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 (Cambridge AS & A Level Physics 9702 – Topic 7.3)

Learning Objective

Syllabus reference: 7.3 – Use the expression

\[

f{o}= \frac{f{s}\,v}{\,v \pm v_{s}\,}

\]

to determine the frequency heard by a stationary observer when the source of sound moves through a medium.

1. Conceptual Overview

• The Doppler effect is a change in the observed frequency; the source continues to emit the same frequency \(f_{s}\).
• For sound a material medium (air, water, etc.) is required – all speeds are measured with respect to the medium, not directly between source and observer.
• When the source moves, the spacing between successive wave‑crests (the wavelength) is altered; this altered wavelength together with the constant wave speed \(v\) gives the new observed frequency.

2. Derivation – Moving Source, Stationary Observer

1. Source emits waves of frequency \(f{s}\) (period \(T=1/f{s}\)). In a stationary medium the wavelength would be \(\lambda = v/f_{s}\).
2. If the source moves with speed \(v{s}\) toward the observer, during one period it advances a distance \(v{s}T\). The second crest is therefore emitted closer to the first.
3. Effective wavelength in the direction of motion:

   \[

   \lambda' = \lambda - v_{s}T

   = \frac{v}{f{s}} - \frac{v{s}}{f_{s}}

   = \frac{v - v{s}}{f{s}}.

   \]

   For motion away from the observer the source retreats, giving \(\lambda' = (v+v{s})/f{s}\).

4. The observer measures frequency from the wave speed \(v\) and the altered wavelength:

   \[

   f_{o}= \frac{v}{\lambda'}=

   \begin{cases}

   \displaystyle\frac{f{s}\,v}{v - v{s}} & \text{source approaching},\\[8pt]

   \displaystyle\frac{f{s}\,v}{v + v{s}} & \text{source receding}.

   \end{cases}

   \]

5. Both cases are written compactly as

   \[

   f{o}= \frac{f{s}\,v}{\,v \pm v_{s}\,}.

   \]

   The upper sign (‑) is used for an approaching source; the lower sign (+) for a receding source.

3. Sign‑Convention Summary

Caption: Upper sign (‑) = source moving toward the observer; lower sign (+) = source moving away from the observer.

4. Worked Examples

Example 1 – Approaching Source

Problem: A police siren emits \(f{s}=800\;\text{Hz}\). The speed of sound in air is \(v=340\;\text{m s}^{-1}\). The siren moves toward a stationary listener at \(v{s}=30\;\text{m s}^{-1}\). Find the frequency heard.

1. Source is approaching → use “−”.
2. Insert values:

   \[

   f_{o}= \frac{800 \times 340}{340 - 30}= \frac{272\,000}{310}\approx 877\;\text{Hz}.

   \]

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

Example 2 – Receding Source

Problem: A train horn emits \(f{s}=500\;\text{Hz}\). The train travels away from a stationary observer at \(v{s}=20\;\text{m s}^{-1}\). Take \(v=340\;\text{m s}^{-1}\). Find the observed frequency.

1. Source is receding → use “+”.
2. Insert values:

   \[

   f_{o}= \frac{500 \times 340}{340 + 20}= \frac{170\,000}{360}\approx 472\;\text{Hz}.

   \]

3. Interpretation: the pitch heard is lower (472 Hz) because the source is moving away.

5. Comparison with a Moving Observer (Reference Only)

When the source is stationary and the observer moves, the observed frequency is

\[

f{o}= f{s}\left(1 \pm \frac{v_{o}}{v}\right),

\]

where \(v_{o}\) is the observer’s speed (positive when moving toward the source). This relation lies outside the core requirement of syllabus 7.3 and is provided only for completeness.

6. Common Misconceptions

• “The Doppler effect depends only on the relative speed.” – For sound the relevant speeds are measured with respect to the medium, not directly between source and observer.
• “The sign is always negative for an approaching source.” – The compact formula uses “‑” for approach and “+” for recession; mixing them up reverses the result.
• “The source frequency changes.” – The emitted frequency \(f_{s}\) remains constant; only the wavelength in the medium is altered.
• “The Doppler shift is the same for light.” – Light does not require a medium; its shift follows a relativistic formula and is treated later in the optics/quantum sections of the syllabus.

7. Summary Checklist (Exam‑Ready)

1. Identify whether the source is moving toward or away from the stationary observer.
2. Choose the correct sign in \(v \pm v_{s}\) (‑ for approach, + for recession).
3. Confirm the condition \(v_{s}

4. Substitute the known values into

   \[

   f{o}= \frac{f{s}v}{\,v \pm v_{s}\,}.

   \]

5. Calculate \(f{o}\) and state whether the pitch is higher or lower than \(f{s}\).

8. Practice Questions

1. A train horn emits \(f_{s}=500\;\text{Hz}\). The train travels at \(20\;\text{m s}^{-1}\) toward a listener. Take \(v=340\;\text{m s}^{-1}\). Find the observed frequency.
2. A fireworks explosion produces \(f{s}=1000\;\text{Hz}\). The source moves away from a stationary observer at \(15\;\text{m s}^{-1}\). With \(v=340\;\text{m s}^{-1}\), determine \(f{o}\).
3. Explain why the Doppler shift for sound differs from that for light, emphasizing the role of the medium.