use the expression fο = f sv / (v ± vs) for the observed frequency when a source of sound waves moves relative to a stationary observer

Doppler Effect for Sound Waves

Learning Objective

Use the expression

\$fo = \frac{fs\,v}{v \pm v_s}\$

to determine the observed frequency when a source of sound moves relative to a stationary observer.

1. Conceptual Overview

The Doppler effect describes the change in frequency (or pitch) of a wave for an observer moving relative to the source of the wave. For sound, the effect is most noticeable when either the source or the observer (or both) are moving at speeds comparable to the speed of sound in the medium.

2. Derivation of the Formula for a Moving Source

Assume a stationary observer and a source moving directly towards or away from the observer.

1. Let \$f_s\$ be the emitted frequency of the source.
2. The wavelength produced while the source moves with speed \$v_s\$ is altered because successive wave fronts are emitted from different positions.
3. When the source moves towards the observer, the effective wavelength \$\lambda'\$ is reduced:

   \$\lambda' = \lambda - \frac{vs}{fs} = \frac{v}{fs} - \frac{vs}{fs} = \frac{v - vs}{f_s}\$

4. The observer measures frequency \$f_o = \dfrac{v}{\lambda'}\$, giving

   \$fo = \frac{v}{\dfrac{v - vs}{fs}} = \frac{fs\,v}{v - v_s}\$

5. If the source moves away, the sign of \$v_s\$ is reversed, leading to

   \$fo = \frac{fs\,v}{v + v_s}\$

Both cases can be written compactly as

\$fo = \frac{fs\,v}{v \pm v_s}\$

where the upper sign (−) is used when the source approaches the observer and the lower sign (+) when it recedes.

3. Sign Convention Summary

4. Applying the Formula – Worked Example

Problem: A police siren emits a sound of \$fs = 800\ \text{Hz}\$. The speed of sound in air is \$v = 340\ \text{m s}^{-1}\$. The siren moves towards a stationary listener at \$vs = 30\ \text{m s}^{-1}\$. What frequency does the listener hear?

1. Identify the correct sign: source is approaching → use “−”.
2. Insert values into the formula:

   \$f_o = \frac{800 \times 340}{340 - 30} = \frac{272000}{310} \approx 877\ \text{Hz}\$

3. Interpretation: The listener perceives a higher pitch (877 Hz) than the emitted 800 Hz.

5. Comparison with a Moving Observer

When the observer moves while the source remains stationary, the observed frequency is given by

\$fo = fs\left(1 \pm \frac{v_o}{v}\right)\$

where \$v_o\$ is the observer’s speed (positive when moving towards the source). This note focuses on the moving‑source case, but the two formulas are analogous.

6. Common Misconceptions

• “The Doppler effect depends on the relative speed only.” – For sound, the medium matters; the speed of the source and observer are taken with respect to the medium, not each other.
• “The sign is always negative for an approaching source.” – The sign convention in the formula \$v \pm v_s\$ uses “−” for approach and “+” for recession; mixing them up leads to inverted results.
• “Frequency changes for the source, not the wavelength.” – The source’s emitted frequency \$f_s\$ remains constant; the wavelength in the medium changes, which the observer interprets as a frequency shift.

7. Summary Checklist

• Identify whether the source is moving towards or away from the observer.
• Choose the correct sign in \$v \pm v_s\$ (− for approach, + for recession).
• Insert known values for \$fs\$, \$v\$, and \$vs\$ into \$fo = \frac{fs v}{v \pm v_s}\$.
• Calculate \$fo\$ and interpret whether the pitch is higher or lower than \$fs\$.

8. Practice Questions

1. A train horn emits \$f_s = 500\ \text{Hz}\$. The train travels at \$20\ \text{m s}^{-1}\$ towards a listener. Take \$v = 340\ \text{m s}^{-1}\$. Find the observed frequency.
2. A fireworks display produces a sound of \$fs = 1000\ \text{Hz}\$. The explosion occurs \$150\ \text{m}\$ away and the sound source moves away from a stationary observer at \$15\ \text{m s}^{-1}\$. Determine \$fo\$.
3. Explain why the Doppler shift for sound is different from that for light, referencing the role of the medium.