# Homework Help: Number of waves from oscillating source in one period

1. Jul 12, 2018 at 12:23 PM

### {???}

1. The problem statement, all variables and given/known data
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?

2. Relevant 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.

3. 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?

Last edited: Jul 12, 2018 at 1:53 PM
2. Jul 12, 2018 at 4:29 PM

### kuruman

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.

3. Jul 12, 2018 at 5:45 PM

### haruspex

The sign change is handled by the cos function.
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?

4. Jul 12, 2018 at 6:09 PM

### Delta²

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.

Last edited: Jul 12, 2018 at 10:15 PM
5. Jul 12, 2018 at 6:16 PM

### kuruman

We so as @haruspex already pointed out.

6. Jul 12, 2018 at 6:44 PM

### haruspex

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?

7. Jul 12, 2018 at 10:31 PM

### Delta²

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.
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. $T=2\pi\sqrt\frac{m}{k}$

8. Jul 13, 2018 at 12:34 AM

### haruspex

They can't, which is why 2πf√(m/k), the number emitted each cycle, is also the number received.
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.

9. Jul 13, 2018 at 2:09 AM

### haruspex

It is not, and that gives the wrong answer as explained above.

10. Jul 13, 2018 at 2:42 AM

### Delta²

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.
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.

11. Jul 13, 2018 at 3:08 AM

### haruspex

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.
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.

12. Jul 13, 2018 at 9:19 AM

### Delta²

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.
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.

13. Jul 13, 2018 at 5:18 PM

### haruspex

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.
$t'=t+\frac{r-\frac{v_0}{\omega}\sin(\omega t)}{v_s}$, $dt'=dt(1-\frac{v_0}{v_s}\cos(\omega t))$.
$f'=\frac{v_s}{v_s-v_0\cos(\omega t)}f$.
$f'.dt'=f.dt$.

It is not correct up to there. Please stop ignoring the points I have raised.
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.

Last edited: Jul 13, 2018 at 5:27 PM
14. Jul 13, 2018 at 8:45 PM

### haruspex

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.

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.

Last edited: Jul 13, 2018 at 9:05 PM
15. Jul 13, 2018 at 10:21 PM

### Tom.G

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

16. Jul 13, 2018 at 10:42 PM

### haruspex

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.
Simple. It moves at speed v in alternating directions.
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.

Last edited: Jul 13, 2018 at 11:05 PM
17. Jul 13, 2018 at 10:44 PM

### haruspex

Correct.
You are confusing the oscillation of the mass (emitter) with the oscillation of a presumed diaphragm within the emitter to create the sound.

18. Jul 13, 2018 at 11:33 PM

### Delta²

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.

Last edited: Jul 14, 2018 at 12:14 AM
19. Jul 14, 2018 at 12:34 AM

### haruspex

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.

20. Jul 14, 2018 at 12:58 AM

### Delta²

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?)

21. Jul 14, 2018 at 1:36 AM

### haruspex

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.
Yes.

22. Jul 14, 2018 at 1:47 AM

### haruspex

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.

23. Jul 14, 2018 at 2:30 AM

### Delta²

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.

Last edited: Jul 14, 2018 at 2:50 AM
24. Jul 14, 2018 at 4:03 AM

### haruspex

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.

25. Jul 14, 2018 at 4:06 AM

### haruspex

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