Number of waves from an oscillating source in one period

[haruspex]

So you admit to trying to solve the transient problem?

Not sure what you mean by that. If you mean partial periods then no, and I do not understand what makes you think that.
Do you or do you not understand that:
- it is not possible for the integral over multiple whole periods at the observer to diverge ever further from what the emitter actually generated?
- there is a time delay before the wavefronts reach the observer, so f'(t) is not equal to F(t); rather, F(t+D(t))=f'(t)?

I see no point in looking further at it

It is a most interesting and subtle problem. I really think you would find the investment rewarding.

 

[Delta2]

… Makes no difference. The time delay makes no difference, and the number of emitted waves (pulses) = the number of received pulses..
…

That's what i thought too but after reading posts #27 and #34 (where we deal with a simpler problem) I concluded that it is exactly what makes the difference. The time delay creates some sort of "pseudo relativistic " effect which is hard for me to explain but I ll try to:

The half period during which the source travels closer to the stationary observer corresponds to something less than a half period for the stationary observer. What i mean by the word correspond is that the waves generated by the source in this half period will be completely received by the stationary observer not in a half period but in a time interval smaller than a half period.(if the speed of wave was infinite, then those two time intervals would be the same obviously, but it is the finite speed and the time delay that creates this mismatch). So though the frequency get larger and larger during this half period, the time interval that these frequencies last is smaller than a half period. So we have larger frequencies but multiplied by a smaller half period.On the next half period where the source moves away from the observer, we have smaller frequencies but multiplied by "bigger" half period. The sum of smaller half period+bigger half period equals exactly one period for the stationary observer.

EDIT: After reading what I wrote I admit it isn't so well explained, I ll write later a post where I ll write some math as well, atm I am busy and I cant.

 

[haruspex]

I mean that you're trying to solve the problem with the oscillator starting from rest

No, no! It is the "steady state" behaviour, as though the oscillations have been going on forever.

Where is your answer?

The answer that the OP understood must be right at the end of post #1: the same number as emitted each period, $2\pi f\sqrt{\frac mk}$.

 

[haruspex]

I find it wholly irrelevant and confusing

They are not difficult questions.

1. If the integral over one whole period gives more wavefronts received than emitted, will the integral over the next whole period produce a different discrepancy or the same?
2. Is there a time delay between the frequency observed for a fragment of the signal by a local stationary observer and that same fragment at a remote observer? Does that mean F and f' are related as f'(t)=F(t+D(t)) for some delay function D?

 

[kuruman]

It seems to me that this problem can be solved by considering that we have two time scales, one is the spring-mass system that repeats its motion every $T=2\pi\sqrt{m/k}$ time interval and one is the sound source that "emits peaks" every $\tau=1/\bar f$ time interval. The latter has a variable rate. We want to know how many peaks are emitted in one period $T$. Let $N$ be that number. Now imagine a clock that ticks regularly every $T/N$ time units and a second clock that ticks every time a peak is detected by the receiver. Clearly, the two clocks will not tick synchronously. The time difference between ticks of the same order is$$\delta(t) =\frac{T}{N}-\frac{1}{\bar f} =\frac{T}{N}-\frac{v_s-v_0\cos\omega t}{v_s f} $$and the time-averaged difference is$$\bar {\delta}=\frac{\int_0^T {\delta (t')dt'}}{\int_0^T {dt'}}$$This average $\bar {\delta}$ must be zero because the second clock sometimes runs faster and sometimes slower, but at the end of one period, $T$, as @haruspex pointed out in #6, all the emitted peaks must be received. Stated differently, regardless of which clock one uses, $N$ ticks mark the passage of one period $T$. On the right side, the cosine term integrates to zero and one is left with a very simple equation to solve for $N$ and obtain OP's "obvious answer".

 

[andrewkirk]

I haven't absorbed all the posts in this thread but it seems to me that the answer does not require any integration, and only involves counting wavefronts, which is what @haruspex seems to be suggesting.

I assume $v_s$ irefers to the speed of sound in the medium and $v_0$ is the maximum speed of the emitter. It has been assumed that $v_s>v_0$, so the emitter should never be passed by, and should never pass, a wavefront emitted earlier. The number of waves observed by the stationary observer in a period is the number of wavefronts that pass the observer in that period.

Consider a period that starts with the emitter closest to the observer, at its furthest point from equilibrium, and ends when the emitter returns there next time. That is a full period of the oscillation, call it T1. Let the period of the emission be T2. Then the emitter will emit T2 / T1 wavefronts during that period, and they will all pass the observer in succession, in a period of length T1, with no extra wavefronts in between them. So the observer counts T2 / T1 wavefronts in time T1 and infers an average period of T2.

 

[haruspex]

the answer does not require any integration,

Thanks for confirming that, but the intriguing question is why the integral at the start of post#1 gives the wrong answer.

 

[kuruman]

Thanks for confirming that, but the intriguing question is why the integral at the start of post#1 gives the wrong answer.

I think it has to do with the setup of the integral itself. To get an integral, usually one starts with a differential equation and adds differences. For example, a mass element $dm=\lambda(x)dx$ gives the total mass of a rod by adding all such mass elements over the length of the rod. Implicit in this derivation is that $dx$ can be made continuously as small as one pleases. OP's integral implicitly assumes that $dN=\tilde f dt$. Here, $dt$ cannot be as small as one pleases because it cannot be smaller than $1/f$, the period of the wave. In fact $dt$ must be several wave periods long for the idea of the number of wave fronts within that time interval to make sense. In short, there is graininess in $N$ which is not a continuous function of time, therefore one has to count as @andrewkirk suggested. In post #47, I got around the problem of having to write an expression for $dN$ by considering an average of $\delta(t)$ which is a continuous function of time.

 

[haruspex]

I think it has to do with the setup of the integral itself. To get an integral, usually one starts with a differential equation and adds differences. For example, a mass element $dm=\lambda(x)dx$ gives the total mass of a rod by adding all such mass elements over the length of the rod. Implicit in this derivation is that $dx$ can be made continuously as small as one pleases. OP's integral implicitly assumes that $dN=\tilde f dt$. Here, $dt$ cannot be as small as one pleases because it cannot be smaller than $1/f$, the period of the wave. In fact $dt$ must be several wave periods long for the idea of the number of wave fronts within that time interval to make sense. In short, there is graininess in $N$ which is not a continuous function of time, therefore one has to count as @andrewkirk suggested. In post #47, I got around the problem of having to write an expression for $dN$ by considering an average of $\delta(t)$ which is a continuous function of time.

No, I don't think it has anything to do with the graininess. In my analysis I assumed f was arbitrarily large yet arrived at the simple equality between the wavefronts emitted each oscillation and the wavefronts received in the same time interval. Or equivalently, you could suppose 1/f is a multiple of the oscillation time, T.

 

[Delta2]

Thanks for confirming that, but the intriguing question is why the integral at the start of post#1 gives the wrong answer.

Because that integral has the right interval of integration but wrong integrand. The correct integrand is not $\tilde{f}(t)dt$ but it is $\tilde{f}(t)dt'$. That is the observer "sees" the frequency $\tilde{f}(t)$ between the times t' and t'+dt' and not between the time t and t+dt.

For the definition of t' and dt' one should look at your post #16. Jut to say that you use the symbol f' there, where I believe you should use $\tilde{f}$ to keep in touch with the symbolism used in OP.

 

[atyy]

Thanks for confirming that, but the intriguing question is why the integral at the start of post#1 gives the wrong answer.

I would guess that the equation in the OP is wrong because of the issue brought up by @rude man in post #9. When the frequency is changing, the instantaneous frequency is given by the rate of change of phase. The equation should count the total phase change, and then divide that by 2π to get the number of waves.

Edit: Hmm, but @rude man says the proper equation using the instantaneous phase gives the same result?

 

[andrewkirk]

Thanks for confirming that, but the intriguing question is why the integral at the start of post#1 gives the wrong answer.

I think I can see two problems with the reasoning in the OP.

(1) The third equation line in section 3 reads:

$$
N=\int_0^T\tilde{f}(t)\,\mathrm{d}t=f\int_{-\frac{\pi}{\omega}}^\frac{\pi}{\omega}\frac{v_s\,\mathrm{d}t}{v_s-v_0\cos\omega t}.
$$

There is no justification for the change in the integral limits, since the integration variable has not changed. The limits should still be 0 and T, where T is also equal to $2\pi/\omega$.

The limits upon the change of integrator to $\theta$ should be 0 and $\omega T = 2\pi$, not $-\pi$ and $\pi$ as shown.

Then upon change of integrator to $\phi$ the limits should become 0 and $\pi$. This range includes a point where cos is zero, while the range in the OP ($(-\pi/2,\pi/2)$) does not, a fact which will be important to the next observation.

(2) The sixth equation line in section 3 reads:

$$N ...
=\frac{2v_sf}{\omega}\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\frac{\sec^2\phi\,\mathrm{d}\phi}{\sec^2\phi[(v_s-v_0)\cos^2\phi+(v_s+v_0)\sin^2\phi]}
=\frac{2v_sf}{\omega}\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\frac{\sec^2\phi\,\mathrm{d}\phi}{(v_s-v_0)+(v_s+v_0)\tan^2\phi}.
$$
This step involves dividing both numerator and denominator by $\cos^2\phi$, which is zero at a point in the correct integration range (see point (1)), in which case the operation is an invalid division by zero.

Given those problems, I see no reason to expect the reasoning to lead to a correct answer.

 

[haruspex]

you use the symbol f' there, where I believe you should use $\bar {f}$ to keep in touch with the symbolism used in OP.

The OP used both, f' in a generic equation at the start. I stuck with that to simplify typing.

Maybe limiting to second order as I have done gives the wrong answer.

No, second order is enough to see that the integral cannot give the answer that over each whole oscillation the emitter and observer count the same number.

Given those problems, I see no reason to expect the reasoning to lead to a correct answer.

As mentioned above, you don't need to go through the exact integration to determine that the integral must be wrong. The second order approximation is adequate to show that.

I encourage everyone on this thread to consider the simplified version I described:
A fly flies back and forth a distance L at constant speed v, buzzing with frequency f.
Using c for the speed of sound, the observer at one end hears frequency $\frac c{c-v}f$ when the fly is approaching and frequency $\frac c{c+v}f$ when the fly is moving away.
Each traverse takes time L/v, so we may be tempted to write that the total number of wavefronts heard over one lap is $\frac{Lf}v(\frac c{c-v}+\frac c{c+v})=\frac{2c^2Lf}{v(c^2-v^2)}$. This is essentially the same approach as in the integral in post #1. Clearly (I would have thought) the answer should be $\frac{2Lf}v$.

If we look at it from the hearer's perspective the error becomes clear. The periods during which the two frequencies are heard are not equal. The high frequency interval is shorter because the first wavefront in that train had further to travel than the last, while in the other direction the interval is lengthened.
When this is taken into account the discrepancy disappears.

I believe the integral in post #1 falls into the same trap, but it is less obvious what is going on (and harder to fix) because of the smooth change in velocity.

 

[haruspex]

Congratulations, you have just disproved your formula of $\frac{2Lf}v$. "Clearly (I would have thought)"? How "clearly"? when you just showed the opposite and buttressed my 2nd order approximation formula.

No, what I can disprove is the $\frac{2c^2Lf}{v(c^2-v^2)}$ formula.
As I wrote,

"If we look at it from the hearer's perspective the error becomes clear. The periods during which the two frequencies are heard are not equal. The high frequency interval is shorter because the first wavefront in that train had further to travel than the last, while in the other direction the interval is lengthened.
When this is taken into account the discrepancy disappears."
​

In the higher frequency wavetrain, the first wavefront has to travel L to reach the listener, but the last has no distance to travel. Therefore the duration of that frequency for the listener is L/v-L/c, Multiplying that by the perceived frequency yields Lf/v. Hardly surprising because that is the number of wavefronts the emitter generated in that train.
Correspondingly, in the other half lap, the wavetrain at the lower frequency lasts L/c+L/v for the listener, and again carries Lf/v wavefronts.

So we see that it is unsafe to integrate the frequencies perceived by a stationary observer using the emitter's sense of when the wavefronts that will be heard at those frequencies were generated. You have to use the arrival times at the observer.
This is the flaw in the integral in post #1, but it is much harder to see exactly why in the case of a smoothly varying velocity.

 

[haruspex]

It would certainly surprise Mr. Doppler, whose entire theory is based on the fact that the number of pulses received is NOT equal to the number generated by the source if the source is moving with respect to the observer.

That is for unidirectional movement, measuring both over the same time interval.
In the buzzing fly problem, I split the complete lap period into unequal intervals at the receiver.
When the fly is moving towards the observer, the observer hears a higher frequency but over a time interval less than half a lap.

 

[vela]

Has anyone really evaluated the integral?

I evaluated the OP's integral using Mathematica and got the same result as the OP did, and your integral is essentially the same form.

There is no justification for the change in the integral limits, since the integration variable has not changed. The limits should still be 0 and T, where T is also equal to $2\pi/\omega$.

The integrand is periodic with period $T$, so integrating from $0$ to $T$ is the same as integrating from $-T/2$ to $T/2$. These limits have the advantage of sidestepping the singularity when the cosine vanishes in the later steps. Anyway, the most straightforward way I found to do the integral by hand is to shift the limits as the OP did and then use the substitution $z = \tan \frac\theta 2$.

Of course, evaluating the integral is pointless because it's not the correct expression in the first place for the reason @haruspex has explained.

 

[haruspex]

This thread needs to be summarized

Good idea. Here's mine.

Let the frequency observed remotely (r from middle of oscillation) at time t' be F(t').
The number of waves received in an interval is therefore ∫F(t').dt'.

Let the amplitude of the oscillation be A=v₀/ω.
At time t, the emitter is at A sin(ωt). An adjacent stationary observer, on the side towards the remote observer, hears frequency $\bar f(t)=\frac{c}{c-v(t)}f$, where c is the speed of sound.
That tone will be heard by the remote observer at time $t+\frac{r-A\sin(\omega t)}c$.
Defining t' to be that, we have:
$F(t')=\bar f(t)$
$dt'=d(t+\frac{r-A\sin(\omega t)}c)=dt-\frac {A\omega}c\cos(\omega t)dt=dt(1-\frac{v_0}c\cos(\omega t))=dt\frac{c-v(t)}c$.
$\int_{t'=a'}^{b'} F(t').dt'=\int \bar f(t).dt\frac{c-v(t)}c=\int _{t'=a'}^{b'}f.dt$.
Note that the final integral is wrt t but the bounds are in terms of t'. However, an increase of $\frac{2\pi}{\omega}$ in one corresponds to the same change in the other, so over one such period we have $\int _{t'=a'}^{a'+2\pi/\omega}f.dt =\int _{t=a}^{a+2\pi/\omega}f.dt =\frac{2\pi f}{\omega}$.

 

[rude man]

Good idea. Here's mine.

Let the frequency observed remotely (r from middle of oscillation) at time t' be F(t').
The number of waves received in an interval is therefore ∫F(t').dt'.

Let the amplitude of the oscillation be A=v₀/ω.
At time t, the emitter is at A sin(ωt). An adjacent stationary observer, on the side towards the remote observer, hears frequency $\bar f(t)=\frac{c}{c-v(t)}f$, where c is the speed of sound.
That tone will be heard by the remote observer at time $t+\frac{r-A\sin(\omega t)}c$.
Defining t' to be that, we have:
$F(t')=\bar f(t)$
$dt'=d(t+\frac{r-A\sin(\omega t)}c)=dt-\frac {A\omega}c\cos(\omega t)dt=dt(1-\frac{v_0}c\cos(\omega t))=dt\frac{c-v(t)}c$.
$\int_{t'=a'}^{b'} F(t').dt'=\int \bar f(t).dt\frac{c-v(t)}c=\int _{t'=a'}^{b'}f.dt$.
Note that the final integral is wrt t but the bounds are in terms of t'. However, an increase of $\frac{2\pi}{\omega}$ in one corresponds to the same change in the other, so over one such period we have $\int _{t'=a'}^{a'+2\pi/\omega}f.dt =\int _{t=a}^{a+2\pi/\omega}f.dt =\frac{2\pi f}{\omega}$.

Thank you haruspex. Having vela support your view, plus the fact that the problem answer is much more likely to be the simple one you came up with rather than my next-to-impossible-to-solve one, forces me to reconsider even though I still can't imagine where I erred. I take some comfort that you too seem to wonder why my integral is wrong. Please let us know if you figure that out. Meantime I'll try to fathom your summary and go on from there.

 

[haruspex]

you too seem to wonder why my integral is wrong. Please let us know if you figure that out.

Yes, it did stump me for a while. I was quite convinced by the OP's argument in favour of the "obvious" result at the end of post #1, so the question for me was where did the algebra go wrong? I tried the approximation for low speed, and could see that the initial integral was already flawed.

Then I realized it might not be safe to integrate the observer's delayed measure of the frequency using the emitter's view of when those wavefronts were generated. That's why I thought up the simpler constant speed example.
In that, we see that the emitter considers the two motions, left-to-right and right-to-left, as taking equal time; but the remote observer hears the higher frequency for less time than the lower frequency.

After that, it was a matter of considering exactly what the observer would hear and when. I was quite relieved when the "right" result emerged.
Quite a subtle trap.

 

[rude man]

I think the point I missed is that the frequency formula can be misinterpreted since the sound waves are not increased in number but just compressed or expanded in one oscillatory cycle. Thus the instantaneous frequency changes but the total number of emitted pulses = the number received over a cycle. Given that, the correct formula 2pi f/w follows immediately. 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. Sound right?

PS I've deleted almost all my posts in this thread and regret having possibly misled the OP, although no one has heard of him/her since his/her opening post.

 

[haruspex]

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.

 

[Delta2]

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.

 

[rude man]

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.

(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.

Look at @haruspex 's post 43. If you count the received + and - pulses then at the end you should receive the same number of pulses as the number sent out by the source in the time the fly goes +u, then -u, then stops. Why? Because your net u = 0 so your total count had better be the same as the source's as if it had never moved. But, using the frequency formula it's not, as post 43 showed. It's really a matter of understanding the physics before forging ahead with a formula that's inappropriate to the situation. As I unfortunately did.

 

[Delta2]

@rude man , what's the formula for $f$' if $u(t)$ is time varying? I suppose 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.

Which makes me wonder, if $w(t)$ is the waveform of the source, what is the waveform seen by the observer if the source is moving with speed $u(t)$. If $u(t)=u$ is constant I think the stationary observer sees the waveform as $w(\frac{c}{c-u}t)$ (with the extra assumption that the waveform doesn't diminish due to distance from the source).

 

[haruspex]

what's the formula for f' if u(t) is time varying?

Can generalise the equation in post #47.
If the source is at offset x(t) at time t then the wavefronts it emits at that time will reach the observer at time t+(r-x)/c and be heard as having frequency $\frac c{c-\dot x}f$.
If the heard frequency is F(t) then $F(t+\frac{r-x}c)=\frac c{c-\dot x}f(t)$.

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.

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]

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.

An FM signal has theoretically infinite spectrum of frequencies. Which does the observer hear and you denote as $F(t)$??

In order for me to fully understand this we first must find the waveform $W(t')$ that the observer sees and then do Fourier transform of $W(t')$. And the interesting question for me is :

If the source emits waveform $w(t)$, and is moving with speed $u(t)$, what is the waveform $W(t')$ that the observer sees?

Is it $W(t')=w(\frac{c}{c-u(t')}t')$? If u(t)=u constant then I think this is correct but if u(t) is time varying what is the equation that relates w(t),u(t) and W(t') I am not sure at all...

I agree that W(t') going to look like an FM signal where in the role of the carrier is the frequency of the moving source(if the source emits a pure sinusoid) and in the role of the modulating signal is the velocity u(t) of the moving source.

 

[rude man]

I deleted my last post, was questionable derivation. Will continue to work on it.

 

[Delta2]

Maybe it is $W(t')=w(\int_0^{t'}\frac{c}{c-u(t)}dt)$. I don't have a proof for it , I just guessed it.

If it is correct and we assume that $w(t)=\sin{2\pi f_ct}$ that is the source emits a sine wave, then $W(t')=\sin{(2\pi f_c \int_0^{t'}\frac{c}{c-u(t)}dt)}$ so the phase for the observer is $\Phi(t')=2\pi f_c \int_0^{t'}\frac{c}{c-u(t)}dt$ and the instantaneous frequency defined as $\frac{d\Phi}{dt}=2\pi f_c \frac{c}{c-u(t)}$ which is very convenient.

 

[Delta2]

Yes ok thanks @rude man , no need to analyse the terms sin(ab sin(…)) or cos(ab sin(…)) in terms of Bessel functions...

I have my doubts about these two lines

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

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 $x'(t)=\sin{(2\pi\frac{c}{c-u}t)}$

 

[rude man]

Yes ok thanks @rude man , no need to analyse the terms sin(ab sin(…)) or cos(ab sin(…)) in terms of Bessel functions...

But if you don't you don't see all the harmonics.

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 $x'(t)=\sin{(2\pi\frac{c}{c-u}t)}$

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'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) ).