9.1 Doppler effect for stationary and moving sources

Understanding the Doppler Effect

The Doppler effect describes how the observed frequency of a sound wave changes when the source and the observer are moving relative to each other. It's the reason an ambulance siren sounds higher-pitched as it approaches you and lower-pitched after it passes. Beyond everyday experience, this effect is central to technologies like radar speed detection, weather tracking, and medical ultrasound.

Doppler Effect and Sound Perception

Sound travels outward from a source in all directions as a series of wavefronts. When the source and observer are both stationary, the observer receives wavefronts at the same rate they're emitted, so the perceived frequency matches the emitted frequency.

Once relative motion enters the picture, that changes:

• If the source and observer are moving toward each other, wavefronts arrive more frequently, and the perceived frequency increases (higher pitch).
• If they're moving apart, wavefronts arrive less frequently, and the perceived frequency decreases (lower pitch).

The size of the frequency shift depends on how fast the source or observer is moving relative to the speed of sound in the medium. A car traveling at 30 m/s produces a much smaller shift than a jet at 300 m/s.

Frequency Calculation for a Moving Observer

When the observer moves but the source is stationary, use:

$f' = f \frac{v \pm v_o}{v}$

• $f'$ = observed (apparent) frequency
• $f$ = frequency emitted by the source
• $v$ = speed of sound in the medium (about 343 m/s in air at 20°C)
• $v_o$ = speed of the observer

Sign convention: Use + when the observer moves toward the source, and − when the observer moves away.

The logic here is straightforward. An observer moving toward the source encounters wavefronts faster than they're being emitted, so the effective wave speed relative to the observer increases. Moving away has the opposite effect.

Example: A stationary siren emits sound at 500 Hz. You're driving toward it at 30 m/s. The speed of sound is 343 m/s.

$f' = 500 \times \frac{343 + 30}{343} = 500 \times \frac{373}{343} \approx 544 \text{ Hz}$

You'd hear a pitch noticeably higher than the actual 500 Hz tone.

Frequency Calculation for a Moving Source

When the source moves but the observer is stationary, the formula changes:

$f' = f \frac{v}{v \mp v_s}$

• $v_s$ = speed of the source

Sign convention: Use − when the source moves toward the observer, and + when it moves away.

Why is the source velocity in the denominator instead of the numerator? A moving source physically compresses the wavefronts ahead of it (shorter wavelength, higher frequency) and stretches them behind it (longer wavelength, lower frequency). This is different from a moving observer, who simply intercepts the same wavefronts at a different rate.

Example: A car horn emits sound at 400 Hz while the car approaches a pedestrian at 25 m/s. Speed of sound is 343 m/s.

$f' = 400 \times \frac{343}{343 - 25} = 400 \times \frac{343}{318} \approx 431 \text{ Hz}$

After the car passes and moves away at the same speed:

$f' = 400 \times \frac{343}{343 + 25} = 400 \times \frac{343}{368} \approx 373 \text{ Hz}$

That jump from 431 Hz to 373 Hz is the characteristic pitch drop you hear as a vehicle passes.

Doppler Effect with Both Source and Observer Moving

When both the source and observer are in motion, combine everything into one formula:

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

The sign rules stay the same as before:

• Numerator (observer): + if moving toward the source, − if moving away
• Denominator (source): − if moving toward the observer, + if moving away

A helpful way to remember: any motion that closes the gap between source and observer increases the observed frequency. Any motion that opens the gap decreases it. In the numerator, closing the gap means adding $v_o$. In the denominator, closing the gap means subtracting $v_s$ (which makes the denominator smaller and the whole fraction larger).

Steps for solving combined Doppler problems:

1. Identify who is the source and who is the observer.
2. Determine each one's speed and direction of motion.
3. Decide the sign for $v_o$: + toward, − away.
4. Decide the sign for $v_s$: − toward, + away.
5. Plug values into the combined formula and solve.

Example: An ambulance (source) emitting a 700 Hz siren moves toward you at 35 m/s. You're driving toward the ambulance at 20 m/s. Speed of sound is 343 m/s.

$f' = 700 \times \frac{343 + 20}{343 - 35} = 700 \times \frac{363}{308} \approx 825 \text{ Hz}$

Both motions close the gap, so the frequency shift is larger than either motion alone would produce.

Applications of the combined formula include traffic flow monitoring (both vehicles and detectors may be moving), Doppler weather radar (tracking storm cells moving relative to a rotating antenna), and medical ultrasound (where the "source" is a transducer and the "observer" is moving blood or a fetal heartbeat, with the reflected signal analyzed for frequency shift).