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A Common, but Terrible, Mistake in Teaching Math and Science
Rick Garlikov

The following is a typical approach to teaching math and science and then testing the students on what they were taught.  Textbooks which have questions at the end of a unit or chapter will sometimes also make the same error, which tends to show up then in students' homework grades.  I will give two examples; one from math, and one from science.  (It has turned out, as you will see below, that the science example I chose is even much more difficult than I had thought when I chose it for being unfairly difficult in the first place.) 

Math Example
Students are taught that distance equals rate times the time, the formula being d = rt.  Some examples are given to show that the longer you drive at some rate or the faster you drive for some time, the further you will go.  Students are also shown that if you know the distance and the rate, you can use the formula to find the time; and vice versa, if you know the distance and the time, you can find the average rate.  So then some problems are worked that are pretty straightforward, where the time and rate are given and the student needs to compute the distance, or where the distance and either the time or rate is given and students need to compute the other. 

Then some examples or some problems might be given where the rate varies during a trip, and the student needs to compute the average rate.  Typically, the problem is something like: John drives for one hour at 60 mph and then drives for an hour at 30 mph.  What is his average rate of speed for the total distance?  The answer is essentially that since he drives one hour at 60 mph, he goes 60 miles in that time, and since he drives the next hour at 30 mph, he goes 30 miles in that time.  Therefore he has driven 90 miles in two hours, so he averaged 45 mph, which is the average of his two speeds.  One can work it out using the formulas because you need to find the total distance divided by the total time, so you first figure out the distance he drives at 60 and the distance he drives at 30, add them together to get the total distance, and then divide that by the total time, 2 hours, to get the average rate he drove.  Some students may have some difficulty with that when it is first presented, but after it is explained and some more problems of that sort are given, many or most students will be able to do them.  So they practice on things like: a plane flies 200 mph for an hour with the wind, and then it flies 100 mph against the wind for an hour.  What is its average speed.  The answer will be 150 mph. 

So far so good.  But then on the test or in the chapter questions in a textbook, the following problem might be given: John drives 100 miles at 60 miles an hour but then comes to a long patch of highway construction and drives the next 100 miles at 30 mph.  What is his average rate of speed.  Students then go through the same process they have "learned" and they figure the answer to be 45 mph because he has gone the same distance at the two different rates, so the average rate for the whole distance must be the average of the two rates, just as it has been in all the problems they have worked in the book and in class.  Or there might be a variation to the problem, such as:  John needs to average 60 mph over a two mile course in order to qualify for a race.  After the first mile, he develops some sort of engine problem that allowed him only to average 30 mph for the second mile.  How fast would he have had to drive the first mile in order to qualify at the average of 60 mph?  The answer seems fairly obviously to be 90 mph. 

Both those answers are wrong, however.  I will explain that in a minute, but the typical rationale for the book's or the teacher's changing the form of the problem is to see whether, or how many, students really understand the formula d = rt and understand the concept of average rates, etc.  The view of the teacher or the text is that they have only added a slightly different "wrinkle" in order to see how the students adapt what they have learned to take it into account and solve the problem.  Normally, few students, if any can adapt. 

There will sometimes be some students who will get the answers right, but most will get them wrong and be very surprised and confused.  Moreover, many of those students will then become discouraged, figuring they just don't understand math and are no good at it.  That is a bad deduction on their part.  What is really true is that they were taught incorrectly and they were the victim of a trick that the teacher does not realize is an unfair trick.  The teacher and textbook writer will know the question is a trick question, but they will think it is fair because they mistakenly think they are only testing for what ought to be understood.  I will come back to this.  But first the science example. 

Science Example
The teacher or text will explain about the Doppler effect, which is the fact that the pitch of something like a siren or a train whistle gets higher and higher as a train or police car approaches you at a high rate of speed with its siren/whistle blaring and then passes you, at which point the pitch immediately sounds lower.  Most people have heard an example of the Doppler effect at some point.  The teacher or text may then give the reasons for why the phenomenon occurs -- that as the train approaches the sound waves emitted by the whistle approach you closer and closer together so they sound like the pitch of different whistles with higher and higher frequencies, which is a function of the number of waves per second that reach your ear.  Then numerous sample problems are computed where students get practice figuring out pitches from wavelengths, etc.  Then, on the test or for a homework assignment, the teacher will ask the innocent looking question, in order, he thinks, to test student understanding, "How does the pitch of a tornado siren on a building change as you drive fast toward that building and then pass the building?" 

Most students, if not all, will say that the Doppler effect comes into play and the pitch will get higher and higher as you approach the building, and then become immediately lower as you pass it just as it does when a siren on a police car approaches you.  That answer will be in part wrong; there will be a slight increase in pitch which, if noticeable at all, will be much different from the relative increase in pitch when the source of sound is moving toward you rather than your moving toward it.  While the two different situations may seem to be symmetrical, they are not if you analyze or think about them more minutely or more deeply.  But with no reason to have analyzed the two different situations before the test or before doing their homework, most students will fail to see the components of these two situations are different.  And unfortunately when they find out they were mistaken, many students will then draw the mistaken conclusion that they just must not be any good at science, and that they don't understand it.  So they become discouraged and lose interest, as with the math case above. 

[I give an explanation later in this essay of the difference the way sound changes when you move toward a stationary source of sound and the way it changes when a source of sound moves toward you when you are stationary, relative to the air.  I made a mistake though in using the example of the car above, because for the difference in the two situations to be noticeable to the ear, the closing (i.e., approaching) speeds in the two different situations need to be much faster than automobile speeds.  This was pointed out to me in an e-mail from physicist Birger Bjerkeng of Norway. I enclose his explanation in an appropriate context below. The general point of my using this example is not affected (and is perhaps even enhanced by the complexity involved), but I have got the details not quite right because I used an example that was difficult for me to fully comprehend intuitively instead of just mathematically.]

Two Common Sense Example
Suppose you live in the suburbs and have a nice home, even if relatively small, with a relatively decent front and back lawn.  Suppose you buy a carton of eggs at the supermarket and take them home and take one of the eggs either up to the roof of your house and drop it on the lawn, or you simply take one of the eggs out of the carton and throw it up in the air as high as you can, perhaps even over your house or over a tall tree, and let it come down onto the ground?  What will happen to the egg? Well, the obvious answer based on our experience with eggs is that the eggs will splat; they will break into a bunch of pieces and make a mess in the grass.  That answer is wrong.  Most of the time, unless one has a really weak or bad or already cracked egg, the egg will not break, at least not with just one such throw or drop. 

Now, of course, this is difficult to believe.  I have even done it for students and for myself and for my kids, numerous times, and each time, though it has worked almost every time before, I expect it not to work, and I expect the egg to crush itself on impact.  It doesn't happen -- as long as the egg lands on grass, just regular lawn grass, and does not hit a bare spot of dirt or a stone in the grass.  It turns out that the egg is a tremendously strong shape as long as forces are distributed in certain ways on it, and the grass apparently helps distribute the force of impact in those ways.  I believe, but am not certain, that the egg shape is the starter pattern, or is similar to the design, of the huge geodesic domes that serve as sports arenas. 

Does the fact that most adults would mistakenly think that the egg would break mean they do not understand science or that they would not be any good at it?  Probably not.  Most scientists in various fields, and perhaps even many physicists, would likely answer incorrectly.  It is not a familiar phenomena and it is rather surprising, particularly given that it takes very little for an egg to break if accidentally dropped in the kitchen.  And you don't seem to have to hit an egg very hard against a bowl or with a knife or spoon to break it.  That an egg would survive a fall that would kill a person or at least break both his legs, seems quite unimaginable.  Yet, if we watched eggs get dropped onto lawns, we would know they don't usually break.  Would that make us good at science?  No.  It only means we have experienced the phenomena and know what happens, though we may have no understanding why. 

The second example is: imagine that the earth is its current size (roughly 24,000 miles around) but that it is smooth like a ball instead of having mountains and valleys, or high and low planes, etc.  Then imagine that we could tie a ribbon around the equator and pull it tight, so that all around the globe it is tight against the ground. Then imagine that we splice in, at some place, one additional yard of ribbon (36 inches), so it makes a little loop, at that point.  But we don't want a loop or any bulges or uneven places in the ribbon, so we go all around the globe pulling out the slack and making the ribbon so it is evenly near or off the ground the same distance everywhere.  Will the ribbon then be very high off the ground or would it be raised up above the ground such a minute amount you wouldn't even notice it.  Remember, we are dispersing 36 inches of slack (less than an 18 high loop) over a distance of 24,000 miles.  Would anyone likely trip over the ribbon as they walked across the equator? 

The intuitive answer is that it would not even be noticeable and the ribbon would be practically an infinitesimal distance above the ground.  The actual answer, however, is that the ribbon would be nearly 6 inches off the ground all the way around.  Again, however, I will come back to this and what it means that almost everyone gets it wrong. 

Returning to the Math and Science Examples
What has happened in the math and science examples above is that a change is made in a crucial part of the problem without students normally having had any experience or any reason to believe the change is crucial.  In many cases the student does not even notice there was a change in the form of the problem at all.  If you do only problems where the times traveled at different rates are always the same, you tend not to notice that the test problem is different because it is talking about traveling the same distances at different rates.  Even if you do notice it is different, you will not likely think the difference is significant in any way.  However, it is.   The average rate traveling for one hour at 60 miles per hour and one hour at 30 mile per hour is 45 miles per hour.  But the average rate traveling 100 miles at 60 mph and another 100 miles at 30 mph is not 45 mph.  If you drive 100 miles at 60 mph, it will take you 1 hour and 40 minutes.  It takes 3 hours and 20 minutes to drive 100 miles at 30 mph.  That means it would take you 5 hours to drive 200 miles under those conditions.  The average rate then will be 200 divided by 5, which is 40 mph. 

The results of the car race question are as counter-intuitive as the egg case and the circumference of the earth case.  Since 60 mph is one mile a minute, to qualify at an average of 60 mph for a two mile distance, you must complete the entire two miles in two minutes or less.  But if you only drive 30 mph for one of those miles, you use up the full two minutes just driving that one mile.  You still have another mile to go, but you have no time left to do it.  That means no matter how fast you drive the other mile, short of  going infinitely fast, you cannot average 60 for both miles together. 

Under homework or test conditions, most students have no reason to see the difference in the assigned problem as being crucial to the way it has to be worked out.  Often they do not even notice there is a difference in the way the problem is stated.  Under test conditions in particular, even if they notice there is a difference in the problem and even if they notice that difference might be crucial, they are not likely to have the presence of mind or the time to figure out what they need to do in order to take that difference into account. 

In the sound wave case, the problem is similar, and perhaps even harder.  There is a difference in the rates at which sound waves reach your ear if you are still, relative to the air with the sound moving toward you from the way they reach your ear if the source of the sound is still, relative to the air, and you are moving toward it. The demonstration and rationale for that difference, however, is complicated and not particularly easy to see without a great deal of thought and scrutiny. 

More importantly, science is not simply an a priori enterprise, and that means it is impossible to deduce knowledge reliably in science only from a theory.  Science is empirical.  Deduction in science has to do with trying to figure out an explanation for known phenomena and with then trying to figure out testable consequences that would confirm or deny the theory by predicting how other phenomena will behave.  The history of science is littered with failed predictions based on deductions, thus causing a revision or elaboration of the theory or a hunt for some intervening cause which was unsuspected until the prediction did not pan out as expected. 

Therefore to give students a theory or explanation and some examples of the phenomena to which it is known to apply, and then to ask them to predict how other phenomena will turn out when they apply the theory as they understand it to that phenomena -- and then to grade them down when their predictions are mistaken -- is not to be teaching them science at all.  It is grading them on the basis of their not being omniscient.  It is grading them on not being able to predict the right answer from insufficient teaching and insufficient evidence or experience. 

Surely no one is lacking scientific ability because s/he might miss the egg question.  And no student or adult is necessarily devoid of mathematical skill and knowledge just because s/he does not think to use math, particularly the circumference/radius ratio, to figure out the earth/ribbon problem.  Yet teachers will grade students down in science or in math because a student does not realize during a test or while trying to finish chapter questions from a textbook that something applies in a way s/he had no reason to suspect it would. 

There is even at least one, by now, fairly well-known trick math question that is very difficult for mathematicians to figure out but easy for those who do not know nearly as much math.  It is a favorite problem to trick unsuspecting math professors with. 

Two trains start out simultaneously, 750 miles apart on the same track, heading toward each other. The train in the west is traveling 70 mph and the train in the east is traveling 55 mph. At the time the trains begin, a bee that flies 300 mph starts at one train and flies until it reaches the other, at which time it reverses (without losing any speed) and immediately flies back to the first train, which, of course, is now closer. The bee keeps going back and forth between the two ever-closer trains until it is squashed between them when they crash into each other. What is the total distance the bee flies? 

The computationally extremely difficult, but psychologically logically apparent, solution is to "sum an infinite series". Mathematicians tend to lock into that method. The easy solution, however, is that the trains are approaching each other at a combined rate of 125 mph, so they will cover the 750 miles, and crash, in 6 hours. The bee is constantly flying 300 mph; so in that 6 hours he will fly 1800 miles. (One mathematician is supposed to have given the answer immediately, astonishing a questioner who responded how incredible that was "since most mathematicians try to sum an infinite series." The mathematician responded with astonishment of his own, "but that is what I did.") 

It is not that mathematicians do not know how to solve this problem the easy way; it is that it is constructed in a way to make them not think about the easy way.  Although teachers who put a "wrinkle" or "twist" into a test question are not constructing problems in order to trick students in the way the bee problem is intentionally constructed to trick mathematicians, it has the same result.  And it is less fairly a reflection on the student who gets a grade than it is on how well the student was taught by the teacher who unfairly does not get a grade. 

Some students will get these trick problems correct on occasion.  When they do, the interesting question is not why the other students did not get it, but why the ones who did knew how to do that.  Often it will be because they worked a problem like it some time, though that problem did not occur as an assignment or a class problem.  In a relaxed moment of reflection, a student might have noticed that all the homework rate-time problems with variable rates involved equal time intervals and became curious about working out an equal rate interval (or unequal time or rate intervals) with the time to think it through.  Perhaps a student interested in auto racing will have seen a circumstance similar to the question, or perhaps a student will have taken auto trips with his/her family and noticed that when they stopped or slowed down, that it seemed to throw off their average speed or time in a way disproportionate to just the change in rate, and they may have looked into it mathematically to try to figure out why.  Perhaps someone once even drove at a fast clip toward a tornado siren and noticed the pitch did not change as much as s/he might have anticipated, and began to wonder why that might be, and trying to work it out on paper. 

Perhaps a teacher, thinking he is giving a clear hint, even points out in class one day that it should be noticed the time intervals were all equal in these variable rate problems, not the distance intervals.  While most students might have merely thought "Ok" and simply taken that as a fact to be noticed and then ignored, one or two students might have taken it as a sign that something else perhaps happened when the distance intervals were the same for the various rates, and they might have sat down to figure out what that would be, why, and how to calculate it. 

Perhaps a student has a parent, aunt or uncle, or an older sibling who knows this sort of thing and has explained it to him/her.  Perhaps a student found an old exam (and answers) somewhere and studied it.  There are any of a number of ways a student might know how to do any of these problems without meaning the student is good at math or science or better at math or science than the students who did not get it right. 

It is therefore wrong to give grades based on such questions.  What is far better is to pique student interest and teach them more by going over such questions in class so that they see the subject is deeper, more complex, and perhaps more interesting and of more consequence than they previously had recognized.  In other words the "trick" questions, or questions with a "new wrinkle" can be used to teach, but not to test with, until after the material has been thoroughly taught and digested by those students who are seriously trying to learn. 

This work is available here free, so that those who cannot afford it can still have access to it, and so that no one has to pay before they read something that might not be what they really are seeking.  But if you find it meaningful and helpful and would like to contribute whatever easily affordable amount you feel it is worth, please do do.  I will appreciate it. The button to the right will take you to PayPal where you can make any size donation (of 25 cents or more) you wish, using either your PayPal account or a credit card without a PayPal account.


This problem is essentially the question of how much a particular circle's radius changes when you add 36 inches to its circumference, because the difference between (1) the distance from the center of the earth to the ribbon and (2) the center of the earth to the ground, will be the height above the ground that the ribbon would be if it were supported equally high all around the globe. Hence it is the difference between the length of the radius to the ribbon and the length of the radius to the ground. 

Since any radius is equal to the circumference divided by two times pi, the radius from the earth's center to the ribbon will be: 
(24,000 miles + 36 inches)/2pi 

This is the same as: 
(24,000miles/2pi) + 36 inches/2pi 

But 24,000miles/2pi is the radius of the earth to begin with, so it is the distance to the ground from the center of the earth. 

That leaves the ribbon to be a height of 36inches/2pi above the ground. 

Since pi is a little more than 3, then 2pi is a little more than 6, which makes 36 inches divided by 2pi a little less than 6 inches. 

That means the ribbon will be almost six inches off the ground all the way around. 

That seems almost impossible to believe, but it is true. In fact, one can generalize the math to show that whenever you change the circumference of any circle, no matter how large or how small, by any amount, you will change the radius of that circle by a little less than 1/6 that amount. Adding (or subtracting) 36 inches to (or from) the circumference of a dime or the circumference of the universe changes the radius of either by just under 6 inches. (Return to text.

I have used some problems that are relatively difficult for adults because they exemplify the point better. When adults look at problems of this sort that are difficult for young students but not for adults, they often think there is no problem and that the students ought to be able to "see" what looks easy and obvious to them, as adults after years of practice.  For example, a teacher might teach students how to subtract two digit numbers where borrowing is necessary, but on a test stick in a three-digit number or, after teaching subtraction using a zero, stick in a particularly difficult subtraction for beginners, such as 1000 - 909.  It does not take much to make a test very difficult for students of any age, if you give them problems that require even the slightest modification of what you were teaching them -- if you have not taught them about the modifications or taught them to think about modifications that might occur. (Return to text.

This one is difficult to explain, in part because although I can do it mathematically, I can't quite "see" why it works the way it does.  If we were going to do it totally correctly it would have to involve frequencies of waves and their relationship to pitch, but to simplify it a little, I will talk about frequencies or intervals of a short beeping sound. The principles I am interested in will be the same for both cases. 

Consider a source that emits brief beeping sounds one second apart.  In a nutshell, there is a difference in (1) the time interval between hearing those two sounds if the source of the sound is approaching you from a distance and (2) the time interval between hearing those two sounds if you were approaching their source from a distance.  That is because, I think it will boil down to, when the source of sound is approaching you it [meaning each successive sound] has less ground to cover at the speed of sound, but when you are approaching it, it [each successive sound] has less ground to cover at your speed. It can normally cover more ground at the speed of sound then it can cover traveling a car or train's speed. [As you will see later in the letter from Birger Bjerkeng from Norway, I did not state this part in green italics correctly. Or at least not precisely or clearly.  He states it more precisely but I cannot quite "see" it intuitively his way.  I think I "see" it the way I have said it, but my statement of it may not be helpful to anyone else even if I am seeing it right. The following mathematical explanation is correct, but seeing or stating its meaning intuitively is not easy -- at least not for me. Each reader who is so inclined will likely have to work this out for him/herself in order to see intuitively or verbally what is occurring.  In fact, after receiving Bjerkeng's letter, I thought about the problem a great deal more and came up with a way of looking at this that I explain later, using automobiles being driven off a moving train toward a stationary you, versus their being driven off a stationary train and your moving toward them. The essential point for this essay, though, is, of course, that this is not something students should reasonably be able to see, for the first time, in an exam situation.  It is not even easy for most of us when we have plenty of time and when we are told it is a different phenomena from the usual Doppler effect.]

[A second reader, Ethan Skyler, has written also about the sentences in green italics, offering his explanation.  I have put my exchange with him as a separate, supplemental web page that may be helpful after you have read this page.]

As an example, take a fast moving source, say one moving half the speed of sound at approximately 185 yards per second.  We will remain still relative to the air as the source comes toward us.  Remember, the source of the sound we hear beeps every second.  At some point in time we hear the first beep.  The second beep will have been sounded at the source one second after the first beep, but it will therefore go off when the source is 185 yards closer to us, which means it takes 1/2 second less time to reach us than the first beep did, and so we will hear the second beep 1/2 second after the first beep. 

But now consider that the source is still, relative to the air, and that we are moving toward it at 185 yards per second.  At some point we hear the first beep.  Will we still hear the second beep a half second later?  No, because half a second later we will only be 92.5 yards closer to the source then when we heard the first beep, instead of 185 yards closer to it.  And that does not get us close enough to the source to hear the sound then; it will not yet have got to where we are. 

It turns out that we will hear it 2/3 of a second after we hear the first beep.  At that time we will be approximately 123 yards closer to the source of the sounds than we were to it when it emitted its first beep.  Sound travels 123 yards in one third of a second, so it travels 1/3 second less than the other sound to reach us.  The sound will take 1/3 less time to reach us than the first sound did, and so it will reach us there. 

As you can see, this is fairly complicated and confusing, and you have to be very careful how you say it and how you think about it.  It is not something a student could be expected to see or do off the top of his head, particularly during an exam. And not being able to see it or do it intuitively or immediately should not mean one has no understanding of science or no ability to do it. 

Since I received the e-mail from Dr. Bjerkeng, I have examined this problem more carefully and have some additional observations and comments.  I am trying to see what it is about the two different situations that makes them or causes them to work out differently mathematically.  I know they work out differently, and I know how to compute the difference, but I cannot quite see why it works out that way.  There are numerous phenomena that you can compute differently and get the same answers.  But this one gives different answers, and I cannot quite see what it is about the physical aspects of the two different situations that makes for such different mathematical (and actual) differences. 

The difference in elapsed times between hearing the first and second sound in the two cases is because the second case involves combining the speed of sound and the speed of the receiver after the second sound has been made, whereas the first case involves just the speed of sound after the second sound has been made, but from a distance that is closer because of the speed of the sound source. What I am not quite able to see, but hope to someday is why that difference in the two cases makes the answers come out different in the way they do. 

I think part of it is this: in the case where the source is moving, the velocity of each successive beep comes toward you the whole time at the speed of sound, because you are not moving, though each beep starts at a point closer to you than the previous beep started.  That is what makes it arrive closer to the previous beep in time than it was sent out.  However, in the case where the observer is moving and the sound comes from the same spot each time, the closing speed of the sound to the observer is the combination of the speed the sound is traveling and the speed the observer is traveling.  The observer is moving toward the sound after it has been emanated and while it is traveling toward the observer.  In the case where the source is moving and the observer is stationary, the closing speed of the sound to the observer is always simply the speed of sound.  I presume it is this difference in the phenomena that causes the difference in the result, but I cannot see exactly how it operates to make the relationships between the two cases be exactly what they are. 

Plus, I presume that this phenomena is exactly the same if we were talking about throwing and catching successive balls -- that the intervals between the catching of balls thrown some specific constant interval apart would be different depending on whether 1) the thrower moves closer (to a stationary receiver) at a constant velocity between throws, or whether 2) the thrower stays stationary and throws the balls to a receiver who is moving toward him at that (same) constant velocity.  The time difference is not likely detectable at normal throwing speeds, but still it will exist.  [A day after having written these last two paragraphs, and working further with the idea in them, I have finally solved this problem to my own satisfaction, and I think I can explain it pretty well now. That explanation is at the end of this essay.]

Understanding this kind of a problem in the way I am asking it is important to me because I do not seem to be able to do physics or much math without understanding the concepts behind the calculations.  Many mathematicians and physicists, seem quite comfortable just looking at the math, and seem to be able to tell what is the "right math" to use when looking at complex phenomena.  I, on the other hand, tend to try to view things conceptually in order then to figure out how to do the math, but when I do physics, I seem often to focus on the wrong concepts or I see the concepts incorrectly, and then misapply the math.  When I have it right conceptually, I am fine, but in those cases where I get it wrong conceptually, it is not good.  But I don't have a way of telling the difference just on my own, unless I do experiments and find out that my conceptual/mathematical scheme does not comply with the empirical results. 

Physicists and mathematicians however seem not to have this difficulty with known phenomena, just with new phenomena.  I am always having to re-invent each wheel however, if all I have is the math to go by instead of the conceptual understanding of it.  I do not think I am alone in this regard, and that is in part why I think that it is unfair to ask students for the first time on a test to be able to recognize variations that are likely to be different from what they have been taught and have been working with.  It is good to do that in class or in a test that does not count against them, but it is not fair to expect someone to see, either mathematically or conceptually, that just a slight change in the form of the problems they have been working can make a large difference in the results or in the proper way to solve the problem. 

(Return to text.

The following is from Birger Bjerkeng: 
"[O]n one point I think your discussion should be revised. It concerns the example of the Doppler effect. You are right that the case of stationary source and moving receiver is different from the case of moving source and stationary receiver. Your discussion of the numerical examples where the speed of the source or the receiver is half the speed of sound is also quite right. Under more normal circumstances, however, the difference is much smaller, and hardly noticable. 

As a physicist, I found your qualitative argument a bit counterintuitive, and your discussion prompted me to work through the two cases. That lead to the following two formulas, which I am sure you have also derived already [he is being kindly over-generous to me in that assumption --RG] : 

        For sound source moving at speed v towards a stationary receiver: 
                        f' = f/(1-v/u) 

        For receiver moving at speed v towards a stationary source: 
                        f' = f*(1+v/u) 

In both cases: 
        f = the true frequency of the sound source 
        u = the speed of sound relative to air 
        f' = the received frequency. 

In normal "everyday" circumstances, v will be much less than u, and in that case, the relative change will be about the same. For instance, driving at a normal car speed of about 18.5 yards per second (about 60 km/h) would mean that f' = f*1.053 for moving source, and f'=f*1.05 for moving receiver.  Thus, your statement that the moving receiver will hear "a slight increase in pitch which, if noticable at all, will be much different from the relative increase in pitch when the sound source is moving..." is not correct for normal circumstances, like observations from a moving car. The frequency is the ratio between wave velocity and wavelength (distance between pulses or peaks). When the source of the sound is approaching you, the wave velocity through the air and relative to you is the same, and unaffected by the movement of the source, but the wavelength is decreased in the direction towards you. The velocity/wavelength ratio increases. If you move through the air towards a sound source, the wavelength is unchanged, but the waves approach you at higher velocity (relative to you), and the velocity/wavelength ratio again increases. In both cases there is a frequency increase, but the effect of a given velocity is different. For small changes the difference is negligible, but for high velocities the outcome is quite different. Mathematically it is the case of comparing adding and subtracting a certain percentage. If you add or subtract 5 % to some number, the ratio between the largest and smallest numbers are almost the same, close to 1,05. However, if you add or subtract 50 %, the ratio of the largest and smallest number is 1.5 or 2.0, respectively.

Only when the speed v is comparable to u, as in your example, will the difference be really large.

The derivation of the formulas can be done as follows:
The sound consists of moving pressure waves, with peaks emitted at a frequency f, i.e. with a time interval t=1/f between emitted peaks (or sound pulses).

During this interval, a sound source with velocity v will have moved a distance v*t between emitted pulses, while the first pulse will have moved a distance u*t from the point where it was emitted. The distance between the two pulses in front of the source is L1 = (u-v)*t=(u-v)/f. This is the wavelength in this direction. The frequency is the ratio between velocity and wavelength, since the wave propagates with velocity u through air, the actual frequency of the sound in the direction in front of the moving sound source will be f’ = f*u /(u-v).

For a stationary sound source emitting with frequency f, the sound wave will have wavelength L2=u/f. An observer moving towards the source will meet pressure peaks travelling in the other direction with velocity u, so the velocity of the observer relative to the sound waves is u+v. The observer will thus pass peaks with time interval t2=L2/(u+v)=(u/f)/(u+v). The frequency is the inverse of the time interval between peaks, so in this case f’=f*(u+v)/v.

As the speed v becomes equal to u in size, the difference becomes qualitative, and not just quantitative: 
For a moving source and a stationary observer: if v=u, f' goes to infinity (Wavelength becomes zero, "supersonic flight"), and if v=-u, f' is reduced by 1/3. 

For stationary sound source and a moving observer: if v=u, f' increases by a factor 2, and if v=-u, f' becomes 0, so the observer can not hear any sound, because he/she moves with the sound waves (as surfers are not hit by the waves, but follows them). 

The formulas above can be rewritten: 
        For sound source moving: 
                        u/f' = (u-v)/f = L1 
        For receiver moving: 
                        (u+v)/f' = u/f = L2 

L1 and L2 are true wave lengths of the propagating sound through the stationary air. 

The difference between the two cases arises from the fact that the movement affects different aspects of the sound wave perception of the observer.  For the moving sound source, the sound wave is objectively changed, independent of the observer, in that the actual wavelength is contracted due to movement of the source. Since the propagation velocity through the air is unchanged, the frequency is actually changed in the direction of movement. This actual change, an increase in front of the moving source, and a decrease behind it, is heard by stationary observers. They hear the “true sound” propagating through air.

With stationary sound source and moving observer, the sound wave is of cause not actually changed, and the observed frequency change is just an impression, created by the observer’s movement. The effective wave velocity relative to the observer is increased when the observer moves towards the source, and reduced when the observer moves away from it.

The discussion above assumes that either the source or the observer is stationary relative to the air. In a real-life situation, for instance if you drive along a road, and meet a police car or an ambulance with sirens, both the source and the observer are moving, and then both effects come into play.  If there is a wind in the direction between source and observer, this will affect the speed of both the sound source and the observer equally relative to air.

For the general case (with constant velocities), where the sound source moves with velocity v1, and the observer with opposing velocity v2, both relative to air, the actual wavelength will be (u-v1)/f, while the effective velocity of the observer compared to the sound waves is (u+v2). The observed frequency will then be:


If the source and observer is at a fixed distance from each other, with air moving at speed v from source to observer, then v1=v-, and v2=v; the observed frequency is unchanged. This is intuitively obvious; with fixed distance, and constant velocities, each pulse will take the same time from source to observer.

Best regards
Birger Bjerkeng (return to text

Eureka!! Finally after knowing about this problem for about two years, I finally see, to my own understanding why and how it is that the asymmetry in the two cases (one where the source of sound is moving, and the other where the receiver of the sound is moving) causes the results to be different.  I see what makes the difference happen.  I will explain first with a particular example, and then show the more general explanation based on it. 

Suppose we have automobiles driving in the following manners in two different situations:  (I will use the unit miles for distance and miles per hour for time because I am an American, but for those of you who use kilometers, just use kilometers with the same numbers -- you don't need to convert miles into kilometers because the length of the unit used below does not matter.  So where I talk about 80 m/hour, just think in terms of 80 km/hour. The concept will work out the same.): 

In both situations two or more of the cars will be driving or pointing (when parked) west and one of the cars will be driving or pointing east. All the cars will be in their proper lanes, so the ones that meet will simply pass each other, not collide.  Here are the two situations: 

1) First situation -- the case of the "moving source and stationary receiver": the car in the west that is pointing east does not move at all, but is parked, waiting for the cars coming from the east to get to it. Here is the way the eastern automobiles begin their journey.  They are on a train that is going west.  The train is traveling 40 m/hr.  Car thieves surreptitiously board the train and start to steal the automobiles.  The automobiles are chained up in such a way that they have to be unchained one at a time, and it takes one hour to unchain each one after the previous one has been unchained.  As soon as they unchain each automobile, a thief immediately drives it off the train (using a ramp they have designed for the purpose) and heads west at 80 m/hr on a road next to the railroad track.  Since the train is only doing 40mph and the car is going 80, the car soon leaves the train far behind it. 

So let's look at what this means for any two successive cars.  They will be 40 miles apart, because in the hour it took to unchain the second one after the first one took off, the first car will have driven 80 miles, but the train will have traveled 40 miles.  So the second car will be 40 miles behind the first, driving at the same speed as the first, 80 m/hr.  So it will stay 40 miles behind the first the whole rest of the trip.  That means that 1/2 hour after the first car passes the stationary car that remained in the west, the second car will also pass it, because it will take it 1/2 hour at 80 m/hr to drive the 40 miles that was between it and the first car when the first car passed the stationary car.  The same will be true for each following car the thieves have got off the train; they will each be 40 miles behind the one in front of it, each driving 80 miles/hour.  The cars will have started one hour apart from each other -- the time it took to unchain them -- but they will have started on the roadway only 40 miles behind each other, and will thus only be a half hour (30 minutes) driving distance at their speed behind each other. 

2) Second situation -- case of the "moving receiver and stationary source": the car in the west will be on a train headed east at 40 m/hr.  The train on which the eastern cars start out will not be moving, but will be at a siding.  The car thieves still need one hour to unchain each automobile after they have freed the one in front of it.  As before, each time they get one loose, they drive it off the train and start heading west at 80 m/hr. 

Now, look at what is happening in this case. Each stolen car will be 80 miles behind the one in front of it because they were stolen one hour apart, and the one in front will have driven 80 miles in that hour.  Since all the cars which are headed west are moving at the same speed, they will stay 80 miles apart. 
However, when the first car passes the train coming east from the west, the train traveling east will be going 40 m/hr toward the next car that is 80 miles back.  Since the train is going 40 m/hr east and the next car is driving 80 m/hr west, their combined speeds at which they are closing toward each other is 120 m/hr.  Since they have 80 miles to go to meet, it will take them 80/120 of an hour or 2/3 of an hour, which is 40 minutes. 

In the first case, the distance between the two cars driving west is half the distance it is between the two cars driving west in the second case.  But they are closing on the stationary, western, car at 80 m/hr.  In the second case, the two cars driving east are 80 miles apart, but they are closing on the now moving, western, car (sitting on a train) at a combined closing speed of 120 m/hr. 

If we designate, just for the sake of simplicity here, 80 m/hr to be the "maximum (legal) speed of the cars", then in the first case --the case of the moving source-- the (stationary) eastward-pointed car and the second westward-heading car will be closing toward each other at the maximum speed and will only be half the distance apart that the maximum speed allows them to travel in an hour.  So they will be half an hour apart, passing the stationary car.  But in the second case, the eastern-heading car and the second westward-heading car will be the full distance apart that the maximum driving speed allows them to travel in one hour, but they will be traveling toward each other at a combined rate of speed that is 1 and 1/2 times the maximum legal speed rate.  Hence the two times to compare are: 
Situation one -- moving source: half the distance at max speed = 1/2 distance/(full speed) = 1/2 time unit 
Situation two -- moving receiver: full distance at 1.5 max speed = full distance/(1.5 speed) = 2/3 time unit 

To generalize the situations: 
Distance between the two successive cars:
Situation one -- moving source: the distance between the cars will be the time, t, between their release, multiplied by the difference in the speed of the car and the speed of the train, so it will be t(Vcar - Vtrain) 
Situation two -- moving receiver: the distance between the two successive cars will be t multiplied by the speed of the cars, so it will be t(Vcar) 

Closing speed of the eastward-heading car and the second westward-heading car will be:
Situation one -- moving source: the speed of the westward-heading car (since it is the only one moving), so it is Vcar
Situation two -- moving receiver: combined speeds of the two cars, so it is (Vcar + Vtrain)

Hence, the closing times of the two cases will be, the closing distances divided by the closing speeds, which will be:
Situation one -- moving source: t(Vcar - Vtrain)/Vcar 
Situation two -- moving receiver: t(Vcar)/(Vcar + Vtrain)

In situation one -- moving source -- as long as the train is going slower than the cars, the time the closing time will be less than the time between the release of the cars.  When the train is moving at the same rate as the cars, the closing time will be zero, because the cars will be traveling side by side.  When the train is not moving at all, the closing time will be the same as the release interval time since all the cars will be released from the same place and travel the same distance at the same time.  When the train is moving faster than the cars, the closing time will be "negative" which only means in this case that the cars will be passed in the opposite order they were let off the train.  That is because each successive car will actually be let out in front of the previously car because the train moves ahead of each car when it is released, since it goes faster.  This is like a conductor's throwing people off a train at each successive stop when he finds out they don't have tickets to a final destination.  The guy thrown off last will be closest to the final destination; the guy thrown off first will be furthest. 

In situation two -- moving receiver -- when the train is not moving at all, the closing time will be the same as the interval of the release time, since all the cars will be traveling the same distance at the same speed. Whether the train is moving slower than the cars or faster, the closing times will be positive and the cars will be passed in the order they were let off, and the time will be less in all those cases than the time interval in which they were let off. 

I believe, that as long as the train is moving slower than the cars, the closing interval of the moving source situation will be smaller or quicker than the closing interval of the moving receiver situation.  In the case where the train is not moving, the closing intervals will be the same (which is the same as the release interval time of the cars in both situations). But suppose we have a number of different cases where the train is going faster than the cars.  As you go from case to case, each time increasing the train's constant speed relative to the car's constant speed, I believe that: 

in the moving source case, the time interval will be larger between successive cars (as the train goes faster and faster relative to the cars), but the cars will be passed in reverse order from what they were let off the train. 

in the moving receiver case, though, the time interval will shrink as the train goes faster and faster, but the cars will still be passed in the order they were released from the train. 

Some comments on the car-train case and the sound case: 
I have chosen speeds and time intervals in the car-train case that give a noticeable result.  If we used a release interval of the cars of only a second or two, the difference in intervals between passing the successive cars in  the moving source case on the one hand and the moving receiver case on the other would still be proportionally the same, but since they would be fractions of a second, they would not be particularly noticeable unless one were measuring time very closely. 

I have presumed in the car-train case that we have the cars and the trains moving at speeds relative to the ground. In the case of sound and pitch, the cases were done with speeds of the source and the receiver relative to the air. There might be situations where there is a moving medium of some sort involved as well, as in river boat cases or in moving air cases involving sound.  But I have disregarded that added variable here. 

When speaking about audible pitch in the sound case, it involves frequencies, which are, or are related to, the time intervals between wavelength crests or pulses or some such similar "places" on waves, none of which I really comprehend very well.  But it takes a proportional increase in frequency to attain the next higher note on the scale, and that may involve a substantial frequency change as one goes higher and higher on the musical scale.  And moreover some ears are more discerning than others as to what pitch or fequency they are hearing.  So I have avoided bringing hearing or pitch into the discussion on my part of it, since I am not familar with most of that material, and cannot remember the little I once knew and thought I understood about it.  I have only concentrated on interval differences involving the receiving of different objects or impulses launched at a constant time interval apart under the two conditions of moving source of their launch, and moving platform receiving them.  Frequencies and audible pitches are based on that, but to me at this point are a secondary part of the issue. (Return to text.