine), and at the other end a detector (your eyes). Regardless of what the tape speed may be (as long as it is constant), if the source punches one hole per second, you will detect one hole per second. Depending on the tape speed, these holes will be further apart or closer together (wavelength), but they will still be passing your eyeball at one hole per second (frequency).

Good analogy Paul. And if the "detector" is moving toward the tape punch it/he will see the holes more frequently (so the frequency will be percieved to increase); or, if away from the tape punch less frequently (so the frequency will be percieved to decrease); or, if away at the same speed as the tape speed, at no frequency at all!

**Paul Elliott** · 15-03-2013, 12:14

Quote:

Originally Posted by svHyLyte

or, if away at the same speed as the tape speed, at no frequency at all!

And if if you go away from the source faster than the speed of the tape (speed of sound), the music plays backwards until you reach the beginning of the song.

I like the punched-tape analogy. Thanks, David! (Although here I'm mixing up paper tape, magnetic tape, and sound.)

jongleur · 15-03-2013, 12:19

So, if the emitter is moving at the speed of sound, mach I, the forward waves will be perceived by the emitter as not moving at all.

As someone asked, where would the waves be 'buffered'? Are they buffered and then spent in a boom or whip crack? Is that a different frequency for the listener, or is it the sum of all frequencies?

Andrew Troup · 15-03-2013, 12:23

It's interesting looking at the nature of the posts.

Some people clearly think that assertion is sufficient, and some of the asserters seem to think assertions have explanatory power.

Others have no interest in assertion but are very interested in explanation.

The asserters tend to get impatient and angry with the explainers.

**Paul Elliott** · 15-03-2013, 12:34

Quote:

Originally Posted by jongleur

So, if the emitter is moving at the speed of sound, mach I, the forward waves will be perceived by the emitter as not moving at all.

As someone asked, where would the waves be 'buffered'? Are they buffered and then spent in a boom or whip crack? Is that a different frequency for the listener, or is it the sum of all frequencies?

Let's ignore all those difficult real-world effects like turbulance, and assume isotropic radiators (infinitely small and radiating equally in all directions) and all that simplifying stuff, I think that the pressure variations (sound) in the forward direction would effectively cancel each other out. The energy delivered on one half-cycle would be returned on the alternate half-cycle, and so no power would be coupled. That's just in the straight-ahead direction. In all other directions, the pressure waves would propagate out at the speed of sound. Here's a nice image I just found:

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