A siren on a pole by a straight road sounds a steady note of 510 Hz…

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Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A siren on a pole by a straight road sounds a steady note of $510\ \text{Hz}$ . Far along the road stands a large flat wall, facing the siren. A cyclist rides along the road at a steady speed, away from the siren and towards the wall, as in the figure. The cyclist hears the sound coming straight from the siren and its echo from the wall, and counts $6.0$ beats per second. The speed of sound in still air is $340\ \text{m/s}$ .

A straight road seen from the side. On the left a siren on a pole sounds a note of 510 Hz. Far to the right stands a large flat wall facing the siren. Between them a cyclist rides towards the wall, shown by an arrow pointing right. Wavy lines show sound going from the siren to the cyclist, and an echo coming back from the wall to the cyclist.

What is the speed of the cyclist?
What frequency does the cyclist hear in the sound that comes straight from the siren?
What frequency does the cyclist hear in the echo?
On another day a steady wind of $10\ \text{m/s}$ blows along the road, from the siren towards the wall. The cyclist rides at the same speed. What frequency does the cyclist now hear in the echo?
How many beats per second does the cyclist hear in the wind?

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. The siren and the wall are both at rest, and both send the cyclist sound of frequency $f$ . The cyclist moves away from one and towards the other. In a wind, use the speed of the sound over the ground in each direction.

Step 1. Speed (a). Moving away from the siren the cyclist hears $f\dfrac{c - u}{c}$ ; moving towards the wall, $f\dfrac{c + u}{c}$ . The beat frequency is $\dfrac{2fu}{c}$ , so

\displaystyle u = \frac{6.0\ \text{Hz} \times 340\ \text{m/s}}{2 \times 510\ \text{Hz}} = 2.0\ \text{m/s}

Step 2. The two notes (b, c). $f_d = 510\ \text{Hz} \times \dfrac{338\ \text{m/s}}{340\ \text{m/s}} = 507.0\ \text{Hz}$ and $f_e = 510\ \text{Hz} \times \dfrac{342\ \text{m/s}}{340\ \text{m/s}} = 513.0\ \text{Hz}$ .

Step 3. The echo in the wind (d). Between two bodies at rest a wind does not change the frequency, so the wall still sends back $510\ \text{Hz}$ . The echo travels against the wind at $330\ \text{m/s}$ over the ground, and the cyclist meets it at $332\ \text{m/s}$ :

\displaystyle f_e' = 510\ \text{Hz} \times \frac{332\ \text{m/s}}{330\ \text{m/s}} = 513.09\ \text{Hz}

Step 4. Beats in the wind (e). The direct sound now travels with the wind at $350\ \text{m/s}$ , and the cyclist runs from it, so it passes the cyclist at $348\ \text{m/s}$ : $f_d' = 510\ \text{Hz} \times \dfrac{348\ \text{m/s}}{350\ \text{m/s}} = 507.09\ \text{Hz}$ . The beats are $513.09\ \text{Hz} - 507.09\ \text{Hz} = 6.01\ \text{Hz}$ .

Answer: (a) $2.0\ \text{m/s}$ . (b) $507.0\ \text{Hz}$ . (c) $513.0\ \text{Hz}$ . (d) $513.09\ \text{Hz}$ . (e) $6.01\ \text{Hz}$ .

Check. The wind moves each note by about $0.1\ \text{Hz}$ , in the same direction, so the beats hardly change. To first order the beat frequency is $2fu/c$ whatever the wind.

Watch out. Do not add the wind to the cyclist's speed. The wind changes the speed of the sound, and the listener's formula is $f(c_{\text{ground}} \pm u)/c_{\text{ground}}$ .

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