The Doppler Shift Formula describes how the observed frequency of a wave changes when the source or the observer is in motion relative to each other. It is expressed as \[ f’ = f \left( \frac{v + v_o}{v – v_s} \right) \] where \( f’ \) is the observed frequency, \( f \) is the source frequency, \( v \) is the wave speed, \( v_o \) is the observer’s velocity, and \( v_s \) is the source’s velocity. This concept is covered in NCERT Class 11 Physics, Chapter 15 (Waves). It is also a frequently tested topic in JEE Main and NEET. This article covers the formula, its derivation, a complete formula sheet, three solved examples, CBSE exam tips, and common mistakes to avoid.
Key Doppler Shift Formulas at a Glance
Quick reference for the most important Doppler effect formulas used in CBSE and competitive exams.
- 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 in wavelength: \( \Delta \lambda = \lambda \dfrac{v_s}{v} \)
- Relativistic Doppler formula (light): \( f’ = f \sqrt{\dfrac{c + v}{c – v}} \)
What is the Doppler Shift Formula?
The Doppler Shift Formula quantifies the apparent change in frequency (or wavelength) of a wave as perceived by an observer who is moving relative to the wave source. This phenomenon is called the Doppler Effect, named after Austrian physicist Christian Doppler who described it in 1842.
In NCERT Class 11 Physics, Chapter 15 (Waves), the Doppler effect is introduced as a fundamental property of all waves, including sound and light. When a source moves towards an observer, the observed frequency is higher than the actual frequency. When it moves away, the observed frequency is lower.
The Doppler Shift Formula applies to sound waves in everyday situations such as a moving ambulance siren or a speeding train. It also applies to light waves in astronomy, where it helps scientists determine whether stars are moving towards or away from Earth. Understanding this formula is essential for CBSE Class 11 board exams and competitive exams like JEE Main and NEET.
Doppler Shift Formula — Expression and Variables
The general Doppler Shift Formula for sound waves is:
\[ f’ = f \left( \frac{v + v_o}{v – v_s} \right) \]
Here, the sign convention is critical. Velocities directed from the observer towards the source are taken as positive for \( v_o \), and velocities directed from the source towards the observer are taken as positive for \( v_s \).
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( f’ \) | Observed (apparent) frequency | Hertz (Hz) |
| \( f \) | Actual (source) frequency | 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 |
| \( \Delta f \) | Frequency shift (\( f’ – f \)) | Hertz (Hz) |
| \( \Delta \lambda \) | Wavelength shift | metre (m) |
Derivation of the Doppler Shift Formula
Consider a source S emitting sound of frequency \( f \) and wavelength \( \lambda = v/f \). The source moves towards a stationary observer O at speed \( v_s \).
Step 1: In one time period \( T = 1/f \), the source moves a distance \( v_s T \) closer to the observer.
Step 2: The effective wavelength received by the observer is compressed:
\[ \lambda’ = \lambda – v_s T = \frac{v}{f} – \frac{v_s}{f} = \frac{v – v_s}{f} \]
Step 3: The observer receives waves at speed \( v \), so the observed frequency is:
\[ f’ = \frac{v}{\lambda’} = \frac{v}{(v – v_s)/f} = f \left( \frac{v}{v – v_s} \right) \]
Step 4: Incorporating observer motion \( v_o \) (observer moving towards source), the relative speed of wave with respect to observer increases to \( v + v_o \), giving the general formula:
\[ f’ = f \left( \frac{v + v_o}{v – v_s} \right) \]
Complete Waves Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| General Doppler Shift Formula | \( f’ = f \left( \dfrac{v + v_o}{v – v_s} \right) \) | f=source freq, v=wave speed, 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 |
| Doppler Wavelength Shift | \( \Delta \lambda = \lambda \dfrac{v_s}{v} \) | λ=wavelength, v_s=source speed | m | Class 11, Ch 15 |
| Wave Speed Formula | \( v = f \lambda \) | v=wave speed, f=frequency, λ=wavelength | m/s | Class 11, Ch 15 |
| Speed of Sound in Air (at 0°C) | \( v = 332 \text{ m/s} \) | Standard value at 0°C | m/s | Class 11, Ch 15 |
| Relativistic Doppler (Light, approaching) | \( f’ = f \sqrt{\dfrac{c + v}{c – v}} \) | c=speed of light, v=relative speed | Hz | Class 12 / JEE |
| Mach Number | \( M = \dfrac{v_s}{v} \) | v_s=source speed, v=sound speed | Dimensionless | Class 11, Ch 15 |
Doppler Shift Formula — Solved Examples
Example 1 (Class 9-10 Level): Moving Ambulance
Problem: An ambulance siren emits sound at a frequency of 700 Hz. The ambulance moves towards a stationary person at 20 m/s. The speed of sound in air is 340 m/s. Find the frequency heard by the person.
Given: \( f = 700 \) Hz, \( v_s = 20 \) m/s, \( v_o = 0 \), \( v = 340 \) m/s
Step 1: Since the observer is stationary and the source moves towards the observer, use:
\( f’ = f \left( \dfrac{v}{v – v_s} \right) \)
Step 2: Substitute the values:
\( f’ = 700 \times \left( \dfrac{340}{340 – 20} \right) = 700 \times \dfrac{340}{320} \)
Step 3: Calculate:
\( f’ = 700 \times 1.0625 = 743.75 \) Hz
Answer
The frequency heard by the person is 743.75 Hz, which is higher than the source frequency because the ambulance is approaching.
Example 2 (Class 11-12 Level): Both Source and Observer Moving
Problem: A train whistle has a frequency of 500 Hz. The train moves towards a person at 30 m/s. The person runs away from the train at 10 m/s. The speed of sound is 340 m/s. Calculate the frequency heard by the person.
Given: \( f = 500 \) Hz, \( v_s = 30 \) m/s (towards observer), \( v_o = 10 \) m/s (away from source), \( v = 340 \) m/s
Step 1: Apply the general Doppler Shift Formula. Since the observer moves away from the source, \( v_o \) is subtracted in the numerator:
\( f’ = f \left( \dfrac{v – v_o}{v – v_s} \right) \)
Step 2: Substitute values:
\( f’ = 500 \times \left( \dfrac{340 – 10}{340 – 30} \right) = 500 \times \dfrac{330}{310} \)
Step 3: Simplify:
\( f’ = 500 \times 1.0645 \approx 532.3 \) Hz
Step 4: Verify the sign convention. The source approaches (denominator decreases, increasing \( f’ \)). The observer recedes (numerator decreases, slightly reducing \( f’ \)). Net effect is still an increase, which is consistent.
Answer
The frequency heard by the person is approximately 532.3 Hz.
Example 3 (JEE/NEET Level): Wavelength Shift and Reflected Sound
Problem: A car moves towards a stationary wall at 20 m/s while sounding a horn of frequency 400 Hz. The speed of sound is 340 m/s. (a) Find the frequency of sound reflected by the wall and heard by the car driver. (b) Find the beat frequency heard by the driver between the direct sound and the reflected sound.
Given: \( f = 400 \) Hz, \( v_s = 20 \) m/s (car towards wall), \( v = 340 \) m/s
Step 1: The wall acts as a stationary observer. The frequency received by the wall (acting as observer):
\( f_{wall} = f \left( \dfrac{v}{v – v_s} \right) = 400 \times \dfrac{340}{340 – 20} = 400 \times \dfrac{340}{320} = 425 \) Hz
Step 2: The wall reflects this frequency as a new source at \( f_{wall} = 425 \) Hz. Now the car (driver) acts as an observer moving towards this stationary “source” (the wall) at 20 m/s:
\( f’ = f_{wall} \left( \dfrac{v + v_o}{v} \right) = 425 \times \dfrac{340 + 20}{340} = 425 \times \dfrac{360}{340} \)
Step 3: Calculate:
\( f’ = 425 \times 1.0588 \approx 450 \) Hz
Step 4: Beat frequency = difference between reflected frequency and direct frequency:
\( f_{beat} = f’ – f = 450 – 400 = 50 \) Hz
Answer
(a) The frequency of reflected sound heard by the driver is 450 Hz. (b) The beat frequency heard is 50 Hz.
CBSE Exam Tips 2025-26
- Memorise sign conventions first: In CBSE 2025-26 exams, sign errors are the most common reason for losing marks. Always write: numerator uses \( + v_o \) when observer moves towards source, and denominator uses \( – v_s \) when source moves towards observer.
- Draw a diagram for every problem: We recommend sketching the positions of source and observer with arrows showing direction. This prevents sign mistakes instantly.
- Learn all four special cases: CBSE frequently tests the individual cases (only source moving, only observer moving). Memorise each case as a separate formula variant.
- Use 340 m/s as the standard speed of sound: Unless the question specifies a different medium or temperature, use \( v = 340 \) m/s for air at room temperature in your calculations.
- Check your answer logically: If the source approaches the observer, \( f’ > f \). If it recedes, \( f’ < f \). Always verify your numerical answer against this logic before writing it.
- Practise beat frequency problems: In 2025-26 board exams, Doppler effect is often combined with beats. Our experts suggest practising at least five such combined problems.
Common Mistakes to Avoid
- Reversing the sign convention: Many students write \( v + v_s \) in the denominator when the source moves towards the observer. This is wrong. The correct denominator is \( v – v_s \) when the source approaches. The wave is compressed, so the effective wavelength decreases.
- Confusing source and observer velocities: Students sometimes substitute \( v_o \) in the denominator and \( v_s \) in the numerator. Always remember: the numerator carries \( v_o \) and the denominator carries \( v_s \).
- Applying the formula to supersonic speeds: The standard Doppler formula is only valid when \( v_s < v \) (subsonic source). If \( v_s \geq v \), a shock wave (sonic boom) is produced. The formula breaks down and gives a negative or infinite value.
- Forgetting the medium for light: The classical Doppler formula applies to sound. For light (electromagnetic waves), the relativistic formula \( f’ = f \sqrt{(c+v)/(c-v)} \) must be used. Do not apply the sound formula to optical Doppler problems.
- Using the wrong value of wave speed: The speed \( v \) in the formula is always the speed of the wave in the medium, not the speed of the source or observer. Students sometimes confuse \( v \) with \( v_s \) when the values are close.
JEE/NEET Application of Doppler Shift Formula
In our experience, JEE aspirants encounter the Doppler Shift Formula in two to three questions per year in JEE Main. NEET also tests this concept in the context of sound waves and medical applications such as ultrasound imaging.
Pattern 1: Reflected Sound and Beat Frequency
JEE Main frequently combines the Doppler effect with the concept of beats. A moving source reflects sound off a stationary wall. The driver hears two frequencies simultaneously: the direct sound and the reflected sound. The beat frequency equals the difference between these two frequencies. This was demonstrated in Example 3 above. Students must apply the Doppler formula twice in such problems.
Pattern 2: Source Moving at an Angle
JEE Advanced tests a more complex scenario where the source moves at an angle \( \theta \) to the line joining source and observer. In this case, only the component of velocity along the line connecting them is used:
\[ f’ = f \left( \frac{v}{v – v_s \cos\theta} \right) \]
This formula is not in NCERT but appears in JEE Advanced. Students should practise it using the concept of resolved velocity components.
Pattern 3: Doppler Effect in Light (Red Shift and Blue Shift)
NEET and JEE Advanced both test the concept of red shift and blue shift of light. When a star moves away from Earth, its light is red-shifted (lower frequency, longer wavelength). When it approaches, the light is blue-shifted (higher frequency, shorter wavelength). The formula used is the relativistic Doppler formula:
\[ f’ = f \sqrt{\frac{c + v}{c – v}} \quad (\text{source approaching}) \]
In our experience, JEE aspirants who master the sign convention for sound Doppler problems find the conceptual leap to light Doppler problems much easier. We recommend studying both together for maximum efficiency.
FAQs on Doppler Shift Formula
We hope this comprehensive guide to the Doppler Shift Formula has helped you build a strong conceptual foundation. For more Physics formulas, explore our Complete Physics Formulas Hub. You may also find these related articles useful: Uniform Circular Motion Formula, Refractive Index Formula, and Electric Field Formula. For the official NCERT syllabus, refer to the NCERT official website.