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

Before you start, one refresher so every symbol here is earned:

Build the formula from a picture (so nothing here is a black box)

Everything on this page rests on one equation, so let us earn it visually before testing it.

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

Observer alone (spacing fixed at $λ = v / f$ , arrival speed becomes $v + v_{o}$ ): $f^{'} = \frac{v + v _{o}}{λ} = \frac{v + v _{o}}{v / f} = f \frac{v + v _{o}}{v}$

Source alone — see s02. In one period $T$ the source moves $v_{s} T$ toward you before releasing the next crest, so each new crest starts $v_{s} T$ nearer than the last: $λ^{'} = λ - v_{s} T = \frac{v}{f} - \frac{v _{s}}{f} = \frac{v - v _{s}}{f}, f^{'} = \frac{v}{λ ^{'}} = f \frac{v}{v - v _{s}}$

That single line is what every question below probes.

Oblique motion — the general case (why $cos θ$ appears)

The master formula assumed everyone moves straight along the line joining source and observer. What if a car drives past on a road offset from you?

Intuition Only the toward-component counts

In s03 the source velocity $v_{s}$ points along the road, but the crests that reach you only care about how fast the source closes the line of sight to your ear. Using $θ_{s}$ (the angle between the source's velocity and that line of sight, as defined above), that closing rate is the projection $v_{s} cos θ_{s}$ . Likewise, if the observer moves at an angle $θ_{o}$ to the line of sight, only $v_{o} cos θ_{o}$ of its speed changes the crest-arrival rate. Anything sideways neither compresses nor stretches the crests heading your way. $f^{'} = f \frac{v + v _{o} c o s θ _{o}}{v - v _{s} c o s θ _{s}}$ Head-on means $θ_{s} = θ_{o} = 0$ , $cos 0 = 1$ → full master formula. Straight across (the instant of closest approach) means the relevant angle is $9 0^{\circ}$ , $cos 9 0^{\circ} = 0$ → no shift at that instant, even at high speed. Every "perpendicular = no shift" claim below is this $cos 9 0^{\circ} = 0$ .

Why the formula dies at $v_{s} \geq v$ — the Mach cone

Intuition Crests piling into a wall

Trace s04. Each dot is where the source was when it emitted a still-expanding circular crest. When $v_{s} < v$ (left) the circles nest neatly and a listener ahead simply gets a high, finite $f^{'}$ . When $v_{s} = v$ (middle) every crest touches the source's current position — they all pile onto one flat front, so $λ^{'} \to 0$ and the formula's denominator $v - v_{s} \to 0$ , forcing $f^{'} \to \infty$ . That divergence is the maths screaming that ordinary Doppler no longer applies. When $v_{s} > v$ (right) the crests can never get ahead; their common tangent is a cone. $α$ is the half-angle of that cone — the angle between the source's direction of travel and the cone's edge, marked in s04. Why $sin α = v / v_{s}$ : in a time $t$ a crest emitted at the start has spread outward a distance $v t$ (its radius), while the source has advanced $v_{s} t$ along its path. Those two distances form a right triangle — the crest's radius $v t$ is the side opposite the half-angle $α$ , and the source's travel $v_{s} t$ is the hypotenuse (from the emission point to the source's current position). So $sin α = opposite / hypotenuse = v t / (v_{s} t) = v / v_{s}$ . This is the Mach cone of a sonic boom. Beyond this speed you must abandon $f^{'} = f v / (v - v_{s})$ entirely.

True or false — justify

Motion of the source changes the wavelength in the medium

True. The source travels

v_{s} T

between emitting crests, so successive crests are born closer together (in front) or farther apart (behind):

λ^{'} = (v \mp v_{s}) / f

Motion of the observer changes the wavelength in the medium

False. The observer doesn't touch the crests already laid down;

λ

is fixed. The observer only changes the speed at which those crests sweep past,

v \pm v_{o}

A moving source changes the actual speed of the sound waves in the air

False. The medium sets the wave speed

v

; the source only sets where each crest is emitted from. Speed stays

v

, spacing changes.

Source approaching at

30

m/s gives the same

f^{'}

as observer approaching at

30

m/s

False. Different mechanisms — source sits in the denominator (wavelength), observer in the numerator (arrival speed). They only approximately agree at low speeds.

If both source and observer move toward each other, the pitch shift is larger than either effect alone

True. Numerator grows (

+ v_{o}

) and denominator shrinks (

- v_{s}

), so the fraction climbs more than either change by itself.

When source and observer move together at the same velocity, there is no Doppler shift

True. The distance between them never changes, so crests neither pile up nor stretch;

f^{'} = f

The Doppler formula for sound is symmetric between source and observer, just like for light

False. Sound has a medium as referee, so swapping roles gives different numbers. Light has no medium and depends only on relative velocity — Doppler effect of light is symmetric.

If the observer moves away faster than the wave speed, they hear nothing

True. Crests can never catch up to an observer fleeing at

v_{o} > v

; the numerator

v - v_{o}

goes negative, which physically means no wavefronts ever reach them.

As a source's speed approaches the wave speed, the heard frequency approaches infinity

True. Denominator

v - v_{s} \to 0

, so

f^{'} \to \infty

: all crests pile onto one front — the onset of a shock wave, see Sonic boom and shock waves.

A tailwind blowing from source to observer (

w > 0

) raises the pitch heard

False. Wind shifts

v \to v + w

in both numerator and denominator, so for a still source and observer the factor cancels and pitch is unchanged. Wind mostly bends and delays sound, not its pitch.

Spot the error

"Observer runs away from a still source at

v_{o}

, so

f^{'} = f \frac{v}{v - v _{o}}

Wrong line. The observer's speed is always in the numerator:

f^{'} = f \frac{v - v _{o}}{v}

. Putting

v_{o}

in the denominator is the source's slot.

"Source recedes at

v_{s}

, so use

- v_{s}

f^{'} = f \frac{v}{v - v _{s}}

Sign is backwards. Receding must lower pitch, so it needs

+ v_{s}

f^{'} = f \frac{v}{v + v _{s}}

. Always sanity-check against 'approach raises, recession lowers'.

"With wind speed

w

toward the observer, use

v \to v + w

but also change

v_{o}

v_{o} + w

Only the wave speed

v

absorbs the wind. The source and observer speeds

v_{s}, v_{o}

are measured relative to the ground and stay as-is; double-counting the wind is the trap.

"Source moving at twice the sound speed just gives a very high

f^{'}

from the formula."

The formula is invalid for

v_{s} \geq v

. The denominator turns negative/zero — physically the source outruns its own crests and forms a Mach cone (s04); no ordinary Doppler frequency exists.

"Source and observer both moving right at the same speed, so plug

+ v_{o}

top and

- v_{s}

bottom."

The signs must reflect approach, not just motion. Moving the same direction at equal speed means neither approaches the other: this is receding-from-behind for one and approaching-from-front for the other — they cancel to

f^{'} = f

, not a shift.

"Doppler shift depends on the distance between source and observer."

It depends on their relative velocity along the line of sight, not the separation. A source at any distance moving toward you at

v_{s}

gives the same

f^{'}

"A car passing on a nearby road shifts steadily as it goes by."

No — the shift depends on

cos θ_{s}

(s03). It is high while approaching, passes smoothly through zero at closest approach (

θ_{s} = 9 0^{\circ}

), then goes low. The change is a continuous swoop, not a step.

Why questions

Why is

v_{o}

in the numerator but

v_{s}

in the denominator?

They act on different quantities: the observer changes the crest-arrival speed (top of

f^{'} = speed / λ

), the source changes the wavelength

λ

(bottom). Different physics, different line.

Why does the medium — not the source — decide the wave speed

v

A wave is a disturbance passing through the medium's particles; how fast that disturbance travels depends on the medium's stiffness and density, not on how the source moved. See Wave speed in a medium.

Why does moving the source toward you shorten the wavelength?

In one period

T

the source advances

v_{s} T

toward you before emitting the next crest, so the gap between crests shrinks by exactly

v_{s} T

λ^{'} = λ - v_{s} T

(see s02).

Why can a moving observer never change the wavelength already in the air?

The crests are fixed in space once emitted; the observer merely samples them faster or slower. Wavelength is a property of the wave pattern, independent of who looks at it. This rests on Wavelength and frequency relation.

Why does an ambulance siren drop in pitch as it passes you?

Approaching, the toward-component

v_{s} cos θ_{s}

compresses wavelength (higher

f^{'}

); as it passes,

cos θ_{s}

falls smoothly through zero, so

f^{'}

glides down through the true

f

; receding, the component reverses and stretches wavelength (lower

f^{'}

). The pitch changes continuously, sweeping from high to low with no instant jump — it merely changes fastest right near the passing point.

Why does only

v_{s} cos θ_{s}

matter for an off-axis source?

Only the component of velocity along the line of sight compresses the crests heading toward you; the perpendicular part slides the source sideways without changing crest spacing on your line (s03).

Why do sound and light Doppler formulas differ in structure?

Sound needs a medium as an absolute referee, so source and observer motions enter asymmetrically. Light has no medium — only relative velocity matters (and relativity enters), making it symmetric; contrast with Doppler effect of light.

Why is 'relative velocity' not enough to get the sound Doppler shift right?

Because two motions with the same relative velocity (source-moves vs observer-moves) give different

λ

-vs-arrival-speed effects. You must know the velocity of each relative to the medium, per Relative velocity.

Why do two nearby Doppler-shifted sirens produce a throbbing sound?

Their slightly different heard frequencies interfere, producing a slow amplitude pulsation at the difference frequency — that is Beats.

Edge cases

Source and observer both stationary

v_{s} = v_{o} = 0

, so

f^{'} = f

. No motion through the medium means no compression and no arrival-speed change.

Observer moving exactly at the wave speed toward the source

f^{'} = f (v + v) / v = 2 f

. Crests arrive twice as often because the observer sweeps into them at double speed — perfectly finite, unlike the source case.

Observer fleeing at exactly the wave speed (

v_{o} = v

)

f^{'} = f (v - v) / v = 0

. Crests exactly keep pace and never overtake; the observer meets no wavefronts and hears silence.

Source approaching at exactly the wave speed (

v_{s} = v

)

Denominator

\to 0

f^{'} \to \infty

. All crests emitted onto one wall of air — this is the shock front, boundary of Sonic boom and shock waves (s04).

Source velocity perpendicular to the line of sight

At that instant

cos θ_{s} = 0

, so

f^{'} = f

momentarily (transverse case for sound), even though the source is moving fast. Only the along-the-line speed enters.

Both move toward each other at equal speed

u

f^{'} = f \frac{v + u}{v - u} > f

: still a net rise, and larger than either moving alone, because top and bottom both push

f^{'}

up.

Observer chasing a faster-receding source

If source is ahead and recedes at

v_{s}

while you chase at

v_{o} < v_{s}

f^{'} = f \frac{v + v _{o}}{v + v _{s}} < f

: recession still wins, a small drop, since you can't quite close the gap.

Supersonic source,

v_{s} > v

The Mach cone forms (s04) with half-angle

sin α = v / v_{s}

; the ordinary formula is void — a listener hears nothing until the shock front arrives, then a boom.

Wind blowing from observer to source (

w < 0

, headwind for the sound)

Replace

v \to v + w = v - ∣ w ∣

in both lines. For a still source and observer the factor still cancels — no pitch change — but the effective wave speed is now slower, referenced to air drifting against the sound.

Build the formula from a picture (so nothing here is a black box)

Oblique motion — the general case (why cosθ appears)

Why the formula dies at vs​≥v — the Mach cone

True or false — justify

Spot the error

Why questions

Edge cases

Oblique motion — the general case (why $cos θ$ appears)

Why the formula dies at $v_{s} \geq v$ — the Mach cone