Number of waves from an oscillating source in one period

[Delta2]

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.

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..

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) ).

I think the post with the analysis with the $\lambda_i$ was correct, you shouldn't have deleted it.

 

[rude man]

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..

Agreed. I did goof big-time by integrating u(t) up to time t.

OK I took another shot at it, to wit:

Scenario: we have a speaker with speaker cone displacement x(t) relative to the speaker body.. The speaker body moves with velocity u(t) relative to the air.. A stationary microphone picks up the moving speaker output which is the speaker output x(t) modulated by the motion u(t). I think this setup has to be clearly defined.

Let x(t) = x₀ cos(2πf₀t) = x₀ cos(2πct/λ₀), c = speed of sound. This defines λ₀.

Fundamental Doppler equation is λ = λ₀ - uT
where T = λ₀/c and λ is λ₀ reduced by uT after time T. u is assumed constant during time T. u > 0 for speaker body approaching the microphone.

We can remove the requirement on u(t) = constant during T by writing λ = λ₀ - $\int_T u(t) \, dt$

Thus λ = λ₀ - $\int_T u(t) \, dt$ = λ₀ -$\int_t^{t+T} u(t') \, dt'$ with t' a dummy variable.

Assume u(t) = u₀ cos(ω₀t).

Then $\int_T u(t) \, dt$ = (u₀/ω₀)[sin(ω₀T + ω₀t) - sin(ω₀t)]

and if we exploit the fact that ω₀T << 1 (speaker cone frequency >> speaker body frequency),

$\int_T u(t) \, dt$ = u₀T cos(ω₀t).

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.

Letting 2πf₀t = a, u₀T/λ₀ = b, z = ab and θ = ω₀t,
x'(t) = x₀' [cos(a) cos(z cosθ) + sin(a) sin(z cosθ)]

= x₀'[cos(a) {J₀(z) + $2\sum_{k=1}^\infty (-1)^k J_{2k}(z) cos(2kθ)$}

+ sin(a) { $2\sum_{k=0}^\infty (-1)^k J_{2k+1}(z) cos(2k+1)θ$}]

with J(z) the Bessel function of the 1st kind,
showing that all even and odd harmonics of ω₀ are present.

I'm hoping I got it this time.

 

[rude man]

I think the post with the analysis with the $\lambda_i$ was correct, you shouldn't have deleted it.

Miraculous recovery! (I usually make a .doc file for each post, for just such an eventuality!)

Kind of late now as the OP seems to have lost interest, but, following my disastrous posts, most of which I have deleted, I decided to finally investigate the Doppler effect more seriously.

It turns out that, for any movement of the source, if the source returns to where it started from, then the number of pulses in that cycle seen by the observer = the number sent by the transducer. Of course, if the observer is distant from the source, there will be a time delay, but that delay does not alter this statement.

Proof:

The basicDoppler relation is λ_i = λ₀ - u_iT .... (1)
where λ_i = i'th spatial distance between two adjacent pulses as seen by observer,
λ₀ = same distance when u = 0,
Δλ_i = λ_i - λ₀
T = one period of the transducer output
u_i = velocity during time interval i of duration T; u can be > or ≤ 0.
f₀ = transducer output frequency = 1/T.

(1) assumes that u_i is constant for the time interval T; but more generally we can let u vary in time, then
λ_i = λ₀ - $\int_T u(t) \, dt$ with u = average u between i and i+1.

Let NT = the period of kinetic oscillation of the mass (the source) and
x(t) = instantaneous position of the source, with x(0) = 0.
So x(t) = $\int_0^t u(t') \, dt'$ and
$\sum_{i=0}^N$ Δλ_i = $\int_0^T u(t) \, dt + \int_T^{2T} u(t) \, dt + ... \int_{(N-1)T}^{NT} u(t) \, dt = \int_0^{NT} u(t) \, dt$ = x(NT) - x(0)
= net distance x traveled by source in time NT.

This proves that if net x = 0 i.e x returns to its starting place so that x(NT) = x(0) , then irrespective of any interim variation of positive or negative x or u,
$\sum_{i=0}^N$ Δλ_i = 0 over that time span NT, which implies
Σ λ_i = Σ λ₀ and the number of transducer pulses = no. of observed pulses in time NT.

So, warning: your textbook formula f' = f⋅(c/(c-u)) is valid only for u constant. It's λ that varies linearly with u, not f. Wish I had understood sooner.

 

[Delta2]

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.

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 $g^{-1}(g(t))=t$.

 

[rude man]

Can you perhaps rewrite what you doing here, I think there are some errors with the parentheses involved.

Yes, what a mess - seems like I never get it right. Thanks a lot for helping me out here.
I edited post 62. Thanks again. I really should use LaTex more. BTW sexy new avatar!

If we take $\lambda (t)=\lambda_0-u_0T\cos(\omega_0t)$ then you say that $x'(t)=x_0'\cos{\frac{2\pi ct}{\lambda(t)}}$?

Actually, I think that part was OK. Wasn't it?

 

[Delta2]

Actually, I think that part was OK. Wasn't it?

Nope I don't agree (I am not sure though), see my post #64 I edited it .

You essentially say that if the waveform of the source is $x(t)$ then the observer sees the waveform $x(\frac{ct}{\lambda (t)})$, I say its $x(g^{-1}(t))$

Suppose $x(t)$ and $x'(t)$ are as you defined in post 62 (with the difference that we assume that the amplitude $x_0$ doesn't diminish with distance from source, so it is $x_0=x_0'$), and $g(t)$ as defined in post 64. Then, as I see it, the microphone diaphragm displacement will be $x(t)$ at time $t'=g(t)$ , so it will be $x'(t')=x(t)$ that is

$x'(g(t))=x(t)$ . we can see that

$x(g^{-1}(t))=x'(g(g^{-1}(t))=x'(t)$.

 

[rude man]

OK. I see that you considered the retardation time (time for sound to go from speaker to microphone. Which is what I think @haruspex did. But I never considered that time. I saw nothing wrong with setting the microphone a few wavelengths (or less) in front of the speaker, making the delay time 0.

So if we assume no attenuation as you did, which is fine, then if x = x₀cos(2πct/λ₀) we get x' = x₀cos(2πct/λ). The only difference between x and x' is that x' is x wavelength-modulated by u, with essentially no time delay.

 

[rude man]

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 $g^{-1}(g(t))=t$.

What is R? I have a tough time with this one.

I understand x(t). But why is x' all of a sudden a function of (time)^-1? Why not retardation time? So if u=0 we get x'(t) = x(t-τ), τ = delay time?

Anyway, as I said, I saw no reason to invoke any retardation time. The expression λ = λ₀ - uT does not imply any time shifting.

 

[haruspex]

But why is x' all of a sudden a function of (time)^-1?

@Delta² is using g^-1 to mean the inverse function of g, not 1/g. If g(t)=t+τ Then g^-1(t)=t-τ.

 

[vela]

Anyway, as I said, I saw no reason to invoke any retardation time. The expression λ = λ₀ - uT does not imply any time shifting.

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.

 

[rude man]

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.

So do you agree with my post 62 or not? Assuming R=0?

EDIT: sorry if I sounded importunate. Didn't mean to. Happens occasionally I'm afraid

But I really did want to know if you agreed or not. And I still find it difficult to see how R enters into the situation. If u is constant the formula f' = f c/(c-u) surely applies, and it doesn't contain any R.