Objective: Understand how the motion of a sound source relative to a stationary observer changes the frequency we hear.
When a sound source moves, the waves it creates are compressed in front of it and stretched behind it. This changes the wavelength and therefore the frequency you hear. 🎧🚗
| Scenario | Formula |
|---|---|
| Source moving towards observer | \$f' = f \frac{v}{v - v_s}\$ |
| Source moving away from observer | \$f' = f \frac{v}{v + v_s}\$ |
Picture a runner (the source) shouting a whistle while sprinting towards you. The sound waves in front of the runner are squeezed together, so you hear a higher pitch. When the runner turns and moves away, the waves stretch out, giving a lower pitch. This is exactly what happens with any moving source of sound.
Use the second formula: \$f' = 500 \times \frac{340}{340+20} \approx 500 \times \frac{340}{360} \approx 472\,\text{Hz}\$. The pitch drops.
🔍 Remember: The sign in the denominator changes with direction: minus for approaching, plus for receding.
🧮 Quick check: If \$v_s\$ is small compared to \$v\$, the change in frequency is small.
📚 Practice: Work through problems with different \$v_s\$ values to see the trend.
1️⃣ If a source moves at \$30\,\text{m/s}\$ towards you and emits \$400\,\text{Hz}\$, what frequency do you hear? (Use \$v = 340\,\text{m/s}\$.)
2️⃣ What happens to the observed frequency if the source moves faster?
💡 Answer 1: \$f' = 400 \times \frac{340}{340-30} \approx 400 \times \frac{340}{310} \approx 439\,\text{Hz}\$.
💡 Answer 2: The observed frequency increases when approaching, decreases when receding.
When a sound source moves relative to a stationary observer, the waves get compressed or stretched, changing the frequency you hear. The key formula and the sign convention are all you need to solve exam questions on this topic.