Both source and observer moving — when is there no apparent change

Question

Both the source of sound and the observer are moving. Under what condition is there no apparent change in frequency (no Doppler effect)?

Solution — Step by Step

When both source and observer move, the apparent frequency is:

f' = f \cdot \frac{v \pm v_o}{v \mp v_s}

Where:

$f$ = actual frequency of source
$v$ = speed of sound
$v_o$ = speed of observer
$v_s$ = speed of source
Upper sign (+ numerator, − denominator): observer moving toward source / source moving toward observer
Lower sign: moving away

For no apparent change in frequency: $f' = f$

f = f \cdot \frac{v \pm v_o}{v \mp v_s}

1 = \frac{v \pm v_o}{v \mp v_s}

v \mp v_s = v \pm v_o

This gives: $\pm v_o = \mp v_s$ , which simplifies to $v_o = v_s$

But we must also consider the direction of motion.

For $f' = f$ , we need:

\frac{v + v_o}{v + v_s} = 1 \quad \text{or} \quad \frac{v - v_o}{v - v_s} = 1

Both give $v_o = v_s$ . But the physical situation that makes this meaningful is:

The source and observer move in the same direction with the same speed.

When both move in the same direction at the same speed, their relative velocity is zero — the distance between them doesn’t change, so no Doppler effect occurs.

There is no apparent change in frequency when the source and observer move in the same direction with the same speed.

The key insight: Doppler effect depends on the relative velocity between source and observer, not their individual speeds. If their relative velocity is zero (moving together), the frequency heard equals the frequency emitted.

Answer: When both source and observer move with the same velocity (same speed AND same direction), $f' = f$ — no Doppler shift.

Why This Works

The Doppler effect occurs because the wavelength (spacing of wavefronts) is compressed or stretched depending on relative motion. When a source moves toward you, wavefronts pile up (shorter wavelength, higher frequency). When you move toward the source, you intercept wavefronts faster (higher frequency).

When both move together at the same speed and direction, neither of these effects occurs. The source isn’t catching up to wavefronts, and the observer isn’t intercepting them faster. The situation is identical to both being stationary in a frame moving with them.

Alternative Method — Reference Frame Argument

Transform to the reference frame moving with the observer at speed $v_o$ . In this frame, the observer is stationary. The source’s speed relative to the observer is $v_s - v_o$ (if moving in the same direction).

For $f' = f$ : relative speed of source = 0, so $v_s - v_o = 0$ , giving $v_s = v_o$ .

This confirms: equal speeds in the same direction means no Doppler shift.

Common Mistake

Thinking that any situation with both moving cancels out. The speeds must be equal AND directions must be the same. If source moves at 30 m/s north and observer moves at 30 m/s south, they’re approaching each other — massive Doppler shift, not zero. The condition is same speed in same direction (zero relative velocity).

Question

Solution — Step by Step

Why This Works

Alternative Method — Reference Frame Argument

Common Mistake

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