The Doppler effect

The siren that changes its note

Stand by a road while an ambulance races past with its siren on. As it comes towards you the note is high. The moment it passes, the note drops. The driver hears the same note all the time. The siren itself does not change.

This change of frequency with motion is the Doppler effect. In this lesson we find how large it is, for sound.

A moving source

A source of frequency $f$ sends one crest every $T = 1/f$ seconds. Each crest spreads out at the speed of sound $v$, measured in the air. Now let the source move at speed $v_s$ towards a listener A.

Between two crests the source moves forward by $v_sT$. So the crests in front are closer together. The wavelength in front is

The crests still travel at $v$, so A hears

That is higher than $f$. Behind the source, for listener B, the crests are further apart, and $f' = f\,\dfrac{v}{v + v_s}$, which is lower.

A worked example. An ambulance siren of $700\ \text{Hz}$ moves at $20\ \text{m/s}$.

• Coming towards you: $f' = 700\ \text{Hz} \times \dfrac{340\ \text{m/s}}{320\ \text{m/s}} = 744\ \text{Hz}$.
• Going away: $f' = 700\ \text{Hz} \times \dfrac{340\ \text{m/s}}{360\ \text{m/s}} = 661\ \text{Hz}$.

As it passes, the note drops by about $83\ \text{Hz}$.

A moving listener

Now the source is still and the listener moves towards it at $v_o$. The wavelength does not change. But the listener runs into the crests, so they reach the listener at a speed $v + v_o$ relative to the listener. The frequency heard is

Moving away, it is $f\,\dfrac{v - v_o}{v}$.

Both move

Put the two together. With the source and the listener moving along the line between them,

The sign rule. Use the upper sign when that body moves towards the other one. Moving towards always raises the frequency. Check each sign with that idea.

• The listener's speed goes on the top. Towards the source: $+v_o$.
• The source's speed goes on the bottom. Towards the listener: $-v_s$.
• All speeds are measured relative to the air.

Notice that the two cases are not the same. A source moving at $20\ \text{m/s}$ towards you raises $700\ \text{Hz}$ to $744\ \text{Hz}$. You moving at $20\ \text{m/s}$ towards a still source hear $741\ \text{Hz}$. For speeds small compared with $v$ they are nearly equal: $\dfrac{\Delta f}{f} \approx \dfrac{v_s}{v}$.

A steady wind along the line just changes the speed of the crests. Use $v + w$ in place of $v$ when the wind blows from the source towards the listener.

Push the source faster. The circles crowd together in front of it, and the frequency at A rises. At B it falls. Now let A run towards the source as well. The frequency at A rises a little more.

Sound bounced off a moving car

A detector at rest sends sound of frequency $f$ towards a car that comes towards it at $u$. The echo comes back with a higher frequency. Work it in two steps.

1. The car is a listener moving towards the source. It receives $f_1 = f\,\dfrac{v + u}{v}$.
2. The car sends that sound back, as a source moving towards the detector. The detector hears $f_2 = f_1\,\dfrac{v}{v - u}$.

Together, $f_2 = f\,\dfrac{v + u}{v - u}$. A bat uses this. It hears how much its own call has changed, and knows how fast it is closing in on an insect or a wall.

Practise with new numbers

Try it in Python

The program finds what you hear from a $700\ \text{Hz}$ siren, coming and going, at four speeds.

Common mistakes

• Swapping the signs. Moving towards raises the frequency. Check the answer is higher.
• Using the speed between the two bodies. The rule needs each speed relative to the air.
• Treating a moving source like a moving listener. The formulas differ. The source changes the wavelength. The listener changes how fast the crests arrive.
• Doing the echo in one step. The reflector is first a listener, then a source.

Remember these

• Moving source: $f' = f\,\dfrac{v}{v \mp v_s}$, minus when it comes towards you.
• Moving listener: $f' = f\,\dfrac{v \pm v_o}{v}$, plus when you move towards it.
• Both: $f' = f\,\dfrac{v \pm v_o}{v \mp v_s}$, upper signs for towards.
• Echo from a reflector at speed $u$ towards you: $f\,\dfrac{v + u}{v - u}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A whistle of $600\ \text{Hz}$ comes towards you at $34\ \text{m/s}$. What do you hear?

2. While the whistle comes towards you at a steady speed, does the note you hear keep rising?

3. You and a still source are $100\ \text{m}$ apart. You walk away from it. Does the frequency you hear rise or fall?

4. Why does an echo from a car coming towards you change in two steps?