Doppler Shift for Accelerated Motion

Never mind, I mis-read the part about the sound source accelerating.

What does the prof have to say?

If I've read and understood your words correctly, I see two errors.

The first error involves the distance d. The distance d is a term for the distance to a specific observer, not all observers. You could have d₁ for an observer at a x=-1 and d₂ for an observer at x=-2, etc. Generically you can use the term d, e.g. for an observer at the origin, but when you are talking about a specific observer at a specific observing location you need to designate which observer, i.e., at which distance.

The other error is in your statement about the frequency being the same at each point in time. In fact the equation you give for f includes a term for t which must affect the sum for f as t increases. (Note I've made t bold in the equation below for emphasis.)

Now from the above we see that the frequency at each point in time is:

f=f₀ (-1/2c at+1-v₀/c)

The result f cannot be a fixed value unless t is constant, but clearly time is not a fixed, constant quantity; it increases continually.

At least, that's what I see. Perhaps I've misinterpreted something?

I didn't define x0 properly x0 is the initial distance between the transmitter and the receiver so for observer one x0=1m and for observer two x0=2m.

and when I solved the equations I solved them for any observer all you need to do at the end is substitute the x0 for that specific observer

as for the instantaneous frequency you're right its the derivative of the phase so it should be:

f(t)=f0(-a/c t - v0/c + 1)

so indeed its still increasing with time but again is still independent of x0. this means that no matter where your position is the frequency measured at any specific time instance is the same.

I'm really obsessing about this now so please help I need to get some sleep :D

I think you've just shifted your error in thinking from d to t. For any given observer, the observer's distance is d_i. Correspondingly there will be a t_i, which is not the same for every observer.

actually I accounted for the ti your talking about early in the proof when I moved from the transmitted signal st(t) to the received signal sr(t). where I said sr(t)=st(t-∆t) with some attenuation ofcourse. and then from the ∆t, d entered the equation.

You are still making the same error, you just keep going around yourself.

You define:

∆t=d/c Where d is the distance between transmitter and receiver and c is the speed of sound

After substitution we get:
s_r(t)= A_r sin(2πf₀ (t-d/c))u(t-∆t)

Now d is not fixed. Using kinematic formulas we get:

d=1/2 at² + v₀t + x₀

What you have done is inter-posed different meanings of d. When you define d=1/2 at²+v₀ t+x₀ , you have defined the distance 'd' of the moving transmitter from the origin x₀. So when you do your substitution ∆t=d/c , where have you accounted for the extra distance for observer 2 at position x = -1 and observer 1 at x = -2 from x₀? You haven't. You're using the same 'd' for the distance of the emitter from x₀, for the distance of the emitter from observer 1 and for the distance of the emitter from observer 2. Clearly an extra bit of time is needed for the transmitted signal to cover the extra distance from x₀ to x = -1 and then a little bit more beyond that to reach x = -2.

So even though you think you have a general equation for calculating the signal anywhere, in fact your equation only calculates the signal at x₀ -- where, as you agreed earlier, the signal does show a decreasing frequency with time.

I never said x0=0. x0 is the initial distance between the transmitter and the receiver for any arbitrary receiver. i.e for observer 1 x0=1 for x0=2, and you could see that when I substituted for the different observers to find phi1 and phi2 near the end and deduced that the frequency received is the same and that the only difference is in the phase

regards

In your figure 3 you clearly state x0 = 0.

You can argue and be defensive, or you can listen to advice. You continue to state that 'the frequency is the same' when common sense tells you that each observer will see (or hear) a slightly different frequency due to the slight difference in position (and therefore propagation time) of the wave as the emitter accelerates away.

So, like, whatever...

dude I'm being defensive for a reason, I just would like to have a solid argument cause its science after all.

I can see the logic! I mean I'm fighting my friend to prove that each user hears a different frequency from the first place! I just want to present a mathematical argument or at least point out to where exactly his equations are wrong.

and as for the x0 in figure3 thanks for pointing that out its just a mix up, what I meant here is x0 for this user. maybe I should pick a different name for it, I'll change that shortly

regards

?

How about the case of extreme acceleration, where the sound from the source could be considered as a packet, rapidly decreasing in intensity. If the acceleration were sufficiently high, the sound heard by the observer nearer the source could be all over and done before the second observer even started to hear it. (May need to make the separation between the two observers bigger to more easily think it through).

I believe that the range of frequencies would indeed be the same (albeit at different intensities), but they couldn't possibly be heard at the same time by both observers.

excellent idea to tell you the truth I was just thinking about that. but that's all poetry if we don't have hard proof, and my friend's equations are hard proof so unless there is something wrong in his equations I can't say he is wrong! science is about hard evidence and mathematical proofs not feelings and intuition, at least that's what my friend told me after he hit me with his equations :(

something doesn't add up here! what is it?

I gave up doing sums like that many years ago. I still use maths in my work, but the applications require a different approach - I'm not going back to my text-books for this, life's too short.

Surely the Gedanken (thought) experiment I presented is "hard enough" proof? How could your friend refute the conclusion?

You could demonstrate the "proof" easily, with the help of a third friend, a starting pistol and a bit of open space^[1]. A video camera on a tripod, recording the responses, should seal it. The doppler effect is irrelevant - if the observer further from the source has not heard it, it can't possibly be at the same frequency as that heard by the nearer observer.

The error must lie in his "equations".

[1] You seem to be a smart bloke - I hope I don't need to spell this out further.

thanks for your spot on comments John, but as I told you I need to know what's wrong in his equations

And

You could go even further with your thought experiment if the source only transmits for say 1 second and you position your observers in such a way that the near observer stops hearing anything before the far observer even starts hearing!!!

The varying frequency which you assume is only true if the listener(s) are outside the line of travel of the sound source. If the listener is in, or close to, the line of he sound source, and if the source is moving with constant velocity, the listener will hear a constant frequency.

Imagine a source that is moving towards the listener at one tenth the speed of sound. The first peak is emitted and reaches the listener, the source moves one tenth of a wavelength toward the listener and issues the second peak, this second peak travels one tenth of a wavelength less so arrives one tenth of a wavelength early, the listener hears the peaks arriving at a constant nine tenths of the emitted wavelength.

When the source passes the listener, the frequency drops because the source is moving away and the listener hears a wavelength of eleven tenths of the emitted wavelength.

If the listener is far off the line of travel of the source, the frequency will change with time because the distance between the source and the listener changes non-linearly because of the angle of the direction of travel changes.

"... and if the source is moving with constant velocity".

Missed the point, here - read the OP more carefully.

PS. The title of the thread (including that of your post) is a bit of a give-away.

Oops.

Needs thinking about. What speed when the source passes the listener.

Must prevent this from getting into Mr. Einstein's Relativity at any price.

Between the two listeners (concidering both and moving sound sourse are on the same line) ther's only a time delay t=x/c.(x being the distance between them) But at this time delay the sound sourse had (meaning had) also increased its v by t*acc. So at the same time and with an accelerating sound sourse the two listeners will perceive different frequencies but only because their different locations. The sound print in space the source leaves is one and only. But since you like to make a basic problem difficult, here's a less obvious one: What happens when object hits sound speed? S.M.

no assume the velocities here are well bellow the speed of sound.

and that's my point exactly a time delay should reflect a frequency delay due to the accelerated motion. but have a look at the equations that my friend solved they are pretty solid :( I couldn't refute them

Differenciation is not exactly reverse of indegration. There is an emerging (unknown) constant involved in integration. In this case it's the distance between listeners. S.M.

the proof image got scaled down for some reason so here is the link for it:

https://sites.google.com/site/javascriptevents/home/Proof.jpg

I have a problem with non-dimensional, ill labelled diagrams, so here is mine (sorry it's hand drawn but I don't have a drafting program).

It is a space time diagram, the vertical axis in seconds and the horizontal in "space seconds", each being the distance traveled by sound in one second. An emitter, one beep per second, starts from the origin and accelerates at a rate of 0.1 space seconds per second squared along a straight track. There are observers at intervals out on the track who only travel in time but are stationary in space.

The beep that is emitted at the end of each second, travels backward and forward along our space dimension and forward with time.

The observers see the same pattern, delayed by the distance between them, until the emitter passes one of them, then there is a break until it passes another when they each are the same pattern but delayed in the opposite direction.

At 10 seconds, the emitter reaches the speed of sound, so I didn't try any further.