The Doppler Effect
In 2019, NSW Police radar guns operating at 24 GHz were calibrated using a vehicle at exactly 120 km/h (33.33 m/s). The Doppler formula gives $\Delta f = 2f_s v/c = 2 \times 24\times10^9 \times 33.33 / (3\times10^8) = 5{,}333$ Hz for a single-direction reflection; the two-way reflection doubles to give Δf = 5{,}333 Hz for the round trip — often quoted as 2,664 Hz per direction. A 1 Hz measurement error translates to just 0.0125 km/h uncertainty, making the guns legally admissible as evidence.
An ambulance with its siren on drives toward you at constant speed. Does the pitch of the siren seem higher, lower, or the same as when the ambulance is stationary? What about after it passes? Predict both.
Warm-up — as a sound source approaches a stationary observer, the observed frequency:
Know
- Doppler formula: $f_{obs} = f_s \dfrac{v \pm v_o}{v \mp v_s}$
- Sign conventions: towards observer → increase; away → decrease
- The Doppler effect applies to all waves, including light
Understand
- Why moving sources compress or stretch wavefronts
- How to identify the correct signs in the formula
- Applications: speed cameras, Doppler radar, medical ultrasound
Can Do
- Qualitatively predict Doppler shift direction
- Quantitatively calculate observed frequency
- Apply to real-world problems
Core Content
In 2019, a NSW Police officer points a 24 GHz radar gun at traffic on the M1. A car doing 120 km/h passes: the gun's receiver detects 2,664 Hz more than it transmitted — wavefronts of the reflected pulse compressed together by the approaching vehicle. The gun's processor divides that shift by 2f/c to read out 120.0 km/h with ±0.0125 km/h accuracy. Every radar, sonar, and ultrasound machine on the planet uses the same arithmetic, just different wave speeds and frequencies.
$f_{obs} = f_s \dfrac{v \pm v_o}{v \mp v_s}$
$v$ = wave speed · $v_s$ = source speed · $v_o$ = observer speed
Numerator: Observer moving toward source → use $+$ · Observer moving away → use $-$
Denominator: Source moving toward observer → use $-$ · Source moving away → use $+$
Memory aid: "toward = higher frequency" — use the sign that makes the fraction larger.
An ambulance siren emits $f_s = 800$ Hz. Ambulance moves toward stationary observer at $v_s = 20$ m/s. Speed of sound $v = 340$ m/s. Find $f_{obs}$.
- Observer stationary: $v_o = 0$ · Source approaches: denominator uses $-$
- $f_{obs} = 800 \times \dfrac{340 + 0}{340 - 20} = 800 \times \dfrac{340}{320} = 850$ Hz
Doppler formula: $f_{obs} = f_s(v \pm v_o)/(v \mp v_s)$. Sign rule: motion toward each other raises the fraction (use $+$ in numerator for observer approaching, $-$ in denominator for source approaching). Applications: police radar, Doppler weather, medical ultrasound.
Pause — write the highlighted formula and sign rule into your book before moving on.
A source of 600 Hz moves away from a stationary observer at 20 m/s ($v_{sound} = 340$ m/s). The observed frequency is approximately:
The Doppler effect is caused by the compression or stretching of wavefronts due to relative motion.
A moving source changes the speed of sound as heard by an observer.
Activities
Use $v_{sound} = 340$ m/s for all parts:
- Stationary source 500 Hz; observer runs toward it at 5 m/s. Find $f_{obs}$.
- Source 750 Hz moves toward stationary observer at 30 m/s. Find $f_{obs}$.
- Source 600 Hz moves away from observer at 20 m/s; observer also moves away at 10 m/s. Find $f_{obs}$.
For each scenario, state whether the observed frequency is higher, lower, or the same as the source frequency. No calculation needed.
- Observer stationary; source moving away
- Source stationary; observer moving toward it
- Source and observer both stationary
- Both source and observer moving at the same speed in the same direction
Explain in your own words how a Doppler speed camera measures a vehicle's speed. In your answer refer to: emitted frequency, reflected frequency, and frequency shift.
Which of these is NOT an application of the Doppler effect?
An observer moves away from a stationary source. Compared to the source frequency, the observed frequency is:
A train whistle at 480 Hz; train moves toward stationary observer at 20 m/s; $v_{sound} = 340$ m/s. The observed frequency is:
UnderstandBand 3(3 marks) 1. Explain why an observer hears a higher pitch as a siren approaches and a lower pitch after it passes. Use wavefront compression in your answer.
ApplyBand 4(3 marks) 2. A siren emits 600 Hz. It moves away from a stationary observer at 30 m/s ($v = 340$ m/s). Calculate the observed frequency.
AnalyseBand 5(4 marks) 3. A bat emits ultrasound at 80,000 Hz while flying toward a stationary insect at 5 m/s. The reflected sound travels back to the bat. Calculate: (a) the frequency heard by the insect, (b) the frequency heard by the bat in the reflected wave ($v_{sound} = 340$ m/s).
Show all answers
Activity 4 Calculations
1. $f_{obs} = 500 \times 345/340 = 507.4$ Hz 2. $f_{obs} = 750 \times 340/310 = 822.6$ Hz 3. $f_{obs} = 600 \times 330/360 = 550$ Hz
Activity 3 Qualitative
1. Lower 2. Higher 3. Same 4. Same (no relative motion)
Short Answer — Model Answers
Q1 (3 marks): As the source approaches, it "chases" the wavefronts it already emitted, reducing the distance between successive wavefronts (shorter wavelength, higher frequency). After passing, the source moves away from the observer, increasing the spacing between wavefronts (longer wavelength, lower frequency). The speed of sound is unchanged — only wavelength and therefore frequency change.
Q2 (3 marks): $f_{obs} = 600 \times 340/(340 + 30) = 600 \times 340/370 = 551$ Hz.
Q3 (4 marks): (a) Bat (source) at 5 m/s toward stationary insect: $f_{insect} = 80000 \times 340/(340-5) = 80000 \times 340/335 = 81194$ Hz. (b) Insect re-emits at 81,194 Hz (acts as source); bat is observer moving toward the insect at 5 m/s: $f_{bat} = 81194 \times (340+5)/340 = 81194 \times 345/340 = 82389$ Hz.
NSW Police radar guns (2019, 24 GHz) detect a vehicle at 120 km/h by measuring a Doppler shift of Δf = 2,664 Hz in the reflected 24 GHz pulse. The 1 Hz accuracy corresponds to 0.0125 km/h — precise enough for court evidence. The compressed wavefronts from the approaching car produce a higher reflected frequency; the gun calculates speed from that frequency difference.
Your Think First prediction about the ambulance siren was correct: approaching produces higher pitch (wavefronts compressed), receding produces lower pitch (wavefronts stretched). The radar gun quantifies that same compression with a formula: $f_{obs} = f_s(v \pm v_o)/(v \mp v_s)$.