NORTHERN CAPE DEPARTMENT OF EDUCATION
PHYSICAL SCIENCES
GRADE 12
PHYSICS
TERM II
DOPPLER EFFECT
G. Izquierdo Rodríguez
&
G. Izquierdo Gómez
2020
Ontwikkel deur Prof. G Izquierdo Rodríguez & G Izquierdo Gómez; Gemodereer, Vertaal &
Saamgestel deur W. Henning
Bladsy I
TABLE OF CONTENTS
Lesson 1: Definition of Doppler Effect. Explanation of Doppler effect. Doppler effect equation.
Applications of Doppler effect.
2
Lesson 2: Doppler effect with light. Red shifts and blue shifts.
10
Lesson 3: Revision exercises.
20
SELF-ASSESSMENT…………………………………………………………………………
30
MEMORAMDUM FOR SELF-ASSESSMENT……………………………………………………31
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Page 1
Lesson 1: Definition of Doppler Effect. Explanation of Doppler Effect. Doppler
Effect equation. Applications of Doppler Effect.
Objective:
Learners must be able to:


State the Doppler Effect as the change in frequency (or pitch) of the sound
detected by a listener because the sound source and the listener have different
velocities relative to the medium of sound propagation.
Explain (using appropriate illustrations) the change in pitch observed when a
source moves toward or away from a listener.
𝑣±𝑣
Solve problems using the equation 𝑓𝐿 = ( 𝐿 ) 𝑓𝑠 when EITHER the source or the

listener is moving.
State applications of the Doppler Effect.

𝑣±𝑣𝑠
Introduction:
From your experience you have noticed that if a police car is parked in the side of the
highway, sounding its 1000 Hz siren and if you are also parked in the highway, you will
hear the same frequency.
But if there is relative motion between you and the police car, either toward or away
from each other, you will hear a different frequency (pitch). For example if you are
driving toward the police car at 120 km·h-1you will hear a higher frequency (1096 Hz an
increase by 96 Hz).
If you are driving away from the police car at the same speed, you will hear a lower
frequency (904 Hz a decrease of 96 Hz).
Development:
According to what we saw in the example above when a source of sound is moving with
respect to an observer or an observer is moving with respect to the source, the pitch of
the sound heard by the observer is different from the pitch of the source.
The change in frequency caused by the relative motion is called the Doppler Effect.
The effect was proposed (although not full workout) in 1842 by the Austrian physicist
Johanm Christian Doppler.
It was tested experimentally in 1845 by Buys Bullot in Holland.
The Doppler effect is the change in frequency (or pitch) of the sound detected by
a listener because the sound source and the listener have different velocities
relative to the medium of sound propagation.
The change in pitch observed when a source of sound moves toward or away from a
listener
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First experiment:
When the source and the listener are both stationary the frequency emitted by the
source equal to the frequency observed by the listener.
Second experiment:
When the bird is flying the wavelength in front of the bird decreases and the wavelength
behind the bird increases.
Science frequency and wavelength are inversely proportional as speed of sound is
constant (v=fλ), the frequency in front of the bird increases and the frequency behind
the bird decreases.
When the source moves toward the listener, the frequency increases. When the
source moves away from the listener, the frequency decreases.
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Explanation:
When the bird approaches the listener the sound waves emitted by the bird are
compressed in front of the bird; more sound waves reach the listener per second and
the pitch (frequency) appears to be higher than the sound emitted by the source (bird).
The opposite occurs when the bird flies away.
Third experiment:
When the listener moves towards the source, the frequency observed increases.
When the listener moves away from the source, the frequency observed decreases.
We can conclude that:
The frequency heard by the observer is higher than the frequency of the source
when the source moves towards the observer and lower when the source moves
away from the source.
When the observer moves then the effective speed of sound changes. The
wavelength remains fixed, but as the observer approaches the source, the higher
effective speed of sound produces a higher frequency. When moving away, the
lower speed of sound produces a lower perceived frequency.
When both the observer and the source move the effects combine, apraoching higher
frequency and moving away lower frequency.
The frequency of the wave sound heard by the listener can be calculated by the
following equation:
𝑣 ± 𝑣𝐿
𝑓𝐿 = (
)𝑓
𝑣 ± 𝑣𝑠 𝑠
If the source is moving away from the observer the speed of the waves relative to
the listener is Vsound+ Vsource.
If the source is moving towards the observer the speed of the waves relative to
the listener is Vsound- Vsource.
If the listener is moving towards the source, then speed of the waves relative to
the listener is Vsound + Vlistener.
If the listener is moving away from the source, then the speed of the waves
relative to the listener is Vsound - Vlistener.
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EXAMPLE 1:
A man is standing on a pavement when he hears an ambulance approaching. The siren
of the ambulance is emitting a wave sound with a frequency of 500 Hz.
The ambulance is moving with constant speed of 30 m·s-1. Calculate the frequency of
the sound the man hears. Take the speed of the sound in air as 340 m·s -1.
Solution:
𝑣 ± 𝑣𝐿
𝑓𝐿 = (
)𝑓
𝑣 ± 𝑣𝑠 𝑠
fL  (
340  0
)  500
340  30
f L  1,10  500
f L  550 Hz
Applications of the Doppler effect
The Doppler effect holds not only for sound waves but also for electromagnetic waves,
including microwaves, radio waves, and visible light.
Police use the Doppler effect with microwaves to determine the speed of a car, a radar
unit emits microwaves of certain frequency f toward the oncoming car. The microwaves
reflected from the metal portions of the car back to the radar unit have a higher
frequency f ' due to the motion of the car relative to the radar unit.
The radar unit translates the difference between f’and f into the speed of the car, which
is then displayed for the operator to see.
Ultrasonic waves (ultrasound) are sound waves with a frequency greater than 20 000
Hz (the upper limit of human hearing).
These waves can be used in medicine to determine the direction of blood flow. The
device, called a Doppler flow meter, sends out sound waves. The sound wave scan
travel through skin and tissue and will be reflected by moving objects in the body (like
blood).
The reflected waves return to the flow meter where its frequency (received frequency) is
compared to the transmitted frequency. Because of the Doppler effect, blood that is
moving towards the flow meter will change the sound to a higher frequency and blood
that is moving away from the flow meter will cause a lower frequency.
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Ultrasound can be used to determine whether blood is flowing in the right direction in
the circulation system of unborn babies, or identify areas in the body where blood flow is
restricted due to narrow veins. The use of ultrasound equipment in medicine is called
sonography or ultrasonography.
SUMMARY:
The Doppler Effect is the change in frequency (or pitch) of the sound detected
by a listener because the sound source and the listener have different velocities
relative to the medium of sound propagation.
The frequency heard by the observer is higher than the frequency of the source
when the source moves towards the observer and lower when the source moves
away from the source.
When the observer moves then the effective speed of sound changes. The
wavelength remains fixed, but as the observer approaches the source, the higher
effective speed of sound produces a higher frequency. When moving away, the
lower speed of sound produces a lower perceived frequency.
The frequency of the wave sound heard by the listener can be calculated by the
following equation. f L  (
vsound  vL
)  fS
vsound  vS
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REVISION EXERCICES
QUESTION 1
Four possible options are provided as answers. Choose the correct option.
1.1
A stationary source produces a sound of frequency 400 Hz. A listener is moving
towards the source of sound with a speed of 20 m.s -1. If the speed of the sound
is 340 m.s-1 the listener hears sound with a frequency of:
A
400 Hz
B
376 Hz
C
424 Hz
D
800 Hz
1.2
An ambulance is moving at a speed of 28 m.s-1 and its siren emits sound waves
of frequency 700 Hz. If the speed of the sound in air is 340 m·s-1, when the
ambulance approaches to the hospital, a nurse standing on the entrance of the
hospital, hears a sound with a frequency :
A
644Hz
B
763Hz
C
700 Hz
D
350 Hz
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1.3 The figure below shows a fire engine moving at constant speed of 100 km·h-1 with
its siren on, the frequency of the sound produced by the siren is 700 Hz. A girl standing
next to the road observes how the fire engine approaches her.
Which one of the following frequency vs time graphs shows the frequency of the sound
observed (heard) by the girl:
QUESTION 2
A police car is moving at a speed of 20 m·s-1 and its siren emits a sound of 1000 Hz.
What frequency is heard by a stationary driver when the police car is:
2.1.
2.2.
moving away from the driver.
approaching to the driver
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SOLUTIONS:
QUESTION 1
1.1
(C)
fL= (
fL  (
v sound  vlistener
) x fsource
v sound  v source
340  20
)  400  424 Hz
340
1.2
v sound  vlistener
) x fsource
v sound  v source
(B)
fL= (
fL  (
340  0
) x700
340  28
fL= 763 Hz
1.3
(D)
QUESTION 2:
2.1
fL= (
v sound  vlistener
) x fsource
v sound  v source
fL  (
340  0
) x1000
340  20
FL= 944,4 Hz
v sound  vlistener
) x fsource
v sound  v source
2.2
fL= (
fL  (
340  0
) x1000
340  20
fL= 1062,5 Hz
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-Lesson 2: Doppler Effect with light. Red shifts and blue shifts.
Objective:
Learners must be able to:


Explain red shifts and blue shifts using the Doppler Effect.
Use the Doppler Effect to explain why we conclude that the universe is
expanding.
Baseline assessment (oral questions)
1. State the Doppler Effect.
The Doppler Effect is the change in frequency (or pitch) of the sound detected by
a listener because the sound source and the listener have different velocities
relative to the medium of sound propagation.
2. Write down ONE application of Doppler Effect in medicine.
ANY ONE
 It is used (in flow meters) in medical science to measure:
- the speed and direction (velocity) of blood flow.
- movement of the heart of a foetus.
 To find the rate of blood flow (Doppler scanning)
 To see the unborn child (Ultra sound scanning)
 To hear the heart of a foetus (Ultra sound scanning)
 It is used in medical sonography to generate images (and sounds) of
blood.
 To detect blood clotting (Doppler ultrasound test)
flowing
Introduction:
Light is a wave and earlier you learnt how you can study the properties of one wave and
apply the same ideas to another wave. The same applies to sound and light. We know
the Doppler Effect is relevant in the context of sound waves when the source is moving
relative to the observer, or the observer is moving relative to the source.
Therefore, in the context of light (EM waves), the frequency of observed light should be
different to the emitted frequency when the source of the light is moving relative to the
observer.
Development:
The pattern formed when a beam of light is broken up into its component frequencies
(colours) is called spectrum. For example when white light passes through a prism a
visible spectrum is observed (seven colours).
Spectrum is the distinctive pattern of wavelengths emitted by a source of light.
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Spectra can be classified into two main types: emission spectra and absorption
spectra.
The emission spectra are produced by objects at high temperature, for example the Sun
(figure below).
Absorption spectrum:
When white light travels through a cold, low pressure gas the atoms of the gas absorb
photons of light with specific frequencies. These specific wavelengths and frequencies
will not be seen after a spectroscope has been applied. Dark or black lines will appear
where the coloured lines appeared in the emission spectra and this explains the
formation of the absorption spectra (figure below).
The dark lines correspond to the frequencies of light that have been absorbed by the
gas.
Each star has a specific absorption spectrum. When we compare the absorption
spectrum of a star to that of our Sun (which we considered to be at rest), we can
determine the relative motion of that star with respect to our Sun.
Red shifts and blue shifts
A frequency shift of light in the visible spectrum could result in a change of colour which
could be observable with the naked eye. There will still be a frequency shift for
frequencies of EM radiation we cannot see.
We can apply all the ideas that we learnt about the Doppler effect to light. When talking
about light we use slightly different terminology to describe what happens. If you look at
the colour spectrum then you will see that blue light has a shorter wavelength than red
light.
Since for light, c = fλ, shorter wavelength equals higher frequency as speed of light is
constant. Relative to the middle of the visible spectrum (approximately green light)
longer wavelengths (or lower frequencies) are redder and shorter wavelengths (or
higher frequencies) are bluer.
So we call shifts towards longer wave lengths ”red shifts” and shifts towards shorter
wavelengths ”blue shifts”.
A shift in wavelength implies that there is also a shift in frequency. Longer wavelengths
of light have lower frequencies and shorter wavelengths have higher frequencies.
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From the Doppler Effect we know that when the source moves towards the observer
any waves they emit that you measure are shifted to shorter wavelengths (blue-shifted).
If the source moves away from the observer, the shift is to longer wavelengths (red
shifted).
Suppose we have three similar stars, A, B and C with identical absorption spectra. A is
at rest and B and C are moving relative to star A.
Absorption spectrum of star B
moving towards star A. (Blue Shift)
Absorption spectrum of star A at rest
Absorption spectrum of star C
moving away from star A. (Red
Shift)
We can conclude that:
1.
2.
If the light source is moving away from the observer (positive velocity) then the
observed frequency is lower and the observed wavelength is greater (redshifted).
If the source is moving towards the observer (negative velocity), the observed
frequency is higher and the wavelength is shorter (blue-shifted).
Stars emit light, which is why we can see them at night. Galaxies are huge collections of
stars.
An example is our own Galaxy, the Milky Way, of which our sun is only one of the
billions of stars!
Using large telescopes like the Southern African Large Telescope (SALT) in the Karoo,
astronomers can measure the light from distant galaxies.
The spectrum of light can tell us what elements are in the stars in the galaxies because
each element has unique energy levels and therefore emits or absorbs light at particular
wavelengths.
These characteristic wavelengths are called spectral lines because the lines show up as
discrete frequencies in the spectrum of light from the star.
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If these lines are observed to be shifted from their usual wavelengths to shorter
wavelengths, then the light from the galaxy is said to be blue shifted. If the spectral lines
are shifted to longer wavelengths, then the light from the galaxy is said to be red shifted.
If we think of the blue shift and red shift in Doppler Effect terms, then a blue shifted
galaxy would appear to be moving towards us (the observers) and a red shifted galaxy
would appear to be moving away from us.
Edwin Hubble (20 November 1889 - 28 September 1953) measured the Doppler shift of
a large sample of galaxies. He found that the light from distant galaxies is red shifted
and he discovered that there is a proportionality relationship between the red shift and
the distance to the galaxy. Galaxies that are further away always appear redder shifted
than nearby galaxies.
Remember that a red shift in Doppler terms means a velocity of the light source directed
away from the observer. So why do all distant galaxies appear to be moving away from
our Galaxy? None of them seem to be moving towards us.
The reason is that the universe is expanding! Some of the galaxies will be moving in
our direction but more slowly than the space between us and them is expanding. The
expansion is so large that it is the primary effect that we observe. The primary reason
the light is red shifted isn’t actually because all of the Doppler effect, it is red shifted
because the space is expanding, the waves are being stretched out. If the Doppler
Effect were a larger effect then some of the galaxies would still be blue shifted (just less
than if space were not expanding).
You might think that this means we are at the centre of the universe. This isn’t correct
the situation will look the same from every galaxy because space is expanding in all
directions.
Demonstration 1:
Get a balloon and draw some dots on it with a marker. As you blow the balloon up all
the dots get further away from all the other dots. The dots represent galaxies in a twodimensional, expanding universe (the balloon surface).
In the following picture the bottom vertex represents the beginning of time, the flat
surface represents space. As you move up through the panels you are moving later in
time and the expansion of the flat surface shows the expansion of the universe. The
galaxies shown on the surface get further away from each other just because of the
expansion of space.
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SUMMARY:
The Doppler Effect is the change in frequency (or pitch) of the sound detected
by a listener because the sound source and the listener have different velocities
relative to the medium of sound propagation.
The frequency heard by the observer is higher than the frequency of the source
when the source moves towards the observer and lower when the source moves
away from the source.
When the observer moves then the effective speed of sound changes. The
wavelength remains fixed, but as the observer approaches the source, the higher
effective speed of sound produces a higher frequency. When moving away, the
lower speed of sound produces a lower perceived frequency.
The frequency of the wave sound heard by the listener can be calculated by the
following equation. f L  (
vsound  vL
)  fS
vsound  vS
The Doppler Effect can be observed in all types of waves, including ultrasound,
light and radio waves.
Sonography makes use of ultrasound and the Doppler Effect to determine the
direction of blood flow.
Light is emitted by stars. Due to the Doppler Effect, the frequency of this light
decreases and the stars appear redder than if they were stationary. This is called
a red shift and means that the stars are moving away from the Earth. This is one
of the reasons we can conclude that the Universe is expanding.
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REVISON EXERCISE
QUESTION 1 (Question 6 Physics, November 2014):
1.1
The siren of a stationary ambulance emits a note of frequency 1130 Hz. When
the ambulance moves at a constant speed, stationary observer detects a
frequency that is 70 Hz higher than that emitted by the siren.
1.1.1 State the Dppler effect in words.
1.1.2 Is the ambulance moving towards or away from the observer?
1.1.3 Calculate the speed at which the ambulance is travelling.
Take the speed of sound in air as 343 m∙s-1.
1.2
(2)
(2)
(5)
A study of spectral lines obtained from various stars can provide valuable
information about the movement of the stars.
The two diagrams below represent different spectral lines of an element.
Diagram 1 represents the spectrum of the element in a laboratory on Earth.
Diagram 2 represents the spectrum of the same element from a distant star.
Is the star moving towards or away from the Earth?
Explain the answer by referring to the shifts in the spectral lines
in the two diagrams above.
(2)
[11]
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QUESTION 2: (QUESTION 6 September 2014 NC)
A whistle of a train emits sound waves at a frequency of 555 Hz. A person (listener)
standing next to the train track hears a sound of the whistle with wavelength of 0,71 m.
Assume that the speed of sound in air is 340 m·s-1.
2.1
Is the train approaching or moving away a person (listener)? Show how you
arrived at the answer.
(4)
2.2
State the Doppler effect in words.
(2)
2.3
Calculate the speed of the train.
(5)
2.4
Write down TWO practical applications of the above phenomenon in the
medical field.
(2)
2.5
Using large telescopes like the Southern African Large Telescope (SALT) in
the Karoo, astronomers can measure the light from distant galaxies. In 1929
Edwin Hubble found that the light from distant galaxies is redshifted.
What does the redshifts tell us about the Universe?
(2)
[15]
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SOLUTIONS:
QUESTION 1:
1.1.1 The Doppler Effect is the change in frequency (or pitch) of the sound detected
by a listener because the sound source and the listener have different velocities
relative to the medium of sound propagation
1.1.2 The ambulance is moving towards the observer
1.1.3
v
v
f L  ( sound L )  f S
vsound  vS
v
v
f L  ( sound L )  f S
vsound  vS
1200  (
342  0
) 1130
343  vS
vS  20,01m  s 2
1.2
From the diagram we can see that spectral lines of the element obtained from the star
are sifted slightly to the blue end of the spectrum. That is they are blue shifted.
We know that the blue end is the higher frequency end. From this we can infer that the
frequency detected is higher than the frequency emitted by the element.
Thus, according to the Doppler Effect, the blue-shift in the spectrum implies that the star
is moving towards the Earth.
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QUESTION 2:
v  f 
2.1
340= f (0,71)
f= 478,87 Hz
The frequency of the sound detected by the person (listener) is smaller than
the frequency of the source (𝑓𝐿 < 𝑓𝑆 ) then the train is moving away from the
listener.
(4)
2.2
The change in pitch or frequency of the sound detected by an observer
because the sound source and the observer have different velocities with
respect to the medium of sound propagation. 
OR
The Doppler effect is the change in the observed frequency of a wave when
the source or the detector moves relative to the transmitting medium. 
OR
The (apparent) change in the frequency of a wave when there is a relative
motion between the source of the wave and an observer. 
(2)
2.3
POSITIVE MARKING FROM 2.1
Option 1:
v

v
f l   sound listener  f s 
 vsound  vsource 
 340  0 
  x 555
 478,87  
 340  vsource 
Vs = 54,05 m·s-1
Option 2:


vsound
 f s 
f l  
 vsound  vsource 


340
 555
 478,87  
 340  vsource 
Vs = 54,05 m·s-1
(5)
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Option 3:
fs

fL 
vs
1
v
555
 478,87 

vs
1

340
vs = 54,05 m·s-1 
2.4
ANY TWO
To find the rate of blood flow (Doppler scanning)
(2)
To see the unborn child (Ultra sound scanning)
To hear the heart of a foetus (Ultra sound scanning) 
2.4
The universe is expanding
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(2)
[15]
Page 19
Les 3: Revision exercises.
Objective:
Learners must be able to:






State the Doppler Effect as the change in frequency (or pitch) of the sound
detected by a listener because the sound source and the listener have different
velocities relative to the medium of sound propagation.
Explain (using appropriate illustrations) the change in pitch observed when a
source moves toward or away from a listener.
𝑣±𝑣𝐿
Solve problems using the equation 𝑓𝐿 =
𝑓𝑠 when EITHER the source or the
𝑣±𝑣𝑠
listener is moving.
State applications of the Doppler Effect.
Explain red shifts and blue shifts using the Doppler Effect.
Use the Doppler Effect to explain why we conclude that the universe is
expanding.
Introduction:
Summary:
The Doppler Effect is the change in frequency (or pitch) of the sound detected
by a listener because the sound source and the listener have different velocities
relative to the medium of sound propagation.
The frequency heard by the observer is higher than the frequency of the source
when the source moves towards the observer and lower when the source moves
away from the source.
When the observer moves then the effective speed of sound changes. The
wavelength remains fixed, but as the observer approaches the source, the higher
effective speed of sound produces a higher frequency. When moving away, the
lower speed of sound produces a lower perceived frequency.
The frequency of the wave sound heard by the listener can be calculated by the
following equation. f L  (
vsound  vL
)  fS
vsound  vS
Remember that
 If the source is moving away from the observer the speed of the waves
relative to the listener is vsound + vsource.
 If the source is moving towards the observer the speed of the waves
relative to the listener is vsound - vsource.
 If the listener is moving towards the source, then speed of the waves
relative to the listener is vsound + vlistener.
 If the listener is moving away from the source, then the speed of the
waves relative to the listener is vsound - vlistener.
The Doppler effect can be observed in all types of waves, including ultrasound,
light and radio waves.
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Sonography makes use of ultrasound and the Doppler Effect to determine the
direction of blood flow.
Development:
Example 1:
A whistle of a locomotive emits sound wave of a frequency of 2000 Hz. A stationary
listener hears a sound of 1920 Hz. The speed of sound in air is 340 m·s-1.
1.1.
1.2.
1.3.
Name and state the wave phenomenon that takes places.
Is the locomotive approaching to the listener or moving away from the listener?
Explain your answer.
Calculate the speed of the locomotive.
Solution:
1.1
Doppler Effect is the change in frequency (or pitch) of the sound detected by a
listener because the sound source and the listener have different velocities
relative to the medium of sound propagation.
OR
Doppler Effect is the change of the frequency of the waves detected by a
detector (listener), as a consequence of the relative movement between the
source and the detector.
Or
The Doppler Effect is the apparent change in frequency and wavelength of a
wave when the observer and the source of the wave move relative to each other.
1.2
1.3.
The locomotive is moving away from the listener, because the frequency heard
by the listener is smaller than the frequency of the source.When the locomotive
moves away from the listener the sound waves emitted are spread out and the
wavelength appears longer behind the locomotive as the speed of sound is
constant the frequency heard is lower; less sound waves reach the listener per
second and the pitch (frequency) appears to be less than the sound emitted by
the source (locomotive).
v
v
fL= ( sound listener ) x fsource
v sound  v source
fL = (
340  0
) x 2000
340  v source
(1920x340) +1920vsource=340x2000
652800+1920Vsource=680000
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1920 V= 680000-652800
1920V= 27200
V= 14,17 m.s-1
REVISON EXERCISES
QUESTION 1:
The siren of a police car produces a sound of frequency 420 Hz. A man sitting next to
the road notices that the pitch of the sound changes as the car moves towards and then
away from him.
1.1
Write down the name of the above phenomenon.
1.2
Assume that the speed of sound in air is 340 m∙s-1. Calculate the frequency of
the sound of the siren observed by the man, when the car is moving towards him
at a speed of 16 m·s-1.
1.3
The police car moves away from the man at constant velocity, then slows down
and finally comes to rest.
1.3.1 How will the observed frequency compare with the original frequency of
the siren when the police car moves away from the man at constant
velocity? Write only GREATER THAN, SMALLER THAN or EQUAL TO.
1.3.2 How will the observed frequency change as the car slows down whilst
moving away? Write only INCREASES, DECREASES or REMAINS THE
SAME.
QUESTION 2
A boy made a video recording of a car race. An analysis of a section of the recording
shows that one of the cars was travelling away from him and that the frequency of the
sound of the engine as it was recorded, was 95 % of the actual frequency of the sound
emitted by the engine. The actual frequency of the sound of the engine was 800 Hz.
2.1
2.2
2.3
Name and state the wave phenomenon that takes places.
How fast was the car travelling at that stage?
Describe one application of this phenomenon in medicine.
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QUESTION 3: (Question 7 June 2014 NC)
The siren of a stationary police car emits sound waves of frequency of 620 Hz. A
stationary listener watches the police car approaching him at constant velocity on a
straight road. Assume that the speed of sound in air is 340 m·s-1.
3.1
How does the wavelength of the sound waves heard by the listener compare
to the wavelength of the sound produced by the siren of the police car when it
approaches the listener? Only write down LONGER THAN, SHORTER THAN
or EQUAL TO. Explain your answer.
(4)
3.2
Name the phenomenon observed in QUESTION 3.1.
(1)
3.3
Calculate the wavelength of the sound waves detected by the stationary
listener if the police car moves toward him at a speed of 110 km·h -1.
(6)
3.4
How will the answer to QUESTION 3.3 change if the police car moves away
from the listener at 120 km·h-1? Write down INCREASES, DECREASES or
REMAINS THE SAME.
(1)
3.5
Write down ONE application of Doppler effect in medicine.
(1)
[13]
QUESTION 4(Question 4 Feb – March 2011)
The whistle of a train emits sound waves of frequency 2 000 Hz. A stationary listener
measures the frequency of these emitted sound waves as 2 080 Hz. The speed of
sound in air is
340 m·s-1
4.1
4.2
4.3
4.4
Name the phenomenon responsible for the observed change in
frequency.
(1)
Is the train moving AWAY FROM or TOWARDS the stationary listener? (1)
Calculate the speed of the train.
(4)
Will the frequency observed by a passenger, sitting in the train, be GREATER
THAN, EQUAL TO or SMALLER THAN 2 000 Hz? Explain the answer.
(2)
[8]
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QUESTION 5:
A locomotive moving at a constant speed emits a wave sound with it whistle. A man
that stands on the side of the railroad hears a sound of frequency 1840, Hz when the
locomotive moves away from him and a girl that stands on the side of the railroad hears
a sound of frequency 2200 Hz when the locomotive approaches her.
D2
S
D1
5.1 Name the phenomenon that describes the change in the frequency heard by the
man and the girl.
5.2 Explain with the aid of diagram, why the man hears a sound with lower frequency
and the girl hears a sound with higher frequency.
5.3 Calculate the speed of the locomotive.
5.4 Calculate the frequency at which the whistle of the locomotive emits the sound
waves.
5.5 Describe applications of this phenomenon in medicine.
QUESTION 6:
The following are the spectra of the same gas found in three different stars.
6.1 If star B is used as the reference star (stationary star) which star is moving towards
star B?
Give a reason.
6.2 If star B is used as the reference star (stationary star) which star is moving away
from star B? Give a reason.
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SOLUTIONS
QUESTION 1:
1.1
Doppler Effect
1.2
Car approaching:
fL 
v
v  vL
fs
fs OR fL 
v  vs
v  vs
=(
340
) (420)
340  16
 fL = 440,74 Hz
1.3.1. Smaller than
1.3.2. Increases
QUESTION 2:
2.1 Doppler Effect:
Doppler Effect is the change in frequency (or pitch) of the sound detected by a listener
because the sound source and the listener have different velocities relative to the
medium of sound propagation.
OR
Doppler Effect is the difference between the value of the frequency of the waves
detected by an observer/listener and the value of the frequency emitted by the source,
as a consequence of the relative movement between the source and the
observer/listener.
 v  vL 
 xf s
2.2 f L  
 v  vs 
 330  0 
 x 800
0.95 x800 = 
330

v
s 

 330  0 
 x 800
760 = 
 330  v s 
(330 + vs ) x 760 = 330 x 800
330 x 760 +760 vs= 264000
760vs= 264000 – 250800
760vs= 13 200
vs= 17, 37 m∙s-1
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2.3 To measure the speed at which the blood is flowing. This enable doctors to
determine areas where blood vessels have been narrowed resulting in an increase
blood flow rate.
Or
It can also be used to monitor the heartbeat of a foetus in the womb.
QUESTION 3:
3.1.
Lower than
As the speed of sound is constant 
wavelength is inversely proportional to frequency
and frequency is higher.
OR
Lower than
When v is constant 
1
λ
f

and frequency is higher.
(4)
OR
Lower than
As frequency heard is higher
the wavelength is smaller because when the speed of sound is constant
wavelength and frequency are inversely proportional.
3.2
Doppler effect
(1)
3.3
Option 1:
v

v
f l   sound listener  f s 
 vsound  vsource 
(6)
 340  0 
fl  
  x 620
 340  30.56 
fl  681,23 Hz
v  f 
340  681,23 
  0,50 m
Option 2:


vsound
 f s 
f l  
 vsound  vsource 
340


fl  
  x 620
 340  30.56 
f l  681,23 Hz
v  f 
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340  681,23 
  0,50 m
Option 3:
fs

fL 
vs
1
v
555

fl 
30.56
1

340
f l  568,87 Hz
v  f 
340  681.23 
  0,50 m
3.4
Increases
(1)
3.5
ANY ONE

It is used (in flow meters) in medical science to measure:
the speed and direction (velocity) of blood flow.
movement of the heart of a foetus.

To find the rate of blood flow (Doppler scanning)

To see the unborn child (Ultra sound scanning)

To hear the heart of a foetus (Ultra sound scanning)

It is used in medical sonography to generate images (and sounds) of
flowing blood.

To detect blood clotting (Doppler ultrasound test)
QUESTION 4:
4.1 Doppler Effect.
4.2 Towards.
4.3
4.4. Equal to (2 000 Hz)
The passenger moves at the same velocity as the train. / There is no difference in
velocity of the passenger relative to the train/ the passenger is as rest with respect
to the train.
Total 8 marks
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(1)
[13]
QUESTION 5:
5.1
Doppler Effect. The man hears a smaller frequency and the girl hears a greater
frequency
5.2
The girl hears a greater frequency because the speed of sound is constant but the
wavelength of the sound decreases and according to the following equation v=f.λ
F is inversely proportional the wavelength when v is constant. The man hears a smaller
frequency because the wavelength is bigger.
5.3 The frequency of the source is the same for both
v
v
f Listener  sound listener  f source
vsound  vsource
fs 
f listener
vsound  vlistener
vsound  vsource
f L1
f L2

v  v L1 v  v L 2
v  vs
v  vs
 v  vL 2 
 v  vL1 
  f L 2  

f L1  
 v  vs 
 v  vs 
 340 
 340 
  2200

1840
340

v
340

v
s 
s 


625600 748000

340  vs 340  vs
625600 (340+Vsource)=748000(340- Vsource)’
212704000+ 625600Vsource=254320000-748000Vsource
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625600Vsource+748000Vsource=254320000-212704000
625600Vsource+748000Vsource=41616000
(625600+748000) V= 41616000
1373600V= 41616000
Vsource=30,3 m.s--1≈ 30 m.s-1
f Listener 
vsound  vlistener
 f source
vsound  vsource
 v
 vsource 

f s  f L  sound
 vsound  v Listener 
 340  30,3 

 340 
fsource= 1840 
fsource≈ 2004 Hz
0r
 v
 vsource 

f s  f L  sound
 vsound  v Listener 
 340  30,3 
 ≈2004 Hz
 340 
fsource= 2200 
5.5
Doppler Effect with ultrasound waves can be used in medicine for example to measure
the speed at which the blood is flowing.
This enable doctors to determine areas where blood vessels have been narrowed
resulting in an increase in blood flow rate,it can also be used to monitor the heartbeat.
QUESTION 6:
6.1 Star A, since it is blueshifted.
6.2 Star C since it is a red shifted.
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SELF-ASSESSMENT
QUESTION 1
An ambulance approaches a tunnel in a mountain at a constant speed of 20 m∙s-1. The
siren of the ambulance emits sound waves having a wavelength of 0,30 m. Take the
speed of sound in air as 340 m∙s-1.
1.1
State the Doppler effect in words.
1.2
Calculate the frequency of the sound waves emitted by the siren as heard by
1.3
(2)
the ambulance’s driver.
(3)
Calculate the frequency of the sound waves heard by an observer standing
(4)
near the tunnel entrance.
1.4
How would the answer to QUESTION 1.3 change if the speed of the
ambulance were GREATER THAN 20 m∙s-1? Write down only INCREASES,
DECREASES or REMAINS THE SAME.
1.5
1.6
(1)
The sound from the siren reflects from the mountain back to the ambulance
driver. Calculate the frequency heard by the ambulance driver.
(3)
Write down ONE application of the Doppler Effect in medicine.
(1)
[14]
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MEMORAMDUM FOR SELF-ASSESSMENT
QUESTION 1
1.1
Doppler effect is the change in frequency (or pitch) of the sound detected
by a listener because the sound source and the listener have different
velocities relative to the medium of sound propagation.
OR
The change in the observed frequency when there is relative motion
between the source and the observer.
(2)
1.2
v = fλ 
340 = f(0,30)
fs = 1 133,33 Hz 
1.3
(3)
OPTION 1
𝑓𝐿 = (
𝑣±𝑣𝐿
𝑣±𝑣𝑠
𝑣
) 𝑓𝑠  OR/OF 𝑓𝐿 = (
𝑣±𝑣𝑠
) 𝑓𝑠
 340 
fL  
  1133,33
 340 - 20 
fL = 1204,16 Hz
OPTION 2
fL 
fs

vs
1
v
fL 
1133,33 
20

1
340
(4)
fL = 1204,16 Hz
1.4
Increases/Verhoog
1.5
𝑓𝐿 = ( 𝑣±𝑣𝐿 ) 𝑓𝑠 OR/OF 𝑓𝐿 = ( 𝑣 𝐿) 𝑓𝑠
𝑣±𝑣
(1)
𝑣±𝑣
𝑠
 340  20 
fL  
  1204,16 
 340 
fL = 1274.99 Hz
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OR
𝑓𝐿 = (1 +
𝑣𝑠
)𝑓
𝑣 𝑠
20
𝑓𝐿 = (1 + 340)1204,16
(3)
fL = 1274,99 Hz
1.6
ANY ONE






It is used (in flow meters) in medical science to measure:
the speed and direction (velocity) of blood flow.
movement of the heart of a foetus.
To find the rate of blood flow (Doppler scanning)
To see the unborn child (Ultra sound scanning)
To hear the heart of a foetus (Ultra sound scanning)
It is used in medical sonography to generate images (and
sounds) of flowing blood.
To detect blood clotting (Doppler ultrasound test)
(1)
[14]
©Developed by: G. Izquierdo Rodríguez & G Izquierdo Gómez.
Copyright reserved
Page 32