Back to Course
Doppler Effect
The Doppler Effect: An Overview
The Doppler Effect is the change in Apparent Frequency of a wave when the source and observer are in Relative Motion.
$$f_{obs} \neq f_{source}$$
The observed frequency deviates from the source frequency due to motion.
$f_{obs}$=Frequency heard by observer(Hz)
$f_{source}$=Frequency emitted by source(Hz)
Frequency-Pitch Link: In sound, this is perceived as a change in Pitch.
Wave Fronts: Motion causes Wave Fronts to bunch up or spread out in the medium.
Key Distinction: The source frequency $f$ never changes — only the observed frequency $f'$ shifts due to motion.
Medium Dependence: The Doppler effect for sound depends on motion relative to the medium (air), not just relative motion between source and observer.
Case 1: The Observer is Moving
When an observer moves towards a Stationary Source, they encounter wave crests more frequently, increasing the Modified Frequency.
$$f_A = f \left( \frac{v \pm u_o}{v} \right)$$
The relative speed of the waves changes as the observer moves through the medium.
$v$=Speed of sound in medium(m/s)
$u_o$=Velocity of the observer(m/s)
$f$=Source frequency(Hz)
$u_o$ towards source
→Use $+$ (Higher frequency)
$u_o$ away from source
→Use $-$ (Lower frequency)
Wavelength Stability: Unlike a moving source, the physical wavelength in the medium remains constant.
Relative Velocity: The observer perceives a higher wave speed ($v + u_o$) when moving toward the source.
Proportionality: The fractional frequency shift is $\Delta f / f = u_o / v$ — directly proportional to observer speed.
Limiting Case: When $u_o = 0$, $f' = f$ — no shift. When $u_o = v$ (observer at speed of sound towards source), $f' = 2f$ — frequency doubles.
Case 2: The Source is Moving
A moving source physically alters the wavelength in the medium, causing Wavelength Compression ahead and expansion behind.
$$f_C = f \left( \frac{v}{v \mp u_s} \right)$$
The distance between wave crests changes because the source 'chases' its own emitted waves.
$u_s$=[Source Velocity](m/s)
$v$=Speed of sound(m/s)
$u_s$ towards observer
→Use $-$ in denominator (Increases frequency)
$u_s$ away from observer
→Use $+$ in denominator (Decreases frequency)
Doppler Shift: The change in wavelength is calculated as $\Delta \lambda = u_s / f$.
Wave Speed: The speed of waves ($v$) relative to the medium is unaffected by the source's speed.
Asymmetry: Source approaching gives $f' = f \cdot v/(v - u_s)$ and receding gives $f' = f \cdot v/(v + u_s)$. The approaching shift is always larger than the receding shift for the same speed.
Limiting Case: As $u_s \to v$, the denominator $(v - u_s) \to 0$ and $f' \to \infty$ — the source catches up to its own waves, creating a Shock Wave (sonic boom).
Case 3: Both Source and Observer Moving
When both the source and observer are in motion, both effects combine into a single general formula for the Modified Frequency.
$$f' = f \left( \frac{v \pm u_o}{v \mp u_s} \right)$$
The numerator handles observer motion (changes how fast the observer encounters waves) and the denominator handles source motion (changes the physical wavelength).
$v$=Speed of sound in medium(m/s)
$u_o$=Observer velocity(m/s)
$u_s$=Source velocity(m/s)
Both approaching each other
→$f' = f(v + u_o)/(v - u_s)$ — maximum frequency increase
Both receding from each other
→$f' = f(v - u_o)/(v + u_s)$ — maximum frequency decrease
Same speed, same direction
→$f' = f$ — no shift (relative velocity is zero)
Sign Convention: Use $+u_o$ when observer moves towards source, $-u_o$ when away. Use $-u_s$ when source moves towards observer, $+u_s$ when away.
Memory Aid: Think 'towards = higher frequency'. In the formula: numerator increase or denominator decrease both raise $f'$.
Echo Problems: When sound reflects off a surface (wall, cliff), treat it as two Doppler shifts — the reflecting surface acts as both receiver and re-emitter.
Applications in Technology and Astronomy
The Doppler Effect is a fundamental tool for measuring velocity in Radar, Sonar, and medical imaging.
$$\Delta \lambda = \lambda_{obs} - \lambda_{source}$$
The Doppler shift in wavelength indicates the direction and speed of relative motion.
$\Delta \lambda$=Change in wavelength(m (or nm for light))
$\lambda_{obs}$=Wavelength measured by observer
$\lambda_{source}$=Wavelength emitted at rest
$\Delta \lambda > 0$
→Red Shift — source moving away (wavelength increases)
$\Delta \lambda < 0$
→Blue Shift — source approaching (wavelength decreases)
Red Shift: Distant galaxies moving away exhibit a shift toward longer (red) wavelengths — key evidence for the expanding universe.
Blue Shift: Stars or galaxies approaching Earth have spectra shifted toward shorter (blue) wavelengths.
Radar Speed Trap: Microwaves reflected off a moving car are Doppler-shifted; the shift is doubled because the signal bounces back.
Sonar: Underwater echo-ranging uses sound waves. Doppler detection measures target speed via frequency shift of echoes — used for submarine detection and depth measurement.
Echolocation: Bats and dolphins use Doppler shifts of reflected ultrasound to determine the speed of moving prey.
Blood Flow: Medical ultrasound (5–10 MHz) directed at arteries detects backscattered signal shifts to measure blood flow velocity.
Application Strategy
Worked Examples
Problem Statement
A stationary siren emits sound at 500 Hz. An observer runs towards it at 20 m/s. What frequency does the observer hear? (v = 340 m/s)
Moving observerStationary source
Strategic Toolkit
$f' = f \left( \frac{v + u_o}{v} \right)$
Step 1
Identify Variables
$f = 500$ Hz, $v = 340$ m/s, $u_o = 20$ m/s (towards source). Source is stationary ($u_s = 0$).
Step 2
Select Formula
Observer moving towards a stationary source → use numerator formula with $+u_o$.
$$f' = 500 \times \left( \frac{340 + 20}{340} \right)$$
Step 3
Calculate Result
Add observer speed to sound speed in the numerator.
$$f' = 500 \times \frac{360}{340} = 529.4 \text{ Hz}$$
Problem Statement
A train approaches a platform at 25 m/s while blowing a 1000 Hz whistle. Calculate the frequency heard by a stationary person on the platform. (Speed of sound = 340 m/s)
Moving sourceStationary observer
Strategic Toolkit
$f' = f \left( \frac{v}{v - u_s} \right)$
Step 1
Identify Variables
Source frequency ($f$) = 1000 Hz, Speed of sound ($v$) = 340 m/s, Source velocity ($u_s$) = 25 m/s. Observer velocity ($u_o$) = 0.
Step 2
Select Formula
Since the source is moving towards a stationary observer, we use the formula with $(v - u_s)$ in the denominator.
$$f' = 1000 \times \left( \frac{340}{340 - 25} \right)$$
Step 3
Calculate Result
Simplify the denominator and divide.
$$f' = 1000 \times \frac{340}{315} \approx 1079.4 \text{ Hz}$$
Problem Statement
A galaxy emits light with a Calcium alpha line at 397 nm. An observer on Earth measures this line at 478 nm. Determine if the galaxy is moving towards or away and calculate its speed.
Light Doppler shiftRedshift
Strategic Toolkit
$\lambda_{obs} = \lambda \left( \frac{c + u_s}{c} \right)$
Step 1
Analyze Motion
Since $\lambda_{obs}$ (478 nm) is greater than $\lambda_{source}$ (397 nm), the wavelength has increased. This is a [Red Shift], meaning the galaxy is moving away.
Step 2
Setup Equation
Use the wavelength version of the Doppler formula for a receding source.
$$478 = 397 \left( \frac{3.0 \times 10^8 + u_s}{3.0 \times 10^8} \right)$$
Step 3
Solve for Velocity
Rearrange to isolate $u_s$.
$$\frac{478}{397} = 1 + \frac{u_s}{3.0 \times 10^8}$$
$$u_s = \left( \frac{478}{397} - 1 \right) \times 3.0 \times 10^8 \approx 6.1 \times 10^7 \text{ m/s}$$
Problem Statement
Car P travels at 12 m/s, car Q behind it at 20 m/s sounds its horn. P's driver estimates 830 Hz. What frequency does Q's driver hear? (v = 340 m/s)
Both movingSame direction
Strategic Toolkit
$f' = f \left( \frac{v - u_o}{v - u_s} \right)$
Step 1
Identify Variables
Car P (observer) at $u_o = 12$ m/s, Car Q (source) at $u_s = 20$ m/s. Both travel in the same direction. The observer hears $f' = 830$ Hz. Speed of sound $v = 340$ m/s.
Step 2
Determine Signs
Q is behind P and catching up, so Q approaches P. Source moves towards observer → denominator uses $(v - u_s)$. Observer moves away from source → numerator uses $(v - u_o)$.
$$f' = f \left( \frac{v - u_o}{v - u_s} \right)$$
Step 3
Solve for Source Frequency
Rearrange and substitute.
$$830 = f \times \frac{340 - 12}{340 - 20} = f \times \frac{328}{320}$$
$$f = 830 \times \frac{320}{328} \approx 810 \text{ Hz}$$
Memorize formulas
Observer Moving Formula
Plus for towards, minus for away. Affects the numerator.
Formula
$f' = f \left( \frac{v \pm u_o}{v} \right)$
Source Moving Formula
Minus for towards, plus for away. Affects the denominator.
Formula
$f' = f \left( \frac{v}{v \mp u_s} \right)$
General Doppler Formula
Both source and observer moving. Numerator = observer, denominator = source.
Formula
$f' = f \left( \frac{v \pm u_o}{v \mp u_s} \right)$
Doppler Wavelength Shift
Physical change in distance between wave crests due to source motion.
Formula
$\Delta \lambda = \frac{u_s}{f}$
Review
1
A star is moving away from the Earth at a significant fraction of the speed of light. Which observation is most likely?
Ans
The observed wavelength is longer than the laboratory wavelength.
Review
Tap to reveal
Apply the red-shift trend
When a light source recedes, the wave crests arrive less frequently, so the observed frequency drops and the wavelength increases.
Identify the matching observation
The only option that reflects a longer observed wavelength without changing the speed of light is the one describing a wavelength larger than the laboratory value.
$$\lambda_{obs} > \lambda_{lab}$$
2
An observer moves towards a stationary sound source with velocity $u_o$. If the velocity of sound is $v$ and the source frequency is $f$, what is the percentage increase in frequency?
Ans
$(u_o / v) \times 100$
Review
Tap to reveal
Use the moving-observer formula
For an observer moving toward a stationary source, the apparent frequency is increased by the factor $(v+u_o)/v$.
Convert the increase to a percentage
The fractional increase is $u_o/v$, so multiplying by 100 gives the percentage increase.
$$\frac{f' - f}{f} \times 100 = \frac{u_o}{v} \times 100$$
3
In a radar speed trap, a microwave signal is reflected off a car moving toward the radar. How does the reflected frequency compare to the emitted frequency?
Ans
Reflected frequency is higher because the car acts as a moving source.
Review
Tap to reveal
Track the wave on the way in
The approaching car intercepts wave fronts more quickly than a stationary target, so it receives a higher frequency than the radar emits.
Track the wave on the way back
After reflection, the moving car behaves like a source moving toward the radar, which compresses the reflected waves again and raises the frequency further.
The reflected frequency is higher than the emitted frequency.
4
What physically happens to the wave fronts in the medium when the source moves toward a stationary listener?
Ans
The distance between successive wave fronts decreases in the direction of motion.
Review
Tap to reveal
Keep the wave speed fixed by the medium
The source moving does not change the speed of sound in air; it only changes where each crest is emitted.
See the crest spacing shrink ahead of the source
Because the source advances between emissions, successive wave fronts are released closer together in front of it.
The wavelength decreases in the direction of motion.
5
If a source of frequency 1000 Hz moves away from a stationary observer at a speed equal to half the speed of sound ($u_s = 0.5v$), what frequency is heard?
Ans
667 Hz
Review
Tap to reveal
Substitute the source-receding formula
For a source moving away from a stationary observer, the observed frequency is reduced by the factor $v/(v+u_s)$.
Insert the given speed ratio
With $u_s = 0.5v$, the denominator becomes $1.5v$, so the frequency is divided by 1.5.
$$f' = 1000 \times \frac{v}{v + 0.5v} = \frac{1000}{1.5} \approx 667\ \text{Hz}$$
6
A fire truck with a 600 Hz siren drives towards a stationary pedestrian at 72 km/h. What frequency does the pedestrian hear? (v = 340 m/s)
Ans
637.5 Hz
Review
Tap to reveal
Convert the truck speed
The siren speed must be in m/s before using the Doppler formula, so $72\ \text{km/h} = 20\ \text{m/s}$.
Apply the approaching-source formula
A source moving toward a stationary observer increases the frequency by the factor $v/(v-u_s)$.
$$f' = 600 \times \frac{340}{340 - 20} = 600 \times \frac{340}{320} = 637.5\ \text{Hz}$$
7
Two trains on parallel tracks move towards each other. Train X (source) blows a 600 Hz whistle while moving at 25 m/s. Train Y (observer) approaches at 15 m/s. What frequency does Train Y hear? (v = 340 m/s)
Ans
676.2 Hz
Review
Tap to reveal
Include both Doppler effects
The observer moving toward the source increases the numerator, and the source moving toward the observer decreases the denominator.
Substitute the train speeds
Use $v+u_o = 340+15 = 355$ and $v-u_s = 340-25 = 315$ to evaluate the frequency.
$$f' = 600 \times \frac{355}{315} \approx 676.2\ \text{Hz}$$
8
A source approaches a stationary observer and the apparent frequency is 20% higher than the actual frequency. If $v = 340$ m/s, what is the source speed?
Ans
56.7 m/s
Review
Tap to reveal
Translate the percentage statement
A 20% increase means the observed frequency is $1.2f$, so the Doppler factor must equal 1.2.
Solve the source-speed relation
For an approaching source, $1.2 = v/(v-u_s)$; solving gives the source speed.
$$1.2 = \frac{340}{340-u_s} \Rightarrow 340-u_s = 283.3 \Rightarrow u_s = 56.7\ \text{m/s}$$
9
A car drives at 15 m/s towards a cliff while honking its 500 Hz horn. What is the frequency of the echo heard by the driver? (v = 340 m/s)
Ans
546.2 Hz
Review
Tap to reveal
Apply Doppler shift on the outgoing wave
The cliff acts like a stationary reflector, so the moving car first sends a higher-frequency wave toward it.
Apply the shift again on the return wave
The reflected wave is then heard by the approaching driver, so the observed echo frequency is boosted a second time.
$$f' = 500 \times \frac{340 + 15}{340 - 15} = 546.2\ \text{Hz}$$
10
Why is the Doppler effect for a source moving at speed $u$ NOT the same as for an observer moving at the same speed $u$, even though the relative velocity is the same?
Ans
Because the source motion physically changes the wavelength in the medium, while the observer motion does not.
Review
Tap to reveal
Separate source motion from observer motion
For sound, the source changes the spacing of crests in the medium, while the observer only changes how quickly those crests are intercepted.
Use the medium as the reference frame
Because the air is the wave medium, equal speeds for source and observer do not produce identical effects on frequency.
The source changes wavelength; the observer changes encounter rate.
Question 1 of 10
0:00
1
A star is moving away from the Earth at a significant fraction of the speed of light. Which observation is most likely?