Year 11 PhysicsModule 3⏱ ~40 min5 MC · 3 Short AnswerLesson 12 of 18
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 a one-way shift of $f_s v/c = 24\times10^9 \times 33.33 / (3\times10^8) = 2{,}666$ Hz; because the pulse makes a round trip (gun to car and back), the gun measures twice this, $\Delta f = 2f_s v/c = 5{,}333$ Hz. A 1 Hz measurement error translates to just 0.0125 km/h uncertainty, making the guns legally admissible as evidence.
Today's hook: In 2019, NSW Police radar guns at 24 GHz were calibrated for a test vehicle at 120 km/h. The reflected signal arrives shifted by Δf = 5,333 Hz, the round-trip shift (twice the 2,666 Hz one-way value). A 1 Hz measurement error corresponds to 0.0125 km/h, making these guns accurate enough to be court-admissible evidence. The physics is one equation: $f_{obs} = f_s(v \pm v_o)/(v \mp v_s)$, with moving source compressing wavefronts (higher observed f) or stretching them (lower observed f).
0/5TASKS
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Warm up first
Warm up first
Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.
Before you read, predict
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:
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
Cross-lesson links: L09 (nature of sound) established that sound is a longitudinal wave moving at v ≈ 340 m/s, the wave speed used in all Doppler calculations. L10 (intensity and beats) introduced frequency differences causing audible beats; the Doppler effect is a related frequency shift caused by relative motion rather than two sources. The radar gun application (electromagnetic waves at c) previews Year 12 Module 7 special relativity, where the relativistic Doppler formula applies.
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The Doppler Formula
Core Content
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The Doppler Formula
+5 XP
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 5,333 Hz more than it transmitted, wavefronts of the reflected pulse compressed together by the approaching vehicle. The gun's processor divides that round-trip 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.
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.
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Activity 4, Doppler Calculations
Activities
Activity 4, Doppler Calculations
ApplyBand 4
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 600 Hz moves away from observer at 20 m/s; observer also moves away at 10 m/s. Find $f_{obs}$.
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Activity 3, Qualitative Doppler
Activity 3, Qualitative Doppler
UnderstandBand 3
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
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Activity 2, Speed Camera Application
Activity 2, Speed Camera Application
ExplainBand 4
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:
✓
Show what you have learned
Multiple Choice, Doppler effect
+5 XP
Short Answer, 10 marks
+5 XP
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).
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.
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.
✓
Retrieve, reflect and finish
Check what actually stuck
How did your thinking change?
NSW Police radar guns (2019, 24 GHz) detect a vehicle at 120 km/h by measuring a round-trip Doppler shift of Δf = 5,333 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: $f_{obs} = f_s(v \pm v_o)/(v \mp v_s)$.