Doppler effect for sound waves

1. Overview

The Doppler effect is the apparent change in the observed frequency (and wavelength) of a wave when there is relative motion between the source of the waves and the observer. In the context of Cambridge A-Level Physics (9702), this phenomenon is specifically examined for sound waves where a source moves at a constant velocity relative to a stationary observer.

The fundamental principle underlying the Doppler effect is that the speed of sound ($v$) in a specific medium is determined solely by the properties of that medium (such as temperature and density). Therefore, the speed of the sound waves through the air remains constant regardless of how fast the source is moving. However, because the source is moving while it emits successive wavefronts, the physical distance between those wavefronts (the wavelength) is altered in the direction of motion. This change in wavelength results in a corresponding change in the frequency detected by the observer, which is perceived as a change in pitch.

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Key Definitions

To secure full marks in descriptive questions, use the following precise definitions:

• Doppler effect: The apparent change in the observed frequency (or wavelength) of a wave when there is relative motion between the source and the observer.
• Source frequency ($f_s$): The actual frequency of the sound waves being emitted by the source, measured in hertz (Hz). This is the frequency an observer would hear if they were moving at the same velocity as the source.
• Observed frequency ($f_o$): The frequency of the sound waves detected by a stationary observer, measured in hertz (Hz).
• Source speed ($v_s$): The speed at which the sound source is moving relative to the medium (usually air), measured in $\text{m s}^{-1}$.
• Wave speed ($v$): The speed at which the wavefronts propagate through the medium, measured in $\text{m s}^{-1}$. For sound in air at standard temperature, this is typically between $330 \text{ m s}^{-1}$ and $340 \text{ m s}^{-1}$.

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Content

3.1 Qualitative Understanding: The Physics of Wavefronts

When a sound source is stationary, it emits spherical wavefronts that expand outward at the speed of sound $v$. These wavefronts are concentric circles, meaning the distance between any two successive wavefronts (the wavelength $\lambda$) is the same in all directions.

When the source begins to move, the situation changes:

1. Approaching the Observer: As the source moves toward a stationary observer, it "chases" the wavefronts it has already emitted. Each new wavefront is emitted from a position closer to the previous wavefront than it would be if the source were stationary. This results in the compression of wavefronts in front of the source.

   • The observed wavelength ($\lambda_o$) decreases.
   • Since $v = f \lambda$ and $v$ is constant, a decrease in wavelength leads to an increase in observed frequency ($f_o$).
   • Perception: The pitch of the sound (e.g., a siren) sounds higher than its actual frequency.

2. Receding from the Observer: As the source moves away from a stationary observer, it moves in the opposite direction to the wavefronts it emits. Each successive wavefront is emitted from a position further away from the previous one. This results in the stretching (rarefaction) of wavefronts behind the source.

   • The observed wavelength ($\lambda_o$) increases.
   • Since $v$ is constant, an increase in wavelength leads to a decrease in observed frequency ($f_o$).
   • Perception: The pitch of the sound sounds lower than its actual frequency.

3.2 Mathematical Derivation

The Cambridge syllabus requires an understanding of how the motion of the source leads to the Doppler equation.

1. Consider a source emitting sound with frequency $f_s$. The time between the emission of successive wavefronts is the period $T$, where $T = 1/f_s$.
2. In the time $T$, the first wavefront travels a distance $d_w = vT$.
3. During this same time interval $T$, the source moves a distance $d_s = v_s T$ toward the observer.
4. The observed wavelength ($\lambda_o$) is the physical distance between these two successive wavefronts in the medium: $$\lambda_o = d_w - d_s$$ $$\lambda_o = vT - v_s T = (v - v_s)T$$
5. Substitute $T = 1/f_s$ into the equation: $$\lambda_o = \frac{v - v_s}{f_s}$$
6. The observer detects the frequency $f_o$ based on the wave speed $v$ and this new wavelength $\lambda_o$: $$f_o = \frac{v}{\lambda_o}$$ $$f_o = \frac{v}{(v - v_s)/f_s}$$ $$f_o = \frac{f_s v}{v - v_s}$$

If the source is moving away from the observer, the source distance $d_s$ is added to the wave distance $d_w$, leading to the denominator $(v + v_s)$.

3.3 The General Equation

For a stationary observer and a source moving at speed $v_s$, the observed frequency $f_o$ is given by:

$$\mathbf{f_o = \frac{f_s v}{v \mp v_s}}$$

• The Minus Sign ($-$): Use when the source is approaching the observer. This makes the denominator smaller, resulting in a higher $f_o$.
• The Plus Sign ($+$): Use when the source is moving away from the observer. This makes the denominator larger, resulting in a lower $f_o$.

3.4 Worked Example 1 — Calculating Frequency of an Approaching Source

Question: A racing car emits a constant engine note at a frequency of $850 \text{ Hz}$. The car travels at a constant speed of $45.0 \text{ m s}^{-1}$ toward a stationary spectator. Taking the speed of sound in air to be $340 \text{ m s}^{-1}$, calculate the frequency heard by the spectator.

Strategy:

1. Identify the variables: $f_s = 850 \text{ Hz}$, $v_s = 45.0 \text{ m s}^{-1}$, $v = 340 \text{ m s}^{-1}$.
2. Determine the sign: The car is approaching, so use the minus sign ($-$) to get a higher frequency.

Working:

• Equation: $f_o = \frac{f_s v}{v - v_s}$
• Substitution: $f_o = \frac{850 \times 340}{340 - 45.0}$
• Intermediate step: $f_o = \frac{289000}{295}$
• Calculation: $f_o = 979.66... \text{ Hz}$

Answer: $f_o = 980 \text{ Hz}$ (to 3 significant figures).

3.5 Worked Example 2 — Finding the Speed of the Source

Question: A stationary observer uses a frequency detector to measure the sound from a departing train. The train's whistle has a known frequency of $500 \text{ Hz}$, but the detector records a frequency of $465 \text{ Hz}$. If the speed of sound is $330 \text{ m s}^{-1}$, calculate the speed of the train.

Strategy:

1. Identify variables: $f_s = 500 \text{ Hz}$, $f_o = 465 \text{ Hz}$, $v = 330 \text{ m s}^{-1}$.
2. Determine the sign: The train is departing (moving away), so use the plus sign ($+$).
3. Rearrange the formula to solve for $v_s$.

Working:

• Equation: $f_o = \frac{f_s v}{v + v_s}$
• Substitution: $465 = \frac{500 \times 330}{330 + v_s}$
• Rearrange: $465(330 + v_s) = 165000$
• Expand: $153450 + 465v_s = 165000$
• Isolate $v_s$: $465v_s = 165000 - 153450$
• $465v_s = 11550$
• $v_s = \frac{11550}{465} = 24.838... \text{ m s}^{-1}$

Answer: $v_s = 24.8 \text{ m s}^{-1}$ (to 3 significant figures).

3.6 Worked Example 3 — The Frequency "Jump"

Question: A police car with a siren of frequency $1200 \text{ Hz}$ passes a stationary observer at a constant speed of $30.0 \text{ m s}^{-1}$. Calculate the total change in frequency heard by the observer as the car passes them. (Speed of sound $v = 340 \text{ m s}^{-1}$).

Strategy:

1. Calculate $f_o$ for the approach ($f_{o,app}$).
2. Calculate $f_o$ for the recession ($f_{o,rec}$).
3. Find the difference: $\Delta f = f_{o,app} - f_{o,rec}$.

Working:

• Step 1 (Approach): $f_{o,app} = \frac{1200 \times 340}{340 - 30} = \frac{408000}{310} = 1316.1 \text{ Hz}$
• Step 2 (Recession): $f_{o,rec} = \frac{1200 \times 340}{340 + 30} = \frac{408000}{370} = 1102.7 \text{ Hz}$
• Step 3 (Difference): $\Delta f = 1316.1 - 1102.7 = 213.4 \text{ Hz}$

Answer: $\Delta f = 213 \text{ Hz}$ (to 3 significant figures).

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Key Equations

Equation	Description	Data Sheet?
$f_o = \frac{f_s v}{v \pm v_s}$	Doppler effect for a moving source. Use ($-$) for approach and ($+$) for moving away.	Yes
$v = f \lambda$	Wave equation. Relates speed, frequency, and wavelength.	Yes
$T = \frac{1}{f}$	Period-frequency relationship. Used in derivations.	Yes

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Common Mistakes to Avoid

• ❌ Sign Errors: Using $+$ for an approaching source.
  • ✓ Right: Always perform a "logic check." If the source is approaching, the frequency must increase. This requires a smaller denominator, so you must subtract $v_s$ ($v - v_s$).
• ❌ Confusing $v$ and $v_s$: Swapping the speed of sound with the speed of the object.
  • ✓ Right: $v$ is the speed of the wave (usually $\approx 340 \text{ m s}^{-1}$). $v_s$ is the speed of the source (usually much smaller, e.g., $20 \text{ m s}^{-1}$).
• ❌ Incorrect Units: Using speeds in $\text{km h}^{-1}$.
  • ✓ Right: All speeds must be in $\text{m s}^{-1}$. To convert $\text{km h}^{-1}$ to $\text{m s}^{-1}$, divide by $3.6$.
• ❌ Applying to Moving Observers: Using the formula when the observer is moving and the source is stationary.
  • ✓ Right: The 9702 syllabus specifically states that understanding of a moving observer is not required. If a question describes a moving observer, look closely—it is likely the source that is actually moving relative to the medium, or it is the relative motion between the source and the medium that matters.
• ❌ Wavelength vs Frequency: Stating that the frequency changes because the speed of sound changes.
  • ✓ Right: The speed of sound $v$ is constant in a uniform medium. The frequency changes because the wavelength is physically compressed or stretched by the source's motion.

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Exam Tips

1. The "Pass-By" Scenario: In many exam questions, a source passes an observer. The frequency does not change continuously as it passes; it stays at a constant high frequency ($f_{o,app}$) while approaching and then drops to a constant lower frequency ($f_{o,rec}$) as it moves away.
2. Data Sheet Values: If the speed of sound is not given in the question, check the "Data and Formulae" page at the front of the exam paper. It is usually listed as $330 \text{ m s}^{-1}$ or $340 \text{ m s}^{-1}$.
3. Significant Figures: Pay close attention to the precision of the values provided. If the speed of sound is $340 \text{ m s}^{-1}$ (2 s.f.) and the frequency is $500 \text{ Hz}$ (3 s.f.), your final answer should generally be given to 2 or 3 s.f.
4. Algebraic Rearrangement: Practice rearranging the Doppler formula to solve for $v_s$ or $v$. This is a common source of algebraic errors under exam pressure.
5. Vector Direction: Remember that $v_s$ is the component of the velocity along the line joining the source and the observer. If a source is moving at an angle, the Doppler shift is reduced (though 9702 usually sticks to direct linear motion).
6. Ratio Method: If you are given two different observed frequencies for the same source, you can often set up a ratio: $$\frac{f_{o,1}}{f_{o,2}} = \frac{v \pm v_{s,2}}{v \pm v_{s,1}}$$ This can save time in complex multiple-choice questions.