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

Throughout, unless told otherwise, take the speed of sound in still air as $v = 340$ m/s.

Formula Unpacking the

$\pm$ and $\mp$ — a sign legend The two stacked symbols are not independent guesses. They are a compact way of writing four situations at once. Read them like this:

The top symbol in $\pm$ (i.e. $+ v_{o}$ ) and the top symbol in $\mp$ (i.e. $- v_{s}$ ) both correspond to approach.
The bottom symbol in each ( $- v_{o}$ , $+ v_{s}$ ) both correspond to recession.

The one rule that generates every sign: choose whichever sign makes $f^{'}$ larger when things approach, smaller when they separate. Observer speed $v_{o}$ lives on top (numerator); source speed $v_{s}$ lives on the bottom (denominator). The chart below spells out all four combinations — see the figure

Motion	Observer $v_{o}$ (top)	Source $v_{s}$ (bottom)	Effect on $f^{'}$
toward	$+ v_{o}$	$- v_{s}$	raises pitch
away	$- v_{o}$	$+ v_{s}$	lowers pitch

So "observer toward, source away" reads $f^{'} = f \frac{v + v _{o}}{v + v _{s}}$ : $+$ on top (approach), $+$ on bottom (recede).

Level 1 — Recognition

Can you spot which speed is which, and which sign each takes?

L1.1 — Name the mechanism

A car horn stands still and sounds at $400$ Hz. You cycle toward it. Does the pitch you hear go up or down, and is it the wavelength that changes or the crest-arrival speed?

Recall Solution

Up. You are the observer, and the source is still, so the wavelength floating in the air is untouched — it is the crest-arrival speed that changes. Running into the crests, you meet them faster, so more crests hit your ear per second → higher $f^{'}$ . In the formula, source still means denominator is plain $v$ ; toward means $+ v_{o}$ on top: $f^{'} = f \frac{v + v _{o}}{v} > f .$

L1.2 — Pick the signs

A train sounds its whistle and moves away from a stationary listener at $v_{s} = 25$ m/s. Write the correct signed formula for $f^{'}$ (don't compute yet).

Recall Solution

Observer still ⇒ top is plain $v$ . Source away ⇒ it stretches the wavelength, lowering pitch, so we need the sign that makes $f^{'}$ smaller: that is $+ v_{s}$ in the denominator. $f^{'} = f \frac{v}{v + v _{s}} .$ Check the logic: a bigger denominator means smaller $f^{'}$ = lower pitch, exactly what receding should do. ✓

Level 2 — Application

Plug into the one master formula with correct signs.

L2.1 — Observer toward still siren

A siren fixed to a pole emits $f = 680$ Hz. You drive toward it at $v_{o} = 34$ m/s. Find $f^{'}$ .

Recall Solution

Source still, observer toward ⇒ $+ v_{o}$ on top: $f^{'} = 680 \cdot \frac{340 + 34}{340} = 680 \cdot \frac{374}{340} = 680 \cdot 1.1 = 748 Hz .$ Higher than $680$ ✓ (you are approaching). Compare the wavefront picture

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

: the observer runs into crests, meeting them faster.

L2.2 — Source toward still listener

An ambulance siren at $f = 660$ Hz drives toward you at $v_{s} = 30$ m/s while you stand still. Find $f^{'}$ .

Recall Solution

Observer still ⇒ plain $v$ on top. Source toward ⇒ wavelength squeezed ⇒ $- v_{s}$ in denominator: $f^{'} = 660 \cdot \frac{340}{340 - 30} = 660 \cdot \frac{340}{310} = 660 \cdot 1.09677 \dots = 723.9 Hz .$ The figure

shows the crests bunched ahead of the moving source — a physically shorter wavelength.

L2.3 — Observer moving AWAY from a still source

A loudspeaker fixed to a wall emits $f = 500$ Hz. You walk away from it at $v_{o} = 17$ m/s. Find $f^{'}$ .

Recall Solution

Source still ⇒ denominator plain $v$ . Observer away ⇒ you run with the crests, so they overtake you more slowly ⇒ $- v_{o}$ on top (recession lowers pitch): $f^{'} = f \frac{v - v _{o}}{v} = 500 \cdot \frac{340 - 17}{340} = 500 \cdot \frac{323}{340} = 500 \cdot 0.95 = 475 Hz .$ Lower than $500$ ✓. Notice this is the exact mirror of L2.1 — same mechanism (crest-arrival speed), opposite sign.

L2.4 — Source receding

An ambulance ( $f = 660$ Hz, $v_{s} = 30$ m/s) has now passed you and is driving away; you are still stationary. Find $f^{'}$ .

Recall Solution

Receding source stretches the wavelength ⇒ $+ v_{s}$ in denominator: $f^{'} = 660 \cdot \frac{340}{340 + 30} = 660 \cdot \frac{340}{370} = 660 \cdot 0.918918 \dots = 606.5 Hz .$ The full "pass-by" drop is $723.9 \to 606.5$ Hz — the falling wail you hear as it goes by.

Level 3 — Analysis

Both moving, sign reasoning under pressure, and reflected/relayed sound.

L3.1 — Both approaching

A train ( $f = 500$ Hz) moves toward a cyclist at $v_{s} = 30$ m/s; the cyclist rides toward the train at $v_{o} = 10$ m/s. Find $f^{'}$ .

Recall Solution

Both motions are "toward," both raise the pitch ⇒ $+ v_{o}$ top, $- v_{s}$ bottom: $f^{'} = 500 \cdot \frac{340 + 10}{340 - 30} = 500 \cdot \frac{350}{310} = 500 \cdot 1.129032 \dots = 564.5 Hz .$ Bigger shift than either alone would give ✓.

L3.2 — Chasing a fleeing source

A police car ( $f = 700$ Hz) speeds away from you at $v_{s} = 40$ m/s. You chase it (moving toward it) at $v_{o} = 25$ m/s. Find $f^{'}$ and say whether the pitch rises or falls.

Recall Solution

You move toward the source ⇒ $+ v_{o}$ on top. The source moves away ⇒ $+ v_{s}$ in the denominator. $f^{'} = 700 \cdot \frac{340 + 25}{340 + 40} = 700 \cdot \frac{365}{380} = 700 \cdot 0.960526 \dots = 672.4 Hz .$ Falls slightly. Forecast check: you are chasing but not quite catching a fleeing source, so recession just wins — a small drop, exactly what we got ✓.

L3.3 — Echo off a moving wall

A stationary horn emits $f = 500$ Hz. A wall moves toward the horn at speed $u = 20$ m/s. Find the frequency of the echo returning to the (stationary) horn.

Trick: the wall first hears a Doppler-shifted frequency (it plays the role of a moving observer, so here $v_{o} = u$ ), then re-radiates it as a moving source (now $v_{s} = u$ ). Do it in two hops.

Recall Solution

Hop 1 — the wall is a moving observer approaching the still horn, so its observer speed is $v_{o} = u$ : $f_{1} = f \frac{v + v _{o}}{v} = f \frac{v + u}{v} = 500 \cdot \frac{360}{340} .$ Hop 2 — the wall now acts as a moving source (emitting $f_{1}$ ) approaching the still horn, so its source speed is $v_{s} = u$ : $f^{'} = f_{1} \frac{v}{v - v _{s}} = f_{1} \frac{v}{v - u} = 500 \cdot \frac{v + u}{v} \cdot \frac{v}{v - u} = 500 \cdot \frac{v + u}{v - u} .$ $f^{'} = 500 \cdot \frac{360}{320} = 500 \cdot 1.125 = 562.5 Hz .$ Notice the clean result: for a reflector approaching a still source, the shift is $\frac{v + u}{v - u}$ (both effects stack, once for hearing, once for re-emitting). Here $u$ is just the wall's ground speed, which plays the role of $v_{o}$ in hop 1 and $v_{s}$ in hop 2.

Level 4 — Synthesis

Combine Doppler with wind, with two-tone beats, and solve backwards.

L4.1 — Doppler with a tailwind AND a headwind

A source ( $f = 400$ Hz) is stationary; the observer moves toward it at $v_{o} = 20$ m/s. Still-air $v = 340$ m/s. (a) A wind of $w = 10$ m/s blows from source to observer (toward the observer). Find $f^{'}$ . (b) Now the same wind blows from observer to source (a headwind, $w = - 10$ m/s). Find $f^{'}$ .

Recall Solution

Wind shifts the effective wave speed everywhere: $v \to v + w$ , where $w$ is positive if the air drifts toward the observer and negative if it drifts toward the source. See Relative velocity for why the crest speeds add. Source still ⇒ effective- $v$ on top; observer toward ⇒ $+ v_{o}$ : $f^{'} = f \frac{( v + w ) + v _{o}}{v + w} .$ (a) Tailwind, $w = + 10$ : $f^{'} = 400 \cdot \frac{350 + 20}{350} = 400 \cdot \frac{370}{350} = 422.9 Hz .$ (b) Headwind, $w = - 10$ : $f^{'} = 400 \cdot \frac{330 + 20}{330} = 400 \cdot \frac{350}{330} = 424.2 Hz .$ (Compare no-wind $f^{'} = 400 \cdot 360/340 = 423.5$ Hz. The tailwind inflates both top and bottom, shrinking the fractional shift; the headwind deflates both, enlarging it. The sign of $w$ is the whole story — it never touches $v_{o}$ .)

L4.2 — Beats from an approaching + receding pair

Two identical horns each emit $f = 340$ Hz. One moves toward you at $v_{s} = 2$ m/s, the other moves away from you at $v_{s} = 2$ m/s. You are stationary. How many beats per second do you hear? (See Beats.)

Recall Solution

Approaching horn (source toward ⇒ $- v_{s}$ in denominator): $f_{1} = 340 \cdot \frac{340}{340 - 2} = 340 \cdot \frac{340}{338}$ . Receding horn (source away ⇒ $+ v_{s}$ ): $f_{2} = 340 \cdot \frac{340}{340 + 2} = 340 \cdot \frac{340}{342}$ . $f_{1} = 342.0118 \dots Hz, f_{2} = 337.9532 \dots Hz .$ Beat frequency = $∣ f_{1} - f_{2} ∣$ : $f_{beat} = 342.0118 - 337.9532 = 4.06 Hz .$ So about 4 beats per second — a slow throbbing.

L4.3 — Solve for the unknown speed

A stationary listener hears a horn's true $f = 500$ Hz shifted to $f^{'} = 540$ Hz as it approaches. Find the source speed $v_{s}$ .

Recall Solution

Observer still, source approaching: $f^{'} = f \frac{v}{v - v _{s}} \Rightarrow v - v_{s} = \frac{f v}{f ^{'}} = \frac{500 \cdot 340}{540} .$ $v - v_{s} = 314.8148 \dots m/s \Rightarrow v_{s} = 340 - 314.8148 = 25.19 m/s .$ So the horn moves at about $25.2$ m/s toward the listener.

Level 5 — Mastery

The degenerate and limiting cases — where the formula strains or breaks.

L5.1 — The forbidden speed

A jet carries a $1000$ Hz siren and flies toward a stationary observer. Compute $f^{'}$ at $v_{s} = 300$ m/s, then $v_{s} = 339$ m/s, then explain what happens at $v_{s} = 340$ m/s and beyond.

Recall Solution

Source approaching, observer still: $f^{'} = f \frac{v}{v - v _{s}}$ .

At $v_{s} = 300$ : $f^{'} = 1000 \cdot \frac{340}{40} = 8500 Hz$ .
At $v_{s} = 339$ : $f^{'} = 1000 \cdot \frac{340}{1} = 340000 Hz$ . As $v_{s} \to v = 340$ , the denominator $\to 0$ and $f^{'} \to \infty$ : every crest is emitted essentially at the same place as the previous one arrives, so all wavefronts pile onto a single front. That front is the shock wave / sonic boom — the Doppler formula is no longer valid because there is no longer a well-defined wavelength ahead of the source. For $v_{s} > v$ the source outruns its own sound and the ordinary formula gives a negative frequency (physically meaningless). See Sonic boom and shock waves.

L5.2 — Two symmetric-looking cases that aren't equal

True $f = 400$ Hz, $v = 340$ m/s, one speed of $40$ m/s. (a) Observer moves toward a still source at $40$ m/s. (b) Source moves toward a still observer at $40$ m/s. Compute both $f^{'}$ and explain, physically, why they differ even though "the gap closes at $40$ m/s" in both.

Recall Solution

(a) $f^{'} = 400 \cdot \frac{340 + 40}{340} = 400 \cdot \frac{380}{340} = 447.06 Hz .$ (b) $f^{'} = 400 \cdot \frac{340}{340 - 40} = 400 \cdot \frac{340}{300} = 453.33 Hz .$ Why different: the observer case is additive — it just speeds up crest arrival, top line. The source case is multiplicative on the wavelength — it physically shrinks the spacing, bottom line. Algebraically $\frac{v + u}{v}$ vs $\frac{v}{v - u}$ agree only to first order in $u / v$ ; the source term has extra "compounding" (the shrinking is repeated for every crest), so it always shifts a little more. This asymmetry is exactly why the acoustic Doppler effect (a medium exists) is not symmetric, unlike Doppler effect of light, where no medium exists and only relative velocity matters.

L5.3 — Limiting sanity check (both speeds → 0)

Show that when $v_{o} \to 0$ and $v_{s} \to 0$ , the master formula returns $f^{'} = f$ , and estimate the fractional shift for small equal speeds $v_{o} = v_{s} = u ≪ v$ (both toward each other).

Recall Solution

Directly, $f^{'} = f \frac{v + 0}{v - 0} = f$ — no motion, no shift ✓. For small $u$ (both toward): $f^{'} = f \frac{v + u}{v - u}$ . Using $\frac{v + u}{v - u} \approx (1 + \frac{u}{v}) (1 + \frac{u}{v}) \approx 1 + \frac{2 u}{v}$ , the fractional shift is $\frac{f ^{'} - f}{f} \approx \frac{2 u}{v} .$ Concretely at $u = 5$ m/s: $\frac{2 \cdot 5}{340} = 0.0294 = 2.94%$ . So two objects closing at a combined $10$ m/s produce roughly a $3%$ pitch rise — a clearly audible shift.

The two-hop echo, visually

Recall Why the echo shift is

$\frac{v + u}{v - u}$ Look at the figure: the wall hears an approaching horn (moving observer, factor $\frac{v + u}{v}$ ), then re-emits what it heard while itself approaching the horn (moving source, factor $\frac{v}{v - u}$ ). Multiply the two hops and the $v$ 's cancel: $\frac{v + u}{v} \cdot \frac{v}{v - u} = \frac{v + u}{v - u}$ .

Connections

Wave speed in a medium — sets $v$ and, with wind, $v + w$ .
Wavelength and frequency relation — the $λ = v / f$ behind every step, i.e. the crest-arrival spacing.
Relative velocity — crest-arrival speeds and wind corrections.
Beats — the throbbing tone of L4.2.
Sonic boom and shock waves — the L5.1 breakdown.
Doppler effect of light — the symmetric, medium-free cousin (L5.2).