Corrected exercises: the Doppler effect, moving source, moving observer and radial component (203-SN3-RE)

The locomotive emits one wavefront every $T=\dfrac{1}{500}=2.00$ ms. At rest, each wavefront has time to travel $vT$ before the next one: $\lambda_{0}=\dfrac{343}{500}=0.686$ m. When it moves, it advances by $v_{s}T$ during the same interval, so the gap between two wavefronts AHEAD of it is $vT-v_{s}T$: $\lambda_{M}=\dfrac{343-30}{500}=\dfrac{313}{500}=0.626$ m. Behind it, the two displacements add: $\lambda_{N}=\dfrac{343+30}{500}=\dfrac{373}{500}=0.746$ m. The figure shows it without calculation: the circles crowd together on the side toward which the source is heading.

Once a wavefront has been released, the air knows nothing more about the locomotive: each wavefront moves at $343$ m/s in both directions, at $M$ as at $N$. This is the point to hold on to. The frequency can then be read directly: $f_{M}=\dfrac{v}{\lambda_{M}}=\dfrac{343}{0.626}=547.9$ Hz and $f_{N}=\dfrac{343}{0.746}=459.8$ Hz. Direction check: the source is coming toward $M$, so $M$ must hear a HIGHER pitch than $500$ Hz, and $547.9>500$; it is moving away from $N$, which must hear a lower pitch, and $459.8<500$. Both directions are right.

Nothing changes on the emission side: the wavefronts are those of a stationary source, spaced by $\lambda_{0}=0.686$ m, and this wavelength does not change one bit because a microphone is moving through it. What changes is the speed at which the microphone meets them: $343+30=373$ m/s. It therefore crosses $f'=\dfrac{373}{0.686}=543.7$ Hz of them every second. Direction checked again: it rushes toward the source, so a higher pitch than $500$ Hz.

$547.9$ Hz against $543.7$ Hz: a difference of $4.2$ Hz, or $0.8$ percent, perfectly audible to a musical ear. The two situations are not equivalent because air is not a neutral backdrop: it is the medium that carries the wave and serves as the reference. Moving source: the wavelength itself is made shorter, and the speed of encounter stays at $343$ m/s. Moving observer: the wavelength stays at $0.686$ m and it is the speed of encounter that rises to $373$ m/s. Two different mechanisms, hence two different formulas, $f\dfrac{v}{v-v_{s}}$ on one side and $f\dfrac{v+v_{o}}{v}$ on the other. A student who remembers a single formula and fiddles with the sign is off by $4$ Hz here, which is enough to lose the question; worse, they lose the one idea of the chapter. Remember the sentence: ahead of a moving source, it is the WAVELENGTH that is compressed; for a moving observer, it is the RATE OF ENCOUNTER that increases.

Expected direction: they are closing in, so a pitch higher than $440$ Hz. The axis runs from $A$ to $B$. $A$ moves toward $B$, so $v_{s}=+25.0$; $B$ moves toward $A$, that is against the axis, so $v_{o}=-15.0$. $f'=440\times\dfrac{340+15}{340-25}=440\times\dfrac{355}{315}=495.9$ Hz. This is indeed higher: the direction is respected.

Expected direction: they are moving apart, so a lower pitch. We reorient the axis, still from the source to the observer. $A$ recedes relative to this axis, $v_{s}=-25.0$; $B$ moves away, $v_{o}=+15.0$. $f'=440\times\dfrac{340-15}{340+25}=440\times\dfrac{325}{365}=391.8$ Hz. Lower, direction respected. At the crossing, the passenger therefore hears a drop from $495.9$ to $391.8$ Hz, that is $104.1$ Hz all at once.

Expected direction: $A$ travels faster than $B$ and catches up with it, so the distance is shrinking and the frequency must RISE, but less than in a), since the closing speed is no longer $40$ m/s but $10$ m/s. Axis from $A$ to $B$: $A$ moves toward $B$, $v_{s}=+25.0$; $B$ moves in the same direction as the axis, so it is running away, $v_{o}=+15.0$. $f'=440\times\dfrac{340-15}{340-25}=440\times\dfrac{325}{315}=454.0$ Hz. Between $440$ and $495.9$: consistent with what was expected.

We need $\dfrac{340-v_{o}}{340-25}=1$, so $v_{o}=25.0$ m/s: $B$ must travel at exactly the speed of $A$. The distance between them no longer changes, no wavefront is compressed or stretched from the point of view of $B$, and it hears the emission frequency, even though both trains are speeding along at $90$ km/h through the air. It is a good way to remember that only the RELATIVE motion along the line joining the two matters. As for $390.4$ Hz, that is $440\times\dfrac{315}{355}$: the classmate inverted numerator and denominator. The direction check would have saved them in two seconds, since $390.4$ Hz is LOWER while the trains are closing in. A $10$-mark question lost for an upside-down fraction, and the worst part is that the wrong result looks like the right one: the two values $495.9$ and $390.4$ bracket $440$, and only the position relative to $440$ tells them apart.

The circles in the figure are off-centre: the wavefront emitted $nT$ ago has radius $vnT$ but its centre has drifted by $wnT$. On the side of $M$, the wavefronts are therefore spaced by $(v+w)T$ and on the side of $N$ by $(v-w)T$. Numerically $\lambda_{M}=\dfrac{360}{440}=0.818$ m, $\lambda_{N}=\dfrac{320}{440}=0.727$ m, against $\dfrac{340}{440}=0.773$ m with no wind. The wavelengths ARE different, and yet both microphones hear $440$ Hz: on the side of $M$ the wavefront moves at $360$ m/s over the ground, and $\dfrac{360}{0.818}=440$; on the side of $N$ it moves at $320$ m/s, and $\dfrac{320}{0.727}=440$. The speed and the wavelength have been multiplied by the SAME factor, their ratio does not move. This is the direction check of the chapter: nobody moves relative to anybody, so there is no shift, whatever the wind does.

In the reference frame of the air, the source has velocity $v_{s}-w$ and the observer $v_{o}-w$ along the axis oriented from the source to the observer. The general formula gives $f'=f_{0}\dfrac{v-(v_{o}-w)}{v-(v_{s}-w)}=f_{0}\dfrac{(v+w)-v_{o}}{(v+w)-v_{s}}$: the wind merely replaces $v$ by $v+w$ everywhere, $w$ being counted positive in the direction from the source to the observer. Here $v+w=360$, $v_{o}=0$, $v_{s}=+30.0$, so $f'=440\times\dfrac{360}{330}=480.0$ Hz.

The wind changes sign: $w=-20.0$, so $v+w=320$. $f'=440\times\dfrac{320}{320-30}=440\times\dfrac{320}{290}=485.5$ Hz.

With no wind, $f'=440\times\dfrac{340}{310}=482.6$ Hz. The ranking is therefore: following wind $480.0$ Hz, no wind $482.6$ Hz, headwind $485.5$ Hz. The common saying is wrong twice over. First, when nobody moves, the wind changes NOTHING in the pitch of the sound, as part a) has just established; it only changes the travel time and the intensity received, which gives the impression of a sound that is carried. Second, when the source moves, a following wind REDUCES the shift instead of increasing it, because what counts is the ratio $\dfrac{v_{s}}{v+w}$: by increasing the propagation speed over the ground, the wind leaves the source less time to catch up with its own wavefronts. The difference is modest, $5.5$ Hz between the two extreme winds, but the DIRECTION of the effect is the opposite of intuition, and this is exactly the kind of three-mark question that is lost by answering on gut feeling.

Very far away, $\theta\to 0$ and $\cos\theta\to 1$: the velocity is entirely radial. $f'=450\times\dfrac{340}{340-70}=450\times\dfrac{340}{270}=566.7$ Hz. This is the MAXIMUM value of the whole passage, and it is reached when the train is farthest away, which always comes as a surprise.

At the closest point, the velocity is perpendicular to the line of sight: $\theta=90°$, $\cos\theta=0$, $v_{r}=0$. At that instant the train is neither approaching nor receding from $O$; its distance goes through a minimum and its derivative is zero. So $f'=f_{0}=450$ Hz exactly. No shift at the moment of passing: this is the most counterintuitive result of the chapter, and the one that exams like to ask about.

The figure gives the geometry: the train is $60.0$ m from the closest point and $O$ is $60.0$ m from the track, so the line of sight makes $45°$ with the track. $\cos\theta=\dfrac{60}{\sqrt{60^{2}+60^{2}}}=\dfrac{1}{\sqrt{2}}=0.7071$. The radial component is $v_{r}=70.0\times 0.7071=49.50$ m/s: the train rushes along at $252$ km/h but approaches $O$ at only $49.5$ m/s. $f'=450\times\dfrac{340}{340-49.50}=\dfrac{153000}{290.50}=526.7$ Hz. Between $450$ and $566.7$: the direction is right.

We solve $450\times\dfrac{340}{340-70\cos\theta}=500$, that is $340-70\cos\theta=\dfrac{450\times 340}{500}=306$, so $70\cos\theta=34$ and $\cos\theta=0.4857$. Hence $\theta=60.9°$. The distance to the closest point is read from the right triangle: $\tan\theta=\dfrac{d}{x}$, so $x=\dfrac{60.0}{\tan 60.9°}=\dfrac{60.0}{1.800}=33.3$ m. The train is therefore only about thirty metres from the crossing point when the frequency has already fallen from $566.7$ to $500$ Hz: the whole slide happens over the last few hundred metres, which is why it is heard as a sudden drop and not as a slow descent. The classic trap is to write $\cos\theta=\dfrac{x}{d}$ instead of $\dfrac{x}{\sqrt{x^{2}+d^{2}}}$, or to take the angle at vertex $O$ rather than at vertex $T$: both mistakes give a cosine greater than $1$ as soon as the train is far away, which should raise an alarm immediately.

First stage: the moth is the observer and the bat the source. Axis from the bat to the moth: $v_{s}=+8.00$, $v_{o}=+3.00$ since the moth flees in the direction of the axis. $f_{1}=42000\times\dfrac{340-3}{340-8}=42000\times\dfrac{337}{332}=42632.5$ Hz. Second stage, the roles are swapped: the moth becomes the source of $f_{1}$ and the bat the observer. New axis, from the moth to the bat: the moth is moving away, so $v_{s}=-3.00$, and the bat is approaching, so $v_{o}=-8.00$. $f_{2}=42632.5\times\dfrac{340+8}{340+3}=42632.5\times\dfrac{348}{343}=43254.0$ Hz. Total shift: $43254-42000=1254$ Hz, that is $1.25$ kHz.

Regrouping, $f_{2}=f_{0}\dfrac{(v-v_{m})(v+v_{b})}{(v-v_{b})(v+v_{m})}$. For $v_{b}$ and $v_{m}$ small compared with $v$, each factor is $1$ plus a small term, and keeping only the first order we are left with $f_{2}\approx f_{0}\left(1+\dfrac{2(v_{b}-v_{m})}{v}\right)$, so $\Delta f\approx\dfrac{2\times 42000\times 5.00}{340}=1235$ Hz. Against $1254$ Hz exactly: a $1.5$ percent difference, which is negligible for an estimate but not for a calculation asked to the hundredth. The factor $2$ is the signature of the round trip, and it is the one to remember.

The two now approach each other. First stage: $v_{s}=+8.00$, $v_{o}=-3.00$, $f_{1}=42000\times\dfrac{343}{332}=43391.6$ Hz. Second stage: the moth moves toward the bat, $v_{s}=+3.00$, and the bat moves toward it, $v_{o}=-8.00$: $f_{2}=43391.6\times\dfrac{348}{337}=44807.9$ Hz. Shift $2808$ Hz, that is $2.81$ kHz, more than double that of a). The bat thus has a very clear measurement of the DIRECTION in which its prey is moving, and this is precisely the information it needs to correct its flight path.

With a single application, $42000\times\dfrac{340}{340-5}=42626.9$ Hz, a shift of $627$ Hz. They are off by a factor of exactly $2$, since $2\times 627=1254$. The error is invisible on reading, the result has the right size and the right sign, and yet it costs the entire question. The rule to hold on to: whenever there is a REFLECTION off a moving target, the shift counts double, because the target undergoes the Doppler effect once on receiving and a second time on re-emitting. This same factor $2$ will come back in the medical flowmeter of exercise 9 and in the speed radar of exercise 10, and it holds for all waves, sound or electromagnetic.

The tangential speed is $u=2\pi Rn=2\pi\times 1.00\times 3.00=18.85$ m/s, that is $67.9$ km/h at the end of a one-metre string. This is more than enough for an audible shift, which is why this experiment is done in every laboratory.

The frequency is maximum when the radial component toward $O$ is maximum, that is when the velocity points straight at $O$: $f_{max}=1200\times\dfrac{340}{340-18.85}=\dfrac{408000}{321.15}=1270.4$ Hz. It is minimum when the velocity points straight away: $f_{min}=1200\times\dfrac{340}{358.85}=1137.0$ Hz. The sound therefore oscillates between these two values three times per second, and this wobble in pitch is exactly what you hear when you whirl a sounding object on a string.

Position 2: the velocity points toward $O$, and this is where $f_{max}$ is reached. Position 4: the velocity points away, and this is $f_{min}$. At positions 1 and 3, the velocity is PERPENDICULAR to the line of sight, so $v_{r}=0$ and $O$ hears exactly $f_{0}=1200$ Hz. Now position 1 is the point of the circle CLOSEST to $O$ and position 3 the FARTHEST: the extremum of distance does not coincide with the extremum of frequency, it is offset by a quarter turn. This is the same result as in exercise 4, transposed to circular motion, and it is the favourite trap of the question: we think the sound is highest when the source is closest, whereas that is the moment when it is not shifted at all.

The arithmetic mean gives $\dfrac{1270.4+1137.0}{2}=1203.7$ Hz, not $1200$ Hz: the $3.7$ Hz difference is not a rounding error, it is structural. The reason is that $f_{max}$ and $f_{min}$ are not symmetric about $f_{0}$: the denominator $v-u$ pulls harder upward than $v+u$ pulls downward. The right mean is the HARMONIC mean, $\dfrac{2f_{max}f_{min}}{f_{max}+f_{min}}$, which gives back $f_{0}$ exactly. We can check by substituting: $\dfrac{2\,f_{0}^{2}v^{2}}{(v^{2}-u^{2})}\times\dfrac{v^{2}-u^{2}}{2f_{0}v^{2}}=f_{0}$, with no approximation at all. Remembering this identity is worth remembering for an exam, because it lets you recover both $f_{0}$ and $u$ from just two measurements, without ever having access to the source at rest.

Recession: $f'=1.00\times\sqrt{\dfrac{1-0.600}{1+0.600}}=\sqrt{\dfrac{0.400}{1.600}}=\sqrt{0.250}=0.500$ GHz. Direction check: it is receding, so a shift downward, and $0.500<1.00$. Approach: the radical is inverted, $f'=\sqrt{4.00}=2.00$ GHz. The product is $0.500\times 2.00=1.00=f_{0}^{2}$, which is no accident: the two factors are inverses of each other, a direct consequence of the fact that reversing the direction of motion amounts to changing $\beta$ into $-\beta$. Note in passing the scale: at $0.6\,c$ the frequency is divided or multiplied by two, whereas the locomotive of exercise 1 shifted the pitch by less than ten percent.

$\gamma=\dfrac{1}{\sqrt{1-0.360}}=\dfrac{1}{\sqrt{0.640}}=\dfrac{1}{0.800}=1.25$, so $f'=\dfrac{1.00}{1.25}=0.800$ GHz. And this is where light truly parts company with sound: in acoustics, at the closest point, the radial component is zero and the observer hears EXACTLY $f_{0}$, as exercise 4 established. Here it receives $0.800$ GHz, a redshift of twenty percent although the probe is neither receding nor approaching at that instant. This transverse shift has no classical equivalent: it does not come from the geometry of the wavefronts but from TIME DILATION, the probe's clock ticking more slowly as seen from Earth. It is, incidentally, one of the measurements that confirmed special relativity.

For small $\beta$, $\sqrt{\dfrac{1-\beta}{1+\beta}}\approx(1-\tfrac{\beta}{2})(1-\tfrac{\beta}{2})\approx 1-\beta$, and since $\lambda$ varies inversely with $f$, $\dfrac{\Delta\lambda}{\lambda}\approx\beta=\dfrac{v}{c}$. Here $\beta=\dfrac{30.0}{3.00\times 10^{5}}=1.00\times 10^{-4}$, so $\Delta\lambda=500.0\times 1.00\times 10^{-4}=0.0500$ nm. A twentieth of a nanometre: it is tiny, and yet every radial velocity of a star and the detection of exoplanets come from this measurement. The quantum physics set of this same course applies exactly this formula to the redshift of a galaxy and to Hubble's law.

Because the acoustic asymmetry came entirely from the existence of a medium. Air provides a privileged reference frame, so there is a real physical difference between moving relative to the air and standing still in it: in one case a different wavelength is produced, in the other the wavefronts are met faster. For light, there is no ether, no experiment can tell a moving source from a receding observer, and the principle of relativity even forbids such a distinction from existing. Only one quantity remains, the relative speed, and one formula. A practical corollary for the exam: as soon as a problem mentions light, radar waves or radio waves, stop asking who moves relative to what and keep only the speed of approach or recession.

False, and this is the most common error in the chapter. The speed of sound depends only on the medium, essentially on its nature and its temperature: $343$ m/s in air at $20$ °C, whatever the source does. Once a wavefront has been released, it carries no memory of the motion of its emitter. What is compressed is the WAVELENGTH, because the source has advanced between two successive emissions. Correct statement: the speed of sound is unchanged, the wavelength ahead of the source is shortened, so $f=\dfrac{v}{\lambda}$ increases.

False in acoustics, true for light. Exercise 1 gives the figures: source moving at $30$ m/s toward the observer, $547.9$ Hz; observer moving at $30$ m/s toward the source, $543.7$ Hz. The $4.2$ Hz difference comes from the fact that air is a carrying medium, hence a physically privileged reference frame. Correct statement: in acoustics the two cases give different formulas; it is only for light, for lack of a medium, that only the relative speed counts.

False. The wavefronts are those of a stationary source, regularly spaced by $\lambda_{0}=\dfrac{v}{f_{0}}$ throughout space, and they do not tighten because someone starts running. What changes is the SPEED OF ENCOUNTER, which goes from $v$ to $v+v_{o}$. Correct statement: the wavelength is unchanged, the observer simply crosses more of them per second, hence $f'=f_{0}\dfrac{v+v_{o}}{v}$.

False twice over. If nobody moves, the wind changes nothing at all in the frequency: it lengthens the wavelength over the ground AND speeds up propagation over the ground in the same ratio, as exercise 3 showed, so that $f$ remains $440$ Hz. And if the source moves, a following wind REDUCES the shift instead of increasing it, $480.0$ Hz against $482.6$ Hz with no wind, because the quantity that counts is $\dfrac{v_{s}}{v+w}$. Correct statement: the wind changes the intensity received and the travel time, not the pitch of the sound, except by changing the effective speed of propagation when one of the two parties moves.

False, and this is the most profitable exam trap. At the moment of passing, the velocity of the train is perpendicular to the line of sight, the radial component is zero, and the observer hears EXACTLY the emission frequency, $450$ Hz in exercise 4. The maximum, $566.7$ Hz, was reached much earlier, when the train was still far away and heading almost straight at the observer. Correct statement: the frequency decreases continuously from $f_{0}\dfrac{v}{v-v_{s}}$ to $f_{0}\dfrac{v}{v+v_{s}}$, it passes through $f_{0}$ at the moment of passing, and it is the speed of this crossing, not a jump, that gives the impression of a sudden drop.

The factor $2$ is the one from exercise 5: the red blood cells first receive the wave as moving observers, then send it back as moving sources, so the shift occurs twice. Numerically, $\Delta f=\dfrac{2\times 5.00\times 10^{6}\times 0.450\times 0.500}{1540}=\dfrac{2.25\times 10^{6}}{1540}=1.46\times 10^{3}$ Hz. A shift of $1461$ Hz on a $5$ MHz carrier, that is three ten-thousandths, but above all a value that falls squarely in the audible band.

The instrument measures only $\Delta f$; it deduces $v$ by dividing by $\dfrac{2f_{0}\cos 60.0°}{c}$, that is with the cosine it has been given. The shift actually produced is $\dfrac{2\times 5.00\times 10^{6}\times 0.450\times\cos 65.0°}{1540}=1235$ Hz. The speed displayed is therefore $v_{read}=0.450\times\dfrac{\cos 65.0°}{\cos 60.0°}=0.450\times 0.845=0.380$ m/s, that is an UNDERESTIMATE of $15.5$ percent. Five degrees of error on an angle, and a sixth of the flow disappears: this is why the operator corrects the angle by hand on the image, and not by eye.

The relative error on $v$ is, to first order, $\tan\theta\,\Delta\theta$. At $60°$, $\tan 60°=1.73$ and $\Delta\theta=5°=0.0873$ rad, which gives the $15$ percent found. At $45°$, $\tan 45°=1.00$, so about $8.7$ percent, and the exact calculation, $1-\dfrac{\cos 50°}{\cos 45°}$, gives $9.10$ percent: half the cost for the same clumsiness. The temptation would then be to go down toward $0°$, where the angle error would cost almost nothing. Two reasons forbid it: a vessel is almost parallel to the skin, so a beam at $0°$ would have to enter from the side and would never cross the right depth; and the reflection from the cells is much weaker at grazing incidence. At the other end, $90°$ is catastrophic: $\cos 90°=0$, the shift is ZERO and the instrument displays nothing, whatever the speed of the blood. A vessel seen perpendicularly therefore looks thrombosed although it is flowing normally, and this is a very real diagnostic error. Between $45°$ and $60°$ we keep a clear signal and a sensitivity to the angle that is still reasonable.

Because $1461$ Hz is a note that the ear hears perfectly well. By sending the shift directly to a loudspeaker, the measurement is turned into sound, and the clinician hears the flow: the pitch gives the speed, the rhythm gives the cardiac cycle, and the quality of the sound immediately betrays the turbulence of a stenosis or the reversal of a flow. No numerical display gives this information as quickly. This is not a gimmick, it is the normal way of using the instrument, and it is a direct consequence of the choice of $5$ MHz: a much higher carrier (70 MHz or more) would push the shift above the audible band and be absorbed more strongly by tissue, a lower carrier would penetrate better but resolve less finely.