The Doppler Effect Formula gives the observed frequency of a wave when the source or observer is in motion relative to the medium. Expressed as \ ( f’ = f \left( \frac{v + v_o}{v – v_s} \right) \), it is a core concept in NCERT Class 11 Physics, Chapter 15 (Waves). It also appears regularly in JEE Main, JEE Advanced, and NEET examinations. This article covers the complete formula, derivation, a full physics formula sheet, three progressively difficult solved examples, CBSE exam tips, common mistakes, and JEE/NEET applications.
Key Doppler Effect Formulas at a Glance
Quick reference for the most important Doppler Effect formulas.
- General Doppler formula: \( f’ = f \left( \dfrac{v + v_o}{v – v_s} \right) \)
- Source moving towards stationary observer: \( f’ = f \left( \dfrac{v}{v – v_s} \right) \)
- Source moving away from stationary observer: \( f’ = f \left( \dfrac{v}{v + v_s} \right) \)
- Observer moving towards stationary source: \( f’ = f \left( \dfrac{v + v_o}{v} \right) \)
- Observer moving away from stationary source: \( f’ = f \left( \dfrac{v – v_o}{v} \right) \)
- Doppler shift (change in frequency): \( \Delta f = f’ – f \)
- Apparent wavelength: \( \lambda’ = \dfrac{v – v_s}{f} \)
What is the Doppler Effect Formula?
The Doppler Effect Formula describes how the frequency of a wave changes when the source of the wave and the observer are moving relative to each other. Named after Austrian physicist Christian Doppler, who proposed it in 1842, this phenomenon applies to all types of waves, including sound, light, and water waves.
In NCERT Class 11 Physics, Chapter 15 (Waves), the Doppler Effect is introduced as the apparent change in frequency of a wave due to relative motion between the source and the observer. When the source and observer move closer together, the observer hears a higher frequency. When they move apart, the observed frequency is lower.
The Doppler Effect Formula is mathematically expressed as:
\[ f’ = f \left( \frac{v + v_o}{v – v_s} \right) \]
Here, \( f’ \) is the observed frequency, \( f \) is the source frequency, \( v \) is the speed of sound in the medium, \( v_o \) is the speed of the observer, and \( v_s \) is the speed of the source. The sign convention follows the direction of motion towards the other party as positive. This formula is central to understanding acoustics, radar systems, medical ultrasound, and astrophysics.
Doppler Effect Formula — Expression and Variables
The general Doppler Effect Formula for sound waves is:
\[ f’ = f \left( \frac{v + v_o}{v – v_s} \right) \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( f’ \) | Observed (apparent) frequency | Hertz (Hz) |
| \( f \) | Actual frequency of the source | Hertz (Hz) |
| \( v \) | Speed of sound in the medium | m/s |
| \( v_o \) | Speed of the observer | m/s |
| \( v_s \) | Speed of the source | m/s |
| \( \lambda’ \) | Apparent wavelength | metre (m) |
| \( \Delta f \) | Doppler frequency shift | Hertz (Hz) |
Sign Convention
The sign convention for the Doppler Effect Formula is critical. Motion of the observer towards the source is taken as positive for \( v_o \). Motion of the source towards the observer is taken as positive for \( v_s \) in the denominator (meaning we subtract it, increasing \( f’ \)). Motion away from the other party reverses the sign. Always define the positive direction as from observer to source before substituting values.
Derivation
Consider a source S emitting sound of frequency \( f \) and an observer O. The speed of sound in the medium is \( v \).
Step 1: When both are stationary, the observer receives \( f \) waves per second, each of wavelength \( \lambda = v/f \).
Step 2: If the source moves towards the observer with speed \( v_s \), the wavefronts ahead compress. The apparent wavelength becomes \( \lambda’ = (v – v_s)/f \).
Step 3: If the observer also moves towards the source with speed \( v_o \), the relative speed of waves reaching the observer is \( (v + v_o) \).
Step 4: The observed frequency is then \( f’ = (v + v_o) / \lambda’ \). Substituting \( \lambda’ \):
\[ f’ = f \left( \frac{v + v_o}{v – v_s} \right) \]
This derivation assumes the medium is stationary and the speeds involved are less than the speed of sound.
Complete Physics (Waves) Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| General Doppler Effect | \( f’ = f \left( \dfrac{v + v_o}{v – v_s} \right) \) | f=source freq, v=speed of sound, v_o=observer speed, v_s=source speed | Hz | Class 11, Ch 15 |
| Source towards stationary observer | \( f’ = f \left( \dfrac{v}{v – v_s} \right) \) | v_o = 0 | Hz | Class 11, Ch 15 |
| Source away from stationary observer | \( f’ = f \left( \dfrac{v}{v + v_s} \right) \) | v_o = 0, source receding | Hz | Class 11, Ch 15 |
| Observer towards stationary source | \( f’ = f \left( \dfrac{v + v_o}{v} \right) \) | v_s = 0 | Hz | Class 11, Ch 15 |
| Observer away from stationary source | \( f’ = f \left( \dfrac{v – v_o}{v} \right) \) | v_s = 0, observer receding | Hz | Class 11, Ch 15 |
| Apparent wavelength (source approaching) | \( \lambda’ = \dfrac{v – v_s}{f} \) | v_s = source speed | m | Class 11, Ch 15 |
| Doppler Frequency Shift | \( \Delta f = f’ – f \) | f’=observed freq, f=source freq | Hz | Class 11, Ch 15 |
| Speed of a Wave | \( v = f \lambda \) | f=frequency, λ=wavelength | m/s | Class 11, Ch 15 |
| Speed of Sound in Air (at 0°C) | \( v \approx 332 \) m/s | Standard value at 0°C | m/s | Class 11, Ch 15 |
| Speed of Sound at room temperature | \( v \approx 340 \) m/s | Standard value at ~25°C | m/s | Class 11, Ch 15 |
| Mach Number | \( M = \dfrac{v_{\text{object}}}{v_{\text{sound}}} \) | M=Mach number (dimensionless) | Dimensionless | Class 11, Ch 15 |
Doppler Effect Formula — Solved Examples
Example 1 (Class 9-10 Level): Stationary Observer, Moving Source
Problem: A train horn emits sound at a frequency of 500 Hz. The train moves towards a stationary observer at 20 m/s. The speed of sound in air is 340 m/s. Find the frequency heard by the observer.
Given:
- Source frequency, \( f = 500 \) Hz
- Speed of source, \( v_s = 20 \) m/s (towards observer)
- Speed of observer, \( v_o = 0 \) m/s (stationary)
- Speed of sound, \( v = 340 \) m/s
Step 1: Write the Doppler Effect Formula for a moving source and stationary observer:
\[ f’ = f \left( \frac{v}{v – v_s} \right) \]
Step 2: Substitute the given values:
\[ f’ = 500 \times \left( \frac{340}{340 – 20} \right) = 500 \times \frac{340}{320} \]
Step 3: Simplify the calculation:
\[ f’ = 500 \times 1.0625 = 531.25 \text{ Hz} \]
Answer
The frequency heard by the observer is 531.25 Hz. Since the train is approaching, the observed frequency is higher than the source frequency.
Example 2 (Class 11-12 Level): Both Source and Observer Moving
Problem: A police car siren emits sound at 800 Hz. The car moves towards a person at 30 m/s. The person runs towards the car at 5 m/s. The speed of sound is 340 m/s. Calculate the frequency heard by the person.
Given:
- Source frequency, \( f = 800 \) Hz
- Speed of source, \( v_s = 30 \) m/s (towards observer, so positive in denominator)
- Speed of observer, \( v_o = 5 \) m/s (towards source, so positive in numerator)
- Speed of sound, \( v = 340 \) m/s
Step 1: Apply the general Doppler Effect Formula:
\[ f’ = f \left( \frac{v + v_o}{v – v_s} \right) \]
Step 2: Substitute values:
\[ f’ = 800 \times \left( \frac{340 + 5}{340 – 30} \right) = 800 \times \frac{345}{310} \]
Step 3: Compute the result:
\[ f’ = 800 \times 1.1129 \approx 890.3 \text{ Hz} \]
Step 4: Verify the logic. Both the source and observer move towards each other. The observed frequency must be greater than 800 Hz. Our answer of 890.3 Hz confirms this.
Answer
The frequency heard by the person is approximately 890.3 Hz.
Example 3 (JEE/NEET Level): Finding Source Speed from Observed Frequencies
Problem: A source of sound moves along a straight line. A stationary observer first measures the frequency as 640 Hz when the source approaches, and then as 480 Hz after the source passes and recedes. The speed of sound is 320 m/s. Find the speed of the source and the actual frequency emitted.
Given:
- Frequency when approaching, \( f_1 = 640 \) Hz
- Frequency when receding, \( f_2 = 480 \) Hz
- Speed of sound, \( v = 320 \) m/s
- Observer stationary: \( v_o = 0 \)
Step 1: Write the Doppler formula for both cases.
Approaching: \( f_1 = f \left( \dfrac{v}{v – v_s} \right) \)
Receding: \( f_2 = f \left( \dfrac{v}{v + v_s} \right) \)
Step 2: Divide the two equations to eliminate \( f \):
\[ \frac{f_1}{f_2} = \frac{v + v_s}{v – v_s} \]
Step 3: Substitute numerical values:
\[ \frac{640}{480} = \frac{320 + v_s}{320 – v_s} \]
\[ \frac{4}{3} = \frac{320 + v_s}{320 – v_s} \]
Step 4: Cross-multiply and solve:
\[ 4(320 – v_s) = 3(320 + v_s) \]
\[ 1280 – 4v_s = 960 + 3v_s \]
\[ 320 = 7v_s \implies v_s = \frac{320}{7} \approx 45.7 \text{ m/s} \]
Step 5: Find the actual frequency \( f \) using the approaching case:
\[ f = f_1 \times \frac{v – v_s}{v} = 640 \times \frac{320 – 45.7}{320} = 640 \times \frac{274.3}{320} \approx 548.6 \text{ Hz} \]
Alternatively, using the harmonic mean: \( f = \sqrt{f_1 \times f_2} = \sqrt{640 \times 480} = \sqrt{307200} \approx 554.3 \) Hz (approximate method). The exact method from Step 5 gives 548.6 Hz.
Answer
The speed of the source is approximately 45.7 m/s and the actual emitted frequency is approximately 548.6 Hz.
CBSE Exam Tips 2025-26
- Master the sign convention first. Most marks are lost due to incorrect signs. Always define the positive direction as from observer to source before writing the formula. We recommend drawing a diagram for every problem.
- Memorise all four special cases. CBSE board exams frequently test the four standard cases: source approaching, source receding, observer approaching, and observer receding. Knowing each sub-formula saves time.
- Use ratio method for two-frequency problems. When both the approaching and receding frequencies are given, divide the two Doppler equations. This eliminates the unknown actual frequency and simplifies the algebra significantly.
- Remember the standard speed of sound. Use \( v = 340 \) m/s at room temperature unless the question specifies otherwise. In 2025-26 CBSE papers, the value is usually given in the question.
- Distinguish between frequency and wavelength changes. The Doppler Effect changes the observed frequency. The actual wavelength emitted by the source changes only if the source is moving. Practice questions that ask for apparent wavelength separately.
- Limitations are asked in theory questions. The Doppler Effect Formula is valid only when speeds are much less than the speed of sound and the motion is along the line joining source and observer. State these limitations clearly in 2-mark theory answers.
Common Mistakes to Avoid
- Incorrect sign convention: Students often add \( v_s \) in the numerator instead of the denominator. Remember: \( v_s \) always appears in the denominator of the Doppler Effect Formula. Approaching source decreases the denominator (\( v – v_s \)), increasing \( f’ \).
- Confusing observer and source speeds: \( v_o \) goes in the numerator and \( v_s \) goes in the denominator. Swapping them is a very common error that leads to completely wrong answers.
- Using the wrong value of speed of sound: Do not assume 340 m/s if the problem gives a different medium or temperature. Speed of sound in water is approximately 1500 m/s. Always read the problem carefully.
- Applying the formula when speeds exceed the speed of sound: The standard Doppler Effect Formula breaks down at supersonic speeds. If \( v_s \geq v \), the formula gives a zero or negative denominator, which is physically meaningless.
- Forgetting that the formula applies to the line of motion only: The standard formula assumes motion along the straight line joining source and observer. For motion at an angle, a cosine factor must be included. Ignoring this in advanced problems costs marks in JEE.
JEE/NEET Application of Doppler Effect Formula
In our experience, JEE aspirants encounter the Doppler Effect Formula in at least one question per year, often in the context of multi-concept problems combining waves, relative motion, and sometimes electromagnetism (for light). NEET questions tend to be more straightforward, focusing on direct application of the formula with clear numerical values.
Pattern 1: Two-Frequency Problems (JEE Main Favourite)
JEE Main frequently provides the frequency heard when a source approaches and when it recedes. Students must set up two simultaneous equations using the Doppler Effect Formula and divide them. This eliminates the actual frequency and lets you solve for \( v_s \). Practice this pattern extensively. The key step is:
\[ \frac{f_1}{f_2} = \frac{v + v_s}{v – v_s} \]
Pattern 2: Doppler Effect with Reflection (JEE Advanced)
In some JEE Advanced problems, a sound source moves towards a wall. The wall acts as a secondary source and reflects sound back. The observer (often the original source itself) hears a beat frequency due to the difference between the direct and reflected sounds. Apply the Doppler Effect Formula twice: once for the wave hitting the wall and once for the reflected wave reaching the observer. The beat frequency equals \( f_{\text{reflected}} – f_{\text{direct}} \).
Pattern 3: Medical Ultrasound and Radar (NEET Application)
NEET biology-physics interface questions sometimes mention how ultrasound machines use the Doppler Effect to measure blood flow velocity. The formula used is a modified version:
\[ \Delta f = \frac{2 f v_{\text{blood}} \cos\theta}{v_{\text{sound}}} \]
Here \( \theta \) is the angle between the ultrasound beam and the direction of blood flow. Understanding this conceptually, rather than memorising the formula, is sufficient for NEET 2025-26. Similarly, radar guns use the Doppler Effect with electromagnetic waves to measure vehicle speed.
Limitations of the Doppler Effect Formula
Understanding the limitations is important for both CBSE theory questions and JEE conceptual problems:
- The formula is valid only when the speed of the source or observer is much less than the speed of sound (\( v_s \ll v \) and \( v_o \ll v \)).
- It assumes the medium is stationary. If the medium itself is moving (e.g., wind), corrections must be applied.
- The standard formula applies only when the motion is along the line joining the source and the observer.
- It does not apply when the source speed equals or exceeds the speed of sound (sonic boom regime).
- For light waves, the relativistic Doppler formula must be used instead of the classical formula.
Real-World Uses of the Doppler Effect
- RADAR and SONAR: Speed guns used by traffic police and weather radar systems both rely on the Doppler Effect to measure the speed of objects.
- Medical Imaging: Doppler echocardiography measures blood flow and heart valve function non-invasively.
- Astronomy: The redshift of light from distant galaxies is a Doppler Effect that proves the universe is expanding (Hubble's Law).
- Satellite Communication: Engineers account for Doppler shifts when communicating with moving satellites.
FAQs on Doppler Effect Formula
Explore more related topics on ncertbooks.net. Read our detailed article on the Uniform Circular Motion Formula to strengthen your understanding of wave and motion concepts. You can also revise the Refractive Index Formula for wave optics and the Electric Field Formula for electrostatics. For the complete collection, visit our Physics Formulas hub. For official NCERT resources, refer to the NCERT official website.