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yes.rude man said:The only way to get an actual change in pulses per unit time would be to continuously maintain the mass's velocity in the same direction for that unit of time.
yes.rude man said:The only way to get an actual change in pulses per unit time would be to continuously maintain the mass's velocity in the same direction for that unit of time.
(1) is the basis of the Doppler shift when the source is moving and the observer is at rest in the medium. It says the wavelength λ' is a linear function of u. This is the fundamental basis of the Doppler shift. It's not hard to derive: in time T the source moves uT towards the observer so the next wavelength is curtailed by uT, and there you have it: λ' = λ - uT. The frequency formula derives from this assuming u and f are constant so it's not general.Delta² said:Two things @rude man
1) Can you provide some intuitive /qualitative insight why (1) (and its integral generalization for time varying u) holds? Just to check if I understand it correctly..
2) I didn't understand why the formula for f' doesn't hold if u is time varying.
Can generalise the equation in post #47.Delta² said:what's the formula for f' if u(t) is time varying?
It would be a FM signal. As long as f has a significantly higher frequency than the motion of the mass it is still reasonable to speak of the received frequency as a function of time.Delta² said:we could argue that frequency is not well defined in such a case because the waveform as seen by the observer might not be periodic.
An FM signal has theoretically infinite spectrum of frequencies. Which does the observer hear and you denote as ##F(t)##??haruspex said:It would be a FM signal. As long as f has a significantly higher frequency than the motion of the mass it is still reasonable to speak of the received frequency as a function of time.
x'(t) = sin(2πct/λ(t)) observed signal after doppler shifting
x'(t) = sin(2πct/(λ0 - ∫tu(t′)dt′))'>∫tu(t′)dt′))∫tu(t′)dt′)) as I showed last post
But if you don't you don't see all the harmonics.Delta² said:Yes ok thanks @rude man , no need to analyse the terms sin(ab sin(…)) or cos(ab sin(…)) in terms of Bessel functions...
I've corrected your 1st expression for x'(t). T instead of t. T is the period of one oscillation of the transducer = λ0/c. Also, your 2nd expression for x'(t) is dimensionally incorrect.I have my doubts about these two lines
If we take u(t)=u constant, then the second line gives ##x'(t)=\sin{(2\pi ct\frac{1}{\lambda_0-uT})}## which doesn't seem correct, we know that if u(t) is constant then the observed signal will be ##x'(t)=\sin{(2\pi\frac{c}{c-u}t)}##
Yes you are right at 2nd expression I forgot to put ##f_0=\frac{c}{\lambda_0}## inside the phase. But the first expression if I remember well it was ##\lambda(t)=\lambda_0-\int^tu(t')dt'##, I don't see how you can get something constant like ##uT## from that integral..rude man said:But if you don't you don't see all the harmonics.I've corrected your 1st expression for x'(t). T instead of t. T is the period of one oscillation of the transducer = λ0/c. Also, your 2nd expression for x'(t) is dimensionally incorrect.
I think the post with the analysis with the ##\lambda_i## was correct, you shouldn't have deleted it.I've made numerous math corrections this morning (11 a.m. GMT-7), all of which were parentheses mistakes. But I've deleted recent posts. I'm revising my analysis for x'(t), the microphone diaphragm displacement (I'm modeling the system as a speaker cone with sinusoidal displacement x(t) at frequency f0 = c/T, and a microphone receiver with diaphragm displacement x'(t) ).
Agreed. I did goof big-time by integrating u(t) up to time t.Delta² said:Yes you are right at 2nd expression I forgot to put ##f_0=\frac{c}{\lambda_0}## inside the phase. But the first expression if I remember well it was ##\lambda(t)=\lambda_0-\int^tu(t')dt'##, I don't see how you can get something constant like ##uT## from that integral..
Miraculous recovery! (I usually make a .doc file for each post, for just such an eventuality!)I think the post with the analysis with the ##\lambda_i## was correct, you shouldn't have deleted it.
Can you perhaps rewrite what you doing here, I think there are some errors with the parentheses involved. If we take ##\lambda (t)=\lambda_0-u_0T\cos(\omega_0t)## then you say that ##x'(t)=x_0'\cos{\frac{2\pi ct}{\lambda(t)}}##?rude man said:Let the microphone diaphragm displacement be x'(t) with amplitude x0'.
So x'(t) = x0' cos(2πct/λ0)/(1 - (u0T/λ0) cos(ω0t))
≅ x0' cos[2πf0t(1 + (u0T/λ0)) cos(ω0t)]
since u0T/λ0 << 1.
Yes, what a mess - seems like I never get it right. Thanks a lot for helping me out here.Delta² said:Can you perhaps rewrite what you doing here, I think there are some errors with the parentheses involved.
Actually, I think that part was OK. Wasn't it?Delta² said:If we take ##\lambda (t)=\lambda_0-u_0T\cos(\omega_0t)## then you say that ##x'(t)=x_0'\cos{\frac{2\pi ct}{\lambda(t)}}##?
Nope I don't agree (I am not sure though), see my post #64 I edited it .rude man said:Actually, I think that part was OK. Wasn't it?
What is R? I have a tough time with this one.Delta² said:Can you perhaps rewrite what you doing here, I think there are some errors with the parentheses involved. If we take ##\lambda (t)=\lambda_0-u_0T\cos(\omega_0t)## then you say that ##x'(t)=x_0'\cos{\frac{2\pi ct}{\lambda(t)}}##?
My thoughts for this problem took me to a completely different route. According to my reasoning (which I ll post in a later post ), if we define as ##g(t)## the retarded time that the wavefront reach the observer, that is
##g(t)=t+\frac{R-\int_0^t u(t')dt'}{c}## then the waveform that the observer sees is :
##x(g^{-1}(t))##
where ##x(t)## is the waveform that the source emits and ##g^{-1}(t)## is the inverse of ##g(t)## such that ##g^{-1}(g(t))=t##.
Actually, it does. It's hidden in the last term.rude man said:Anyway, as I said, I saw no reason to invoke any retardation time. The expression λ = λ0 - uT does not imply any time shifting.
So do you agree with my post 62 or not? Assuming R=0?vela said:Actually, it does. It's hidden in the last term.
Let's assume the position of the mass is given by ##x(t)##, the source emits a signal ##y(t) = Y \cos(2\pi f t)##, and the observer is located at ##x=R##.
At time ##t=0##, crest 0 is emitted at ##x=0##, and it subsequently propagates toward the observer with speed ##c##. We can describe its position at ##x_0(t) = ct## for ##t\ge 0##.
At time ##t=T=1/f##, crest 1 is emitted from the position ##x(T)##. Its position is subsequently given by ##x_1(t) = x(T) + c(t-T)## for ##t\ge T##.
At time ##t=2T##, crest 2 is emitted, and its position is subsequently given by ##x_2(t) = x(2T) + c(t-2T)## for ##t \ge 2T##.
In general, for ##t \ge nT##, the position of the crest ##n## is given by ##x_n(t) = x(nT) + c(t-nT)##.
The distance ##\lambda_k## between crest ##k-1## and crest ##k## is
$$\lambda_k = x_{k-1}(t) - x_{k}(t) = cT - [x(kT) - x((k-1)T)] = \lambda - \int_{(k-1)T}^{kT} v(t')\,dt',$$ where ##\lambda=cT## is the un-Doppler-shifted wavelength and ##v(t)## is the velocity of the mass. This is essentially the same expression you derived in an earlier post.
So when does crest ##n## arrive at the observer? Just set ##x_n(t_n) = R## and solve for ##t_n##. When we do this, we get
$$t_n = nT + \frac Rc - \frac{x(nT)}{c}.$$ Since ##t=nT## is the time the crest was emitted, this is @Delta²'s expression for the retarded time.
One way of looking at the problem is that the movement of the source distorts the wave. The changing position of the source as crests are emitted causes them to be squeezed together or stretched apart. This modulated wave then propagates to the observer. The other way is to see the changes in the wave as a result of different propagation times. Compared to the time a crest emitted at ##x=0## would take to reach the observer, a crest is emitted from ##x>0## arrives a little early, and one from ##x<0## arrives a little late.