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
7.3.1 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)
- Doppler effect is the change in frequency of a wave received by an observer moving relative to the source of waves.
- You may have noticed a change in pitch of the note heard when an emergency vehicle passes you while sounding its siren. The pitch is higher as the vehicle approaches you, and lower as it moves away (recedes). This is an example of the Doppler effect; you can hear the same thing if a train passes at speed while sounding its whistle.
- Above Figure shows that the frequency of waves emitted by a stationary source is same in all directions. Thus, the observer at rest perceives frequency f.
- When the source starts moving towards the observer, shorter wavelength waves are moving towards the observer and therefore, high frequency is observed.
- Similarly, when the source starts to move away from the observer, wavelength becomes longer and the received frequency if less than the actual frequency emitted from the source.
- In nutshell, waves get compressed when source moves towards the observer and expand when the source starts moving away from the observer.
Doppler Effect in light:
- A similar kind of phenomena is observed in electromagnetic or light waves.
- Red shift: When the observer is moving away from the source, the frequency of the waves is reduced. Thus, a shift towards red side of the spectrum is observed ( due to increase in wavelength).
- Blue shift: When the observer is moving towards the source, the frequency of the waves is increased. Thus, a shift towards blue side of the spectrum is observed ( due to decrease in wavelength).
Applications of Doppler effect:
- Radar Guns: It detects the change in the speed of the vehicle from the change in frequency of the reflected radio waves.
- Measuring speed of galaxies in astronomy: The Doppler shift of light from stars and galaxies give information about the speed and direction of their movement.
- Sound navigation and Ranging (SONAR): It is used by ships to detect underwater objects (like submarines) by detecting the change in frequency.
7.3.2 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 frequency and wavelength observed by an observer will change according to the speed vs at which the source is moving relative to the stationary observer.
- Considering the Figure given in the previous section, the wavelength of sound emitted by a stationary source is given as λo = \(\frac{v}{fs}\), where v is the wave speed and fs is the source frequency.
- When the source starts moving with a velocity vs, the wavelength of the signal received by the observer changes to:
Moving away from observer: \(\lambda=f_{s}\frac{v}{v+v_{s}}\) (longer wavelength)
- Therefore, the frequency received by the observer will be given as:
Moving away from observer: \(f_{o}=f_{s}\frac{v}{v + v_{s}}\)
- Trick to remember: Source moving closer: Frequency received will be high (small denominator, v – vs)
- Source moving away: Frequency received will be low (large denominator, v + vs)
- The general formula of Doppler effect is given as:
\(f_{o}=f_{s}\frac{v\pm v_{o}}{v\pm v_{s}}\)
Here, vo is the speed of the observor.
- Case I: Source is moving towards the observer at rest.
\(f_{o}=f_{s}\frac{v}{v- v_{s}}\)
- Case II: Source is moving away from the observer at rest.
\(f_{o}=f_{s}\frac{v}{v+ v_{s}}\)
- Case III: Observor moving away from a stationary source.
\(f_{o}=f_{s}\frac{v- v_{o}}{v}\)
- Case IV: Observor moving towards a stationary source.
\(f_{o}=f_{s}\frac{v+ v_{o}}{v}\)
Question
A train with a whistle that emits a note of frequency 800 Hz is approaching a stationary observer at a speed of 60 m\(s^{-1}\).Calculate the frequency of the note heard by the observer.
Answer/Explanation
Speed of sound in air =330 m\(s^{-1}\)
Step 1 Select the appropriate form of the Doppler equation. Here, the source is approaching the observer so we choose the minus sign:
\(f_{o}=f_{s}\frac{v}{v- v_{s}}\)
Step 2 Substitute values from the question and solve:
fo = \(\frac{800\times 330}{330-60}\) = 978 Hz
So, the observer hears a note whose pitch is raised significantly, because the train is travelling at a speed that is a significant fraction of the speed of sound.
Question
Two cars A and B are moving toward each other at a speed of 140 m/s. If the frequency of the horn emitted by A is 800 Hz, then what is the apparent frequency of the horn heard by the passenger sitting in car B?
Answer/Explanation
The velocity of sound in air is 360 m/s.
\(f_{o}=f_{s}\frac{v+ v_{o}}{v- v_{s}}\)
\(f_{o}=800(\frac{360+ 140}{360- 140})\)Hz
= 1818 Hz.
12.5 The Doppler effect for sound waves
You may have noticed a change in pitch of the note heard when an emergency vehicle passes you while sounding its siren. The pitch is higher as the vehicle approaches you, and lower as it moves away (recedes). This is an example of the Doppler effect; you can hear the same thing if a train passes at speed while sounding its whistle.
Figure 12.11 shows why this change in frequency is observed. It shows a source of sound emitting waves with a constant frequency $f_{\mathrm{S}}$, together with two observers $\mathrm{A}$ and $\mathrm{B}$.
- If the source is stationary (Figure 12.11a), waves arrive at $\mathrm{A}$ and $\mathrm{B}$ at the same rate, and so both observers hear sounds of the same frequency $f_{\mathrm{s}}$.
- If the source is moving towards $\mathrm{A}$ and away from B (Figure 12.11b), the situation is different. From the diagram, you can see that the waves are squashed together in the direction of A and spread apart in the direction of $B$.
Observer A will observe, or detect, waves whose wavelength is shortened. More wavelengths per second arrive at A, and so A observes a sound of higher frequency than $f_{\mathrm{s}}$. Similarly, the waves arriving at B have been stretched out and $\mathrm{B}$ will observe a frequency lower than $f_{\mathrm{s}}$.
Figure 12.11: Sound waves (green lines) emitted at constant frequency by a a stationary source, and $\mathbf{b}$ a source moving with speed $v_{\mathrm{s}}$. The separation between adjacent green lines is equal to one wavelength.
An equation for observed frequency
There are two different speeds involved in this situation. The source is moving with speed $v_{\mathrm{s}}$. The sound waves travel through the air with speed $v$, which is unaffected by the speed of the source. (Remember, the speed of a wave depends only on the medium it is travelling through.)
The frequency and wavelength observed by an observer will change according to the speed $v_{\mathrm{s}}$ at which the source is moving relative to the stationary observer. Figure 12.12 shows how we can calculate the observed wavelength $\lambda_0$ and the observed frequency $f_0$.
The wave sections shown in Figure 12.12 represent the $f_{\mathrm{s}}$ wavelengths emitted by the source in $1 \mathrm{~s}$. Provided the source is stationary (Figure 12.12a), the length of this section is equal to the wave speed $v$. The wavelength observed by the observer is simply:
$
\lambda_0=\frac{v}{f_{\mathrm{s}}}
$
The situation is different when the source is moving away (receding) from the observer (Figure 12.12b).
In $1 \mathrm{~s}$, the source moves a distance $v_{\mathrm{s}}$. Now the section of $f_{\mathrm{s}}$ wavelengths will have a length equal to $v+$ $v_{\mathrm{s}}$.
Figure 12.12: Sound waves, emitted at constant frequency by a a stationary source, and b a source moving with speed $v_{\mathrm{S}}$ away from the observer (that is, the person hearing the sound).
The observed wavelength is now given by:
$
\lambda_0=\frac{\left(v+v_{\mathrm{s}}\right)}{f_{\mathrm{s}}}
$
The observed frequency is given by:
$
f_0=\frac{v}{\lambda_0}=\frac{f_{\mathrm{s}} \times v}{\left(v+v_{\mathrm{s}}\right)}
$
where $f_0$ is the observed frequency, $f_{\mathrm{S}}$ is the frequency of the source, $v$ is the speed of the wave and $v_{\mathrm{S}}$ is the speed of the source relative to the observer.
This shows us how to calculate the observed frequency when the source is moving away from the observer. If the source is moving towards the observer, the section of $f_{\mathrm{s}}$ wavelengths will be compressed into a shorter length equal to $v-v_{\mathrm{S}}$, and the observed frequency will be given by:
$
f_0=\frac{v}{\lambda_0}=\frac{f_{\mathrm{s}} \times v}{\left(v-v_{\mathrm{s}}\right)}
$
We can combine these two equations to give a single equation for the Doppler shift in frequency due to a moving source:
observed frequency, $f_0=\frac{f_{\mathrm{s}} \times v}{\left(v \pm v_{\mathrm{s}}\right)}$
where the plus sign applies to a receding source and the minus sign to an approaching source. Note these important points:
The frequency fs of the source is not affected by the movement of the source.
The speed v of the waves as they travel through the air (or other medium) is also unaffected by the movement of the source.
Note that a Doppler effect can also be heard when an observer is moving relative to a stationary source, and also when both source and observer are moving. There is more about the Doppler effect and light in Chapter 31.
WORKED EXAMPLE
3 A train with a whistle that emits a note of frequency $800 \mathrm{~Hz}$ is approaching a stationary observer at a speed of $60 \mathrm{~m} \mathrm{~s}^{-1}$.
Calculate the frequency of the note heard by the observer.
speed of sound in air $=330 \mathrm{~m} \mathrm{~s}^{-1}$
Step 1 Select the appropriate form of the Doppler equation. Here the source is approaching the observer so we choose the minus sign:
$
f_0=\frac{f_{\mathrm{s}} \times v}{\left(v-v_{\mathrm{s}}\right)}
$
Step 2 Substitute values from the question and solve:
$
\begin{aligned}
f_0 & =\frac{800 \times 330}{(330-60)} \\
& =\frac{800 \times 330}{270} \\
& =978 \mathrm{~Hz} \approx 980 \mathrm{~Hz}
\end{aligned}
$
So, the observer hears a note whose pitch is raised significantly, because the train is travelling at a speed that is a significant fraction of the speed of sound.
Question
10 A plane’s engine emits a note of constant frequency $120 \mathrm{~Hz}$. It is flying away from a stationary observer at a speed of $80 \mathrm{~m} \mathrm{~s}^{-1}$. Calculate:
a the observed wavelength of the sound received by the observer
b its observed frequency.
(Speed of sound in air $=330 \mathrm{~m} \mathrm{~s}^{-1}$.)