Siren frequency heard by observer when an ambulance passes

[songoku]

Homework Statement

An ambulance siren emits a sound with a single frequency f. The ambulance travels towards, passes close to, and then travels away from a stationary observer. Which statement describes the frequency of the sound detected by the observer as the ambulance passes the observer?
A. equal to f and decreasing
B. equal to f and increasing
C. greater than f and constant
D. less than f and constant

Relevant Equations

Doppler Effect

I am not sure I understand the question.

Based on "The ambulance travels towards, passes close to, and then travels away from a stationary observer", I would answer greater than f then less than f.

If based on "as the ambulance passes the observer", I would say less than f and constant (option D)

But the answer is (A).

What is my mistake?

Thanks

 

[kuruman]

What is my mistake?

Your mistake is that you probably did this in your head. Write down the relevant equation instead of just "Doppler effect", which is not even an equation, and apply it to the given situation. I am sure you will find your mistake.

 

[Orodruin]

I would rather say it is an issue of the problem being possible to interpret in two different ways:

1. The ambulance passing “close to” the observer means that we can use the approximation that the ambulance passes through the same location as the observer to good accuracy. The frequency will have to be given infinitesimally after passage to be well defined.

2. The approximation cannot be made and you have to take the angle between direction of motion and direction to the ambulance into account.

The listed answer is compatible with 2 whereas your response is compatible with 1.

 

[kuruman]

1. The ambulance passing “close to” the observer means that we can use the approximation that the ambulance passes through the same location as the observer to good accuracy.

My interpretation of "close to" is "close but not in such a manner as to run over the observer." There is a distance of closest approach. We are to answer the question when the ambulance is exactly at that distance.

 

[Orodruin]

My interpretation of "close to" is "close but not in such a manner as to run over the observer." There is a distance of closest approach. We are to answer the question when the ambulance is exactly at that distance.

Which is:

2. The approximation cannot be made and you have to take the angle between direction of motion and direction to the ambulance into account.

1 is not necessarily about running the observer over, but about the passage being so fast that the approximation of a step function is a good approximation to the frequency change during the passage. What is a "good approximation" is of course always up for discussion and a matter of observational accuracy.

 

[kuruman]

I think the questions are "at the point of closest approach, (a) what is the value of the function and (b) what is the sign of the slope?"

I have observed what is described with a train blowing its siren while I being stopped at a railroad crossing. It's one continuous sound with changing frequency and intensity. The longitudinal component of the train's velocity relative to me is a continuous function of the siren's radial distance from me and goes through zero at the point of closest approach.

 

[pbuk]

Based on "The ambulance travels towards, passes close to, and then travels away from a stationary observer", I would answer greater than f then less than f.

But the question is not "Which statement describes the frequency of the sound detected by the observer as the ambulance travels towards, passes close to, and then travels away from a stationary observer" so that is the not an answer to the question that is asked.

If based on "as the ambulance passes the observer", I would say less than f and constant (option D)

Then you would be wrong. Before the ambulance passes the observer it is travelling towards them and so the observed frequency is greater than f. After the ambulance passes the observer it is travelling away from them and so the observed frequency is less than f. At the instant that the ambulance passes the observer it is neither travelling towards them nor away from them and so the observed frequency is equal to f and decreasing.

 

[pbuk]

1 is not necessarily about running the observer over, but about the passage being so fast that the approximation of a step function is a good approximation to the frequency change during the passage.

Maybe but both the value and the gradient of an unqualified step function are undefined at the point of discontinuity and "undefined" is not among the answers.

 

[Orodruin]

Maybe but both the value and the gradient of an unqualified step function are undefined at the point of discontinuity and "undefined" is not among the answers.

Hence why I said:

The frequency will have to be given infinitesimally after passage to be well defined.

I am not arguing that it is the intended interpretation. I am arguing that it is a possible interpretation given the usual fuzziness with implicit assumptions of many physics teachers.

 

[haruspex]

1 is not necessarily about running the observer over, but about the passage being so fast that the approximation of a step function is a good approximation to the frequency change during the passage.

As ever, the right way to treat an idealisation (zero mass, instantaneous impact, inextensible string…) is as the limit of realistic scenarios. The idealisation is valid if the limit (which may involve several parameters varying independently) always exists and is constant.
In the present case, we have the proximity, s, the velocity, v, and the raw frequency, f.
For any non-infinite/infinitesimal combo of those, the answer is A.

 

[Orodruin]

the right way to treat an idealisation (zero mass, instantaneous impact, inextensible string…) is as the limit of realistic scenarios

Distribution theory would like a word …

 

[kuruman]

I do not see the need for a step function or an idealization. Here is the math that accompanies my comments in post #6.

The Doppler equation for a stationary observer relative to the medium is $$f_{\text{obs.}}=\frac{c}{c\mp v_s}f_0$$ where
$f_{\text{obs.}} =~$ the observed frequency;
$f_0 =~$ the emitted frequency;
$c=~$ the speed of sound;
$v_{\text{s}} =~$ the speed of the source relative to the observer; the upper/lower sign is used if the source moves towards/(away from) the observer. "Towards and "away" means along the straight line joining the source and the observer.

We can safely assume that the observer O has heard the siren and is standing on the sidewalk watching the ambulance go by (see diagram on the right). The radial component of the velocity towards/(away from) the observer is $v_{\text{radial}}=v_{\text{s}}\cos\theta.$ The Doppler equation becomes $$f_{\text{obs.}}=\frac{c}{c- v_s\cos\theta}f_0.$$ Note how the cosine in the denominator automatically takes care of all that needs to be taken care of.

 

[Orodruin]

In those variables you cannot see how the step function of timr becomes the limit as the distance of closest approach goes to zero. This is because it is obscured by the fact that in the corresponding limit, $\theta$ approaches a step function in time.

Again, I am not saying there is a need for making that approximation. I am saying that making such an approximation based on the wording in the problem would not be very different from making other implicitly assumed approximations in other problems at the same level. And if you do, then you may mess up the answer relative to the intended one.

The maths here are absolutely trivial and were never in doubt.

 

[haruspex]

Distribution theory would like a word …

Consider this example.
A mass hangs from an inelastic, inextensible, massless string. What is the acceleration of the mass immediately after the string is cut?
Here we have four idealisations, the fourth being "immediately after".
In any realistic scenario, the string has mass, some elasticity and the acceleration can only be judged after some very short interval. Plug in unknowns for those and take limits in some order, or simultaneously according to set ratios, or whatever .
If we let the time interval go to zero first, the answer for the acceleration is zero. If it goes to the limit last, the answer is g. Ratios linking the limit processes can produce any value in between.

Does distribution theory help here?

 

[Orodruin]

Consider this example.
A mass hangs from an inelastic, inextensible, massless string. What is the acceleration of the mass immediately after the string is cut?
Here we have four idealisations, the fourth being "immediately after".
In any realistic scenario, the string has mass, some elasticity and the acceleration can only be judged after some very short interval. Plug in unknowns for those and take limits in some order, or simultaneously according to set ratios, or whatever .
If we let the time interval go to zero first, the answer for the acceleration is zero. If it goes to the limit last, the answer is g. Ratios linking the limit processes can produce any value in between.

Does distribution theory help here?

It helps in the sense that it gives you an answer.

For any non-infinite/infinitesimal combo of those, the answer is A.

That really doesn’t matter though if the accuracy of the experiment cannot tell that particular case ofA from the approximation of the idealization. Then the idealization is sufficiently accurate.

I could talk all day about the mass increase of an object that is heated by two degrees. That doesn’t make it measurable and constant mass is a sufficiently good approximation.

 

[haruspex]

It helps in the sense that it gives you an answer.

Namely?

 

[Orodruin]

Namely?

The idealized differential equation is in t and therefore the solution is a distribution on the time axis. For t=0 the distribution in itself does not take a value but for any time ”after” it is g.

 

[Orodruin]

But either way, your example is just making my point: That the result may very well depend on the assumptions made.

As the closest approach distance tends to zero, the frequency converges pointwise to a step function.

 

[haruspex]

The idealized differential equation is in t and therefore the solution is a distribution on the time axis. For t=0 the distribution in itself does not take a value but for any time ”after” it is g.

It seems you have simply taken the other idealisation limits first. There is no justification for that.

 

[Orodruin]

It seems you have simply taken the other idealisation limits first. There is no justification for that.

No, I have solved the differential equation in the idealized case. It is your formulation with ”almost” that suggests time goes last (you did not say ”almost ideal string”). You’ll have to decide what you really want to say here related to the OP because so far all you have said is ”it might depend on the idealization”, which is exactly what I started by saying.

 

[Tom.G]

I do not see the need for a step function or an idealization. Here is the math that accompanies my comments in post #6.

The Doppler equation for a stationary observer relative to the medium is $$f_{\text{obs.}}=\frac{c}{c\mp v_s}f_0$$ where
$f_{\text{obs.}} =~$ the observed frequency;
$f_0 =~$ the emitted frequency;
$c=~$ the speed of sound;
$v_{\text{s}} =~$ the speed of the source relative to the observer; the upper/lower sign is used if the source moves towards/(away from) the observer. "Towards and "away" means along the straight line joining the source and the observer.

View attachment 353176We can safely assume that the observer O has heard the siren and is standing on the sidewalk watching the ambulance go by (see diagram on the right). The radial component of the velocity towards/(away from) the observer is $v_{\text{radial}}=v_{\text{s}}\cos\theta.$ The Doppler equation becomes $$f_{\text{obs.}}=\frac{c}{c- v_s\cos\theta}f_0.$$ Note how the cosine in the denominator automatically takes care of all that needs to be taken care of.

As the closest approach distance tends to zero, the frequency converges pointwise to a step function.

@Orodruin, wouldn't your statement require the Cos function to be a step function?
That seems somewhat improbable.

 

[haruspex]

so far all you have said is ”it might depend on the idealization”

No, I wrote in post #10 that the answer is unambiguously A.
My example in post #14 shows that sometimes it depends on the idealisation.

 

[Orodruin]

No, I wrote in post #10 that the answer is unambiguously A.
My example in post #14 shows that sometimes it depends on the idealisation.

But as I said already, which you completely ignored, the function converges pointwise to the step function. This means that at some point in the limit it is indistinguishable from going to a constant after the passage because in physics we need to care about experimental accuracy,

 

[Orodruin]

@Orodruin, wouldn't your statement require the Cos function to be a step function?
That seems somewhat improbable.

I already addressed this:

In those variables you cannot see how the step function of timr becomes the limit as the distance of closest approach goes to zero. This is because it is obscured by the fact that in the corresponding limit, θ approaches a step function in time.

 

[haruspex]

But as I said already, which you completely ignored, the function converges pointwise to the step function. This means that at some point in the limit it is indistinguishable from going to a constant after the passage because in physics we need to care about experimental accuracy,

The question asked is which of A to D applies. For every noninfinite/nonzero set of numbers you plug in the answer is A. The limit of the sequence A, A, A, …. is A.

 

[Orodruin]

The question asked is which of A to D applies. For every noninfinite/nonzero set of numbers you plug in the answer is A. The limit of the sequence A, A, A, …. is A.

Yes, but that is because you are assuming that any change is detectable, which - as I said - is not going to be the case, so that is not a very physical answer.

Here is a plot:

The input data are:
$v = 150$ km/h
$d = 1$ m
$c = 340$ m/s
$f_0 = 700$ Hz
The x-axis has units of seconds.
The orange line is the observed frequency in Hz.
The blue line is the number of wavelengths passing by the observer since $t = -1$ s.

We can observe that, yes, the orange line is continuous, it is equal to $f_0$ at $t = 0$, and it decreases. This is the purely mathematical answer, but physics is an empirical science. We need to ask ourselves if this is detectable. The frequency goes to a constant essentially 0.1 s after the passage. If we assume the "detector" to be a human ear, this is most likely way too short to distinguish from an instant change in the frequency.

Even if we do not assume a human ear, but a more ideal detector, there is this graph:

The x-axis is still time in seconds and the graph shows the relative change in frequency during one full oscillation. To meaningfully talk about the frequency of a wave, the frequency should not change appreciably over one period - or ideally several periods, meaning that this graph should be well below 1, which again, it is not for times around the passage time $t = 0$.

Could you detect the very small changes in frequency after $t = 0.1$ s? If you have an extremely sensitive detector, perhaps, but for most practical measurements the frequency - where meaningfully defined - is going to be effectively constant.

Edit: Put somewhat differently: Within all of the idealisations made, you have not taken into account the idealisation of the observer that should detect the frequency change, but simply assumed that this observer can measure frequency with infinite precision over an infinitesimal period of time.

 

[jbriggs444]

A. equal to f and decreasing
B. equal to f and increasing
C. greater than f and constant
D. less than f and constant

Clearly, all four possibilities are incorrect.

At the time of closest approach, the currently perceived sound will have been emitted from a point prior to that. It will be doppler shifted to a higher frequency. This eliminates everything but choice C.

However (barring a collision with the hapless listener) the perceived frequency will be strictly monotone decreasing at all times. Thie eliminates everything other than choice A.