Doppler effect — all cases - source moving, observer moving, both: Worked examples

Before we use the formula, let's build it from a picture, so it stops being a black box.

Building the formula from wavefronts

The heart of the Doppler effect is one question: how many crests hit your ear each second? Frequency heard means exactly that — crests per second.

Look at the picture below. On the left the source sits still; it paints crests as a set of evenly spaced rings, each one a distance $λ = v / f$ apart. On the right the source is moving: each new crest is born from a spot slightly ahead of the last, so the rings bunch up in front and spread out behind.

Figure s02 — two ring pictures side by side. Left: a still source with evenly spaced blue circles (spacing = wavelength). Right: a moving source whose circles bunch up (red) in front and spread out (green) behind. Caption asks: how many crests reach your ear per second?

Figure — Doppler effect — all cases - source moving, observer moving, both

Now let's turn each knob into a term.

The observer term (numerator). Suppose the rings are fixed with spacing $λ$ and travel through the air at speed $v$ . If you stand still, rings sweep past you at speed $v$ , so you meet $v / λ = f$ of them per second — no shift. If you now walk into them at speed $v_{o}$ , the rings approach you at the combined speed $v + v_{o}$ (your closing speed on something coming at you is the sum of the two speeds — see Relative velocity). You now meet $(v + v_{o}) / λ$ crests per second. The observer's speed adds to the closing speed, so it lands on top as $v + v_{o}$ .

Figure s03 — the observer term. Vertical blue crest lines with fixed spacing $λ$ drift right at $v$ ; an orange observer on the right runs left (into them) at $v_{o}$ ; a green box states the closing speed is $v + v_{o}$ , which becomes the numerator.

The source term (denominator). Now hold the observer still and move the source toward them at $v_{s}$ . In one period $T = 1/ f$ the source emits a crest, then travels $v_{s} T$ forward before emitting the next. So the next crest starts $v_{s} T$ closer than it otherwise would: the spacing in front shrinks from $λ = v T$ to $λ^{'} = v T - v_{s} T = (v - v_{s}) T = \frac{v - v _{s}}{f} .$ The rings still travel at $v$ (the medium, not the source, sets wave speed — see Wave speed in a medium). So you meet $v / λ^{'} = f \cdot v / (v - v_{s})$ crests per second. The source's forward motion shrinks the denominator to $v - v_{s}$ .

Figure s04 — the source term. The source at "emit 1" (orange) has moved a distance $v_{s} T$ to "emit 2" (red); the two crest fronts it painted are only $λ^{'} = (v - v_{s}) T$ apart (green double-arrow), which becomes the denominator.

Stacking them. The observer changes the closing speed (top); the source changes the spacing (bottom). Because these two knobs act on different parts of the ratio $\frac{closing speed}{spacing}$ — one on the numerator, one on the denominator — turning both at once simply replaces both parts at the same time: $f^{'} = \frac{closing speed of crests}{spacing you meet} = \frac{v + v _{o}}{( v - v _{s} ) / f} = f \frac{v + v _{o}}{v - v _{s}} .$ Why this step? We are allowed to substitute both corrected pieces into the single fraction $\frac{closing speed}{spacing}$ precisely because they are independent: the observer's motion never touches the ring spacing, and the source's motion never touches the closing speed. Independent effects multiply/divide into the same ratio without interfering, so we may plug in the observer-corrected numerator ( $v + v_{o}$ ) and the source-corrected denominator ( $(v - v_{s}) / f$ ) together. That is the master formula for approach. Reverse either motion and its sign flips — which gives the general form and, crucially, tells us why the sign rule works.

Formula The only formula on this page

$f^{'} = f \frac{v \pm v _{o}}{v \mp v _{s}}$

$f$ = true source frequency (crests emitted per second), $f^{'}$ = frequency heard.
$v$ = wave speed in the medium (still air $\approx 340$ m/s). The medium is the referee: all motion is measured against it.
$v_{o}$ = observer's speed, $v_{s}$ = source's speed.
Reading the $\pm$ and $\mp$ : the two symbols are chosen independently — the top sign is fixed only by what the observer does, the bottom sign only by what the source does. They are NOT locked together; you set each one on its own using the two lines below.
Top sign (observer, $v_{o}$ ): $+$ if the observer moves toward the source, $-$ if away.
Bottom sign (source, $v_{s}$ ): $-$ if the source moves toward the observer, $+$ if away.
One rule that produces both: choose each sign so that moving-together raises the pitch. On top, "toward" adds ( $+ v_{o}$ , bigger numerator). On the bottom, "toward" subtracts ( $- v_{s}$ , smaller denominator). Both raise $f^{'}$ , for opposite algebraic reasons.

The words "top" and "bottom" mean the numerator and denominator — the numbers above and below the dividing line. That is the entire vocabulary we need.

Recall Aside: what if the motion is

not head-on? (off-axis case) Everything above assumed motion straight along the source–observer line. If the observer or source moves at an angle, only the component along the line joining them does the squeezing.

Figure s05 — the off-axis angles. A source and observer sit at two dots joined by a dashed gray "line of sight." Each has a velocity arrow at an angle to that line: the source's velocity makes angle $φ$ with the line, the observer's makes angle $θ$ . The component along the line — $v_{s} cos φ$ for the source, $v_{o} cos θ$ for the observer — is drawn as a shorter arrow on the dashed line; the sideways (perpendicular) component is drawn faint, labelled "no Doppler shift."

Replace $v_{o} \to v_{o} cos θ$ and $v_{s} \to v_{s} cos φ$ , where (see the figure) $θ$ is the angle between the observer's velocity and the line joining the two, and $φ$ is the angle between the source's velocity and that same line. Head-on motion means the velocity points straight along the line, so $θ = 0$ (or $φ = 0$ ) and $cos = 1$ — the full speed counts. Purely sideways motion means the velocity is perpendicular, $θ = 9 0^{\circ}$ , so $cos 9 0^{\circ} = 0$ and there is no first-order shift. We defer the full off-axis treatment; on this page every example is head-on so $cos = 1$ .

The scenario matrix

Every Doppler problem is one row of the table below. The last column names the example that solves it.

#	Who moves?	Direction	What we expect	Solved in
A	Observer only	toward source	pitch up	Example 1
B	Observer only	away from source	pitch down	Example 2
C	Source only	toward observer	pitch up	Example 3
D	Source only	away from observer	pitch down	Example 4
E	Both	toward each other	biggest up	Example 5
F	Both	source flees, observer chases	small net down	Example 6
G	Zero / degenerate	nobody moves ( $v_{o} = v_{s} = 0$ )	no shift	Example 7
H	Real-world word problem	ambulance passing (before & after)	up then down	Example 8
I	Wind blowing	medium itself drifts	shifts $v$	Example 9
J	Limiting / exam twist	$v_{s} \to v$	formula breaks → shock wave	Example 10

For all examples we reuse: $f = 400$ Hz, $v = 340$ m/s, unless a problem says otherwise. Keeping the numbers constant lets you compare rows directly — a trick worth remembering for exams.

The map below collects the whole sign logic on one card — keep it in view while you work through the cells.

Figure s01 — the sign map. The master formula sits at the top; a blue arrow ties the numerator to the word OBSERVER (toward $\to +$ ), an orange arrow ties the denominator to SOURCE (toward $\to -$ ), a green box states the master rule "moving together raises pitch," and a red line marks the validity boundary $v_{s} < v$ .

The worked examples

Cell A — Observer only, moving toward

Worked example 1 — You sprint toward a parked horn

Horn at rest, $f = 400$ Hz, $v = 340$ m/s. You run toward it at $v_{o} = 20$ m/s.

Forecast: Source still, so wavelength is untouched. You are running into the crests, so you meet them faster → higher pitch. Guess a number between 400 and, say, 430 Hz.

Step 1 — Decide which speeds are non-zero. Source doesn't move: $v_{s} = 0$ , so the bottom is just $v$ . Why this step? Removing zero terms first stops sign mistakes and shows only the observer knob is active.

Step 2 — Pick the observer sign. Moving toward → pitch must rise → choose $+ v_{o}$ on top. Why this step? From the derivation, running into the rings adds to their closing speed ( $v + v_{o}$ ).

Step 3 — Substitute. $f^{'} = 400 \cdot \frac{340 + 20}{340} = 400 \cdot \frac{360}{340} = 423.5 Hz$ Why this step? Now it is pure arithmetic; the physics decisions are already made.

Verify: $423.5 > 400$ ✓ (pitch up, as forecast). Units: Hz on both sides since the $v$ -ratio is dimensionless ✓.

Cell B — Observer only, moving away

Worked example 2 — You run away from the same horn

Same horn, but now you flee at $v_{o} = 20$ m/s.

Forecast: You are outrunning the crests, so they catch you slower → pitch drops. Expect just under 400 Hz.

Step 1 — Zero terms. $v_{s} = 0$ , bottom is plain $v$ . Why this step? The source still isn't moving, so it can't touch the wavelength; only the observer knob is live, exactly as in Example 1.

Step 2 — Observer sign. Moving away lowers pitch → choose $- v_{o}$ on top. Why this step? Fleeing the rings subtracts from their closing speed ( $v - v_{o}$ ); it is the mirror image of Example 1.

Step 3 — Substitute. $f^{'} = 400 \cdot \frac{340 - 20}{340} = 400 \cdot \frac{320}{340} = 376.5 Hz$ Why this step? Plug the chosen sign into the observer-only formula; the physics is settled, only arithmetic remains.

Verify: $376.5 < 400$ ✓. Notice the up-shift in Ex 1 was $+ 23.5$ and the down-shift here is $- 23.5$ — equal in size for the observer case, because $v_{o}$ enters linearly on top. That symmetry is special to the observer; the source will break it (see Examples 3 & 4).

Cell C — Source only, moving toward

Worked example 3 — A train horn charges at you (you stand still)

Train sounds $f = 400$ Hz while moving toward you at $v_{s} = 30$ m/s. You are stationary.

Forecast: The source chases its own waves, squeezing the wavelength in front. Shorter wavelength at fixed wave speed → higher pitch. Expect a shift bigger than an observer moving at the same speed, because the denominator shrinks.

Step 1 — Zero terms. $v_{o} = 0$ , top is plain $v$ . Why this step? You aren't moving, so the closing speed of the rings is just their medium speed $v$ ; only the source knob is active.

Step 2 — Source sign. Source toward → pitch up → choose $- v_{s}$ on the bottom (smaller denominator = bigger fraction). Why this step? From the derivation, the source moving toward you shrinks the ring spacing to $(v - v_{s}) / f$ .

Step 3 — Substitute. $f^{'} = 400 \cdot \frac{340}{340 - 30} = 400 \cdot \frac{340}{310} = 438.7 Hz$

Verify: $438.7 > 400$ ✓. Compare to a hypothetical observer at 30 m/s: that would give $400 \cdot \frac{370}{340} = 435.3$ Hz. The source shift (438.7) is bigger — proof the two mechanisms differ. ✓

Cell D — Source only, moving away

Worked example 4 — The train has passed; horn recedes

Same train, $f = 400$ Hz, now moving away at $v_{s} = 30$ m/s; you stand still.

Forecast: Wavelength behind the source is stretched → lower pitch. Because $v_{s}$ is on the bottom, the drop should be smaller in size than the rise in Example 3 (division is not symmetric).

Step 1 — Zero terms. $v_{o} = 0$ , top plain $v$ . Why this step? You're stationary, so the rings sweep past at their medium speed $v$ ; only the source knob is turned.

Step 2 — Source sign. Away → pitch down → make the bottom bigger → $+ v_{s}$ . Why this step? Retreating, the source paints each crest farther than the last, stretching the spacing to $(v + v_{s}) / f$ .

Step 3 — Substitute. $f^{'} = 400 \cdot \frac{340}{340 + 30} = 400 \cdot \frac{340}{370} = 367.6 Hz$ Why this step? Insert the receding sign into the source-only formula; the larger denominator drives $f^{'}$ below 400.

Verify: $367.6 < 400$ ✓. Rise in Ex 3 was $+ 38.7$ ; drop here is only $- 32.4$ . Asymmetric, exactly because $v_{s}$ lives in the denominator. This is why an ambulance's pitch falls more suddenly as it approaches than the gentle sag afterwards.

Cell E — Both move, toward each other

Worked example 5 — Head-on: you run at the charging train

$f = 400$ , $v = 340$ , observer toward at $v_{o} = 20$ , source toward at $v_{s} = 30$ .

Forecast: Both effects raise pitch; they stack. Expect the largest shift so far — above 438.7 Hz of Example 3.

Step 1 — Both non-zero. Neither speed is zero, so both slots are used. Why this step? Both knobs are now turned, so we cannot drop either term.

Step 2 — Signs. Observer toward → $+ v_{o}$ (top). Source toward → $- v_{s}$ (bottom). Both chosen to raise pitch. Why this step? Closing-speed rises and spacing shrinks — the two mechanisms of the derivation act together.

Step 3 — Substitute. $f^{'} = 400 \cdot \frac{340 + 20}{340 - 30} = 400 \cdot \frac{360}{310} = 464.5 Hz$

Verify: $464.5$ exceeds both single-cause results (423.5 and 438.7) ✓. Sanity: shift $= + 64.5$ , close to the sum of the two individual shifts ( $23.5 + 38.7 = 62.2$ ) — they combine, with a small extra from the multiplication. ✓

Cell F — Both move, source flees while observer chases

Worked example 6 — You chase a fleeing siren

Source moves away at $v_{s} = 30$ ; you chase it (i.e. move toward the source) at $v_{o} = 20$ .

Forecast: Two opposing effects. Chasing raises pitch a bit; the source fleeing lowers it. Recession sits on the bottom and tends to dominate → expect a small drop below 400 Hz.

Step 1 — Both non-zero. Why this step? Both observer and source move, so both terms survive; we only need their directions.

Step 2 — Signs. Observer moving toward source → $+ v_{o}$ (top, raises). Source moving away → $+ v_{s}$ (bottom, lowers). Why this step? Careful: "toward source" is what defines the $+$ on top even though the source itself runs off. The direction of the observer's motion relative to the source fixes $v_{o}$ 's sign; the source's own direction fixes $v_{s}$ 's sign.

Step 3 — Substitute. $f^{'} = 400 \cdot \frac{340 + 20}{340 + 30} = 400 \cdot \frac{360}{370} = 389.2 Hz$ Why this step? The two effects land on opposite lines with opposite pitch influence; arithmetic decides which wins.

Verify: $389.2 < 400$ ✓ — small drop, as forecast, because the fleeing source (bottom) outweighs your chase (top). If you ran faster, $v_{o}$ could overtake and lift $f^{'}$ back above 400. Try $v_{o} = 50$ : $400 \cdot \frac{390}{370} = 421.6$ Hz > 400. ✓

Cell G — Degenerate: nobody moves

Cell H — Real-world word problem

Worked example 8 — Ambulance passing you: before and after

An ambulance siren sounds $f = 400$ Hz, driving at $v_{s} = 30$ m/s along a straight road. You stand on the pavement. What pitch do you hear as it approaches, and after it passes? (Take the moments when it is far up/down the road so its motion is nearly straight at you / straight away.)

Forecast: Approaching → high, receding → low, with a sudden fall as it passes. This is the classic "neeeee-oowww."

Step 1 — Approaching phase. Source toward, observer still. This is exactly Cell C: $f_{approach}^{'} = 400 \cdot \frac{340}{340 - 30} = 438.7 Hz$ Why this step? Far up the road the velocity is nearly straight at you, so $cos φ \approx 1$ and the head-on source-only formula applies.

Step 2 — Receding phase. Source away, observer still. Exactly Cell D: $f_{recede}^{'} = 400 \cdot \frac{340}{340 + 30} = 367.6 Hz$ Why this step? One physical event splits into two matrix cells — recognising that is the exam skill.

Step 3 — Size of the jump. $Δ f = 438.7 - 367.6 = 71.1 Hz$ Why this step? The audible drama is the difference between the two phases, so we subtract the receding pitch from the approaching one.

Verify: The pitch drops by about $71$ Hz "instantly" as the ambulance passes — a musical interval of roughly $438.7/367.6 \approx 1.19$ , close to a minor third. Matches the real siren swoop ✓. Note the true 400 Hz is never heard except at the single instant of closest approach (velocity toward you is momentarily zero there — the off-axis $cos = 0$ moment).

Cell I — Wind: the medium itself moves

Worked example 9 — A tailwind blows from source to observer

Source toward observer at $v_{s} = 30$ ; you stand still ( $v_{o} = 0$ ); a wind blows at $w = 10$ m/s from the source toward you.

Forecast: The air (the referee) drifts toward you, so crests ride the wind and effectively travel faster in the medium. Replace $v$ by $v + w$ everywhere it represents wave-in-medium speed. The frequency shift from the source should soften a little because a faster $v$ means $v_{s}$ is a smaller fraction of it.

Step 1 — Correct the wave speed. Wind toward observer adds to wave speed: $v_{eff} = v + w = 350$ m/s. Why this step? The medium sets the wave speed relative to the ground. If the whole body of air drifts toward you at $w$ , a crest riding on it travels at $v$ through the air plus $w$ of the air, so $v + w$ relative to the ground.

Step 2 — Keep $v_{o}$ and $v_{s}$ as ground speeds. Do not add $w$ to $v_{o}$ or $v_{s}$ . Why this step? Wind only carries the rings; it does not push the ambulance's wheels or your feet. The observer's and source's speeds were given relative to the ground, and the ground hasn't moved, so those numbers stay the same. Only the wave, which lives in the drifting air, feels the wind — that is why the $w$ correction attaches solely to $v$ .

Step 3 — Apply the master formula with $v_{eff}$ . Observer still, source toward: $f^{'} = 400 \cdot \frac{v _{eff}}{v _{eff} - v _{s}} = 400 \cdot \frac{350}{350 - 30} = 400 \cdot \frac{350}{320} = 437.5 Hz$ Why this step? With the corrected referee speed in place, the ordinary source-only formula does the rest.

Verify: Without wind (Example 3) we got $438.7$ Hz; with tailwind $437.5$ Hz — a slightly smaller shift, as forecast, because the larger $v_{eff}$ dilutes the ratio ✓.

Cell J — Limiting case: source reaches the speed of sound

Worked example 10 — What if

$v_{s} \to v$ ? The formula's breaking point Push the source speed up toward the wave speed: $v_{s} \to 340$ m/s, observer still.

Forecast: The denominator $v - v_{s} \to 0$ , so $f^{'} \to \infty$ . Physically infinite frequency is nonsense — something must break. It does: the wavefronts pile onto each other.

Step 1 — Watch the denominator shrink. For $v_{s} = 300, 330, 339$ : $f^{'} = 400 \cdot \frac{340}{40} = 3400, 400 \cdot \frac{340}{10} = 13600, 400 \cdot \frac{340}{1} = 136000 Hz$ Why this step? Numbers make the blow-up visceral: as $v_{s} \to v$ , the ring spacing $(v - v_{s}) / f \to 0$ , so $f^{'}$ rockets to infinity.

Step 2 — Interpret physically. At $v_{s} = v$ every crest is emitted at the same forward edge as the previous one arrived — they superpose into a single wall of pressure. That wall is a shock wave; for $v_{s} > v$ it forms the cone of a sonic boom. Why this step? The math says " $\infty$ "; the physics says the assumption (crests spread out as separate rings) no longer holds. See Sonic boom and shock waves.

Verify: $f^{'} \to \infty$ as $v_{s} \to v^{-}$ ✓, and the formula is simply invalid for $v_{s} \geq v$ — you must switch models. This is the one row of the matrix the master formula cannot fill; knowing that boundary is itself the answer.

Recall Self-test: name the cell

"You cycle away from a stationary bell." Which cell, and does pitch rise or fall? ::: Cell B — observer only, moving away → pitch falls ( $- v_{o}$ on top). "A drone flies straight over you sounding a tone." Which cells, in what order? ::: Cell C then Cell D (approach up, recede down) — the Example 8 story. "A jet at exactly Mach 1." Which cell, and what does the formula do? ::: Cell J — denominator $\to 0$ , $f^{'} \to \infty$ , formula invalid; shock wave forms.

Connections

Parent: Doppler — all cases — the derivation these examples exercise.
Wave speed in a medium — why wind changes $v$ (Cell I), not $v_{s}$ .
Wavelength and frequency relation — $λ = v / f$ behind every step.
Relative velocity — the observer-frame crest speed in Cells A, B, F.
Sonic boom and shock waves — the $v_{s} \geq v$ limit (Cell J).
Doppler effect of light — the symmetric, medium-free cousin.
Beats — what two of these shifted tones make together.