ARCHETYPE: Doppler-Effect Numericals

In every CBSE, ICSE, JEE and NEET paper that touches waves, there is a Doppler question waiting — an ambulance, a train, a bat, a reflecting wall. It is worth two to four marks, it is entirely mechanical once you have a system, and it is one of the most reliable places on the paper to lose those marks to a single flipped sign. Students do not fail Doppler problems because the physics is hard; they fail because they improvise the signs under time pressure and the improvisation goes wrong. This article is the problem gym for the Doppler effect in sound. The concept article next door built the master formula and proved it by counting wavefronts; here the setup is the whole point. Every Doppler-for-sound problem, however it is dressed up, is the same four-step drill: name the medium's sound speed, decide who is moving and whether toward or away, assign the numerator and denominator signs by the rule that approach raises pitch, and compute. Get that reflex automatic and the whole family collapses into fill-in-the-blanks. We will run the drill on the two board staples — a source that approaches then recedes, and an observer that moves past a fixed source — and then climb to the two questions that separate the top scorers: a source and observer both moving through a wind, and the archetype that catches almost everyone, the reflected wall, where the shift happens twice and produces an audible beat. By the end you should be able to look at any Doppler problem and know, before you write anything, exactly which signs the answer wears.

Key takeaways

Every Doppler-for-sound problem is one drill: pick the medium's $v$ , decide who moves and toward/away, assign $\dfrac{v \pm v_o}{v \mp v_s}$ by "toward raises pitch," compute.
The observer's speed goes in the numerator ( $+$ toward), the source's speed in the denominator ( $-$ toward). Never swap them.
Ambulance staple: $640\ \mathrm{Hz}$ siren at $20\ \mathrm{m/s}$ , $v = 343$ — approaching $679.6\ \mathrm{Hz}$ , receding $604.7\ \mathrm{Hz}$ , an audible jump of $74.9\ \mathrm{Hz}$ on passing.
Wind enters by replacing $v$ with $v \pm w$ everywhere at once — a tailwind and a headwind change both terms together.
The reflected-wall archetype is TWO Doppler shifts, not one: a source at $500\ \mathrm{Hz}$ moving at $2\ \mathrm{m/s}$ toward a wall hears its own echo at $505.917\ \mathrm{Hz}$ , a beat of $\Delta f = 2fu/(v-u) = 5.917\ \mathrm{Hz}$ .
A moving reflector gives the same beat formula — the physics behind speed traps and Doppler blood-flow.
Key idea: solve every Doppler numerical with one fixed procedure — medium speed, who moves, sign by 'toward is higher', compute — and treat every reflection as two shifts applied in turn.

The core insight

There is only one Doppler-for-sound problem, solved by one procedure: identify the medium's sound speed, decide for the source and the observer separately whether each moves toward or away, place the observer's speed in the numerator and the source's in the denominator with signs fixed by "approach raises pitch," and — whenever sound reflects — apply that procedure twice.

Where this lives today

The reflected-wall beat is not a textbook curiosity; it is a working instrument. A police speed gun fires a beam at your car and reads the beat between the outgoing and reflected waves — the moving-reflector formula $\Delta f = 2fu/(v \mp u)$ , with light instead of sound and a relativistic version of the same double-shift logic. A Doppler ultrasound in a hospital does the identical thing with sound: it bounces megahertz pulses off red blood cells and reads the beat to measure how fast, and which way, blood flows through a valve — the same two-shift chain you drilled in Example 4. Weather radar reads the double shift off raindrops to map wind inside a storm; bats read it off fleeing insects. Every one of these is the reflected-wall archetype, and every one is solved by the same discipline: two shifts in turn, and the beat is the difference.

The setup: one procedure for every problem

In an archetype the bookkeeping is the physics, so learn the bookkeeping cold. The master formula, derived in the concept article by counting compressed wavefronts, is

f' = f\,\frac{v \pm v_o}{v \mp v_s} \qquad (1)

where: $f$ — the emitted frequency; $f'$ — the observed frequency; $v$ — the speed of sound in the medium, in the medium's frame; $v_o$ — the observer's speed; $v_s$ — the source's speed. The four-step procedure that turns any word problem into a number:

Pick $v$ . The speed of sound in the medium, stated by the problem — usually $340$ , $343$ (at $20\ \mathrm{^\circ C}$ ) or $330\ \mathrm{m/s}$ . Fix it once and never change it mid-problem.

Decide who moves, and toward or away. Draw the arrow. A moving observer and a moving source are handled by different terms, so label each separately — this is where sound differs from light and where marks are lost.

Assign signs by "toward is higher." The observer's speed sits on top: $+v_o$ if the observer moves toward the source, $-v_o$ if away. The source's speed sits on the bottom: $-v_s$ if the source moves toward the observer, $+v_s$ if away. Whatever combination makes an approach come out higher in pitch is correct.

Compute, then sanity-check. Evaluate the fraction, then ask: is this a net approach? Then $f'$ must exceed $f$ . If it doesn't, a sign is wrong.

Figure

The sign map for Doppler problems: the observer speed sits in the numerator and the source speed in the denominator, and moving toward always raises the pitch

The double-shift archetype: a reflecting wall

The hardest Doppler problem on any Indian paper is the reflected wall, and it is hard for exactly one reason: the shift happens twice. Sound leaves a moving source, reaches a wall, and comes back — and each leg is a separate Doppler event that you must handle in turn. Treat it as a single shift and you will be wrong every time.

Figure

A source moving toward a wall at speed u: the wall first hears a raised frequency as a stationary observer, then re-radiates it, and the still-moving source hears that echo raised a second time — two shifts producing a beat

Derivation AThe reflected-wall double shift and its beat

The goal is the frequency a source hears back from a wall it is approaching, and the beat that results — the archetype numbers $f_{\text{refl}} = 505.917\ \mathrm{Hz}$ and $\Delta f = 5.917\ \mathrm{Hz}$ for $f = 500\ \mathrm{Hz}$ , $u = 2\ \mathrm{m/s}$ , $v = 340\ \mathrm{m/s}$ . The strategy is to break the round trip into two ordinary Doppler problems: first the wall acting as a stationary observer of an approaching source, then the wall acting as a stationary source whose sound is met by the still-approaching original source, now an observer. Doing them in order — and never trying to combine them into one shift — is the whole skill.

Step 1. First shift: the wall hears the approaching source. The wall is a stationary observer; the source approaches at $u$ , so use the source-toward denominator: $f_1 = f\,\dfrac{v}{v - u} = 500\times\dfrac{340}{338} = 502.96\ \mathrm{Hz}$ . The wall "hears" a raised frequency.

Step 2. Reflection. A wall reflects whatever frequency strikes it, re-radiating $f_1$ as if it were a stationary source sitting at the wall. No new shift happens at the wall itself — it simply becomes a source of frequency $f_1$ .

Step 3. Second shift: the moving source hears the echo. The original source is still moving toward the wall at $u$ , so it is now an observer approaching a (stationary) source of frequency $f_1$ . Use the observer-toward numerator: $f_{\text{refl}} = f_1\,\dfrac{v + u}{v} = 502.96\times\dfrac{342}{340}$ . Chaining the two, $f_{\text{refl}} = f\,\dfrac{v}{v-u}\cdot\dfrac{v+u}{v} = f\,\dfrac{v + u}{v - u} = 500\times\dfrac{342}{338} = 505.917\ \mathrm{Hz}$ .

Step 4. The beat. The source now hears two tones at once: its own direct $500\ \mathrm{Hz}$ and the echo at $505.917\ \mathrm{Hz}$ . They beat at the difference, $\Delta f = f_{\text{refl}} - f = 5.917\ \mathrm{Hz}$ . In closed form, $\Delta f = f\left(\dfrac{v+u}{v-u} - 1\right) = \dfrac{2 f u}{v - u} = \dfrac{2\times500\times2}{338} = 5.917\ \mathrm{Hz}$ — a slow, throbbing beat you could count on a stopwatch. Verified: 500·340/338 = 502.96 Hz, 500·342/338 = 505.917 Hz, and 2·500·2/338 = 5.917 Hz are recomputed in verify_hs2-doppler-problems.py.

Go deeper · Derivation B · 1 minA variant: the moving reflector gives the same beat

Derivation BA variant: the moving reflector gives the same beat

The goal is to show that a stationary source whose sound reflects off a moving object produces the same beat formula $\Delta f = 2fu/(v-u)$ — the physics behind a police speed trap and Doppler blood-flow imaging. The strategy is again two shifts in turn, but with the roles reversed from the wall problem: this time the object is first a moving observer, then a moving source, while the detector sits with the original source. Watching the same $\Delta f$ fall out of the reversed geometry is the point — it is why one formula covers both.

Step 1. First shift: the moving object receives the sound. The source is fixed and emits $f$ ; the object moves toward it at $u$ , so the object is an observer approaching, and receives $f_1 = f\,\dfrac{v + u}{v}$ .

Step 2. Reflection. The object re-radiates $f_1$ , but now it is a moving source — it is still travelling toward the detector at $u$ as it reflects.

Step 3. Second shift: the detector hears the moving echo. The detector sits beside the original source (stationary observer); the reflecting object approaches it at $u$ as a source, so $f_{\text{det}} = f_1\,\dfrac{v}{v - u} = f\,\dfrac{v+u}{v}\cdot\dfrac{v}{v-u} = f\,\dfrac{v+u}{v-u}$ — identical to the reflected-wall result.

Step 4. Same beat. Therefore $\Delta f = f_{\text{det}} - f = \dfrac{2 f u}{v - u}$ , exactly as before. With $f = 500$ , $u = 2$ , $v = 340$ it is again $5.917\ \mathrm{Hz}$ . Whether the source chases a wall or a target flees a fixed source, the double shift wears the same formula — which is why an ultrasound machine can read blood speed from the beat of its own echo. Verified: the moving-reflector chain reproduces f(v+u)/(v−u) = 505.917 Hz and Δf = 5.917 Hz in verify_hs2-doppler-problems.py.

Common variants and where they trap you

variant	how to handle it	the trap
source approaches then recedes	flip the denominator sign; two calculations	reporting one number when the question wants the jump
observer moves past a fixed source	observer speed in the numerator	putting it in the denominator
both move	fill both terms of (1) at once	collapsing to a single "relative speed"
wind present	replace $v$ by $v \pm w$ everywhere	adding the wind to only one term
reflecting wall or moving reflector	apply Doppler twice, then $\Delta f = 2fu/(v \mp u)$	treating it as one shift
$v_s \to v$	denominator $\to 0$ , $f' \to \infty$	calling the infinity an error

Common trapA reflection is two Doppler shifts, never one

The single most expensive mistake in this whole topic is treating a reflected-wall or moving-reflector problem as a single Doppler event. It is always two: the reflector first receives a shifted frequency (as an observer), then re-emits it (as a source), and each step applies formula (1) in its own right. Skipping a step typically halves the beat — you would get $\Delta f \approx fu/v$ instead of the correct $2fu/(v-u)$ , and lose the whole question. The safe habit: write "shift 1" and "shift 2" as separate lines, compute the intermediate frequency the reflector hears, and only then find what returns. The factor of two in $2fu/(v-u)$ is the fingerprint of the double shift — if it is missing from your answer, you did the problem once instead of twice.

Worked examples

1 · Board level — the ambulance that passes you

An ambulance sounds a siren of frequency $640\ \mathrm{Hz}$ and travels at $20\ \mathrm{m/s}$ along a straight road past a stationary pedestrian. Take the speed of sound as $343\ \mathrm{m/s}$ . Find the frequency heard (a) while the ambulance approaches, (b) after it has passed, and (c) the sudden change in pitch the pedestrian hears at the moment of passing.

Step 1. Only the source moves ( $v_o = 0$ ); $v = 343$ , $v_s = 20$ , $f = 640$ . The observer term is just $v$ .

Step 2. Approaching — source toward, denominator $(v - v_s)$ : $f' = 640\times\dfrac{343}{343 - 20} = 640\times\dfrac{343}{323} = \boxed{679.6\ \mathrm{Hz}}$ .

Step 3. Receding — source away, denominator $(v + v_s)$ : $f' = 640\times\dfrac{343}{343 + 20} = 640\times\dfrac{343}{363} = \boxed{604.7\ \mathrm{Hz}}$ .

Step 4. The jump on passing is the difference: $679.6 - 604.7 = \boxed{74.9\ \mathrm{Hz}}$ . That step down of about $75\ \mathrm{Hz}$ is the "nee-naw" collapsing to a lower pair as the ambulance goes by. Note the emitted $640\ \mathrm{Hz}$ is never actually heard by a bystander until the ambulance is exactly level.

2 · Board level — the cyclist and the factory siren

A factory siren sounds a steady $680\ \mathrm{Hz}$ . A cyclist rides directly toward it at $20\ \mathrm{m/s}$ , passes beneath it, and continues away at the same speed. Take $v = 340\ \mathrm{m/s}$ . What frequency does the cyclist hear (a) while approaching and (b) while receding?

Step 1. Now the source is fixed ( $v_s = 0$ ) and the observer moves; $v = 340$ , $v_o = 20$ , $f = 680$ . The source term is just $v$ , and the observer speed goes in the numerator.

Step 2. Approaching — observer toward, numerator $(v + v_o)$ : $f' = 680\times\dfrac{340 + 20}{340} = 680\times\dfrac{360}{340} = \boxed{720.0\ \mathrm{Hz}}$ .

Step 3. Receding — observer away, numerator $(v - v_o)$ : $f' = 680\times\dfrac{340 - 20}{340} = 680\times\dfrac{320}{340} = \boxed{640.0\ \mathrm{Hz}}$ .

Step 4. Sanity: approach gives $720 > 680$ , recession gives $640 < 680$ — "toward is higher" holds. Contrast this with Example 1: there the same kind of $20\ \mathrm{m/s}$ motion belonged to the source and went in the denominator; here it belongs to the observer and goes in the numerator. Placing the speed correctly is the entire difference between the two board problems.

3 · Competitive level — both moving, in a wind

A train sounding a $500\ \mathrm{Hz}$ horn moves at $30\ \mathrm{m/s}$ toward a car; the car moves toward the train at $20\ \mathrm{m/s}$ . A steady wind of $5\ \mathrm{m/s}$ blows from the train toward the car. Take the still-air sound speed as $340\ \mathrm{m/s}$ . Find the frequency heard in the car, and compare with the no-wind case.

Step 1. Deal with the wind first. Blowing from source toward observer, it adds to the sound speed over the ground: $v_{\text{eff}} = 340 + 5 = 345\ \mathrm{m/s}$ . Use $v_{\text{eff}}$ in both terms.

Step 2. Both approach: observer toward gives numerator $(v_{\text{eff}} + v_o)$ , source toward gives denominator $(v_{\text{eff}} - v_s)$ : $f' = 500\times\dfrac{345 + 20}{345 - 30} = 500\times\dfrac{365}{315}$ .

Step 3. Evaluate: $f' = 500\times\dfrac{365}{315} = \boxed{579.37\ \mathrm{Hz}}$ .

Step 4. No-wind comparison: with $v = 340$ , $f' = 500\times\dfrac{340 + 20}{340 - 30} = 500\times\dfrac{360}{310} = 580.65\ \mathrm{Hz}$ . The tailwind lowers the observed frequency slightly, from $580.65$ to $579.37\ \mathrm{Hz}$ , because adding $5$ to both numerator and denominator pulls the ratio toward $1$ . Adding the wind to only one term — the classic error — would push the answer the wrong way entirely. Timing note: about two and a half minutes. The wind substitution is one line if you do it first; the danger is doing the sign assignment before the wind, then forgetting to update one of the terms. Fix $v_{\text{eff}}$ before you touch the signs.

4 · Competitive level — the reflected-wall beat

A car moving at $2\ \mathrm{m/s}$ straight toward a large wall sounds a $500\ \mathrm{Hz}$ horn. Take $v = 340\ \mathrm{m/s}$ . What frequency does the driver hear reflected from the wall, and what beat frequency results between the direct horn and the echo?

Step 1. This is a double shift — write it as two lines. Shift 1: the wall is a stationary observer, the source approaches at $u = 2$ , so the wall receives $f_1 = 500\times\dfrac{340}{340 - 2} = 500\times\dfrac{340}{338} = 502.96\ \mathrm{Hz}$ .

Step 2. Reflection: the wall re-radiates $f_1$ as a stationary source of $502.96\ \mathrm{Hz}$ .

Step 3. Shift 2: the driver is now an observer approaching that source at $u = 2$ , so hears $f_{\text{refl}} = 502.96\times\dfrac{340 + 2}{340}$ . Chained, $f_{\text{refl}} = 500\times\dfrac{342}{338} = \boxed{505.917\ \mathrm{Hz}}$ .

Step 4. The driver hears the direct $500\ \mathrm{Hz}$ and the reflected $505.917\ \mathrm{Hz}$ together, beating at $\Delta f = 505.917 - 500 = \boxed{5.917\ \mathrm{Hz}}$ , which matches the closed form $\dfrac{2fu}{v-u} = \dfrac{2\times500\times2}{338} = 5.917\ \mathrm{Hz}$ . Almost six beats a second — clearly audible. Timing note: about three minutes. The whole question is guarding against doing one shift instead of two. Budget thirty seconds just to write "shift 1 / shift 2" before any arithmetic; the closed form $\Delta f = 2fu/(v-u)$ is your check, not your first move.

Exam traps

Collapsing to a single relative speed. Sound is medium-borne, so a moving source (denominator) and a moving observer (numerator) enter through different terms and never combine into one relative speed. This is the defining feature of the sound Doppler effect and the most common conceptual error.

Swapping numerator and denominator. The observer's speed belongs on top, the source's on the bottom. A distractor built on the swap gives a plausible-looking but wrong number. The physical reason: the observer changes the closing speed (numerator), the source changes the wavelength (denominator).

Doing the reflection once. The reflected-wall and moving-reflector problems are two shifts. The tell is the factor of two in $\Delta f = 2fu/(v-u)$ ; if your beat has no such factor you applied Doppler only once.

Mixing $v$ values. Choose $330$ , $340$ or $343\ \mathrm{m/s}$ as the problem dictates and keep it for the whole solution. Switching from $343$ in part (a) to $340$ in part (b) is a silent, self-inflicted error.

Wind on one side only. A wind changes the sound speed over the ground, so it enters the numerator and denominator together via $v \to v \pm w$ . Adding it to just one term is wrong and usually shifts the answer in the wrong direction.

Units in km/h. Convert every speed to $\mathrm{m/s}$ before it meets $v$ : $72\ \mathrm{km/h} = 20\ \mathrm{m/s}$ . A stray km/h wrecks the ratio.

Board scope. This numerical archetype targets CBSE, ICSE, JEE-Main, JEE-Advanced and NEET, all of which set quantitative Doppler questions and the reflected-wall beat. AP Physics 2 treats Doppler only qualitatively, and IB frames it conceptually, so a pure numericals drill is aimed at the Indian boards — do the calculation-heavy problems here if that is your exam, and read for the ideas if it is not.

Origins · 1 minLord Rayleigh and how this was discovered

Lord Rayleigh

1842–1919 · Cambridge and Terling

John William Strutt, third Baron Rayleigh, is the author of the two-volume The Theory of Sound (1877–78), the book from which nearly every problem type on this page descends. Rayleigh set the modern wave analysis of sound on a rigorous footing — moving sources, moving observers, reflection, interference and beats all treated with the care that lets a student today write down $f(v+u)/(v-u)$ for a reflecting wall and trust it. He combined this theoretical reach with meticulous measurement: his pursuit of a tiny discrepancy in the density of nitrogen led, with William Ramsay, to the discovery of argon and the 1904 Nobel Prize in Physics. For the Doppler problem-solver, Rayleigh is the source of the systematic bookkeeping that makes these numericals routine rather than improvised.

History

1842

— Christian Doppler, in Prague, predicts that motion shifts the frequency of waves from a source, for light and for sound alike.

1845

— C. H. D. Buys Ballot confirms the pitch shift for sound with trumpeters on a moving railway carriage near Utrecht.

1848

— Hippolyte Fizeau independently extends the principle to light, the Doppler–Fizeau shift.

1877

— Lord Rayleigh's "The Theory of Sound" gives the systematic wave treatment of moving sources, observers and reflectors that underlies these problem types.

1940

s — Continuous-wave Doppler radar reads a target's speed from the double frequency shift of its reflected waves — the moving-reflector archetype in hardware.

Going further

Texts. NCERT Physics Class 11, Ch. 15 ("Waves"), the Doppler section and its exercises · H. C. Verma, Concepts of Physics Vol. 1, "Sound Waves" — especially the reflected-wall and moving-reflector problems · D. C. Pandey and similar JEE/NEET problem books for the full variant set · Lord Rayleigh, The Theory of Sound, for the parent analysis.

Provenance. Standard results. A verifier should recompute: the ambulance $640\times343/323 = 679.6$ , $640\times343/363 = 604.7$ , jump $74.9$ ; the cyclist $680\times360/340 = 720.0$ and $680\times320/340 = 640.0$ ; the wind case $500\times365/315 = 579.37$ against the no-wind $500\times360/310 = 580.65$ ; and the reflected wall $500\times340/338 = 502.96$ , $500\times342/338 = 505.917$ , with beat $2\times500\times2/338 = 5.917$ .

Prerequisites. The Doppler Effect in Sound — the concept, the master formula, and the wavefront derivation this gym drills. Leads to. Doppler Sign Discipline — the one procedure that makes the sign assignment automatic and fast · Echo, SONAR and Speed-of-Sound Problems — the sibling gym for reflection timing, where the round trip returns without a frequency shift.

Go deeper. The relativistic version of the moving-reflector beat — how a radar speed gun works with light rather than sound — belongs to special relativity, where the double shift is real but the formula changes because light has no medium. It is the natural sequel once you have this sound version cold.

How do I decide, quickly, which sign each speed takes?

Use one rule and one rule only: moving toward raises the pitch. Write the formula $f' = f(v \pm v_o)/(v \mp v_s)$ , then for the observer choose the numerator sign that makes an approach increase $f'$ (that is $+v_o$ ), and for the source choose the denominator sign that makes an approach increase $f'$ (that is $-v_s$ ). You never memorise four cases; you memorise "toward is higher" and let it pick the signs. After computing, check the direction of the shift against that rule — if an approaching object came out lower, a sign is flipped. This single discipline is drilled to reflex in the sign-craft technique article.

Why does the reflected wall shift the sound twice?

Because two separate Doppler events happen on the round trip. On the way out, the wall is a stationary observer receiving sound from a source moving toward it, so it hears a raised frequency $f_1$ — that is shift one. The wall then reflects, re-radiating $f_1$ as if it were a stationary source at the wall. On the way back, the original source is still moving toward the wall, so it is now an observer approaching that source and hears $f_1$ raised again — shift two. Chaining the two gives $f(v+u)/(v-u)$ , and the beat against the direct sound is $2fu/(v-u)$ . The factor of two is the signature of the two shifts.

Is the beat frequency the same as the reflected frequency?

No, and confusing them is a common slip. The reflected frequency is what the wall sends back — $505.917\ \mathrm{Hz}$ in the worked example — a full audible tone close to the original. The beat frequency is the difference between that reflected tone and the direct horn, $505.917 - 500 = 5.917\ \mathrm{Hz}$ — a slow throb you hear as the loudness pulsing several times a second, not as a pitch. The question usually asks for the beat, because that is what an instrument (or an ear) actually measures when two close tones overlap. Always read which one is wanted.

Which speed of sound should I use — 330, 340, or 343?

Whichever the problem gives you, and then keep it fixed for the whole solution. Many CBSE and ICSE problems use $330\ \mathrm{m/s}$ ; $340\ \mathrm{m/s}$ is the common room-temperature value; $343\ \mathrm{m/s}$ is the value at $20\ \mathrm{^\circ C}$ used in some texts and in the ambulance problem here. They differ because the speed of sound rises with temperature. The error to avoid is switching between them inside one problem — computing the approach with $343$ and the recession with $340$ silently corrupts the jump. Read the stated value, write it at the top, and use only that.

How do I include a wind without getting lost?

Do the wind first, before any signs. A wind blowing from the source toward the observer is carried by the whole medium, so it raises the sound speed over the ground to $v + w$ ; a headwind gives $v - w$ . Replace $v$ by that single effective speed everywhere in the formula, then assign the source and observer signs exactly as usual with all velocities measured relative to the ground. The mistake to avoid is adding the wind to only one term or treating it as another moving object — it is a property of the medium, so it enters the numerator and denominator together. Fix $v_{\text{eff}}$ at the top and the rest is a routine problem.

The emitted frequency is 640 Hz, but neither approach nor recession gives 640. Is that wrong?

No — that is exactly right, and it is a subtle point worth understanding. A stationary bystander only ever hears the true emitted frequency at the single instant the source is level with them, when its velocity is momentarily neither toward nor away. Before that they hear the raised approaching value ( $679.6\ \mathrm{Hz}$ ), and after it the lowered receding value ( $604.7\ \mathrm{Hz}$ ); the pitch steps essentially straight from one to the other. So the emitted $640\ \mathrm{Hz}$ is a value the moving source knows but the roadside listener barely catches. This is why "the frequency you hear" and "the frequency emitted" are genuinely different numbers in a Doppler problem.