Number of waves from an oscillating source in one period

Homework Statement

A mass $m$ attached to a spring of spring constant $k$ emits sound at frequency $f$, detected by a collinear observer at distance $r$. If the mass has maximum velocity $v_0$, what is the total number of waves the observer detects in one period of oscillation?

Homework Equations

Unless I'm missing something, we should only need the Doppler effect for a moving source:
$$f'=\frac{v_s}{v_s-v_0}f$$
the angular velocity of a spring-mass system
$$\omega^2=\frac{k}{m}$$
and the number of waves being the integral of the frequency over time
$$N=\int_{t_i}^{t_f}f\,\mathrm{d}t$$
This last one I'm not sure about, but it makes sense in the case that $f$ is constant, and seems a natural generalization.

The Attempt at a Solution

The spring-mass system oscillates with velocity given by
$$v(t)=v_0\cos\omega t,\qquad\textrm{where}\qquad\omega^2=\frac{k}{m}.$$
The observer is stationary, and so detects a Doppler-shifted frequency $\tilde{f}$ given by
$$\tilde{f}(t)=\frac{v_s}{v_s-v(t)}f=\frac{v_s}{v_s-v_0\cos\omega t}.$$
The number of waves detected by the observer is simply the integral of frequency over time:
$$N=\int_0^T\tilde{f}(t)\,\mathrm{d}t=f\int_{-\frac{\pi}{\omega}}^\frac{\pi}{\omega}\frac{v_s\,\mathrm{d}t}{v_s-v_0\cos\omega t}.$$
We are assuming $v_0<v_s$, so that the denominator is never zero. We first substitute $\theta=\omega t$, so $\mathrm{d}t=\mathrm{d}\theta/\omega$ and $\theta(\pm\frac{\pi}{\omega})=\pm\pi$:
$$N=\frac{v_sf}{\omega}\int_{-\pi}^\pi\frac{\mathrm{d}\theta}{v_s-v_0\cos\theta}.$$
We now substitute $\theta=2\phi$, so $\mathrm{d}\theta=2\,\mathrm{d}\phi$ and $\phi(\pm\pi)=\pm\frac{\pi}{2}$:
$$N=\frac{v_sf}{\omega}\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\frac{2\,\mathrm{d}\phi}{v_s-v_0\cos 2\phi} =\frac{2v_sf}{\omega}\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\frac{\mathrm{d}\phi}{v_s(\cos^2\phi+\sin^2\phi)-v_0(\cos^2\phi-\sin^2\phi)}$$
$$=\frac{2v_sf}{\omega}\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\frac{\sec^2\phi\,\mathrm{d}\phi}{\sec^2\phi[(v_s-v_0)\cos^2\phi+(v_s+v_0)\sin^2\phi]} =\frac{2v_sf}{\omega}\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\frac{\sec^2\phi\,\mathrm{d}\phi}{(v_s-v_0)+(v_s+v_0)\tan^2\phi}.$$
We now substitute $x=\tan\phi$, so $\mathrm{d}x=\sec^2\phi\,\mathrm{d}\phi$, and $x(\pm\frac{\pi}{2})=\pm\infty$:
$$N=\frac{2v_sf}{\omega}\int_{-\infty}^\infty\frac{\mathrm{d}x}{(v_s-v_0)+(v_s+v_0)x^2}.$$
We finally substitute $x=\sqrt{(v_s-v_0)/(v_s+v_0)}z$, so $\mathrm{d}x=\sqrt{(v_s-v_0)/(v_s+v_0)}\,\mathrm{d}z$:
$$N=\frac{2v_sf}{\omega}\sqrt\frac{v_s-v_0}{v_s+v_0}\int_{-\infty}^\infty\frac{\mathrm{d}z}{(v_s-v_0)+(v_s-v_0)z^2} =\frac{2v_sf}{\omega}\frac{1}{\sqrt{v_s^2-v_0^2}}\int_{-\infty}^\infty\frac{\mathrm{d}z}{1+z^2} =\fbox{\displaystyle\frac{2\pi fv_s}{\sqrt{v_s^2-v_0^2}}\sqrt\frac{m}{k}.}$$

There is also the "obvious" answer
$$n=2\pi f\sqrt\frac{m}{k}.$$
I can see why this answer should be right, and why qualitatively my answer above should be wrong. However, I cannot find any flaw in my quantitative argument. Can anybody help me spot the error in my reasoning?

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kuruman
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For half the period the source is moving towards the observer and for the other half away from the observer. This means a change in sign in the denominator which (it seems) you did not take into account.

haruspex
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The sign change is handled by the cos function.
and the number of waves being the integral of the frequency over time
$$N=\int_{t_i}^{t_f}f\,\mathrm{d}t$$
This last one I'm not sure about, but it makes sense in the case that $f$ is constant, and seems a natural generalization.
You are right to be suspicious.
To make it simpler, consider oscillating at constant speed. In each cycle, how long does the observer hear the higher frequency?

Delta2
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For half the period the source is moving towards the observer and for the other half away from the observer. This means a change in sign in the denominator which (it seems) you did not take into account.
the doppler shifted frequency is ##\tilde {f(t)}=\frac{v_s}{v_s-v_0\cos{\omega t}}f## ,so don't we take the change of sign into account because ##\cos{\omega t}## is positive in half cycle and negative in half cycle?

I also don't see why the obvious answer is correct, it would be correct only if there was no doppler effect or it is negligible. In the limit ##v_0<<v_s## that is doppler effect is negligible ##\sqrt{v_s^2-v_0^2}\approx. \sqrt{v_s^2}=v_s## so both formulas give the same result.

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kuruman
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the doppler shifted frequency is ~f(t=vsvs−v0cosωt\tilde {f(t}=\frac{v_s}{v_s-v_0\cos{\omega t}} ,so don't we take the change of sign into account because cosωt\cos{\omega t} is positive in half cycle and negative in half cycle?
We so as @haruspex already pointed out.

haruspex
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I also don't see why the obvious answer is correct,
One complete cycle would be representative of the long term average. Over long periods, the number of waves emitted cannot diverge from the number received.
Did you consider the question I posed at the end of post #3?

Delta2
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One complete cycle would be representative of the long term average. Over long periods, the number of waves emitted cannot diverge from the number received.
Sorry I cant understand what you trying to say here. The source emits waves and the observer at distance r detects waves. I don't understand why some waves can be lost and not received by the observer.
Did you consider the question I posed at the end of post #3?
If oscillation is at speed ##+-v_0## then the number of waves for the half cycle where the speed is ##+v_0## is ##\frac{v_s}{v_s+v_0}f\frac{T}{2}## and for the other half cycle with speed ##-v_0## is ##\frac{v_s}{v_s-v_0}f\frac{T}{2}## and the total number of waves is the sum of those two. ##T=2\pi\sqrt\frac{m}{k}##

andrewkirk
haruspex
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I don't understand why some waves can be lost and not received by the observer.
They can't, which is why 2πf√(m/k), the number emitted each cycle, is also the number received.
the number of waves for the half cycle where the speed is ...
The source emits the same number in each half cycle. When it is approaching the observer the wavelength is shorter, so the time for that half to be received, first to last, is less than for the other half cycle.
This is why the OP's integral is wrong. For the variable of integration it uses the time of when the wave was emitted instead of its arrival time.
It's a bit like time dilation in relativity.

haruspex
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the integral has to be of the form ∫ dx/(a - b cos wx)
It is not, and that gives the wrong answer as explained above.

Delta2
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They can't, which is why 2πf√(m/k), the number emitted each cycle, is also the number received.
we agree that no waves are lost, still I disagree in that i believe that the observer counts different number of waves because for him the frequency of each wave is different.
The source emits the same number in each half cycle. When it is approaching the observer the wavelength is shorter, so the time for that half to be received, first to last, is less than for the other half cycle.
This is why the OP's integral is wrong. For the variable of integration it uses the time of when the wave was emitted instead of its arrival time.
It's a bit like time dilation in relativity.
Sorry again I don't seem to understand your qualitative arguments. Can you write some math? What integral and what time variable we should use? We are in classical physics here, time variable is the same for all sources and observers, regardless of their relative motion.

haruspex
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we agree that no waves are lost, still I disagree in that i believe that the observer counts different number of waves because for him the frequency of each wave is different.
Over an arbitrary period the counts may be different, but it cannot keep increasing without limit. Where would all those waves be accumulating (or get created)?
If a discrepancy arises over a single complete oscillation, that same discrepancy must arise every oscillation.
What integral and what time variable we should use?
Suppose we position a series of microphones along the oscillation path. As the emitter passes one at time t the microphone will detect frequency ##\bar f## as defined in post #1.
The position of the microphone is ##A\sin(\omega t)## where ##A=v_0/\omega##. That signal will reach the observer at time ##t'=t+(r-A\sin(\omega t))/v_s##. The integral should be dt', not dt.

Delta2
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Over an arbitrary period the counts may be different, but it cannot keep increasing without limit. Where would all those waves be accumulating (or get created)?
If a discrepancy arises over a single complete oscillation, that same discrepancy must arise every oscillation.
Hmmm, your arguments here seem to be kind of correct but still I disagree, its not that new waves are lost somewhere or are created from nowhere, it is the way we count them at the stationary observer that makes the difference. And we count them differently because the frequency perceived at the stationary observer is different.
Suppose we position a series of microphones along the oscillation path. As the emitter passes one at time t the microphone will detect frequency ##\bar f## as defined in post #1.
The position of the microphone is ##A\sin(\omega t)## where ##A=v_0/\omega##. That signal will reach the observer at time ##t'=t+(r-A\sin(\omega t))/v_s##. The integral should be dt', not dt.
so if ##\tilde{f(t)}## is defined as by OP , and ##t'=g(t) ## as defined by you, ##dt'=g'(t)dt## so you say we should calculate the integral ##\int \tilde{f(t)}g'(t)dt##. This is a lot more messier (makes wolfram crash) I doubt it gives that simple obvious answer we are waiting for.

haruspex
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the way we count them at the stationary observer
Eh? There will be no disagreement as to what constitutes a wavefront, and they are neither created no destroyed. The arrival intervals vary, but the difference in the counts only varies over a fixed finite range.
This is a lot more messier
##t'=t+\frac{r-\frac{v_0}{\omega}\sin(\omega t)}{v_s}##, ##dt'=dt(1-\frac{v_0}{v_s}\cos(\omega t))##.
##f'=\frac{v_s}{v_s-v_0\cos(\omega t)}f##.
##f'.dt'=f.dt##.

Correct up to here.
It is not correct up to there. Please stop ignoring the points I have raised.
One alternative making the math much more tractable would be to assume that v0 << vs.
I would not call that making it more tractable. I would call it starting with the wrong equation but magicking away the error with an approximation.
If you apply your approximation but keep second order terms you will find it still produces an accumulating discrepancy between the number emitted and the number received, which is clearly not possible.

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Delta2
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you haven't explained why that number is not the OP's formula.
As I pointed out in post #8, f'(t) is not the frequency observed by the remote observer at time t. Let that frequency be f"(t). It is the frequency observed by a local observer, on the remote observer side of the emitter, at an earlier time. The time difference varies.
## f'(t)=f"(t+\frac{r-\frac{v_0}{\omega}\sin(\omega t)}{v_s})##
A correct integral is f"(t).dt.

As mentioned, the issue may become clearer if you answer my question at the end of post #3.

The OP's formula is correct and will remain so unless and until you have come up with a credible alternative.
Barring a flaw in the subsequent working, the OP's formula is wrong because it produces an impossible answer. The challenge is to figure why it is wrong, and that challenge will remain even if my explanation is incorrect.
Incidentally, the risk of there being a subsequent flaw can be minimised by applying your approximation to the OP's integral but keeping second order terms. As you say, this makes the algebra much easier, but still produces an impossible answer.

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Tom.G
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A mass m attached to a spring of spring constant k emits sound at frequency f, detected by a collinear observer at distance r. If the mass has maximum velocity v0, what is the total number of waves the observer detects in one period of oscillation?
This does NOT say that the mass/spring assemblage is moving WRT the observer, only that the mass is moving, presumably oscillating since the next sentence states "in one period of oscillation".

Cheers,
Tom

haruspex
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The formula for f ' is the formula as seen by the observer,
Not quite. There is a time shift, which varies.
At time t, the emitter is at position ## \frac{v_0}{\omega}\sin(\omega t)##. A stationary observer very close to there, just on the side of the emitter towards the remote observer, will detect frequency f'(t). But that signal will take further time ##\frac{r-\frac{v_0}{\omega}\sin(\omega t)}{v_s}## to reach the remote observer.

If we define t' as the time it reaches the remote observer then the correct integral is ##\int f'(t).dt'##.
For the development from there please see post #16.
What does "oscillation at constant speed" mean??
Simple. It moves at speed v in alternating directions.
explain to us why the formula produces an impossible answer?
I don't know how to make it much clearer than already spelt out in posts #6, #8 and #12.
The wave count over one complete oscillation (to nearest whole number of waves) must be representative of the long term average. If there is a discrepancy then the same discrepancy will occur every oscillation, pushing the two counts ever further apart.
Suppose this discrepancy is such that the observer gets fewer waves than were emitted in the oscillation. Those extra wavefronts must lie between the two. Repeating the discrepancy every oscillation must pile up ever more wavefronts between them.

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haruspex
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only that the mass is moving, presumably oscillating since the next sentence states "in one period of oscillation".
Correct.
You are confusing the oscillation of the mass (emitter) with the oscillation of a presumed diaphragm within the emitter to create the sound.

Tom.G and Delta2
Delta2
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Ok lets get a simpler problem that has constant frequencies and doesn't involve moving both closer and further from the observer.

Suppose we have a source that emits a sinusoidal wave ##y(t)=y_0\sin{2\pi ft}## (1) and moves away at constant velocity ##v_0## from a stationary observer.

An observer that is moving together with the source measures y(t) exactly as (1) says and at a time interval ##\Delta t## measures ##f\Delta t## waves.

A stationary observer at distance ##r## measures ##\tilde {y}(t')=y_0\sin{2\pi \tilde {f}t'}## because the frequency for him is ##\tilde{f}=\frac{c}{c+v_0}f## where ##t'=t-\frac{r}{c}##and this observer at the same time interval ##\Delta t'## measure ##\tilde{f}\Delta t'## waves. We can see that ##\Delta t'=\Delta t## because t and t' differ only by a constant.

So each observer counts different number of waves within the same time interval. Are there new wavefronts created or lost in between the stationary observer and the source? Of course not.

Our problem is more complicated because I believe the stationary observer measures ##\tilde {y}(t')=y_0\sin{2\pi \tilde {f}(t')t'}## where ##\tilde{f}(t)## as defined in the OP, and t' as defined by @haruspex so the wave form is quite different from a sinusoidal wave.

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haruspex
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Are there new wavefronts created or lost in between the stationary observer and the source?
The number of wavefronts in transit between the two keeps increasing because the source and observer are getting ever further apart. In post #1, that distance varies only over a fixed finite range, so the number of wavefronts in transit can only vary over such a range.
Please, please, try to solve my problem at the end of post #3.

Delta2
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The number of wavefronts in transit between the two keeps increasing because the source and observer are getting ever further apart. In post #1, that distance varies only over a fixed finite range, so the number of wavefronts in transit can only vary over such a range.
Please, please, try to solve my problem at the end of post #3.
Well I think I solved that problem at post #7 but you told me the answer is wrong and to be honest I didn't understand why.

Anyway I am not an expert with doppler effect, haven't studied it in detail, I am a mathematician anyway.

But you agree that in the simpler problem I posed in post #23 each observer counts different number of waves for the same time interval, right? (ok I think I got your point now, they count differently because the number of wavefronts keeps increasing because the distance inbetween keeps increasing right?)

haruspex
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If oscillation is at speed ##+-v_0## then the number of waves for the half cycle where the speed is ##+v_0## is ##\frac{v_s}{v_s+v_0}f\frac{T}{2}## and for the other half cycle with speed ##-v_0## is ##\frac{v_s}{v_s-v_0}f\frac{T}{2}## and the total number of waves is the sum of those two. ##T=2\pi\sqrt\frac{m}{k}##
If I add ##\frac{v_s}{v_s+v_0}f\frac{T}{2}## and ##\frac{v_s}{v_s-v_0}f\frac{T}{2}## I get ##\frac{v_s^2}{v_s^2-v_0^2}fT##. This is analogous to the result the OP got in post #1 using the integral ∫f'.dt.
But those two inputs do not represent what the observer hears.
Consider a half cycle where the source moves from -A at time t=0 to +A at time 2A/v, a distance 2A towards the observer.
The first of that train arrives at the observer at time ##\frac{r+A}{v_s}##, and the last at time ##\frac{2A}v+\frac{r-A}{v_s}##.
So those Tf=(2A/v)f wavefronts arrive over a period ##\frac{2A}v-\frac{2A}{v_s}##.
This wavetrain arrives at the observer at frequency ##f'=\frac{v_s}{v_s-v}f##.
##\int f'## for this half cycle is therefore ##(\frac{2A}v-\frac{2A}{v_s})*\frac{v_s}{v_s-v}f=\frac{2A}vf##, which is reassuring since that is the number emitted.
Similarly in the other half cycle, switching signs as necessary.
The blunder would be to integrate using the source's view of when each wavetrain started and ended, i.e. integration limits for each half cycle are not the same.
they count differently because the number of wavefronts keeps increasing because the distance inbetween keeps increasing right?
Yes.

atyy
haruspex
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The formula for the observer remains f' = f(vs - v0(t))/vs.
At time t, the source has velocity v(t). At that time, the remote observer is not hearing f(vs - v(t))/vs.
I think you have to agree that if F(t) is the frequency the observer hears at time t then the number of wavefronts received over interval (a,b) is ∫abF(t).dt. Can I persuade you to stop asserting that F(t)=f'(t) (which it obviously is not) and figure out what the relationship actually is?
Any answer which leads to a discrepancy between the total emitted and the total received that tends to infinity over time.

Delta2
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..
##\int f'## for this half cycle is therefore ##(\frac{2A}v-\frac{2A}{v_s})*\frac{v_s}{v_s-v}f=\frac{2A}vf##, which is reassuring since that is the number emitted.
Similarly in the other half cycle, switching signs as necessary.
The blunder would be to integrate using the source's view of when each wavetrain started and ended, i.e. integration limits for each half cycle are not the same.
I don't understand why you multiply by the time interval ##(\frac{2A}v-\frac{2A}{v_s})##. It's like saying that a half cycle for the moving source corresponds to ##(\frac{2A}v-\frac{2A}{v_s})## time for the stationary observer, I don't think this is correct, time runs the same way in classical physics for all observers and sources regardless of their relative motion.

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haruspex
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We're not integrating over infinity. We're integrating over one period of the oscillator which is T = 2π/ω.
Right, but as I pointed out, whatever happens over one oscillation will be repeated exactly over the next oscillation. If Δn more wavefronts are emitted than received during one oscillation then the same discrepancy will occur during the next oscillation. That would mean that the number of wavefronts in transit between source and observer increases by Δn each oscillation. Since the distance between them never exceeds A+r, that clearly makes no sense.

haruspex
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I don't understand why you multiply by the time interval ##(\frac{2A}v-\frac{2A}{v_s})##.
Because that is the time for which the arrival rate is ##\frac{v_s-v}{v_s}f##.

Delta2
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Because that is the time for which the arrival rate is ##\frac{v_s-v}{v_s}f##.

Ok fine but if ##T>\frac{2A}{v}-\frac{2A}{v_s}## we have to continue keep counting cause we want to know the waves for a period T. In other words the interval of integration is [0,T] cause that's by definition what we are asking for. I have to admit though after all these posts I am not sure what the integrand of integration should be...

haruspex
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Ok fine but if ##T>\frac{2A}{v}-\frac{2A}{v_s}##.
The arrival period ##\frac{2A}{v}-\frac{2A}{v_s}## is for the wavetrain emitted in the half cycle in which the emitter is moving towards the observer. The wavetrain for the other half cycle is spread over time ##\frac{2A}{v}+\frac{2A}{v_s}## at the observer. These add up to 4A/v=T, as they should.

Delta2
haruspex
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Who says there are?
Each oscillation is a carbon copy of the one before. If one produces an increase in the number of in transit wavefronts then they all will. Judging from the last part of post #1 the OP understood that.
Show using the OP's equation that that would be the case.
That's crazy. I am claiming that equation is wrong, so of course I cannot use that equation to get the result I say is right.

Instead of just insisting that equation is correct, how about you try to answer my question in post #28?
What is the received frequency, F(t), at time t?
(I gave that answer in post #18, so you can either agree with it or provide an alternative.)

haruspex
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Show that using what you call the "wrong" equation leads to a difference
Yes, sorry.
As I mentioned, starting with the OP's integral you can take a small v0 approximation keeping terms to the second order. The second order term does not vanish when integrated.
It is the OP's f', same as the transmitted frequency. Obviously.
No, obviously not. The waves do not reach the observer instantly. There is a delay. f'(t) is the frequency observed at time t by an observer adjacent to the emitter, on the side towards the remote observer. What frequency does the remote observer hear at that time?
Do you see any flaw in the equation I wrote for it in post #18?

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haruspex
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N = f(1 + v02/2v2)
So in each oscillation (each integral over one period) (1 + v02/2v2) times as many wavefronts are received as were emitted. Can you not see that is impossible?
Makes no difference.
That is your assumption, but we will get nowhere until you accept that it could be wrong and follow what the equations say.
Do you agree with my equation in post #18 or not?

haruspex
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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
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… 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.

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haruspex
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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.
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
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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".