Interactive physics simulator
Doppler Effect
Explore how relative motion between sound sources and observers shifts the apparent pitch of sound. Visualize moving wavefront compression, supersonic Mach shock cones, and radar double shifts.
Doppler Effect Lab
Interact with the canvas: drag elements to position them. Toggle sound, adjust sliders, and listen to the real-time pitch-shifted frequency.
Live Telemetry
What is the Doppler Effect?
The Doppler Effect (or Doppler shift) is the shift in the observed frequency and wavelength of a wave when there is relative motion between the wave source and the observer. It was first proposed by Austrian physicist Christian Doppler in 1842.
When a sound source moves toward an observer, each successive sound wave crest is emitted from a position closer to the observer than the previous crest. As a result, the distance between arriving wavefronts decreases, causing the wavelength to compress and the observed pitch (frequency) to rise. Conversely, when the source moves away, waves are emitted from progressively further positions, stretching the wavelength and dropping the apparent pitch.
This effect is commonly heard when an ambulance siren passes by: the pitch is high as the ambulance approaches, drops noticeably at the exact instant it passes, and stays low as it recedes.
Sign Conventions
Choosing the correct signs in the Doppler equation is essential for accurate calculations:
| Motion Case | Numerator Sign (\(v \pm v_o\)) | Denominator Sign (\(v \mp v_s\)) |
|---|---|---|
| Approaching | \(+\) (Observer moves toward source) | \(-\) (Source moves toward observer) |
| Receding | \(-\) (Observer moves away) | \(+\) (Source moves away) |
Golden Rule: Approach increases frequency (plus on top, minus on bottom). Recession decreases frequency (minus on top, plus on bottom).
General Equation
Calculates the observed frequency due to relative motion:
Where:
- f' = Observed frequency (hertz, Hz)
- f = Emitted frequency (hertz, Hz)
- v = Velocity of sound waves in the medium (m/s)
- v_o = Velocity of the observer (m/s)
- v_s = Velocity of the source (m/s)
Note: Velocities \(v_o\) and \(v_s\) are measured relative to the stationary medium (usually air).
Solved Examples
An ambulance emitting a (400 ext{ Hz}) siren approaches a stationary pedestrian observer at (30 ext{ m/s}). Calculate the apparent frequency heard by the observer (a) as the ambulance approaches, and (b) after it passes and recedes. (Speed of sound (v = 340 ext{ m/s})).
- Identify the given variables: Source frequency (f = 400 ext{ Hz}), Speed of sound (v = 340 ext{ m/s}), Source speed (v_s = 30 ext{ m/s}), Observer speed (v_o = 0).
- For the approaching ambulance (source moving toward stationary observer), use the formula with a negative sign in the denominator: (f' = f cdot rac{v}{v - v_s}).
- Substitute values for approach: (f' = 400 cdot rac{340}{340 - 30} = 400 cdot rac{340}{310} approx 438.71 ext{ Hz}).
- For the receding ambulance (source moving away from stationary observer), use the formula with a positive sign in the denominator: (f' = f cdot rac{v}{v + v_s}).
- Substitute values for recession: (f' = 400 cdot rac{340}{340 + 30} = 400 cdot rac{340}{370} approx 367.57 ext{ Hz}).
Answer: Approaching: (438.7 ext{ Hz}) | Receding: (367.6 ext{ Hz})
A train passenger is riding on a train traveling at (25 ext{ m/s}) toward a stationary station warning bell that is ringing at a frequency of (600 ext{ Hz}). What frequency does the passenger hear? (Speed of sound (v = 343 ext{ m/s})).
- Identify the given variables: Source frequency (f = 600 ext{ Hz}), Speed of sound (v = 343 ext{ m/s}), Source speed (v_s = 0) (stationary), Observer speed (v_o = 25 ext{ m/s}) (approaching).
- Since the observer is moving toward a stationary source, use the formula with a positive sign in the numerator: (f' = f cdot rac{v + v_o}{v}).
- Substitute the values: (f' = 600 cdot rac{343 + 25}{343} = 600 cdot rac{368}{343}).
- Compute the final frequency: (f' approx 600 cdot 1.0729 approx 643.73 ext{ Hz}).
Answer: Apparent frequency (f' approx 643.7 ext{ Hz})
A sonar speed system emits an ultrasonic tone of (30 ext{ kHz}). A car approaches the sonar rig at a speed of (35 ext{ m/s}). The signal reflects off the car and returns to the detector. What is the frequency of the returning wave received back at the sonar detector? (Speed of sound (v = 340 ext{ m/s})).
- This is a double Doppler shift problem. First, the car acts as a moving observer receiving the signal from the stationary sonar source: (f_{car} = f cdot rac{v + v_{car}}{v}).
- Second, the car reflects the wave, behaving as a moving source emitting at the frequency (f_{car}) back toward the stationary sonar receiver: (f' = f_{car} cdot rac{v}{v - v_{car}}).
- Combine the two shifts: (f' = f cdot left( rac{v + v_{car}}{v} ight) cdot left( rac{v}{v - v_{car}} ight) = f cdot rac{v + v_{car}}{v - v_{car}}).
- Substitute the values: (f' = 30 ext{ kHz} cdot rac{340 + 35}{340 - 35} = 30 cdot rac{375}{305}).
- Compute the final frequency: (f' approx 30 cdot 1.2295 approx 36.89 ext{ kHz}).
Answer: Returning frequency (f' approx 36.89 ext{ kHz}) (shifted by (+6.89 ext{ kHz}))
Common Mistakes
- Reversing Numerator & Denominator Signs: Students frequently mix up where the observer and source speed belong. Remember that the observer is on top (numerator) and the source is on the bottom (denominator).
- Confusing Pitch with Loudness: The Doppler effect shifts the pitch (frequency) of sound, not its volume (loudness). Sound is louder when the source is close simply because of distance and sound decay, which is separate from the Doppler pitch-shift.
- Incorrect Frame of Reference: Doppler speeds must be relative to the air medium. If wind is blowing, the speed of sound relative to the ground changes, which must be factored in.
- Relativistic Speed Errors: Applying the classical sound equation to light waves. Light requires the relativistic formula because light speed \(c\) is constant in all reference frames.
Sonic Boom & Mach Cone
What happens when the source moves faster than sound (\(v_s \ge v\)):
- Sound Barrier: At \(v_s = v\), waves pile up at the front, creating a pressure barrier.
- Mach Cone: When \(v_s > v\), circular wave fronts form a conical shock front. The Mach angle \(\theta\) is:
\sin(\theta) = \frac{v}{v_s} = \frac{1}{M}
- Mach Number (M): Ratio of source speed to sound speed (\(M = v_s / v\)).
- Sonic Boom: The intense double pressure crack heard on the ground when the Mach cone sweeps past.
Practice Questions
Question 1
A police car with its siren active at (500 ext{ Hz}) chases a speeding car. The police car travels at (40 ext{ m/s}) and the speeder travels at (30 ext{ m/s}) in front of it. What frequency is heard by the driver of the speeder? (Speed of sound (v = 340 ext{ m/s})).
View Solution & Answer
Both source and observer are moving in the same direction. The source (police, (v_s = 40 ext{ m/s})) is chasing (moving toward) the observer. The observer (speeder, (v_o = 30 ext{ m/s})) is escaping (moving away from) the source. According to sign conventions, (f' = f cdot rac{v - v_o}{v - v_s}). Substituting the values: (f' = 500 cdot rac{340 - 30}{340 - 40} = 500 cdot rac{310}{300} approx 516.67 ext{ Hz}). The speeder hears a pitch of (516.7 ext{ Hz}).
Question 2
A whistle on a factory roof emits sound at (800 ext{ Hz}). On a windy day, wind blows at (15 ext{ m/s}) from the whistle toward a stationary observer. Does the observer hear a Doppler shift? Explain why or why not.
View Solution & Answer
No, a stationary observer hears no frequency shift due to wind. The wind increases the speed of sound traveling toward the observer ((v + v_{wind})) and stretches the wavelength proportionally ((lambda' = lambda + v_{wind}/f)). Since both speed and wavelength increase by the same factor, the frequency (f = v' / lambda') remains exactly (800 ext{ Hz}). The Doppler effect only occurs when the source or observer moves relative to the medium.
Question 3
An airplane flies horizontally at Mach 1.5 at an altitude of (3000 ext{ m}). Calculate the Mach angle of the shockwave cone. (Speed of sound (v = 340 ext{ m/s})).
View Solution & Answer
The Mach number is (M = 1.5). The Mach angle ( heta) is the half-angle of the shockwave cone and is given by: (sin( heta) = rac{v}{v_s} = rac{1}{M}). Substituting (M = 1.5): (sin( heta) = rac{1}{1.5} = rac{2}{3} approx 0.6667). Therefore, ( heta = arcsin(0.6667) approx 41.81^circ). The Mach angle is (41.8^circ).
Question 4
How is the Doppler effect for light different from sound, and why doesn't it require a medium?
View Solution & Answer
Light is an electromagnetic wave and does not require a physical medium to propagate; it travels through a vacuum. Therefore, the Doppler shift for light depends only on the relative velocity between the source and observer ((u)), not their individual velocities relative to a medium. The relativistic Doppler formula is (f' = f cdot sqrt{ rac{1 - eta}{1 + eta}}) (for recession) where (eta = u/c).
Frequently Asked Questions
What is the Doppler Effect?
The Doppler effect is the change in the observed frequency (or pitch) of a wave when there is relative motion between the wave source and the observer.
What is the general formula for the Doppler Effect?
The general formula for the observed frequency f' is: f' = f · ((v ± v_o) / (v ∓ v_s)), where f is the source frequency, v is the speed of sound in the medium, v_o is the observer's speed, and v_s is the source's speed.
How do you choose the signs in the Doppler formula?
Use the sign convention relative to the sound propagation: the numerator is (v + v_o) if the observer moves toward the source, and (v - v_o) if moving away. The denominator is (v - v_s) if the source moves toward the observer, and (v + v_s) if moving away.
Does the Doppler Effect change the actual speed of sound?
No. The speed of sound (v) depends solely on the properties of the medium (such as temperature, humidity, and density) and remains constant, regardless of the motion of the source or observer.
What happens when a source moves at the speed of sound?
When the source speed equals the speed of sound (v_s = v), the sound waves emitted in the direction of motion pile up at the front, creating an infinitely dense barrier of sound pressure known as the sound barrier.
What is a sonic boom and a Mach cone?
When a source travels faster than sound (supersonic, v_s > v), it outruns its own waves. The circular wavefronts overlap constructively behind it to form a conical shockwave front called a Mach cone. The sudden pressure rise across this cone is heard as a sonic boom.
How does the Doppler Effect work in radar speed guns?
A radar speed gun emits electromagnetic waves of a known frequency towards a moving car. The wave is shifted when it hits the car (moving observer) and shifted again when it reflects back to the gun (moving source). The radar measures this double shift to calculate the car's speed.
What is the difference between blue shift and red shift in light?
In light waves, a blue shift occurs when a light source approaches the observer, shifting the observed light to higher frequencies (shorter wavelengths, toward the blue end). A red shift occurs when a source recedes, shifting light to lower frequencies (longer wavelengths, toward the red end).
Does the Doppler Effect change the loudness of sound?
No, not directly. The Doppler effect specifically describes the shift in frequency (pitch). The changes in sound loudness (intensity) are governed by the inverse square law as the distance between the source and observer changes.
Can the Doppler Effect occur in all types of waves?
Yes. The Doppler effect is a fundamental wave phenomenon that occurs in all wave types, including mechanical waves (like sound and water waves) and electromagnetic waves (like light and radar).