Q.When a piece of iron is heated in a hot flame, it first becomes dull red, then reddish yellow and finally white hot. This phenomenon can be explained by

The Doppler Effect: When the Source Moves

Imagine you're standing by a road. A car with its horn blaring races toward you. The pitch sounds high and urgent. The instant it passes you and speeds away, the pitch drops to a lower, flatter note. The horn itself hasn't changed — the car is playing the same note the whole time. What changed? The motion of the source.

That shift in pitch is the Doppler effect. It happens whenever a wave source moves relative to an observer. Sound waves, light waves, even water ripples — all of them get compressed or stretched by motion.

The Intuition: Why Does the Frequency Change?

Think of a stationary source — say, a speaker — sending out sound waves. Each wave crest leaves the source and travels outward at the speed of sound $v$. The distance between successive crests is the wavelength $\lambda$, and the number of crests leaving per second is the frequency $f$. For a stationary source, the waves are evenly spaced in all directions.

Now imagine the source moves toward you. Between one wave crest and the next, the source itself has moved a little closer to you. So the second crest doesn't have to travel as far to reach you as the first one did. The crests bunch up — the wavelength shortens. Since the wave speed $v$ hasn't changed (sound speed depends on the medium, not the source), a shorter wavelength means more crests hit your ear per second. You hear a higher frequency.

When the source moves away, the opposite happens. Each successive crest is launched from a point farther from you. The crests stretch out — wavelength increases — and fewer crests reach you per second. The frequency drops.

The Precise Statement

Let:

• $f$ = frequency emitted by the source (the "true" frequency)
• $f'$ = frequency heard by the observer
• $v$ = speed of sound in the medium
• $v_s$ = speed of the source (relative to the medium)

Case 1: Source moving toward a stationary observer

The source chases its own waves. In one second, the source emits $f$ waves. But because it moves toward you, those $f$ waves are packed into a distance $v - v_s$ instead of $v$. The observed wavelength is $\lambda' = \frac{v - v_s}{f}$. Since $v = f' \lambda'$, we get:

Because $v - v_s < v$, the denominator is smaller than the numerator, so $f' > f$. The pitch rises.

Case 2: Source moving away from a stationary observer

Now the source recedes. The waves stretch out. In one second, the $f$ emitted waves are spread over a distance $v + v_s$. The observed wavelength is $\lambda' = \frac{v + v_s}{f}$, so:

Here $v + v_s > v$, so $f' < f$. The pitch falls.

What the Formula Tells You

The ratio $\frac{v}{v \mp v_s}$ is always greater than 1 when the source approaches, and less than 1 when it recedes. The faster the source moves, the more extreme the shift. If $v_s$ gets close to $v$ (the source approaches the speed of sound), the denominator $v - v_s$ becomes tiny, and $f'$ becomes huge — the sound gets extremely high-pitched. That's the buildup just before a sonic boom. …