Doppler Effect — The Calculations, Not Just the Idea
Every student can explain the Doppler effect in one line: a siren sounds higher-pitched as it approaches and lower-pitched as it moves away. The marks, though, are lost on the actual substitution — deciding which speed goes in the numerator, which goes in the denominator, and whether a term is added or subtracted. This guide gives the exact formula, the sign convention spelt out plainly, and four fully worked numericals covering every combination examiners ask.
The Doppler effect formula (sound)
Here f is the frequency the source actually emits, f′ is the frequency the observer hears, v is the speed of sound in the medium, vo is the observer's speed and vs is the source's speed. We use v = 340 m/s for sound in air throughout this article — the standard value used at CBSE/JEE level; note that the exact speed of sound rises with temperature (about 331 m/s at 0°C, closer to 346 m/s at 25°C), so some textbooks use slightly different numbers. Always use whichever value your syllabus specifies and stay consistent through the whole problem.
The sign convention — this is where marks are actually lost
| Motion | Effect on the term | Why |
|---|---|---|
| Observer moves toward the source | +vo in the numerator | Observer meets more wavefronts per second → f′ increases |
| Observer moves away from the source | −vo in the numerator | Fewer wavefronts reach the observer per second → f′ decreases |
| Source moves toward the observer | −vs in the denominator | Wavefronts bunch up ahead of the source → f′ increases |
| Source moves away from the observer | +vs in the denominator | Wavefronts spread out behind the source → f′ decreases |
The rule that never fails: whichever situation makes the observer and source get closer together pushes f′ up (numerator bigger, denominator smaller); whichever makes them move apart pushes f′ down. Work out "closer or farther" first, then fill in the signs — do not try to memorise the signs on their own.
Worked example 1 — source approaching a stationary observer
An ambulance siren emits f = 500 Hz. It moves toward a stationary observer at vs = 20 m/s. Source approaching → use −vs in the denominator, and vo = 0 since the observer is stationary.
f′ = f × v / (v − vs) = 500 × 340 / (340 − 20) = 500 × 340 / 320
340 ÷ 320 = 1.0625, so f′ = 500 × 1.0625 = 531.25 Hz
The pitch rises from 500 Hz to about 531 Hz as the ambulance approaches.
Worked example 2 — the same source moving away
Same ambulance, same siren, now moving away from the observer at 20 m/s. Source receding → use +vs in the denominator.
f′ = f × v / (v + vs) = 500 × 340 / (340 + 20) = 500 × 340 / 360
340 ÷ 360 = 0.9444, so f′ = 500 × 0.9444 = 472.2 Hz
The pitch drops below the true 500 Hz once the ambulance has passed — this sudden fall from 531 Hz to 472 Hz as it passes you is the familiar "nee-naw" effect.
Worked example 3 — observer moving toward a stationary source
A person runs toward a stationary temple bell at vo = 5 m/s. The bell rings at f = 300 Hz. Observer approaching → use +vo in the numerator, and vs = 0 since the source is stationary.
f′ = f × (v + vo) / v = 300 × (340 + 5) / 340 = 300 × 345 / 340
345 ÷ 340 = 1.01471, so f′ = 300 × 1.01471 = 304.4 Hz
The shift is small here because the observer's own speed (5 m/s) is tiny compared to the speed of sound — running toward a sound source changes the pitch far less than the source itself moving at the same speed would, because vo only affects the numerator.
Worked example 4 — both source and observer moving toward each other
A train (the source) approaches a person standing on the platform who starts running toward the train. Train speed vs = 30 m/s, person's speed vo = 5 m/s, whistle frequency f = 600 Hz. Both are getting closer together, so both effects push f′ up: +vo in the numerator, −vs in the denominator.
f′ = f × (v + vo) / (v − vs) = 600 × (340 + 5) / (340 − 30) = 600 × 345 / 310
345 ÷ 310 = 1.11290, so f′ = 600 × 1.11290 = 667.7 Hz
Notice how much larger this shift is than example 3 alone — both terms working in the same direction compound the effect.
Common mistakes that cost marks
- Swapping the numerator and denominator terms. vo (observer) always sits in the numerator; vs (source) always sits in the denominator — this never changes, only the sign in front does.
- Choosing the sign before deciding "closer or farther." Work out whether the source and observer are approaching or separating first; the sign follows from that, not the other way round.
- Forgetting to convert units. A speed given in km/h must become m/s (divide by 3.6) before it goes anywhere near the formula — mixing units here is one of the most common silent errors.
- Using the wrong speed of sound. If a question states a value for v (or a temperature to derive it from), use that exact value — do not default to 340 m/s out of habit if the paper has already given you a different number.
- Applying the same formula to light. The Doppler effect for light (used in redshift/blueshift) needs relativistic corrections at high speeds and is not the same expression as the sound formula above — do not mix the two.
Where the Doppler effect appears in exams
| Exam | Typical use |
|---|---|
| CBSE/ICSE Class 11 | Waves chapter — sound Doppler effect, all four motion combinations above |
| JEE Main & Advanced | Numericals combining Doppler shift with reflection off a moving wall or a second observer |
| NEET | Direct single-step substitution questions, usually source-moving-only |
| GATE (Engineering Physics) | Doppler shift applied to radar and ultrasonic flow-measurement problems |
Practise the arithmetic, not just the idea. Every Doppler numerical reduces to one division and one multiplication once the sign is fixed — the Scientific Calculator, the view that opens by default, has the keys you need to check a step like 340 ÷ 320 in one go.
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