Q.Explain Doppler Effect.

The Doppler Effect in Sound – From Intuition to Formula

Imagine you're standing on a railway platform. A train approaches, horn blaring. As it comes toward you, the pitch sounds higher than when it is stationary. The moment it passes and moves away, the pitch drops to something lower. The horn itself hasn't changed — the train driver is pressing the same button. What changed? The relative motion between you and the source.

That shift in pitch is the Doppler Effect.

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Why does the pitch change? The intuition

Sound travels as waves — compressions and rarefactions in air. The frequency you hear is the number of wavefronts hitting your ear per second.

When the source moves toward you, it chases its own sound waves. Each new wave is emitted from a position slightly closer to you than the previous one. The waves get squashed together — the wavelength shortens. Shorter wavelength means higher frequency, so you hear a higher pitch.

When the source moves away from you, each wave is emitted from a position slightly farther. The waves get stretched — wavelength lengthens, frequency drops, pitch falls.

The same logic applies if you move toward or away from a stationary source. Your motion relative to the wavefronts changes how many hit you per second.

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The precise statement

When there is relative motion between a sound source and an observer, the observed frequency $f'$ is different from the emitted frequency $f$. The observed frequency depends on the velocities of the source and observer relative to the medium (air).

The general formula is:

Where:

• $f$ = original frequency emitted by the source
• $v$ = speed of sound in the medium (air, typically $340\ \text{m/s}$)
• $v_o$ = speed of the observer relative to the medium
• $v_s$ = speed of the source relative to the medium

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The sign convention — the only tricky part

The signs are chosen so that approach raises frequency and retreat lowers frequency.

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Worked example

A train horn emits sound at $500\ \text{Hz}$. The train moves at $30\ \text{m/s}$ toward a stationary observer. Speed of sound is $340\ \text{m/s}$. What frequency does the observer hear?

Here $v_o = 0$, $v_s = 30\ \text{m/s}$, source moving toward observer → denominator gets minus.

The pitch is higher, as expected.

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What the formula does NOT cover

• It assumes the source and observer move along the line joining them. If motion is at an angle, only the component along the line matters. …