CIE AS/A Level Physics 7.3 Doppler effect for sound waves Study Notes- 2025-2027 Syllabus
CIE AS/A Level Physics 7.3 Doppler effect for sound waves Study Notes – New Syllabus
CIE AS/A Level Physics 7.3 Doppler effect for sound waves Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on AS/A Level Physics latest syllabus with Candidates should be able to:
- understand that when a source of sound waves moves relative to a stationary observer, the observed frequency is different from the source frequency (understanding of the Doppler effect for a stationary source and a moving observer is not required)
- use the expression \(f_o=f_s \times \frac{V}{V\pm V_s}\) for the observed frequency when a source of sound waves moves relative to a stationary observer
The Doppler Effect (Moving Source and Stationary Observer)
The Doppler Effect is the change in the observed frequency of a wave when the source of the wave is moving relative to a stationary observer. It occurs because the motion of the source causes successive wavefronts to be either compressed (if the source is moving toward the observer) or stretched (if moving away).
Conceptual Understanding:
- When the source moves toward the observer, the wavefronts are closer together, leading to a higher observed frequency (and pitch).
- When the source moves away from the observer, the wavefronts are more spread out, leading to a lower observed frequency (and pitch).
- The speed of the waves through the medium (e.g., sound through air) remains constant.
The Doppler effect arises due to the relative motion between the source and observer, which alters the wavelength reaching the observer, and therefore the frequency heard.
Illustration:
- Source moving toward observer → waves bunch up → shorter wavelength → higher frequency.
- Source moving away from observer → waves spread out → longer wavelength → lower frequency.
Real-life Examples:
- Change in pitch of a passing ambulance siren.
- Whistle of a moving train as it approaches and then recedes.
- Police radar systems measuring vehicle speed (based on frequency shift).
Example
Explain why the sound of a car horn seems higher in pitch as the car approaches you and lower in pitch as it drives away.
▶️ Answer / Explanation
As the car approaches, the sound waves are compressed in front of it, decreasing the wavelength and increasing the frequency heard by the observer — the sound is higher in pitch. As the car moves away, the waves are stretched, increasing the wavelength and decreasing the frequency heard — the sound is lower in pitch.
Mathematical Expression for the Doppler Effect (Moving Source, Stationary Observer)
When a source of sound moves relative to a stationary observer, the observed frequency \( \mathrm{f_o} \) is given by:
\( \mathrm{f_o = f_s \times \dfrac{V}{V \pm V_s}} \)
Where:
- \( \mathrm{f_o} \) = observed frequency
- \( \mathrm{f_s} \) = source frequency (frequency emitted by source)
- \( \mathrm{V} \) = speed of sound in the medium (e.g., air)
- \( \mathrm{V_s} \) = speed of the source
- Use \( +V_s \) when the source is moving away from the observer.
- Use \( -V_s \) when the source is moving toward the observer.
Derivation (Conceptual Overview):
- When the source moves, each wavefront is emitted closer (or farther) from the observer than the previous one.
- This changes the wavelength reaching the observer: \( \mathrm{\lambda’ = \lambda \pm V_s/f_s} \).
- The observer still perceives waves traveling at speed \( \mathrm{V} \), so \( \mathrm{f_o = \dfrac{V}{\lambda’}} \).
- Substituting and simplifying gives \( \mathrm{f_o = f_s \times \dfrac{V}{V \pm V_s}} \).
Key Takeaways:
- Frequency increases when the source approaches (\( -V_s \) in denominator).
- Frequency decreases when the source moves away (\( +V_s \) in denominator).
- Applies only for sound waves with a stationary observer.
Example
A car horn emits a sound of frequency \( \mathrm{f_s = 500 \ Hz} \). If the car moves toward a stationary observer at \( \mathrm{V_s = 20 \ m/s} \) and the speed of sound in air is \( \mathrm{V = 340 \ m/s} \), find the observed frequency.
▶️ Answer / Explanation
Given: \( \mathrm{f_s = 500 \ Hz} \), \( \mathrm{V = 340 \ m/s} \), \( \mathrm{V_s = 20 \ m/s} \).
For a source moving toward the observer:
\( \mathrm{f_o = f_s \times \dfrac{V}{V – V_s}} \)
\( \mathrm{f_o = 500 \times \dfrac{340}{340 – 20}} = 500 \times \dfrac{340}{320} = 531.25 \ Hz} \)
Therefore, the observed frequency is approximately \( \mathrm{531 \ Hz} \).
Example
A train whistle emits a sound of frequency \( \mathrm{f_s = 800 \ Hz} \). The train is moving away from a stationary observer at a speed of \( \mathrm{V_s = 30 \ m/s} \). If the speed of sound in air is \( \mathrm{V = 340 \ m/s} \), calculate the frequency heard by the observer.
▶️ Answer / Explanation
Given: \( \mathrm{f_s = 800 \ Hz} \), \( \mathrm{V = 340 \ m/s} \), \( \mathrm{V_s = 30 \ m/s} \)
Since the source (train) is moving away from the observer, we use the \( +V_s \) sign in the denominator:
\( \mathrm{f_o = f_s \times \dfrac{V}{V + V_s}} \)
\( \mathrm{f_o = 800 \times \dfrac{340}{340 + 30}} = 800 \times \dfrac{340}{370}}
\( \mathrm{f_o = 800 \times 0.919 = 735.2 \ Hz} \)
Therefore, the observer hears a lower frequency of approximately \( \mathrm{735 \ Hz} \).
Interpretation: As the train moves away, each successive wavefront takes slightly longer to reach the observer, increasing the effective wavelength and reducing the perceived frequency — the sound appears lower in pitch.