Waves and Modern Physics 203-SN3-RE • CEGEP, Montreal
Revision sheet: the Doppler effect, the choice of signs and the radial component (203-SN3-RE)
Almost all the marks lost in this chapter are lost on SIGNS, and they all come from the same missing step: the student recites a formula instead of first deciding on the direction. This sheet installs the opposite habit. It sorts the four simple cases into a grid where the expected direction comes before the sign, then moves on to the four situations where the basic formula is no longer enough: the off-axis source, the moving medium, the echo off a moving target and light.
Each trap comes with its price tag, because they do not all cost the same: believing that the speed of sound changes when the source moves costs two marks and all the reasoning that follows, forgetting the factor 2 of an echo costs the whole question, and announcing the maximum frequency at the moment the train passes in front of you costs the favourite question of the exam. It is written for CEGEP students in the Natural Science program (Sciences de la nature).
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The thread of the chapter
What changes is NEVER the speed of the wave in the medium: it is the wavelength produced ahead of a moving source, and the rate of encounter of a running observer. Two different mechanisms, hence two different formulas, and a sign that is chosen by deciding FIRST whether the frequency must go up or down.
•The speed of the wave depends ONLY on the medium and its temperature: 343 m/s in air at 20 °C, 1540 m/s in tissue, 1500 m/s in water. Once released, a wavefront has no memory of its source.
•Moving source: between two emissions, the source advances by vsT, so the wavelength is λ=f0v−vs ahead and f0v+vs behind. It is the WAVELENGTH that is made different.
•Moving observer: the wavefronts are intact, spaced by λ0=f0v everywhere. It is the SPEED OF ENCOUNTER that goes from v to v±vo.
•Direct consequence: the two cases are NOT symmetric. At 30 m/s in air, a moving source gives 547.9 Hz and a moving observer 543.7 Hz for the same f0=500 Hz.
The two wavefronts move at the same speed in both directions; only their SPACING changes, three times tighter ahead of the source than behind it.
Air is not a neutral backdrop, it is the carrying medium and therefore a physically privileged reference frame. All the asymmetry of the chapter comes from there, and it disappears for light, which has no medium.
The general formula: one axis, two algebraic velocities
•Orient an axis from the SOURCE to the OBSERVER. Let vs and vo be the algebraic velocities along this axis, positive in the direction of the axis.
•Then f′=f0v−vsv−vo, and there is not a single sign left to guess: the signs are those of the velocities.
•Built-in check: if vo=vs, the fraction is 1 and f′=f0. Two movers that keep the same distance hear no shift, even at 90 km/h in air.
•The four simple cases in the table below are this same formula with one of the two velocities zero. Nothing more to memorize.
The axis runs from the source to the observer: vs=+25.0 m/s and vo=−15.0 m/s, hence f′=440×340−25340+15=495.9 Hz, higher in pitch as expected.
Writing the sentence of expected direction BEFORE the formula often earns half a mark of rigour, and above all it makes a sign error all but impossible to commit.
The four extensions of the basic formula
•Off axis: only the RADIAL component counts, f′=f0v−vscosθv, where θ is the angle between the velocity and the line of sight.
•Moving medium: a wind w counted positive from the source toward the observer simply replaces v by v+w in the general formula.
•Reflection off a moving target: the target is first an observer, then a source. The shift counts DOUBLE, and for small speeds Δf≈c2f0vcosθ.
•Light: no medium, so a single formula and no asymmetry, f′=f01+β1−β for recession with β=cv. Purely transverse, f′=γf0, whereas acoustics would give exactly f0.
Four moves: project the velocity onto the line of sight, replace v by v+w with the wind, count two shifts on an echo, and γf0 for light seen from the side.
The transverse shift is the only effect in the chapter with no classical equivalent: it comes not from the geometry of the wavefronts but from time dilation.
The rules in table form
Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.
The four simple cases: the direction decides the sign
Read from left to right, never from right to left: read the middle column, say it out loud, and only then copy the formula. A red cell flags a row that does not exist and must be replaced by a real question.
Who moves, relative to the air
The frequency must
The formula, sign included
Source toward you
rise
f′=f0v−vsv
Example: 500×343−30343=547.9 Hz
Source moving away
fall
f′=f0v+vsv
Example: 500×373343=459.8 Hz
You toward the source
rise
f′=f0vv+vo
Example: 500×343373=543.7 Hz, and not 547.9 Hz
You moving away
fall
f′=f0vv−vo
Example: 500×343313=456.3 Hz
RELATIVE speed of 30 m/s
rise
decides nothingrule that does not exist
Example: same 30 m/s, two answers: 547.9 Hz if it is the source that moves, 543.7 Hz if it is you.
What to do: Ask WHO moves relative to the air, then take one of the four rows above.
In acoustics, the relative speed is never enough. It is enough for light, and only for light, for lack of a carrying medium.
When the basic formula is no longer enough
Three words in the problem each trigger one modification, and only one. Spot them before calculating.
What the problem says
The move
The modified formula
a wind w
replace v by v+w
f′=f0(v+w)−vs(v+w)−vo
Example: wind of 20 m/s, source at 30 m/s: 440×330360=480.0 Hz, against 482.6 Hz with no wind.
a distance to the track
project, vscosθ
f′=f0v−vscosθv
Example: at 45°, 70.0×0.707=49.5 m/s: 450×290.5340=526.7 Hz.
an echo, a return
apply TWICE
Δf≈c2f0vcosθ
Example: 5.00 MHz, blood at 0.450 m/s at 60.0° in 1540 m/s: Δf=1461 Hz.
light
a single formula
f′=f01+β1−β
Example: β=0.600 in recession: 1.00 GHz becomes 0.500 GHz.
The modifications add up without interfering: an off-axis echo in a moving medium carries all three at once.
The mistakes that cost marks
These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.
1.Writing the signs from memory, without having decided on the direction
half the marks of the question, and it happens again on every question of the chapter
What not to write
“The two trains are closing in, f′=440×340+25340−15=391.8 Hz.”
What to write
“They are closing in, so f′ must be GREATER than 440 Hz. Axis from the source to the observer: vs=+25.0 m/s, vo=−15.0 m/s, so f′=440×340−25340+15=495.9 Hz.”
391.8 and 495.9 Hz look alike: only their position relative to 440 Hz tells them apart. Two trains closing in: higher, so 495.9 Hz (green).
Why: An inverted fraction gives a result that has the right size and the right look: 391.8 and 495.9 bracket 440, and only the position relative to 440 tells them apart. The direction check is the only thing that separates them, and it takes two seconds.
2.Believing that the speed of sound changes when the source moves
2 marks, and all the reasoning that follows
What not to write
“The locomotive moves at 30 m/s, so ahead of it sound travels at 343+30=373 m/s: that is why the whistle is higher in pitch.”
What to write
“The speed of sound depends only on the air and its temperature, 343 m/s ahead and behind. It is the wavelength that is 500343−30=0.626 m instead of 0.686 m, hence f=0.626343=547.9 Hz.”
To scale, 500 Hz and 30 m/s: wavefronts spaced 0.626 m apart ahead, 0.746 m behind, and all moving at 343 m/s (green); none leaves at 373 m/s (red).
Why: This is the modelling error of the chapter: it sometimes gives the right order of magnitude, and it makes everything after it impossible, since the wind, the radial component and the echo all rest on the fact that v is a property of the MEDIUM.
3.Treating a moving source and a moving observer as symmetric
3 marks, and the one idea of the chapter
What not to write
“Only the relative speed counts, 30 m/s in both cases, so 547.9 Hz whether it is the source or me that moves.”
What to write
“In acoustics the two formulas are different: moving source 500×313343=547.9 Hz, moving observer 500×343373=543.7 Hz. The 4.2 Hz difference comes from the fact that air is the carrying medium.”
Same speed, two answers: 547.9 Hz if the source moves, 543.7 Hz if it is the observer; the gap grows with speed, the curve for the source heading to infinity toward 343 m/s.
Why: The mechanism is not the same: on one side the wavelength is made shorter, on the other it is intact and it is the speed of encounter that rises. Two mechanisms, two formulas, and a perfectly audible difference.
4.Taking the full speed when the source is not coming straight at you
the whole question, and it is the favourite question of the exam
What not to write
“At the moment the train passes in front of me, I hear the maximum, 450×340−70340=566.7 Hz.”
What to write
“At the moment of passing, the velocity is perpendicular to the line of sight: cosθ=0, vr=0 and I hear EXACTLY 450 Hz. The maximum 566.7 Hz was reached well before, when the train was still far away and heading almost straight at me.”
The frequency is already 538.7 Hz at 37°, then exactly 450 Hz at the passing, where the right angle cancels the radial component, then 386.4 Hz.
Why: The distance goes through a minimum at the moment of passing, so its derivative vanishes: at that instant the train is no longer approaching and not yet receding. The extremum of distance and the extremum of frequency never coincide.
5.Forgetting that the shift counts DOUBLE on an echo
a factor 2, so the whole question
What not to write
“The bat is closing in on the moth at 5.00 m/s, so the echo returns at 42000×335340=42627 Hz, a shift of 627 Hz.”
What to write
“The target undergoes the Doppler effect twice, once on receiving as a moving observer, once on re-emitting as a moving source: f1=42632 Hz then f2=43254 Hz, a shift of 1254 Hz, exactly double.”
The moth receives f1=42632 Hz as a moving observer, then sends back as a moving source: f2=43254 Hz, a shift of 1254 Hz (green), double a single shift (red).
Why: The error is invisible on rereading, the result having the right size and the right sign. The trigger word is in the problem: echo, return, radar, sonar, ultrasound imaging, bat. As soon as there is reflection off a moving target, the factor 2 is there.
6.Believing that a following wind makes the sound higher in pitch
3 marks on the discussion question
What not to write
“The wind carries the sound of the siren toward me, so the perceived frequency rises.”
What to write
“Nobody is moving: the wind lengthens the wavelength over the ground AND speeds up propagation over the ground in the SAME ratio, so 0.818360=440 Hz, unchanged. And if the source moves, a following wind REDUCES the shift, 480.0 Hz against 482.6 Hz with no wind.”
Nobody moves: the wind changes nothing, 440 Hz (green). Source at 30 m/s: following wind 480.0 Hz, no wind 482.6, headwind 485.5: the following wind REDUCES the shift.
Why: The wind changes the travel time and the intensity received, which gives the impression of a sound that is carried, but not the pitch. When the source moves, the quantity that decides is v+wvs: a following wind leaves the source less time to catch up with its own wavefronts.
7.Taking the arithmetic mean of the two extreme frequencies
2 marks, for a wrong result off by 3.7 Hz that looks like the right one
What not to write
“fmax=1270.4 Hz and fmin=1137.0 Hz, so f0=21270.4+1137.0=1203.7 Hz.”
What to write
“The two extremes are not symmetric about f0: it is the HARMONIC mean that gives it back, fmax+fmin2fmaxfmin=1200.0 Hz.”
The extremes fall at the bottom and top of the circle; at the two points marked with a right angle, the closest to and the farthest from O, exactly 1200 Hz is heard.
Why: The denominator v−u pulls harder upward than v+u pulls downward. The harmonic identity is exact, with no approximation, and it lets you recover f0 AND the speed from just two measurements, without ever having access to the source at rest.
8.Judging a five-degree error on the angle to be negligible
3 marks, and in a clinic a missed diagnosis
What not to write
“Five degrees out of sixty is an eight percent error on the angle, so under ten percent on the speed: negligible.”
What to write
“The relative error on the speed is tanθΔθ: at 60.0° it reaches 15.5 percent, at 45.0° only 9.1 percent. At 90°, cosθ=0, the shift is ZERO and the instrument displays nothing at all.”
Five degrees of error cost 9.1 percent at 45°, 15.5 percent at 60°, and the error blows up toward 90°; the 8 percent from the rule of three (red) underestimates it.
Why: The sensitivity to the angle goes as tanθ, so it blows up as 90° is approached. This is why protocols require a beam between 45° and 60° and never perpendicular: a vessel seen from the side looks blocked although it is flowing normally.
9.Believing that a misaligned radar can overestimate a speed
2 marks, and the direction of the difference is the only thing the question asks
What not to write
“The beam makes 15.0° with the road, so the instrument is wrong, and it may have credited me with more than the real 108 km/h.”
What to write
“The instrument displays vcosθ, and cosθ≤1 ALWAYS: it reads 30.0×0.966=28.98 m/s, that is 104.3 km/h, so 3.7 km/h LESS. A misaligned radar underestimates, never the reverse.”
The instrument reads the projection vcosθ: 28.98 m/s for 30.0 m/s (green). A projection is always shorter than the vector: a misaligned radar underestimates, never the reverse (red).
Why: Same formula as in the flowmeter, but the opposite stakes: in a clinic the underestimate is a defect that must be corrected, on the road it works in the driver's favour and nobody corrects it. The physics does not decide, the use decides.
Which method to choose
Which formula, according to what the problem describes
Do not look at the theme, look at WHO moves and relative to what.
If a single mover, on the line joining the two → one of the four rows of the table, chosen by the expected direction
Example: siren at 30 m/s toward you: 500×313343=547.9 Hz
check that the source moves relative to the AIR, not relative to you
If both move → axis oriented from the source to the observer, then f′=f0v−vsv−vo
Example: two trains head-on, 495.9 Hz
the signs are those of the velocities, there is no ± left to choose
If the problem gives a distance to the track, an angle or a circular motion → keep only the radial component, vscosθ
Example: train 60 m from the track, seen at 45°: 526.7 Hz
at the closest point, cosθ=0 and exactly f0 is heard
If the problem mentions wind, a draught or a drifting medium → replace v by v+w in the general formula, w counted from the source to the observer
Example: 440×330360=480.0 Hz
if nobody moves, the wind changes NO frequency
If the word echo, return, radar, sonar or ultrasound imaging appears → apply the formula twice, or Δf≈c2f0vcosθ
Example: 5.00 MHz on blood at 0.450 m/s at 60.0°: 1461 Hz
the target is first a moving observer, then a moving source
If it is about light, a radio wave or a spectral line → a single formula, f′=f01+β1−β, and only the relative speed counts
Example: β=0.600 in recession: 1.00 GHz gives 0.500 GHz
stop asking who moves relative to what, there is no medium
If no branch applies, it is almost always because the problem describes a transverse motion: in acoustics the answer is then f0 as it stands, and in relativity it is γf0.
How the answer is expected to be written
A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.
A Doppler calculation that does not lose the sign mark
When to use it: As soon as a problem gives an emission frequency and at least one speed.
1State the expected direction in one sentence, BEFORE writing any formula: the distance between the two is decreasing, so the perceived frequency must be higher.
2Orient an axis from the source to the observer and write the two algebraic velocities with their sign along this axis, saying each time why the sign is what it is.
3Write the general formula in letters, f′=f0v−vsv−vo, then substitute with the units.
4Give the result with the right number of significant figures and its unit.
5Compare with the direction stated in step 1, and write it down: it is the only line that proves to the marker that the sign is not a coincidence.
Concluding sentence
“The two trains are closing in, so the perceived frequency must be greater than 440 Hz. Orienting the axis from the source to the observer, vs=+25.0 m/s and vo=−15.0 m/s, so f′=f0v−vsv−vo=440×340−25340+15=495.9 Hz, well above 440 Hz as expected.”
The trap: Skipping step 1 and replacing it with an afterthought. A direction announced AFTER the calculation is worth nothing: it merely describes the result obtained, right or wrong.
Marking: 1 mark the direction announced, 2 marks the two algebraic velocities justified, 2 marks the formula and the substitution, 1 mark the final comparison.
A speed measurement by echo, written up to the direction of the error
When to use it: Radar, sonar, flowmeter, Doppler ultrasound: any problem where the wave leaves the target again.
1Say explicitly that the target undergoes the shift TWICE, first as a moving observer, then as a moving source, and deduce the factor 2.
2Project the velocity of the target onto the axis of the beam, vcosθ, naming the angle on the diagram.
3Write Δf=c2f0vcosθ and specify which speed c is used, that of sound in the medium crossed or that of light.
4Isolate the quantity asked for and calculate, keeping the significant figures of the data.
5Discuss the direction of the angle error: since cos decreases on [0°,90°], a real angle larger than the set angle always makes the DISPLAYED speed too small.
Concluding sentence
“The blood cells receive the wave as moving observers then send it back as moving sources, hence Δf=c2f0vcosθ=15402×5.00×106×0.450×cos60.0°=1.46×103 Hz.”
The trap: Writing c without saying what it is. In a flowmeter it is 1540 m/s, the speed of ultrasound in tissue; in a radar it is 3.00×108 m/s. Confusing the two gives a result that is wrong by a factor of 2×105.
Marking: 2 marks the factor 2 justified, 1 mark the projection, 2 marks the calculation, 1 mark the direction of the angle error.
Check before you hand in
Five minutes of checking recover more marks than one more problem started in a hurry.
The direction, in one second
Reread your announcement sentence and compare your result with f0. Closing in, it must be larger; moving apart, smaller. No exception, in acoustics or in optics.
A result of 391.8 Hz under a sentence that says “they are closing in” is wrong: the fraction was inverted, 495.9 Hz was needed.
Setting the two velocities equal
In your general formula, mentally replace vo by vs. It must return exactly f0, since two movers that keep the same distance hear no shift.
f0v−vsv−vo does give f0; f0v−vsv+vo gives 440×315365=509.8 Hz, so a sign is wrong.
The order of magnitude of the shift
The relative shift is roughly vvs for a one-way trip, and double that for an echo. Compare it with yours before handing in.
30 m/s in 343 m/s gives 9 percent: on 500 Hz, expect about fifty hertz. A reported shift of 200 Hz signals a unit error or a forgotten factor.
The calculator mode, and cos90°
As soon as a cosθ appears, check DEGREE mode, then test θ=90° in your expression: it must give a ZERO shift, never a maximum.
In radian mode, cos60 is −0.952 and the speed would come out negative: the absurd sign is the only warning you will get.
The product of the two directions, for light
Calculate the frequency in approach and in recession at the same speed: their product must equal f02, the two factors being inverses of each other.
At β=0.600: 0.500×2.00=1.00=f02 with f0=1.00 GHz. A different product exposes a radical that was inverted wrongly.
The typical problem, taken apart
The Doppler flowmeter, from the factor 2 to the direction of the angle error
A probe placed on the skin emits ultrasound at f0=5.00 MHz toward a blood vessel in which blood flows at v=0.450 m/s. The ultrasound travels at c=1540 m/s in tissue and the operator sets the instrument to θ=60.0°.
Calculate the shift measured. Then, the real angle being in fact 65.0°, give the speed the instrument displays and the relative error made.
The beam makes the trip out AND back, and this is where the factor 2 comes from; the angle θ is measured between the direction of the beam and that of the flow, never between the beam and the skin.
Step 1
Expected direction: the blood flows up the beam toward the probe, so the echo returns HIGHER than 5.00 MHz and Δf>0.
Why
This is the step that locks every sign that follows. Here it earns few marks directly, but it makes the most expensive mistake impossible, and it takes one line.
Step 2
Count the shifts: there are TWO. The blood cells receive the wave as moving observers, then send it back as moving sources. Hence Δf=c2f0vcosθ.
Why
The factor 2 is not guessed from the formula, it is justified by the mechanism. A marker gives the marks for the sentence that names the two successive roles of the target, not for the 2 copied down.
Step 3
Project the velocity of the blood onto the beam: vcos60.0°=0.450×0.500=0.225 m/s.
Why
Only the component along the beam brings the target closer to or farther from the probe. It is the same projection as in the train passing alongside, and it is read off the figure, not the formula.
Step 4
Δf=15402×5.00×106×0.225=15402.25×106=1.46×103 Hz.
Why
Three significant figures, like the data. The result falls squarely in the audible band: this is the reason these instruments have a loudspeaker, and this remark often earns a mark of discussion.
Step 5
Real angle 65.0°: the instrument divides the shift actually produced by c2f0cos60.0°, so it displays vread=0.450×cos60.0°cos65.0°=0.450×0.845=0.380 m/s.
Why
The instrument measures only Δf: everything else is a calculation made with the angle it was given. Writing the ratio of the two cosines avoids recalculating two shifts and removes one chance of error.
Step 6
Check and conclusion: cos decreases on [0°,90°], so cos65.0°<cos60.0° and the displayed speed MUST be smaller than the true one. Relative error 0.4500.450−0.380=15.5 percent.
Why
The direction of the error is what the question really asks, and it can be predicted without calculation. The value confirms the estimate tan60°×5° in radians, that is 1.73×0.0873=15 percent.
The conclusion, written out
The shift is 1.46×103 Hz. With five degrees of error on the angle, the instrument displays 0.380 m/s instead of 0.450 m/s, an underestimate of 15.5 percent: a sixth of the flow disappears for five degrees.
The classic mistake on this problem: Forgetting the factor 2 and reporting 730 Hz, or getting the direction of the angle error wrong by believing that a larger real angle makes the speed overestimated.
Learn by heart
•General formula, axis oriented from the SOURCE to the OBSERVER: f′=f0v−vsv−vo, the signs being those of the velocities.
•The direction BEFORE the signs: every approach raises the frequency, every recession lowers it, without exception.
•The speed of the wave depends only on the medium: 343 m/s in air at 20 °C, 1540 m/s in tissue, 1500 m/s in water.
•Moving source: it is the WAVELENGTH that is compressed; moving observer: it is the RATE OF ENCOUNTER that rises. The two cases are NOT symmetric.
•Off axis: only vscosθ counts, and at the closest point EXACTLY f0 is heard.
•Wind w counted from the source to the observer: v becomes v+w. If nobody moves, the wind changes NO frequency.
•Echo off a moving target: DOUBLE shift, Δf≈c2f0vcosθ, and the instrument always underestimates when the angle is badly set.
•Rotating source: f0 is the HARMONIC mean of fmax and fmin, never the arithmetic one.
•Light: no medium, so f′=f01+β1−β for recession, and a transverse shift γf0 with no acoustic equivalent.
Frequently asked questions
How do I know whether to add or subtract in the Doppler effect formula?
You decide on the direction before choosing the sign. Ask yourself whether the distance between the source and the observer is decreasing or increasing. If it is decreasing, the perceived frequency must rise, so choose the sign that makes the fraction greater than one. The safest approach is still to orient an axis from the source to the observer and to put the two velocities on it with their sign.
Does the speed of sound change when the source moves?
No, never. The speed of sound depends only on the medium and its temperature, about 343 metres per second in air at 20 degrees Celsius, ahead of and behind the source. What changes is the wavelength: the source advances between two emissions, so the wavefronts are tighter ahead of it and more spread out behind it.
Why is the frequency not at its maximum at the moment the train passes in front of me?
Because only the component of the velocity directed toward you shifts the frequency. At the moment of passing, the train is moving perpendicular to your line of sight: it is no longer approaching and not yet receding, so you hear exactly the emission frequency. The maximum was reached earlier, when the train was heading almost straight at you.
Why does the Doppler shift count double in a radar or an ultrasound scan?
Because the target undergoes the effect twice. It first receives the wave as a moving observer, so already shifted, then it sends it back as a moving source, which shifts a second time the frequency it has just received. This is where the factor two in the formula comes from, valid for sound as for electromagnetic waves.
Does wind change the pitch of the sound I hear?
No, as long as neither the source nor you is moving. The wind lengthens the wavelength over the ground and speeds up propagation over the ground in the same ratio, so the frequency does not move. It changes the intensity received and the travel time, which gives the impression of a sound that is carried. If the source moves, a following wind even slightly reduces the shift.
Does an error of a few degrees on the angle really matter?
Yes, and much more than you would think. The relative error on the speed is roughly the tangent of the angle times the error expressed in radians: five degrees of error cost 15 percent at 60 degrees, against 9 percent at 45 degrees. At 90 degrees the shift is zero and the instrument displays nothing at all.
Practise it
Corrected exercises: The Doppler effect: moving source, moving observer and radial component, 203-SN3-RE
A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.
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Get in touch for a first session. We work through the Doppler effect of the waves and modern physics course at the level of real tests, from the choice of signs to speed measurements by echo.
