Apply Doppler Effect — Apply C.5 Doppler effect in problem-solving and explanations - IB Physics

SECTION 1

Historical Context & Motivation

You have probably noticed that the pitch of an ambulance siren sounds higher as it approaches you and drops lower once it passes. This everyday experience is the Doppler effect, a phenomenon that connects wave physics to everything from weather forecasts to the expanding universe. Understanding the Doppler effect allows physicists to measure speeds remotely—without ever touching the moving object—and it forms a cornerstone of modern astrophysics, medical imaging, and radar technology.

1842 • Doppler's Proposal

Austrian physicist Christian Doppler published a paper predicting that the observed frequency of a wave changes when the source or observer is in motion. He initially applied the idea to light from stars.

1845 • Buys Ballot's Train Experiment

Dutch meteorologist Christophorus Buys Ballot tested Doppler's theory by placing musicians on a moving train while stationary listeners noted the pitch change, confirming the effect for sound waves.

1929 • Hubble and Redshift

Edwin Hubble used the Doppler shift of light from distant galaxies to show that the universe is expanding. Galaxies moving away from us display redshift—a lower observed frequency of light.

1960s • Doppler Radar & Medical Ultrasound

Engineers applied the Doppler effect to radar systems for tracking weather patterns and aircraft. Around the same time, Doppler ultrasound was developed to measure blood flow velocity in the human body.

The central question behind the Doppler effect is deceptively simple: If a wave source or an observer is moving, what frequency does the observer actually detect? Answering this question precisely is the goal of IB Physics topic C.5, and it unlocks powerful applications across science and engineering.

SECTION 2

Core Principles & Definitions

The Doppler effect rests on a few foundational ideas about how waves propagate through a medium. Before diving into equations, it helps to establish the key concepts clearly. Each of the principles below applies to mechanical waves (such as sound) travelling through a medium like air, and some extend to electromagnetic waves (such as light) with a slightly different treatment.

Wave Speed Is Set by the Medium

The speed of a wave (v) is determined by the properties of the medium it travels through—not by the motion of the source. Sound travels at about 340 m s⁻¹ in air regardless of how fast the siren moves.

Frequency vs. Wavelength

When a source moves toward an observer, the wavefronts bunch together, producing a shorter wavelength and a higher observed frequency. The reverse happens when the source moves away.

Observer Motion Matters Too

If the observer moves toward a stationary source, they encounter wavefronts more frequently. If they move away, they encounter wavefronts less frequently. Both source and observer motion affect the observed frequency.

Blueshift & Redshift

When source and observer approach each other, the observed frequency increases (blueshift). When they move apart, the observed frequency decreases (redshift). These terms originate from the light spectrum.

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KEY TAKEAWAY

Think of running into rain. If you stand still, a certain number of raindrops hit you each second. If you run toward the rain, more drops hit you per second—the "frequency" of impacts goes up. If you run away, fewer drops hit you per second. The rain itself hasn't changed—just your motion relative to it. The Doppler effect works the same way with wavefronts instead of raindrops.

SECTION 3

Visualising the Doppler Effect

The diagram below shows how wavefronts change when a source is in motion. On the left, a stationary source emits circular wavefronts that are equally spaced in all directions. On the right, a moving source compresses the wavefronts ahead of it and stretches them behind it.

Left: A stationary source emits circular wavefronts with equal spacing (constant wavelength) in every direction. Right: A moving source shifts the wavefront centres, compressing waves ahead (higher frequency) and stretching waves behind (lower frequency). Observer A detects a higher pitch; Observer B detects a lower pitch.

Notice that the wave speed itself has not changed—the circles expand at the same rate in both diagrams. What changes is where the source is when it emits each wavefront. Because the source keeps moving forward between emissions, the wavefronts pile up in front and spread apart behind. This asymmetry is the physical origin of the frequency shift.

SECTION 4

Mathematical Framework

The IB Physics syllabus (C.5) provides a single master equation that covers both source and observer motion for sound waves. The key is choosing the correct signs. Let's build up the equation step by step.

The General Doppler Equation for Sound

DOPPLER EFFECT — SOUND

f′ = f × (v ± v_o) / (v ∓ v_s)

Where f′ = observed frequency, f = emitted (source) frequency, v = speed of sound in the medium, v_o = speed of observer, v_s = speed of source.

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Sign Convention (IB Approach)

Use the upper signs (+ in numerator, − in denominator) when source and observer move toward each other. Use the lower signs (− in numerator, + in denominator) when they move apart. A simple check: approaching should give f′ > f, and receding should give f′ < f.

Special Cases

MOVING SOURCE, STATIONARY OBSERVER

f′ = f × v / (v ∓ v_s)

Set v_o = 0. Use − when source approaches (f′ increases) and + when source recedes (f′ decreases).

MOVING OBSERVER, STATIONARY SOURCE

f′ = f × (v ± v_o) / v

Set v_s = 0. Use + when observer approaches and − when observer recedes.

Doppler Effect for Electromagnetic Waves

Light and other electromagnetic waves do not require a medium, so the Doppler equation takes a different form. For speeds much less than the speed of light (v ≪ c), the IB syllabus uses the approximation:

DOPPLER EFFECT — LIGHT (v ≪ c)

Δf / f ≈ Δλ / λ ≈ v / c

Where Δf is the change in frequency, Δλ is the change in wavelength, v is the relative speed between source and observer, and c is the speed of light (3.00 × 10⁸ m s⁻¹).

SECTION 5

Applications & Frequency Shift Breakdown

The Doppler effect appears across a surprisingly wide range of fields. The diagram below maps out four major real-world applications, showing how the same physics principle is used in vastly different contexts. After the diagram, a classification table breaks down the type of wave, the quantity measured, and the information extracted.

Four major applications of the Doppler effect. In each case, the same underlying principle is at work: a frequency or wavelength shift reveals the speed of a moving object.

Application	Wave Type	Quantity Measured	Information Obtained
Astronomical redshift	Visible light / EM	Δλ (wavelength shift)	Recession speed of galaxies
Medical ultrasound	Sound (ultrasonic)	Δf (frequency shift)	Blood flow velocity
Speed radar (police)	Microwaves / EM	Δf (frequency shift)	Vehicle speed
Doppler weather radar	Microwaves / EM	Δf (frequency shift)	Wind speed, storm rotation

Summary of Doppler effect applications by wave type and output.

SECTION 6

Worked Example

Let's walk through a full problem involving a moving source, which is one of the most common Doppler calculations on IB exams.

Ambulance Siren — Moving Source, Stationary Observer

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Step 1 — Read the Problem

An ambulance emits a siren at a frequency of 800 Hz. It drives toward a stationary observer at 25 m s⁻¹. The speed of sound in air is 340 m s⁻¹. What frequency does the observer hear?
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Step 2 — Identify Given Values

f = 800 Hz (emitted frequency), v = 340 m s⁻¹ (speed of sound), v_s = 25 m s⁻¹ (source speed), v_o = 0 (observer is stationary). The source is approaching, so we expect f′ > f.
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Step 3 — Select the Correct Equation

Since the observer is stationary, we use: f′ = f × v / (v − v_s). We use the minus sign in the denominator because the source is approaching, which should increase the observed frequency (smaller denominator → larger result).
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Step 4 — Substitute Values

f′ = 800 × 340 / (340 − 25) = 800 × 340 / 315
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Step 5 — Calculate

f′ = 800 × 1.0794 = 863.5 Hz

f′ ≈ 864 Hz
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Step 6 — Sense Check

The observed frequency (864 Hz) is greater than the emitted frequency (800 Hz), which is correct because the source is approaching. The increase is about 8%, which is reasonable for a source moving at roughly 7% of the speed of sound.

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Exam Tip

Always do a sense check on your answer. Approaching = higher frequency. Receding = lower frequency. If your answer contradicts this, re-check your signs.

SECTION 7

Source Motion vs. Observer Motion

Students sometimes wonder whether it matters if the source moves or the observer moves. For sound waves, these situations are physically different because there is a medium (air) that defines a rest frame. The table below compares the two cases for the same relative approach speed.

Feature	Moving Source	Moving Observer
What changes physically	Wavelength is compressed or stretched	Observer intercepts wavefronts more or less frequently
Equation form	f′ = f × v / (v ∓ $v_s$ )	f′ = f × (v ± $v_o$ ) / v
Same relative speed → same f′?	No—slightly different result	No—slightly different result
Why the difference?	Sound has a medium; the medium defines which frame is at rest	Same reason—air is the reference frame
For light (EM waves)	No medium → only relative motion matters	Same result as moving source at the same relative speed

Comparison of moving-source and moving-observer Doppler scenarios for sound.

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KEY TAKEAWAY

For sound, the medium (air) acts like a referee—it distinguishes between a moving source and a moving observer even if the relative speed is the same. For light, there is no medium, so only the relative velocity between source and observer matters. This distinction is exactly why the IB gives you separate equation forms for sound but a single approximation for light.

SECTION 8

Connection to Relativistic Doppler & Advanced Theory

The Doppler equations you learn in IB Physics work beautifully for everyday speeds—cars, ambulances, and even orbiting planets. However, when objects move at a significant fraction of the speed of light, Einstein's theory of special relativity changes the picture. The table below previews how the classical Doppler formula for light gives way to the relativistic Doppler equation.

Aspect	Classical (IB Level)	Relativistic (University Level)
Speed range	v ≪ c (much less than speed of light)	Any v up to c
Equation for light	Δf/f ≈ v/c	f′ = f × √((1 − β)/(1 + β)), where β = v/c
Time dilation included?	No	Yes—built into the equation via the √ factor
Transverse Doppler effect?	Not predicted (classical = no shift at 90°)	Predicted and experimentally confirmed

Classical vs. Relativistic Doppler comparison.

You do not need the relativistic formula for IB exams, but it's good to know that the approximation Δf/f ≈ v/c breaks down at high speeds. Quasars, jets from black holes, and particles in accelerators all require the full relativistic treatment. If you continue to university physics, the Doppler effect will reappear in the context of special relativity and cosmological redshift.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A car horn sounds at a constant frequency as it drives past you. Describe how the pitch you hear changes as the car approaches, passes, and then moves away. Explain your answer in terms of wavefront compression and stretching.

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As the car approaches, it emits each successive wavefront from a position closer to you. This compresses the wavefronts, shortening the wavelength and increasing the frequency you perceive—the pitch sounds higher. At the instant the car is directly beside you, there is no component of velocity along the line connecting you and the car, so you hear roughly the true emitted frequency. As the car moves away, the wavefronts are stretched (longer wavelength), reducing the observed frequency, so the pitch drops. The abrupt change in pitch at the moment of passing is a hallmark of the Doppler effect.

Question 2: BASIC CALCULATION

A train whistle emits sound at 500 Hz. The train moves away from a stationary observer at 30 m s⁻¹. If the speed of sound is 340 m s⁻¹, calculate the frequency heard by the observer.

Show answer

Use f′ = f × v / (v + v_s) because the source is moving away (denominator increases, so f′ < f). Substituting: f′ = 500 × 340 / (340 + 30) = 500 × 340 / 370 = 500 × 0.9189 ≈ 459 Hz. The observer hears a lower frequency of approximately 459 Hz, consistent with the source moving away.

Question 3: INTERMEDIATE

A police car siren emits sound at 700 Hz and approaches a jogger who is running toward the police car at 3.0 m s⁻¹. The police car travels at 20 m s⁻¹ and the speed of sound is 340 m s⁻¹. What frequency does the jogger hear?

Show answer

Both source and observer are moving toward each other, so use f′ = f × (v + v_o) / (v − v_s). Substituting: f′ = 700 × (340 + 3.0) / (340 − 20) = 700 × 343 / 320 = 700 × 1.0719 ≈ 750 Hz. The jogger hears about 750 Hz, which is higher than 700 Hz as expected when source and observer approach each other. The source motion contributes a larger shift than the observer motion because the source speed is greater.

Question 4: APPLIED

An astronomer observes a spectral line from a distant galaxy. The laboratory wavelength of this line is 656.3 nm, but the observed wavelength is 662.1 nm. Calculate the recession speed of the galaxy. (c = 3.00 × 10⁸ m s⁻¹)

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First find Δλ = 662.1 − 656.3 = 5.8 nm. Using the light approximation Δλ/λ ≈ v/c, we get v = c × Δλ/λ = 3.00 × 10⁸ × (5.8 / 656.3) = 3.00 × 10⁸ × 0.008836 ≈ 2.65 × 10⁶ m s⁻¹. The galaxy is receding at approximately 2.65 × 10⁶ m s⁻¹. This is about 0.88% of the speed of light, which validates our use of the non-relativistic approximation (v ≪ c).

Question 5: CRITICAL THINKING

Two identical speakers each emit sound at 1000 Hz. Speaker A is stationary, while Speaker B moves toward a stationary observer at 17 m s⁻¹. The speed of sound is 340 m s⁻¹. (a) Calculate the frequency heard from Speaker B. (b) What beat frequency does the observer hear when both speakers sound simultaneously? (c) Explain what would happen to the beat frequency if Speaker B accelerated to 34 m s⁻¹.

Show answer

(a) f′ = 1000 × 340/(340 − 17) = 1000 × 340/323 ≈ 1053 Hz. (b) The beat frequency equals the difference between the two heard frequencies: f_beat = 1053 − 1000 = 53 Hz. (c) At 34 m s⁻¹: f′ = 1000 × 340/(340 − 34) = 1000 × 340/306 ≈ 1111 Hz, giving f_beat = 111 Hz. Doubling the source speed more than doubles the beat frequency because the Doppler formula is non-linear—the denominator shrinks, and the relationship between speed and frequency shift is not proportional. This demonstrates that the Doppler effect produces larger frequency changes at higher speeds.

SUMMARY

Lesson Summary

Summary

The Doppler effect describes how the observed frequency of a wave changes when there is relative motion between the source and the observer. For sound, a moving source compresses wavefronts ahead (higher f′) and stretches them behind (lower f′), while a moving observer intercepts wavefronts at a different rate. The general equation f′ = f × (v ± v_o) / (v ∓ v_s) handles both cases, with sign choices determined by whether source and observer approach or recede.

For electromagnetic waves at low speeds (v ≪ c), the approximation Δf/f ≈ Δλ/λ ≈ v/c connects redshift and blueshift to the relative speed of source and observer. Applications span astronomy (Hubble's expanding universe), medical imaging (Doppler ultrasound), and radar technology (weather and speed detection). Remember: approaching → higher frequency, receding → lower frequency—always verify your answer against this rule.