Doppler Effect Observed Frequency Under Special Conditions

[Thinker8921]

v_w= Velocity of Wave
v_m= Velocity of Wave Source
f_w= Frequency of the Wave
f_d= Frequency of the Wave relative to Detector as Wave Source is also moving
q= Infinity

For the wave frequency detected by a detector at rest as the wave source* is moving towards it, there is:
f_d=( v_wf_w )/( v_w-v_m )
Now when v_m=v_w, the frequency is undefined.
However if you get v_w to get closer and closer to v_m, while still being a little greater than v_m (to keep things positive for now), it is clear that f_d approaches q.

Keeping that idea in mind, the question is this:
When this wave source meets with the detector, what frequency wave will be detected (by the detector) at that very instant?

*To clear any confusions, the wave is being emitted by the wave source at all times.

 

[BruceW]

Unfortunately, the equation you've used assumes that the velocity of the wave is greater than the velocities of the source or detector.
So when [itex]v_m[\itex] gets close to [itex]v_w[\itex], then the equation no longer works.
It has some information on wikipedia's page on the doppler effect.

 

[Thinker8921]

Thank-you for that, I found it on Wikipedia too:
"The above formula works for sound wave if and only if the speeds of the source and receiver relative to the medium are slower than the speed of sound."
So, there is a sonic boom where the sound waves collide. This causes a great 'thump' of sound.

So now what does the detector detect? (Same conditions apply, and I would like to stress, at that very instant).
What rule should be used for the v_w=v_m condition?
Also, it seems reasonable to think that the great sound produced during the sonic boom is detected only (even a little bit) after the wave source is past the detector. Because it takes some time for the sound waves to collide and merge right?

Also, Wikipedia said it works only for the sound wave?? Is this correct? Because then the infinity frequency thing applies to other waves (such as light) as well?

 

[BruceW]

The equation wiki gives at the top is derived from the fact that the velocity of a signal sent from the source is constant, and this velocity is much greater than the velocities of source or observer. As long as these are true, then it applies to any kind of wave. But when any of the speeds involved are near the speed of light, you must account for special relativity, so the equation will look different.
Another problem is whether sound will still act like a linear wave when the source is at high velocities. When something is described as a wave, it is often only a mathematical approximation, so you have to think about what is physically happening.
Sound is a wave of pressure. So when the source is traveling near the speed of sound, the source keeps traveling forward, not letting pressure to drop and fall in a wave pattern. In other words, I don't think a sound wave would form in the forward direction.

 

[Thinker8921]

...not letting pressure to drop and fall in a wave pattern. I don't think a sound wave would form in the forward direction.

I don't quite understand this. What exactly do you mean by the pressure dropping and falling? I get the next part about how the sound wave will not extend forwards (with respect to the wave source). This means that at one point, the sound waves all merge to became a sonic wave, but the question still remains about the wave frequency detected at that meeting point.

 

[BruceW]

sorry, I meant rise and fall. A sound wave is characterised by pressure rising and falling with a given frequency. But If the source is going at almost the speed of the wave, then the air molecules are all rushing backwards (from the reference frame of the source). So the pressure cannot rise and fall in the way characteristic of a wave (in the forward direction).
By the way, I'm just speculating here, don't take me as an expert on the subject.
In front of the source, it will still be trying to make a wave, but the air travels past it so quickly that no waves can form. So I'd say that when the detector and source collided, there would be no sound detected. After their collision, some sound may catch up.

 

[Thinker8921]

sorry, I meant rise and fall. A sound wave is characterised by pressure rising and falling with a given frequency.

Yes that makes sense. Although that's the first time I've heard of pressure rise and fall in a sound wave. Usually, I say compressions and rarefactions of the moving particles.

But If the source is going at almost the speed of the wave, then the air molecules are all rushing backwards (from the reference frame of the source).

Doesn't this also apply if v_w=v_m as well if the waves have not already combined together to form the sonic wave.

So I'd say that when the detector and source collided, there would be no sound detected. After their collision, some sound may catch up.

You say some sound may catch up. Why is it not certain?

So the question remains. What is the frequency of the wave detected when v_m=v_w?

 

[BruceW]

You said in the first post that the source is emitting waves. Assuming we're talking about sound waves, I'm imagining a jet with a speaker at the front. As the jet gets close to the speed of sound, the speaker is no longer able to create forward-travelling sound waves.
When the jet is traveling at the speed of sound, all sound is emitted backwards, so if you were listening after the jet had passed overhead, you would hear the sonic boom, and then you'd hear the sound created by the jet smacking into the air. The air traveling over the jet is very chaotic, so any sound made by the speaker would be destroyed.
So the frequency of the wave you detect will have no dependence on the frequency of sound emitted by the speaker. You would detect a range of frequencies due to the jet smashing through the air.

 

[Thinker8921]

Thank-you, that explains it well.

 

[BruceW]

no problem. The theory behind sound waves has always been difficult to grasp for me, but I reckon my last post was more-or-less right