The Doppler effect Book page 381 - 386 copycgrahamphysicscom 2016
Doppler effect
bull Definition The apparent shift in frequency if there is relative motion between an observer and the source
bull A stationary source emits a sound of constant frequency f
bull A moving source changes the wavelength at which sound is emitted by its movement
The Doppler Effect
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The apparent shift in frequency
bull Source approaching constant pitch of increasing loudness
bull Source passes you instant change in pitch
bull Source moving away decreasing in loudness
bull High pitch is constant ndash then abrupt change to constant low pitch
bull We do not have a sound of increasing pitch and then decreasing pitch
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Change in pitch or frequency
bull Smaller wavelength bull Increased frequency
bull Larger wavelength bull Decreased frequency
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Doppler Effect If we can figure out what the change in the wavelength is we also know the change in the frequency
Observer at rest
λ =119907
119891=
119907∆119905
119891∆119905
∆119889 = 119907∆119905 minus 119907119904∆119905 Because of the motion of the source the number of waves occupying the distance d is equal to ∆119889
The new wavelength λprime =119907∆119905minus119907119904∆119905
119891∆119905=
119907minus119907119904
119891
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Frequency heard by stationary observer and moving source
bull 119891prime =119907
119907+119907119904119891 = 119891
1
1+119907119904119907
for source moving away from observer
bull 119891prime =119907
λprime=
119891119907
119907minus119907119904
bull 119891prime =119907
119907minus119907119904119891 = 119891
1
1minus119907119904119907
for source moving towards O
bull 119907 = speed of emitted sound
bull 119907119904 =speed of source
λprime =119907 minus 119907119904
119891
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Frequency heard by moving observer from stationary source
bull If the observer is moving with respect to the source things are a bit different
bull The wavelength remains the same but the wave speed is different for the observer
bull λ = unchanged
bull 119907prime changes
O 119907119904119900119906119899119889 = 119907119900119887119904119890119903119907119890119903 + 119907119904119900119906119899119889
λ =119907119900119887119904119890119903119907119890119903+119907119904119900119906119899119889
119891 and 119891prime =
119907prime
λ
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Moving observer
Towards source
bull 119891prime =119907119900+119907119904119900119906119899119889
119907119904119900119906119899119889119891
=119907119900+119907119904119900119906119899119889
119907119904119900119906119899119889times 119891
bull 119891prime =119907119900
119907+ 1 119891
bull 119907 = speed of sound
Away from source
bull 119891prime =119907119904119900119906119899119889minus119907119900
119907119904119900119906119899119889119891
=119907119904119900119906119899119889minus119907119900
119907119904119900119906119899119889times 119891
bull 119891prime =119907119900
119907minus 1 119891
Change in frequency
bull ∆119891 = 119891prime minus 119891 =119907119900
119907+ 1 119891 minus 119891 = 119891 +
119907119900
119907119891 minus 119891 =
119907119900
119907119891
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For light waves bull Relative velocity between source and observer
bull Speed of light does not depend on speed of source
bull All observer measure speed of light = c
bull 119907119904 ≪lt 119888 hence apparent shift in frequency is the same for O moving or S moving
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Light waves
Source moving towards observer
bull ∆119891 = 119891prime minus 119891 =119907119900
119907119891 =
119907
119888119891
bull This expression was derived from binomial expansion which is not required to reproduce
Source moving away from observer
bull ∆119891 = 119891prime minus 119891 =minus119907119900
119907119891
= minus119907
119888119891
These two equations are equivalent to
∆λ = λ119907
119888
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Example bull A source emits sound of f=440Hz It moves in a straight line towards
a stationary observer with a speed of 30m119904minus1
bull An observer hears sound of frequency f = 484Hz
bull Find the speed of sound in air Solution
bull S moves towards observer 119891prime =119907
119907minus119907119904119891
bull 119891 = 440119867119911 119891prime = 484119867119911 119907119904 = 30119898119904minus1
bull 119891prime =119907
119907minus119907119904119891 = 119891
1
1minus119907119904119907
bull 1 minus119907119904
119907=
119891
119891prime
bull 1 minus119891
119891prime=
119907119904
119907
bull 119907 = 1199071199041
1minus119891
119891prime
= 30 times1
1minus440
484
=30
0091= 330119898119904minus1
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Example bull A radio signal from a galaxy is measured to be f =
139x109Hz The same signal from a source in the laboratory has f = 142x109Hz Suggest why the galaxy is moving away from earth and calculate its recession speed away from Earth
Solution bull Frsquo of moving source lt than f source is moving away less f longer λ
bull ∆119891 = minus119907
119888119891
bull 119907 =minus119888∆119891
119891=
minus30times108times 139minus142 times109
142x109
bull = 634 times 106119898119904minus1
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Using the Doppler effect bull Can be used to measure speed of galaxies
bull To measure speed in general
bull Source = transmitter this can be sound or EM waves
bull Frequency is constant
bull Waves are incident on reflector which is moving towards the source at speed v
bull Reflected waves are detected by receiver placed next to transmitter
bull 119907 ≪lt 119888 where c is the speed of wave from transmitter
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Continued
Reflector is moving observer
bull Observer is moving towards stationary source
bull ∆119891 = 119891prime minus 119891 =119907
119888119891
Receiver is stationary observer
bull Stationary observer watches source moving toward him
bull ∆119891 = 119891primeprime minus 119891prime =119907
119888119891rsquo
bull Mathematical manipulation
bull 119891prime minus 119891 =119907
119888119891
119891primeprime minus 119891prime =119907
119888119891rsquo
119891prime minus 119891 + 119891primeprime minus 119891prime =119907
119888119891 +
119907
119888119891rsquo
119891primeprime minus 119891 =119907
119888119891 + 119891prime
bull 119891prime is observer moving tw source
+
copycgrahamphysicscom 2016
continued bull 119891prime is observer moving toward source
119891prime minus 119891 =119907
119888119891
119891prime = 119891 +119907
119888119891 = 119891 1 +
119907
119888
bull 119891primeprime minus 119891 =119907
119888119891 + 119891prime =
119907
119888119891 + 119891 1 +
119907
119888
bull 119891primeprime minus 119891 =119907
119888119891 1 + 1 +
119907
119888=
119907
119888119891 2 +
119907
119888
= 2119891119907
119888+
1199072
1198882119891
bull But since vltltltc 1199072
1198882 can be ignored
bull 119891primeprime minus 119891 =2119907
119888119891 or short ∆119891 =
2119907
119888119891
bull If v = c we must use the full Doppler equation bull For EM radiation we always consider situations were vltltc
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Example bull The speed of sound in blood is 1500 times 103 m 119904minus1
Ultrasound of frequency 100 MHz is reflected from blood flowing in an artery The frequency of the reflected waves received back at the transmitter is 105 MHz Estimate the speed of the blood flow in the artery
Solution
bull Since vltltc we can use ∆119891 =2119907
119888119891
bull 119891primeprime minus 119891 = 105 minus 100 119872119867119911 =2119907
1500 times 103 times 10 times 106
bull 005 times 106 = 133334119907 119907 = 375m 119904minus1
bull We have assumed that ultra sound is incident at perpendicular direction to the blood flow
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Example bull Judy is standing on the platform of a station A high speed train is
approaching the station in a straight line at constant speed and is sounding its whistle As the train passes by Judy the frequency of sound emitted by the whistle as heard by Judy changes from 640Hz to 430Hz Find a) the speed of the train Solution
bull A)
bull (speed of sound = 330m119904minus1)
bull 119891prime(1 minus119907119904
119907) = 119891prime 1 +
119907119904
119907
bull 640 minus 640119907119904
119907= 430 + 430
119907119904
119907
bull 210 = 1070 119907119904
119907 119907119904 =
210
1070times 330 = 6477119898119904minus1
Towards Judy Away from Judy
119891prime = 640119867119911
119891prime = 1198911
1 minus119907119904119907
119891 =119891prime
1
1 minus119907119904119907
119891 = 119891prime 1 minus119907119904
119907
119891prime = 430119867119911
119891prime = 1198911
1 +119907119904119907
119891 =119891prime
1
1 +119907119904119907
119891 = 119891prime 1 +119907119904
119907
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hellipcontinued
bull b) the frequency of the sound emitted by the whistle as heard by a person on the train (speed of sound = 330m119904minus1)
bull Solution
bull 119891 = 119891prime 1 minus119907119904
119907= 640 times 1 minus
6477
330= 514119867119911
copycgrahamphysicscom 2016
Doppler effect
bull Definition The apparent shift in frequency if there is relative motion between an observer and the source
bull A stationary source emits a sound of constant frequency f
bull A moving source changes the wavelength at which sound is emitted by its movement
The Doppler Effect
copycgrahamphysicscom 2016
The apparent shift in frequency
bull Source approaching constant pitch of increasing loudness
bull Source passes you instant change in pitch
bull Source moving away decreasing in loudness
bull High pitch is constant ndash then abrupt change to constant low pitch
bull We do not have a sound of increasing pitch and then decreasing pitch
copycgrahamphysicscom 2016
Change in pitch or frequency
bull Smaller wavelength bull Increased frequency
bull Larger wavelength bull Decreased frequency
copycgrahamphysicscom 2016
Doppler Effect If we can figure out what the change in the wavelength is we also know the change in the frequency
Observer at rest
λ =119907
119891=
119907∆119905
119891∆119905
∆119889 = 119907∆119905 minus 119907119904∆119905 Because of the motion of the source the number of waves occupying the distance d is equal to ∆119889
The new wavelength λprime =119907∆119905minus119907119904∆119905
119891∆119905=
119907minus119907119904
119891
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Frequency heard by stationary observer and moving source
bull 119891prime =119907
119907+119907119904119891 = 119891
1
1+119907119904119907
for source moving away from observer
bull 119891prime =119907
λprime=
119891119907
119907minus119907119904
bull 119891prime =119907
119907minus119907119904119891 = 119891
1
1minus119907119904119907
for source moving towards O
bull 119907 = speed of emitted sound
bull 119907119904 =speed of source
λprime =119907 minus 119907119904
119891
copycgrahamphysicscom 2016
Frequency heard by moving observer from stationary source
bull If the observer is moving with respect to the source things are a bit different
bull The wavelength remains the same but the wave speed is different for the observer
bull λ = unchanged
bull 119907prime changes
O 119907119904119900119906119899119889 = 119907119900119887119904119890119903119907119890119903 + 119907119904119900119906119899119889
λ =119907119900119887119904119890119903119907119890119903+119907119904119900119906119899119889
119891 and 119891prime =
119907prime
λ
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Moving observer
Towards source
bull 119891prime =119907119900+119907119904119900119906119899119889
119907119904119900119906119899119889119891
=119907119900+119907119904119900119906119899119889
119907119904119900119906119899119889times 119891
bull 119891prime =119907119900
119907+ 1 119891
bull 119907 = speed of sound
Away from source
bull 119891prime =119907119904119900119906119899119889minus119907119900
119907119904119900119906119899119889119891
=119907119904119900119906119899119889minus119907119900
119907119904119900119906119899119889times 119891
bull 119891prime =119907119900
119907minus 1 119891
Change in frequency
bull ∆119891 = 119891prime minus 119891 =119907119900
119907+ 1 119891 minus 119891 = 119891 +
119907119900
119907119891 minus 119891 =
119907119900
119907119891
copycgrahamphysicscom 2016
For light waves bull Relative velocity between source and observer
bull Speed of light does not depend on speed of source
bull All observer measure speed of light = c
bull 119907119904 ≪lt 119888 hence apparent shift in frequency is the same for O moving or S moving
copycgrahamphysicscom 2016
Light waves
Source moving towards observer
bull ∆119891 = 119891prime minus 119891 =119907119900
119907119891 =
119907
119888119891
bull This expression was derived from binomial expansion which is not required to reproduce
Source moving away from observer
bull ∆119891 = 119891prime minus 119891 =minus119907119900
119907119891
= minus119907
119888119891
These two equations are equivalent to
∆λ = λ119907
119888
copycgrahamphysicscom 2016
Example bull A source emits sound of f=440Hz It moves in a straight line towards
a stationary observer with a speed of 30m119904minus1
bull An observer hears sound of frequency f = 484Hz
bull Find the speed of sound in air Solution
bull S moves towards observer 119891prime =119907
119907minus119907119904119891
bull 119891 = 440119867119911 119891prime = 484119867119911 119907119904 = 30119898119904minus1
bull 119891prime =119907
119907minus119907119904119891 = 119891
1
1minus119907119904119907
bull 1 minus119907119904
119907=
119891
119891prime
bull 1 minus119891
119891prime=
119907119904
119907
bull 119907 = 1199071199041
1minus119891
119891prime
= 30 times1
1minus440
484
=30
0091= 330119898119904minus1
copycgrahamphysicscom 2016
Example bull A radio signal from a galaxy is measured to be f =
139x109Hz The same signal from a source in the laboratory has f = 142x109Hz Suggest why the galaxy is moving away from earth and calculate its recession speed away from Earth
Solution bull Frsquo of moving source lt than f source is moving away less f longer λ
bull ∆119891 = minus119907
119888119891
bull 119907 =minus119888∆119891
119891=
minus30times108times 139minus142 times109
142x109
bull = 634 times 106119898119904minus1
copycgrahamphysicscom 2016
Using the Doppler effect bull Can be used to measure speed of galaxies
bull To measure speed in general
bull Source = transmitter this can be sound or EM waves
bull Frequency is constant
bull Waves are incident on reflector which is moving towards the source at speed v
bull Reflected waves are detected by receiver placed next to transmitter
bull 119907 ≪lt 119888 where c is the speed of wave from transmitter
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Continued
Reflector is moving observer
bull Observer is moving towards stationary source
bull ∆119891 = 119891prime minus 119891 =119907
119888119891
Receiver is stationary observer
bull Stationary observer watches source moving toward him
bull ∆119891 = 119891primeprime minus 119891prime =119907
119888119891rsquo
bull Mathematical manipulation
bull 119891prime minus 119891 =119907
119888119891
119891primeprime minus 119891prime =119907
119888119891rsquo
119891prime minus 119891 + 119891primeprime minus 119891prime =119907
119888119891 +
119907
119888119891rsquo
119891primeprime minus 119891 =119907
119888119891 + 119891prime
bull 119891prime is observer moving tw source
+
copycgrahamphysicscom 2016
continued bull 119891prime is observer moving toward source
119891prime minus 119891 =119907
119888119891
119891prime = 119891 +119907
119888119891 = 119891 1 +
119907
119888
bull 119891primeprime minus 119891 =119907
119888119891 + 119891prime =
119907
119888119891 + 119891 1 +
119907
119888
bull 119891primeprime minus 119891 =119907
119888119891 1 + 1 +
119907
119888=
119907
119888119891 2 +
119907
119888
= 2119891119907
119888+
1199072
1198882119891
bull But since vltltltc 1199072
1198882 can be ignored
bull 119891primeprime minus 119891 =2119907
119888119891 or short ∆119891 =
2119907
119888119891
bull If v = c we must use the full Doppler equation bull For EM radiation we always consider situations were vltltc
copycgrahamphysicscom 2016
Example bull The speed of sound in blood is 1500 times 103 m 119904minus1
Ultrasound of frequency 100 MHz is reflected from blood flowing in an artery The frequency of the reflected waves received back at the transmitter is 105 MHz Estimate the speed of the blood flow in the artery
Solution
bull Since vltltc we can use ∆119891 =2119907
119888119891
bull 119891primeprime minus 119891 = 105 minus 100 119872119867119911 =2119907
1500 times 103 times 10 times 106
bull 005 times 106 = 133334119907 119907 = 375m 119904minus1
bull We have assumed that ultra sound is incident at perpendicular direction to the blood flow
copycgrahamphysicscom 2016
Example bull Judy is standing on the platform of a station A high speed train is
approaching the station in a straight line at constant speed and is sounding its whistle As the train passes by Judy the frequency of sound emitted by the whistle as heard by Judy changes from 640Hz to 430Hz Find a) the speed of the train Solution
bull A)
bull (speed of sound = 330m119904minus1)
bull 119891prime(1 minus119907119904
119907) = 119891prime 1 +
119907119904
119907
bull 640 minus 640119907119904
119907= 430 + 430
119907119904
119907
bull 210 = 1070 119907119904
119907 119907119904 =
210
1070times 330 = 6477119898119904minus1
Towards Judy Away from Judy
119891prime = 640119867119911
119891prime = 1198911
1 minus119907119904119907
119891 =119891prime
1
1 minus119907119904119907
119891 = 119891prime 1 minus119907119904
119907
119891prime = 430119867119911
119891prime = 1198911
1 +119907119904119907
119891 =119891prime
1
1 +119907119904119907
119891 = 119891prime 1 +119907119904
119907
copycgrahamphysicscom 2016
hellipcontinued
bull b) the frequency of the sound emitted by the whistle as heard by a person on the train (speed of sound = 330m119904minus1)
bull Solution
bull 119891 = 119891prime 1 minus119907119904
119907= 640 times 1 minus
6477
330= 514119867119911
copycgrahamphysicscom 2016
The apparent shift in frequency
bull Source approaching constant pitch of increasing loudness
bull Source passes you instant change in pitch
bull Source moving away decreasing in loudness
bull High pitch is constant ndash then abrupt change to constant low pitch
bull We do not have a sound of increasing pitch and then decreasing pitch
copycgrahamphysicscom 2016
Change in pitch or frequency
bull Smaller wavelength bull Increased frequency
bull Larger wavelength bull Decreased frequency
copycgrahamphysicscom 2016
Doppler Effect If we can figure out what the change in the wavelength is we also know the change in the frequency
Observer at rest
λ =119907
119891=
119907∆119905
119891∆119905
∆119889 = 119907∆119905 minus 119907119904∆119905 Because of the motion of the source the number of waves occupying the distance d is equal to ∆119889
The new wavelength λprime =119907∆119905minus119907119904∆119905
119891∆119905=
119907minus119907119904
119891
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Frequency heard by stationary observer and moving source
bull 119891prime =119907
119907+119907119904119891 = 119891
1
1+119907119904119907
for source moving away from observer
bull 119891prime =119907
λprime=
119891119907
119907minus119907119904
bull 119891prime =119907
119907minus119907119904119891 = 119891
1
1minus119907119904119907
for source moving towards O
bull 119907 = speed of emitted sound
bull 119907119904 =speed of source
λprime =119907 minus 119907119904
119891
copycgrahamphysicscom 2016
Frequency heard by moving observer from stationary source
bull If the observer is moving with respect to the source things are a bit different
bull The wavelength remains the same but the wave speed is different for the observer
bull λ = unchanged
bull 119907prime changes
O 119907119904119900119906119899119889 = 119907119900119887119904119890119903119907119890119903 + 119907119904119900119906119899119889
λ =119907119900119887119904119890119903119907119890119903+119907119904119900119906119899119889
119891 and 119891prime =
119907prime
λ
copycgrahamphysicscom 2016
Moving observer
Towards source
bull 119891prime =119907119900+119907119904119900119906119899119889
119907119904119900119906119899119889119891
=119907119900+119907119904119900119906119899119889
119907119904119900119906119899119889times 119891
bull 119891prime =119907119900
119907+ 1 119891
bull 119907 = speed of sound
Away from source
bull 119891prime =119907119904119900119906119899119889minus119907119900
119907119904119900119906119899119889119891
=119907119904119900119906119899119889minus119907119900
119907119904119900119906119899119889times 119891
bull 119891prime =119907119900
119907minus 1 119891
Change in frequency
bull ∆119891 = 119891prime minus 119891 =119907119900
119907+ 1 119891 minus 119891 = 119891 +
119907119900
119907119891 minus 119891 =
119907119900
119907119891
copycgrahamphysicscom 2016
For light waves bull Relative velocity between source and observer
bull Speed of light does not depend on speed of source
bull All observer measure speed of light = c
bull 119907119904 ≪lt 119888 hence apparent shift in frequency is the same for O moving or S moving
copycgrahamphysicscom 2016
Light waves
Source moving towards observer
bull ∆119891 = 119891prime minus 119891 =119907119900
119907119891 =
119907
119888119891
bull This expression was derived from binomial expansion which is not required to reproduce
Source moving away from observer
bull ∆119891 = 119891prime minus 119891 =minus119907119900
119907119891
= minus119907
119888119891
These two equations are equivalent to
∆λ = λ119907
119888
copycgrahamphysicscom 2016
Example bull A source emits sound of f=440Hz It moves in a straight line towards
a stationary observer with a speed of 30m119904minus1
bull An observer hears sound of frequency f = 484Hz
bull Find the speed of sound in air Solution
bull S moves towards observer 119891prime =119907
119907minus119907119904119891
bull 119891 = 440119867119911 119891prime = 484119867119911 119907119904 = 30119898119904minus1
bull 119891prime =119907
119907minus119907119904119891 = 119891
1
1minus119907119904119907
bull 1 minus119907119904
119907=
119891
119891prime
bull 1 minus119891
119891prime=
119907119904
119907
bull 119907 = 1199071199041
1minus119891
119891prime
= 30 times1
1minus440
484
=30
0091= 330119898119904minus1
copycgrahamphysicscom 2016
Example bull A radio signal from a galaxy is measured to be f =
139x109Hz The same signal from a source in the laboratory has f = 142x109Hz Suggest why the galaxy is moving away from earth and calculate its recession speed away from Earth
Solution bull Frsquo of moving source lt than f source is moving away less f longer λ
bull ∆119891 = minus119907
119888119891
bull 119907 =minus119888∆119891
119891=
minus30times108times 139minus142 times109
142x109
bull = 634 times 106119898119904minus1
copycgrahamphysicscom 2016
Using the Doppler effect bull Can be used to measure speed of galaxies
bull To measure speed in general
bull Source = transmitter this can be sound or EM waves
bull Frequency is constant
bull Waves are incident on reflector which is moving towards the source at speed v
bull Reflected waves are detected by receiver placed next to transmitter
bull 119907 ≪lt 119888 where c is the speed of wave from transmitter
copycgrahamphysicscom 2016
Continued
Reflector is moving observer
bull Observer is moving towards stationary source
bull ∆119891 = 119891prime minus 119891 =119907
119888119891
Receiver is stationary observer
bull Stationary observer watches source moving toward him
bull ∆119891 = 119891primeprime minus 119891prime =119907
119888119891rsquo
bull Mathematical manipulation
bull 119891prime minus 119891 =119907
119888119891
119891primeprime minus 119891prime =119907
119888119891rsquo
119891prime minus 119891 + 119891primeprime minus 119891prime =119907
119888119891 +
119907
119888119891rsquo
119891primeprime minus 119891 =119907
119888119891 + 119891prime
bull 119891prime is observer moving tw source
+
copycgrahamphysicscom 2016
continued bull 119891prime is observer moving toward source
119891prime minus 119891 =119907
119888119891
119891prime = 119891 +119907
119888119891 = 119891 1 +
119907
119888
bull 119891primeprime minus 119891 =119907
119888119891 + 119891prime =
119907
119888119891 + 119891 1 +
119907
119888
bull 119891primeprime minus 119891 =119907
119888119891 1 + 1 +
119907
119888=
119907
119888119891 2 +
119907
119888
= 2119891119907
119888+
1199072
1198882119891
bull But since vltltltc 1199072
1198882 can be ignored
bull 119891primeprime minus 119891 =2119907
119888119891 or short ∆119891 =
2119907
119888119891
bull If v = c we must use the full Doppler equation bull For EM radiation we always consider situations were vltltc
copycgrahamphysicscom 2016
Example bull The speed of sound in blood is 1500 times 103 m 119904minus1
Ultrasound of frequency 100 MHz is reflected from blood flowing in an artery The frequency of the reflected waves received back at the transmitter is 105 MHz Estimate the speed of the blood flow in the artery
Solution
bull Since vltltc we can use ∆119891 =2119907
119888119891
bull 119891primeprime minus 119891 = 105 minus 100 119872119867119911 =2119907
1500 times 103 times 10 times 106
bull 005 times 106 = 133334119907 119907 = 375m 119904minus1
bull We have assumed that ultra sound is incident at perpendicular direction to the blood flow
copycgrahamphysicscom 2016
Example bull Judy is standing on the platform of a station A high speed train is
approaching the station in a straight line at constant speed and is sounding its whistle As the train passes by Judy the frequency of sound emitted by the whistle as heard by Judy changes from 640Hz to 430Hz Find a) the speed of the train Solution
bull A)
bull (speed of sound = 330m119904minus1)
bull 119891prime(1 minus119907119904
119907) = 119891prime 1 +
119907119904
119907
bull 640 minus 640119907119904
119907= 430 + 430
119907119904
119907
bull 210 = 1070 119907119904
119907 119907119904 =
210
1070times 330 = 6477119898119904minus1
Towards Judy Away from Judy
119891prime = 640119867119911
119891prime = 1198911
1 minus119907119904119907
119891 =119891prime
1
1 minus119907119904119907
119891 = 119891prime 1 minus119907119904
119907
119891prime = 430119867119911
119891prime = 1198911
1 +119907119904119907
119891 =119891prime
1
1 +119907119904119907
119891 = 119891prime 1 +119907119904
119907
copycgrahamphysicscom 2016
hellipcontinued
bull b) the frequency of the sound emitted by the whistle as heard by a person on the train (speed of sound = 330m119904minus1)
bull Solution
bull 119891 = 119891prime 1 minus119907119904
119907= 640 times 1 minus
6477
330= 514119867119911
copycgrahamphysicscom 2016
Change in pitch or frequency
bull Smaller wavelength bull Increased frequency
bull Larger wavelength bull Decreased frequency
copycgrahamphysicscom 2016
Doppler Effect If we can figure out what the change in the wavelength is we also know the change in the frequency
Observer at rest
λ =119907
119891=
119907∆119905
119891∆119905
∆119889 = 119907∆119905 minus 119907119904∆119905 Because of the motion of the source the number of waves occupying the distance d is equal to ∆119889
The new wavelength λprime =119907∆119905minus119907119904∆119905
119891∆119905=
119907minus119907119904
119891
copycgrahamphysicscom 2016
Frequency heard by stationary observer and moving source
bull 119891prime =119907
119907+119907119904119891 = 119891
1
1+119907119904119907
for source moving away from observer
bull 119891prime =119907
λprime=
119891119907
119907minus119907119904
bull 119891prime =119907
119907minus119907119904119891 = 119891
1
1minus119907119904119907
for source moving towards O
bull 119907 = speed of emitted sound
bull 119907119904 =speed of source
λprime =119907 minus 119907119904
119891
copycgrahamphysicscom 2016
Frequency heard by moving observer from stationary source
bull If the observer is moving with respect to the source things are a bit different
bull The wavelength remains the same but the wave speed is different for the observer
bull λ = unchanged
bull 119907prime changes
O 119907119904119900119906119899119889 = 119907119900119887119904119890119903119907119890119903 + 119907119904119900119906119899119889
λ =119907119900119887119904119890119903119907119890119903+119907119904119900119906119899119889
119891 and 119891prime =
119907prime
λ
copycgrahamphysicscom 2016
Moving observer
Towards source
bull 119891prime =119907119900+119907119904119900119906119899119889
119907119904119900119906119899119889119891
=119907119900+119907119904119900119906119899119889
119907119904119900119906119899119889times 119891
bull 119891prime =119907119900
119907+ 1 119891
bull 119907 = speed of sound
Away from source
bull 119891prime =119907119904119900119906119899119889minus119907119900
119907119904119900119906119899119889119891
=119907119904119900119906119899119889minus119907119900
119907119904119900119906119899119889times 119891
bull 119891prime =119907119900
119907minus 1 119891
Change in frequency
bull ∆119891 = 119891prime minus 119891 =119907119900
119907+ 1 119891 minus 119891 = 119891 +
119907119900
119907119891 minus 119891 =
119907119900
119907119891
copycgrahamphysicscom 2016
For light waves bull Relative velocity between source and observer
bull Speed of light does not depend on speed of source
bull All observer measure speed of light = c
bull 119907119904 ≪lt 119888 hence apparent shift in frequency is the same for O moving or S moving
copycgrahamphysicscom 2016
Light waves
Source moving towards observer
bull ∆119891 = 119891prime minus 119891 =119907119900
119907119891 =
119907
119888119891
bull This expression was derived from binomial expansion which is not required to reproduce
Source moving away from observer
bull ∆119891 = 119891prime minus 119891 =minus119907119900
119907119891
= minus119907
119888119891
These two equations are equivalent to
∆λ = λ119907
119888
copycgrahamphysicscom 2016
Example bull A source emits sound of f=440Hz It moves in a straight line towards
a stationary observer with a speed of 30m119904minus1
bull An observer hears sound of frequency f = 484Hz
bull Find the speed of sound in air Solution
bull S moves towards observer 119891prime =119907
119907minus119907119904119891
bull 119891 = 440119867119911 119891prime = 484119867119911 119907119904 = 30119898119904minus1
bull 119891prime =119907
119907minus119907119904119891 = 119891
1
1minus119907119904119907
bull 1 minus119907119904
119907=
119891
119891prime
bull 1 minus119891
119891prime=
119907119904
119907
bull 119907 = 1199071199041
1minus119891
119891prime
= 30 times1
1minus440
484
=30
0091= 330119898119904minus1
copycgrahamphysicscom 2016
Example bull A radio signal from a galaxy is measured to be f =
139x109Hz The same signal from a source in the laboratory has f = 142x109Hz Suggest why the galaxy is moving away from earth and calculate its recession speed away from Earth
Solution bull Frsquo of moving source lt than f source is moving away less f longer λ
bull ∆119891 = minus119907
119888119891
bull 119907 =minus119888∆119891
119891=
minus30times108times 139minus142 times109
142x109
bull = 634 times 106119898119904minus1
copycgrahamphysicscom 2016
Using the Doppler effect bull Can be used to measure speed of galaxies
bull To measure speed in general
bull Source = transmitter this can be sound or EM waves
bull Frequency is constant
bull Waves are incident on reflector which is moving towards the source at speed v
bull Reflected waves are detected by receiver placed next to transmitter
bull 119907 ≪lt 119888 where c is the speed of wave from transmitter
copycgrahamphysicscom 2016
Continued
Reflector is moving observer
bull Observer is moving towards stationary source
bull ∆119891 = 119891prime minus 119891 =119907
119888119891
Receiver is stationary observer
bull Stationary observer watches source moving toward him
bull ∆119891 = 119891primeprime minus 119891prime =119907
119888119891rsquo
bull Mathematical manipulation
bull 119891prime minus 119891 =119907
119888119891
119891primeprime minus 119891prime =119907
119888119891rsquo
119891prime minus 119891 + 119891primeprime minus 119891prime =119907
119888119891 +
119907
119888119891rsquo
119891primeprime minus 119891 =119907
119888119891 + 119891prime
bull 119891prime is observer moving tw source
+
copycgrahamphysicscom 2016
continued bull 119891prime is observer moving toward source
119891prime minus 119891 =119907
119888119891
119891prime = 119891 +119907
119888119891 = 119891 1 +
119907
119888
bull 119891primeprime minus 119891 =119907
119888119891 + 119891prime =
119907
119888119891 + 119891 1 +
119907
119888
bull 119891primeprime minus 119891 =119907
119888119891 1 + 1 +
119907
119888=
119907
119888119891 2 +
119907
119888
= 2119891119907
119888+
1199072
1198882119891
bull But since vltltltc 1199072
1198882 can be ignored
bull 119891primeprime minus 119891 =2119907
119888119891 or short ∆119891 =
2119907
119888119891
bull If v = c we must use the full Doppler equation bull For EM radiation we always consider situations were vltltc
copycgrahamphysicscom 2016
Example bull The speed of sound in blood is 1500 times 103 m 119904minus1
Ultrasound of frequency 100 MHz is reflected from blood flowing in an artery The frequency of the reflected waves received back at the transmitter is 105 MHz Estimate the speed of the blood flow in the artery
Solution
bull Since vltltc we can use ∆119891 =2119907
119888119891
bull 119891primeprime minus 119891 = 105 minus 100 119872119867119911 =2119907
1500 times 103 times 10 times 106
bull 005 times 106 = 133334119907 119907 = 375m 119904minus1
bull We have assumed that ultra sound is incident at perpendicular direction to the blood flow
copycgrahamphysicscom 2016
Example bull Judy is standing on the platform of a station A high speed train is
approaching the station in a straight line at constant speed and is sounding its whistle As the train passes by Judy the frequency of sound emitted by the whistle as heard by Judy changes from 640Hz to 430Hz Find a) the speed of the train Solution
bull A)
bull (speed of sound = 330m119904minus1)
bull 119891prime(1 minus119907119904
119907) = 119891prime 1 +
119907119904
119907
bull 640 minus 640119907119904
119907= 430 + 430
119907119904
119907
bull 210 = 1070 119907119904
119907 119907119904 =
210
1070times 330 = 6477119898119904minus1
Towards Judy Away from Judy
119891prime = 640119867119911
119891prime = 1198911
1 minus119907119904119907
119891 =119891prime
1
1 minus119907119904119907
119891 = 119891prime 1 minus119907119904
119907
119891prime = 430119867119911
119891prime = 1198911
1 +119907119904119907
119891 =119891prime
1
1 +119907119904119907
119891 = 119891prime 1 +119907119904
119907
copycgrahamphysicscom 2016
hellipcontinued
bull b) the frequency of the sound emitted by the whistle as heard by a person on the train (speed of sound = 330m119904minus1)
bull Solution
bull 119891 = 119891prime 1 minus119907119904
119907= 640 times 1 minus
6477
330= 514119867119911
copycgrahamphysicscom 2016
Doppler Effect If we can figure out what the change in the wavelength is we also know the change in the frequency
Observer at rest
λ =119907
119891=
119907∆119905
119891∆119905
∆119889 = 119907∆119905 minus 119907119904∆119905 Because of the motion of the source the number of waves occupying the distance d is equal to ∆119889
The new wavelength λprime =119907∆119905minus119907119904∆119905
119891∆119905=
119907minus119907119904
119891
copycgrahamphysicscom 2016
Frequency heard by stationary observer and moving source
bull 119891prime =119907
119907+119907119904119891 = 119891
1
1+119907119904119907
for source moving away from observer
bull 119891prime =119907
λprime=
119891119907
119907minus119907119904
bull 119891prime =119907
119907minus119907119904119891 = 119891
1
1minus119907119904119907
for source moving towards O
bull 119907 = speed of emitted sound
bull 119907119904 =speed of source
λprime =119907 minus 119907119904
119891
copycgrahamphysicscom 2016
Frequency heard by moving observer from stationary source
bull If the observer is moving with respect to the source things are a bit different
bull The wavelength remains the same but the wave speed is different for the observer
bull λ = unchanged
bull 119907prime changes
O 119907119904119900119906119899119889 = 119907119900119887119904119890119903119907119890119903 + 119907119904119900119906119899119889
λ =119907119900119887119904119890119903119907119890119903+119907119904119900119906119899119889
119891 and 119891prime =
119907prime
λ
copycgrahamphysicscom 2016
Moving observer
Towards source
bull 119891prime =119907119900+119907119904119900119906119899119889
119907119904119900119906119899119889119891
=119907119900+119907119904119900119906119899119889
119907119904119900119906119899119889times 119891
bull 119891prime =119907119900
119907+ 1 119891
bull 119907 = speed of sound
Away from source
bull 119891prime =119907119904119900119906119899119889minus119907119900
119907119904119900119906119899119889119891
=119907119904119900119906119899119889minus119907119900
119907119904119900119906119899119889times 119891
bull 119891prime =119907119900
119907minus 1 119891
Change in frequency
bull ∆119891 = 119891prime minus 119891 =119907119900
119907+ 1 119891 minus 119891 = 119891 +
119907119900
119907119891 minus 119891 =
119907119900
119907119891
copycgrahamphysicscom 2016
For light waves bull Relative velocity between source and observer
bull Speed of light does not depend on speed of source
bull All observer measure speed of light = c
bull 119907119904 ≪lt 119888 hence apparent shift in frequency is the same for O moving or S moving
copycgrahamphysicscom 2016
Light waves
Source moving towards observer
bull ∆119891 = 119891prime minus 119891 =119907119900
119907119891 =
119907
119888119891
bull This expression was derived from binomial expansion which is not required to reproduce
Source moving away from observer
bull ∆119891 = 119891prime minus 119891 =minus119907119900
119907119891
= minus119907
119888119891
These two equations are equivalent to
∆λ = λ119907
119888
copycgrahamphysicscom 2016
Example bull A source emits sound of f=440Hz It moves in a straight line towards
a stationary observer with a speed of 30m119904minus1
bull An observer hears sound of frequency f = 484Hz
bull Find the speed of sound in air Solution
bull S moves towards observer 119891prime =119907
119907minus119907119904119891
bull 119891 = 440119867119911 119891prime = 484119867119911 119907119904 = 30119898119904minus1
bull 119891prime =119907
119907minus119907119904119891 = 119891
1
1minus119907119904119907
bull 1 minus119907119904
119907=
119891
119891prime
bull 1 minus119891
119891prime=
119907119904
119907
bull 119907 = 1199071199041
1minus119891
119891prime
= 30 times1
1minus440
484
=30
0091= 330119898119904minus1
copycgrahamphysicscom 2016
Example bull A radio signal from a galaxy is measured to be f =
139x109Hz The same signal from a source in the laboratory has f = 142x109Hz Suggest why the galaxy is moving away from earth and calculate its recession speed away from Earth
Solution bull Frsquo of moving source lt than f source is moving away less f longer λ
bull ∆119891 = minus119907
119888119891
bull 119907 =minus119888∆119891
119891=
minus30times108times 139minus142 times109
142x109
bull = 634 times 106119898119904minus1
copycgrahamphysicscom 2016
Using the Doppler effect bull Can be used to measure speed of galaxies
bull To measure speed in general
bull Source = transmitter this can be sound or EM waves
bull Frequency is constant
bull Waves are incident on reflector which is moving towards the source at speed v
bull Reflected waves are detected by receiver placed next to transmitter
bull 119907 ≪lt 119888 where c is the speed of wave from transmitter
copycgrahamphysicscom 2016
Continued
Reflector is moving observer
bull Observer is moving towards stationary source
bull ∆119891 = 119891prime minus 119891 =119907
119888119891
Receiver is stationary observer
bull Stationary observer watches source moving toward him
bull ∆119891 = 119891primeprime minus 119891prime =119907
119888119891rsquo
bull Mathematical manipulation
bull 119891prime minus 119891 =119907
119888119891
119891primeprime minus 119891prime =119907
119888119891rsquo
119891prime minus 119891 + 119891primeprime minus 119891prime =119907
119888119891 +
119907
119888119891rsquo
119891primeprime minus 119891 =119907
119888119891 + 119891prime
bull 119891prime is observer moving tw source
+
copycgrahamphysicscom 2016
continued bull 119891prime is observer moving toward source
119891prime minus 119891 =119907
119888119891
119891prime = 119891 +119907
119888119891 = 119891 1 +
119907
119888
bull 119891primeprime minus 119891 =119907
119888119891 + 119891prime =
119907
119888119891 + 119891 1 +
119907
119888
bull 119891primeprime minus 119891 =119907
119888119891 1 + 1 +
119907
119888=
119907
119888119891 2 +
119907
119888
= 2119891119907
119888+
1199072
1198882119891
bull But since vltltltc 1199072
1198882 can be ignored
bull 119891primeprime minus 119891 =2119907
119888119891 or short ∆119891 =
2119907
119888119891
bull If v = c we must use the full Doppler equation bull For EM radiation we always consider situations were vltltc
copycgrahamphysicscom 2016
Example bull The speed of sound in blood is 1500 times 103 m 119904minus1
Ultrasound of frequency 100 MHz is reflected from blood flowing in an artery The frequency of the reflected waves received back at the transmitter is 105 MHz Estimate the speed of the blood flow in the artery
Solution
bull Since vltltc we can use ∆119891 =2119907
119888119891
bull 119891primeprime minus 119891 = 105 minus 100 119872119867119911 =2119907
1500 times 103 times 10 times 106
bull 005 times 106 = 133334119907 119907 = 375m 119904minus1
bull We have assumed that ultra sound is incident at perpendicular direction to the blood flow
copycgrahamphysicscom 2016
Example bull Judy is standing on the platform of a station A high speed train is
approaching the station in a straight line at constant speed and is sounding its whistle As the train passes by Judy the frequency of sound emitted by the whistle as heard by Judy changes from 640Hz to 430Hz Find a) the speed of the train Solution
bull A)
bull (speed of sound = 330m119904minus1)
bull 119891prime(1 minus119907119904
119907) = 119891prime 1 +
119907119904
119907
bull 640 minus 640119907119904
119907= 430 + 430
119907119904
119907
bull 210 = 1070 119907119904
119907 119907119904 =
210
1070times 330 = 6477119898119904minus1
Towards Judy Away from Judy
119891prime = 640119867119911
119891prime = 1198911
1 minus119907119904119907
119891 =119891prime
1
1 minus119907119904119907
119891 = 119891prime 1 minus119907119904
119907
119891prime = 430119867119911
119891prime = 1198911
1 +119907119904119907
119891 =119891prime
1
1 +119907119904119907
119891 = 119891prime 1 +119907119904
119907
copycgrahamphysicscom 2016
hellipcontinued
bull b) the frequency of the sound emitted by the whistle as heard by a person on the train (speed of sound = 330m119904minus1)
bull Solution
bull 119891 = 119891prime 1 minus119907119904
119907= 640 times 1 minus
6477
330= 514119867119911
copycgrahamphysicscom 2016
Frequency heard by stationary observer and moving source
bull 119891prime =119907
119907+119907119904119891 = 119891
1
1+119907119904119907
for source moving away from observer
bull 119891prime =119907
λprime=
119891119907
119907minus119907119904
bull 119891prime =119907
119907minus119907119904119891 = 119891
1
1minus119907119904119907
for source moving towards O
bull 119907 = speed of emitted sound
bull 119907119904 =speed of source
λprime =119907 minus 119907119904
119891
copycgrahamphysicscom 2016
Frequency heard by moving observer from stationary source
bull If the observer is moving with respect to the source things are a bit different
bull The wavelength remains the same but the wave speed is different for the observer
bull λ = unchanged
bull 119907prime changes
O 119907119904119900119906119899119889 = 119907119900119887119904119890119903119907119890119903 + 119907119904119900119906119899119889
λ =119907119900119887119904119890119903119907119890119903+119907119904119900119906119899119889
119891 and 119891prime =
119907prime
λ
copycgrahamphysicscom 2016
Moving observer
Towards source
bull 119891prime =119907119900+119907119904119900119906119899119889
119907119904119900119906119899119889119891
=119907119900+119907119904119900119906119899119889
119907119904119900119906119899119889times 119891
bull 119891prime =119907119900
119907+ 1 119891
bull 119907 = speed of sound
Away from source
bull 119891prime =119907119904119900119906119899119889minus119907119900
119907119904119900119906119899119889119891
=119907119904119900119906119899119889minus119907119900
119907119904119900119906119899119889times 119891
bull 119891prime =119907119900
119907minus 1 119891
Change in frequency
bull ∆119891 = 119891prime minus 119891 =119907119900
119907+ 1 119891 minus 119891 = 119891 +
119907119900
119907119891 minus 119891 =
119907119900
119907119891
copycgrahamphysicscom 2016
For light waves bull Relative velocity between source and observer
bull Speed of light does not depend on speed of source
bull All observer measure speed of light = c
bull 119907119904 ≪lt 119888 hence apparent shift in frequency is the same for O moving or S moving
copycgrahamphysicscom 2016
Light waves
Source moving towards observer
bull ∆119891 = 119891prime minus 119891 =119907119900
119907119891 =
119907
119888119891
bull This expression was derived from binomial expansion which is not required to reproduce
Source moving away from observer
bull ∆119891 = 119891prime minus 119891 =minus119907119900
119907119891
= minus119907
119888119891
These two equations are equivalent to
∆λ = λ119907
119888
copycgrahamphysicscom 2016
Example bull A source emits sound of f=440Hz It moves in a straight line towards
a stationary observer with a speed of 30m119904minus1
bull An observer hears sound of frequency f = 484Hz
bull Find the speed of sound in air Solution
bull S moves towards observer 119891prime =119907
119907minus119907119904119891
bull 119891 = 440119867119911 119891prime = 484119867119911 119907119904 = 30119898119904minus1
bull 119891prime =119907
119907minus119907119904119891 = 119891
1
1minus119907119904119907
bull 1 minus119907119904
119907=
119891
119891prime
bull 1 minus119891
119891prime=
119907119904
119907
bull 119907 = 1199071199041
1minus119891
119891prime
= 30 times1
1minus440
484
=30
0091= 330119898119904minus1
copycgrahamphysicscom 2016
Example bull A radio signal from a galaxy is measured to be f =
139x109Hz The same signal from a source in the laboratory has f = 142x109Hz Suggest why the galaxy is moving away from earth and calculate its recession speed away from Earth
Solution bull Frsquo of moving source lt than f source is moving away less f longer λ
bull ∆119891 = minus119907
119888119891
bull 119907 =minus119888∆119891
119891=
minus30times108times 139minus142 times109
142x109
bull = 634 times 106119898119904minus1
copycgrahamphysicscom 2016
Using the Doppler effect bull Can be used to measure speed of galaxies
bull To measure speed in general
bull Source = transmitter this can be sound or EM waves
bull Frequency is constant
bull Waves are incident on reflector which is moving towards the source at speed v
bull Reflected waves are detected by receiver placed next to transmitter
bull 119907 ≪lt 119888 where c is the speed of wave from transmitter
copycgrahamphysicscom 2016
Continued
Reflector is moving observer
bull Observer is moving towards stationary source
bull ∆119891 = 119891prime minus 119891 =119907
119888119891
Receiver is stationary observer
bull Stationary observer watches source moving toward him
bull ∆119891 = 119891primeprime minus 119891prime =119907
119888119891rsquo
bull Mathematical manipulation
bull 119891prime minus 119891 =119907
119888119891
119891primeprime minus 119891prime =119907
119888119891rsquo
119891prime minus 119891 + 119891primeprime minus 119891prime =119907
119888119891 +
119907
119888119891rsquo
119891primeprime minus 119891 =119907
119888119891 + 119891prime
bull 119891prime is observer moving tw source
+
copycgrahamphysicscom 2016
continued bull 119891prime is observer moving toward source
119891prime minus 119891 =119907
119888119891
119891prime = 119891 +119907
119888119891 = 119891 1 +
119907
119888
bull 119891primeprime minus 119891 =119907
119888119891 + 119891prime =
119907
119888119891 + 119891 1 +
119907
119888
bull 119891primeprime minus 119891 =119907
119888119891 1 + 1 +
119907
119888=
119907
119888119891 2 +
119907
119888
= 2119891119907
119888+
1199072
1198882119891
bull But since vltltltc 1199072
1198882 can be ignored
bull 119891primeprime minus 119891 =2119907
119888119891 or short ∆119891 =
2119907
119888119891
bull If v = c we must use the full Doppler equation bull For EM radiation we always consider situations were vltltc
copycgrahamphysicscom 2016
Example bull The speed of sound in blood is 1500 times 103 m 119904minus1
Ultrasound of frequency 100 MHz is reflected from blood flowing in an artery The frequency of the reflected waves received back at the transmitter is 105 MHz Estimate the speed of the blood flow in the artery
Solution
bull Since vltltc we can use ∆119891 =2119907
119888119891
bull 119891primeprime minus 119891 = 105 minus 100 119872119867119911 =2119907
1500 times 103 times 10 times 106
bull 005 times 106 = 133334119907 119907 = 375m 119904minus1
bull We have assumed that ultra sound is incident at perpendicular direction to the blood flow
copycgrahamphysicscom 2016
Example bull Judy is standing on the platform of a station A high speed train is
approaching the station in a straight line at constant speed and is sounding its whistle As the train passes by Judy the frequency of sound emitted by the whistle as heard by Judy changes from 640Hz to 430Hz Find a) the speed of the train Solution
bull A)
bull (speed of sound = 330m119904minus1)
bull 119891prime(1 minus119907119904
119907) = 119891prime 1 +
119907119904
119907
bull 640 minus 640119907119904
119907= 430 + 430
119907119904
119907
bull 210 = 1070 119907119904
119907 119907119904 =
210
1070times 330 = 6477119898119904minus1
Towards Judy Away from Judy
119891prime = 640119867119911
119891prime = 1198911
1 minus119907119904119907
119891 =119891prime
1
1 minus119907119904119907
119891 = 119891prime 1 minus119907119904
119907
119891prime = 430119867119911
119891prime = 1198911
1 +119907119904119907
119891 =119891prime
1
1 +119907119904119907
119891 = 119891prime 1 +119907119904
119907
copycgrahamphysicscom 2016
hellipcontinued
bull b) the frequency of the sound emitted by the whistle as heard by a person on the train (speed of sound = 330m119904minus1)
bull Solution
bull 119891 = 119891prime 1 minus119907119904
119907= 640 times 1 minus
6477
330= 514119867119911
copycgrahamphysicscom 2016
Frequency heard by moving observer from stationary source
bull If the observer is moving with respect to the source things are a bit different
bull The wavelength remains the same but the wave speed is different for the observer
bull λ = unchanged
bull 119907prime changes
O 119907119904119900119906119899119889 = 119907119900119887119904119890119903119907119890119903 + 119907119904119900119906119899119889
λ =119907119900119887119904119890119903119907119890119903+119907119904119900119906119899119889
119891 and 119891prime =
119907prime
λ
copycgrahamphysicscom 2016
Moving observer
Towards source
bull 119891prime =119907119900+119907119904119900119906119899119889
119907119904119900119906119899119889119891
=119907119900+119907119904119900119906119899119889
119907119904119900119906119899119889times 119891
bull 119891prime =119907119900
119907+ 1 119891
bull 119907 = speed of sound
Away from source
bull 119891prime =119907119904119900119906119899119889minus119907119900
119907119904119900119906119899119889119891
=119907119904119900119906119899119889minus119907119900
119907119904119900119906119899119889times 119891
bull 119891prime =119907119900
119907minus 1 119891
Change in frequency
bull ∆119891 = 119891prime minus 119891 =119907119900
119907+ 1 119891 minus 119891 = 119891 +
119907119900
119907119891 minus 119891 =
119907119900
119907119891
copycgrahamphysicscom 2016
For light waves bull Relative velocity between source and observer
bull Speed of light does not depend on speed of source
bull All observer measure speed of light = c
bull 119907119904 ≪lt 119888 hence apparent shift in frequency is the same for O moving or S moving
copycgrahamphysicscom 2016
Light waves
Source moving towards observer
bull ∆119891 = 119891prime minus 119891 =119907119900
119907119891 =
119907
119888119891
bull This expression was derived from binomial expansion which is not required to reproduce
Source moving away from observer
bull ∆119891 = 119891prime minus 119891 =minus119907119900
119907119891
= minus119907
119888119891
These two equations are equivalent to
∆λ = λ119907
119888
copycgrahamphysicscom 2016
Example bull A source emits sound of f=440Hz It moves in a straight line towards
a stationary observer with a speed of 30m119904minus1
bull An observer hears sound of frequency f = 484Hz
bull Find the speed of sound in air Solution
bull S moves towards observer 119891prime =119907
119907minus119907119904119891
bull 119891 = 440119867119911 119891prime = 484119867119911 119907119904 = 30119898119904minus1
bull 119891prime =119907
119907minus119907119904119891 = 119891
1
1minus119907119904119907
bull 1 minus119907119904
119907=
119891
119891prime
bull 1 minus119891
119891prime=
119907119904
119907
bull 119907 = 1199071199041
1minus119891
119891prime
= 30 times1
1minus440
484
=30
0091= 330119898119904minus1
copycgrahamphysicscom 2016
Example bull A radio signal from a galaxy is measured to be f =
139x109Hz The same signal from a source in the laboratory has f = 142x109Hz Suggest why the galaxy is moving away from earth and calculate its recession speed away from Earth
Solution bull Frsquo of moving source lt than f source is moving away less f longer λ
bull ∆119891 = minus119907
119888119891
bull 119907 =minus119888∆119891
119891=
minus30times108times 139minus142 times109
142x109
bull = 634 times 106119898119904minus1
copycgrahamphysicscom 2016
Using the Doppler effect bull Can be used to measure speed of galaxies
bull To measure speed in general
bull Source = transmitter this can be sound or EM waves
bull Frequency is constant
bull Waves are incident on reflector which is moving towards the source at speed v
bull Reflected waves are detected by receiver placed next to transmitter
bull 119907 ≪lt 119888 where c is the speed of wave from transmitter
copycgrahamphysicscom 2016
Continued
Reflector is moving observer
bull Observer is moving towards stationary source
bull ∆119891 = 119891prime minus 119891 =119907
119888119891
Receiver is stationary observer
bull Stationary observer watches source moving toward him
bull ∆119891 = 119891primeprime minus 119891prime =119907
119888119891rsquo
bull Mathematical manipulation
bull 119891prime minus 119891 =119907
119888119891
119891primeprime minus 119891prime =119907
119888119891rsquo
119891prime minus 119891 + 119891primeprime minus 119891prime =119907
119888119891 +
119907
119888119891rsquo
119891primeprime minus 119891 =119907
119888119891 + 119891prime
bull 119891prime is observer moving tw source
+
copycgrahamphysicscom 2016
continued bull 119891prime is observer moving toward source
119891prime minus 119891 =119907
119888119891
119891prime = 119891 +119907
119888119891 = 119891 1 +
119907
119888
bull 119891primeprime minus 119891 =119907
119888119891 + 119891prime =
119907
119888119891 + 119891 1 +
119907
119888
bull 119891primeprime minus 119891 =119907
119888119891 1 + 1 +
119907
119888=
119907
119888119891 2 +
119907
119888
= 2119891119907
119888+
1199072
1198882119891
bull But since vltltltc 1199072
1198882 can be ignored
bull 119891primeprime minus 119891 =2119907
119888119891 or short ∆119891 =
2119907
119888119891
bull If v = c we must use the full Doppler equation bull For EM radiation we always consider situations were vltltc
copycgrahamphysicscom 2016
Example bull The speed of sound in blood is 1500 times 103 m 119904minus1
Ultrasound of frequency 100 MHz is reflected from blood flowing in an artery The frequency of the reflected waves received back at the transmitter is 105 MHz Estimate the speed of the blood flow in the artery
Solution
bull Since vltltc we can use ∆119891 =2119907
119888119891
bull 119891primeprime minus 119891 = 105 minus 100 119872119867119911 =2119907
1500 times 103 times 10 times 106
bull 005 times 106 = 133334119907 119907 = 375m 119904minus1
bull We have assumed that ultra sound is incident at perpendicular direction to the blood flow
copycgrahamphysicscom 2016
Example bull Judy is standing on the platform of a station A high speed train is
approaching the station in a straight line at constant speed and is sounding its whistle As the train passes by Judy the frequency of sound emitted by the whistle as heard by Judy changes from 640Hz to 430Hz Find a) the speed of the train Solution
bull A)
bull (speed of sound = 330m119904minus1)
bull 119891prime(1 minus119907119904
119907) = 119891prime 1 +
119907119904
119907
bull 640 minus 640119907119904
119907= 430 + 430
119907119904
119907
bull 210 = 1070 119907119904
119907 119907119904 =
210
1070times 330 = 6477119898119904minus1
Towards Judy Away from Judy
119891prime = 640119867119911
119891prime = 1198911
1 minus119907119904119907
119891 =119891prime
1
1 minus119907119904119907
119891 = 119891prime 1 minus119907119904
119907
119891prime = 430119867119911
119891prime = 1198911
1 +119907119904119907
119891 =119891prime
1
1 +119907119904119907
119891 = 119891prime 1 +119907119904
119907
copycgrahamphysicscom 2016
hellipcontinued
bull b) the frequency of the sound emitted by the whistle as heard by a person on the train (speed of sound = 330m119904minus1)
bull Solution
bull 119891 = 119891prime 1 minus119907119904
119907= 640 times 1 minus
6477
330= 514119867119911
copycgrahamphysicscom 2016
Moving observer
Towards source
bull 119891prime =119907119900+119907119904119900119906119899119889
119907119904119900119906119899119889119891
=119907119900+119907119904119900119906119899119889
119907119904119900119906119899119889times 119891
bull 119891prime =119907119900
119907+ 1 119891
bull 119907 = speed of sound
Away from source
bull 119891prime =119907119904119900119906119899119889minus119907119900
119907119904119900119906119899119889119891
=119907119904119900119906119899119889minus119907119900
119907119904119900119906119899119889times 119891
bull 119891prime =119907119900
119907minus 1 119891
Change in frequency
bull ∆119891 = 119891prime minus 119891 =119907119900
119907+ 1 119891 minus 119891 = 119891 +
119907119900
119907119891 minus 119891 =
119907119900
119907119891
copycgrahamphysicscom 2016
For light waves bull Relative velocity between source and observer
bull Speed of light does not depend on speed of source
bull All observer measure speed of light = c
bull 119907119904 ≪lt 119888 hence apparent shift in frequency is the same for O moving or S moving
copycgrahamphysicscom 2016
Light waves
Source moving towards observer
bull ∆119891 = 119891prime minus 119891 =119907119900
119907119891 =
119907
119888119891
bull This expression was derived from binomial expansion which is not required to reproduce
Source moving away from observer
bull ∆119891 = 119891prime minus 119891 =minus119907119900
119907119891
= minus119907
119888119891
These two equations are equivalent to
∆λ = λ119907
119888
copycgrahamphysicscom 2016
Example bull A source emits sound of f=440Hz It moves in a straight line towards
a stationary observer with a speed of 30m119904minus1
bull An observer hears sound of frequency f = 484Hz
bull Find the speed of sound in air Solution
bull S moves towards observer 119891prime =119907
119907minus119907119904119891
bull 119891 = 440119867119911 119891prime = 484119867119911 119907119904 = 30119898119904minus1
bull 119891prime =119907
119907minus119907119904119891 = 119891
1
1minus119907119904119907
bull 1 minus119907119904
119907=
119891
119891prime
bull 1 minus119891
119891prime=
119907119904
119907
bull 119907 = 1199071199041
1minus119891
119891prime
= 30 times1
1minus440
484
=30
0091= 330119898119904minus1
copycgrahamphysicscom 2016
Example bull A radio signal from a galaxy is measured to be f =
139x109Hz The same signal from a source in the laboratory has f = 142x109Hz Suggest why the galaxy is moving away from earth and calculate its recession speed away from Earth
Solution bull Frsquo of moving source lt than f source is moving away less f longer λ
bull ∆119891 = minus119907
119888119891
bull 119907 =minus119888∆119891
119891=
minus30times108times 139minus142 times109
142x109
bull = 634 times 106119898119904minus1
copycgrahamphysicscom 2016
Using the Doppler effect bull Can be used to measure speed of galaxies
bull To measure speed in general
bull Source = transmitter this can be sound or EM waves
bull Frequency is constant
bull Waves are incident on reflector which is moving towards the source at speed v
bull Reflected waves are detected by receiver placed next to transmitter
bull 119907 ≪lt 119888 where c is the speed of wave from transmitter
copycgrahamphysicscom 2016
Continued
Reflector is moving observer
bull Observer is moving towards stationary source
bull ∆119891 = 119891prime minus 119891 =119907
119888119891
Receiver is stationary observer
bull Stationary observer watches source moving toward him
bull ∆119891 = 119891primeprime minus 119891prime =119907
119888119891rsquo
bull Mathematical manipulation
bull 119891prime minus 119891 =119907
119888119891
119891primeprime minus 119891prime =119907
119888119891rsquo
119891prime minus 119891 + 119891primeprime minus 119891prime =119907
119888119891 +
119907
119888119891rsquo
119891primeprime minus 119891 =119907
119888119891 + 119891prime
bull 119891prime is observer moving tw source
+
copycgrahamphysicscom 2016
continued bull 119891prime is observer moving toward source
119891prime minus 119891 =119907
119888119891
119891prime = 119891 +119907
119888119891 = 119891 1 +
119907
119888
bull 119891primeprime minus 119891 =119907
119888119891 + 119891prime =
119907
119888119891 + 119891 1 +
119907
119888
bull 119891primeprime minus 119891 =119907
119888119891 1 + 1 +
119907
119888=
119907
119888119891 2 +
119907
119888
= 2119891119907
119888+
1199072
1198882119891
bull But since vltltltc 1199072
1198882 can be ignored
bull 119891primeprime minus 119891 =2119907
119888119891 or short ∆119891 =
2119907
119888119891
bull If v = c we must use the full Doppler equation bull For EM radiation we always consider situations were vltltc
copycgrahamphysicscom 2016
Example bull The speed of sound in blood is 1500 times 103 m 119904minus1
Ultrasound of frequency 100 MHz is reflected from blood flowing in an artery The frequency of the reflected waves received back at the transmitter is 105 MHz Estimate the speed of the blood flow in the artery
Solution
bull Since vltltc we can use ∆119891 =2119907
119888119891
bull 119891primeprime minus 119891 = 105 minus 100 119872119867119911 =2119907
1500 times 103 times 10 times 106
bull 005 times 106 = 133334119907 119907 = 375m 119904minus1
bull We have assumed that ultra sound is incident at perpendicular direction to the blood flow
copycgrahamphysicscom 2016
Example bull Judy is standing on the platform of a station A high speed train is
approaching the station in a straight line at constant speed and is sounding its whistle As the train passes by Judy the frequency of sound emitted by the whistle as heard by Judy changes from 640Hz to 430Hz Find a) the speed of the train Solution
bull A)
bull (speed of sound = 330m119904minus1)
bull 119891prime(1 minus119907119904
119907) = 119891prime 1 +
119907119904
119907
bull 640 minus 640119907119904
119907= 430 + 430
119907119904
119907
bull 210 = 1070 119907119904
119907 119907119904 =
210
1070times 330 = 6477119898119904minus1
Towards Judy Away from Judy
119891prime = 640119867119911
119891prime = 1198911
1 minus119907119904119907
119891 =119891prime
1
1 minus119907119904119907
119891 = 119891prime 1 minus119907119904
119907
119891prime = 430119867119911
119891prime = 1198911
1 +119907119904119907
119891 =119891prime
1
1 +119907119904119907
119891 = 119891prime 1 +119907119904
119907
copycgrahamphysicscom 2016
hellipcontinued
bull b) the frequency of the sound emitted by the whistle as heard by a person on the train (speed of sound = 330m119904minus1)
bull Solution
bull 119891 = 119891prime 1 minus119907119904
119907= 640 times 1 minus
6477
330= 514119867119911
copycgrahamphysicscom 2016
For light waves bull Relative velocity between source and observer
bull Speed of light does not depend on speed of source
bull All observer measure speed of light = c
bull 119907119904 ≪lt 119888 hence apparent shift in frequency is the same for O moving or S moving
copycgrahamphysicscom 2016
Light waves
Source moving towards observer
bull ∆119891 = 119891prime minus 119891 =119907119900
119907119891 =
119907
119888119891
bull This expression was derived from binomial expansion which is not required to reproduce
Source moving away from observer
bull ∆119891 = 119891prime minus 119891 =minus119907119900
119907119891
= minus119907
119888119891
These two equations are equivalent to
∆λ = λ119907
119888
copycgrahamphysicscom 2016
Example bull A source emits sound of f=440Hz It moves in a straight line towards
a stationary observer with a speed of 30m119904minus1
bull An observer hears sound of frequency f = 484Hz
bull Find the speed of sound in air Solution
bull S moves towards observer 119891prime =119907
119907minus119907119904119891
bull 119891 = 440119867119911 119891prime = 484119867119911 119907119904 = 30119898119904minus1
bull 119891prime =119907
119907minus119907119904119891 = 119891
1
1minus119907119904119907
bull 1 minus119907119904
119907=
119891
119891prime
bull 1 minus119891
119891prime=
119907119904
119907
bull 119907 = 1199071199041
1minus119891
119891prime
= 30 times1
1minus440
484
=30
0091= 330119898119904minus1
copycgrahamphysicscom 2016
Example bull A radio signal from a galaxy is measured to be f =
139x109Hz The same signal from a source in the laboratory has f = 142x109Hz Suggest why the galaxy is moving away from earth and calculate its recession speed away from Earth
Solution bull Frsquo of moving source lt than f source is moving away less f longer λ
bull ∆119891 = minus119907
119888119891
bull 119907 =minus119888∆119891
119891=
minus30times108times 139minus142 times109
142x109
bull = 634 times 106119898119904minus1
copycgrahamphysicscom 2016
Using the Doppler effect bull Can be used to measure speed of galaxies
bull To measure speed in general
bull Source = transmitter this can be sound or EM waves
bull Frequency is constant
bull Waves are incident on reflector which is moving towards the source at speed v
bull Reflected waves are detected by receiver placed next to transmitter
bull 119907 ≪lt 119888 where c is the speed of wave from transmitter
copycgrahamphysicscom 2016
Continued
Reflector is moving observer
bull Observer is moving towards stationary source
bull ∆119891 = 119891prime minus 119891 =119907
119888119891
Receiver is stationary observer
bull Stationary observer watches source moving toward him
bull ∆119891 = 119891primeprime minus 119891prime =119907
119888119891rsquo
bull Mathematical manipulation
bull 119891prime minus 119891 =119907
119888119891
119891primeprime minus 119891prime =119907
119888119891rsquo
119891prime minus 119891 + 119891primeprime minus 119891prime =119907
119888119891 +
119907
119888119891rsquo
119891primeprime minus 119891 =119907
119888119891 + 119891prime
bull 119891prime is observer moving tw source
+
copycgrahamphysicscom 2016
continued bull 119891prime is observer moving toward source
119891prime minus 119891 =119907
119888119891
119891prime = 119891 +119907
119888119891 = 119891 1 +
119907
119888
bull 119891primeprime minus 119891 =119907
119888119891 + 119891prime =
119907
119888119891 + 119891 1 +
119907
119888
bull 119891primeprime minus 119891 =119907
119888119891 1 + 1 +
119907
119888=
119907
119888119891 2 +
119907
119888
= 2119891119907
119888+
1199072
1198882119891
bull But since vltltltc 1199072
1198882 can be ignored
bull 119891primeprime minus 119891 =2119907
119888119891 or short ∆119891 =
2119907
119888119891
bull If v = c we must use the full Doppler equation bull For EM radiation we always consider situations were vltltc
copycgrahamphysicscom 2016
Example bull The speed of sound in blood is 1500 times 103 m 119904minus1
Ultrasound of frequency 100 MHz is reflected from blood flowing in an artery The frequency of the reflected waves received back at the transmitter is 105 MHz Estimate the speed of the blood flow in the artery
Solution
bull Since vltltc we can use ∆119891 =2119907
119888119891
bull 119891primeprime minus 119891 = 105 minus 100 119872119867119911 =2119907
1500 times 103 times 10 times 106
bull 005 times 106 = 133334119907 119907 = 375m 119904minus1
bull We have assumed that ultra sound is incident at perpendicular direction to the blood flow
copycgrahamphysicscom 2016
Example bull Judy is standing on the platform of a station A high speed train is
approaching the station in a straight line at constant speed and is sounding its whistle As the train passes by Judy the frequency of sound emitted by the whistle as heard by Judy changes from 640Hz to 430Hz Find a) the speed of the train Solution
bull A)
bull (speed of sound = 330m119904minus1)
bull 119891prime(1 minus119907119904
119907) = 119891prime 1 +
119907119904
119907
bull 640 minus 640119907119904
119907= 430 + 430
119907119904
119907
bull 210 = 1070 119907119904
119907 119907119904 =
210
1070times 330 = 6477119898119904minus1
Towards Judy Away from Judy
119891prime = 640119867119911
119891prime = 1198911
1 minus119907119904119907
119891 =119891prime
1
1 minus119907119904119907
119891 = 119891prime 1 minus119907119904
119907
119891prime = 430119867119911
119891prime = 1198911
1 +119907119904119907
119891 =119891prime
1
1 +119907119904119907
119891 = 119891prime 1 +119907119904
119907
copycgrahamphysicscom 2016
hellipcontinued
bull b) the frequency of the sound emitted by the whistle as heard by a person on the train (speed of sound = 330m119904minus1)
bull Solution
bull 119891 = 119891prime 1 minus119907119904
119907= 640 times 1 minus
6477
330= 514119867119911
copycgrahamphysicscom 2016
Light waves
Source moving towards observer
bull ∆119891 = 119891prime minus 119891 =119907119900
119907119891 =
119907
119888119891
bull This expression was derived from binomial expansion which is not required to reproduce
Source moving away from observer
bull ∆119891 = 119891prime minus 119891 =minus119907119900
119907119891
= minus119907
119888119891
These two equations are equivalent to
∆λ = λ119907
119888
copycgrahamphysicscom 2016
Example bull A source emits sound of f=440Hz It moves in a straight line towards
a stationary observer with a speed of 30m119904minus1
bull An observer hears sound of frequency f = 484Hz
bull Find the speed of sound in air Solution
bull S moves towards observer 119891prime =119907
119907minus119907119904119891
bull 119891 = 440119867119911 119891prime = 484119867119911 119907119904 = 30119898119904minus1
bull 119891prime =119907
119907minus119907119904119891 = 119891
1
1minus119907119904119907
bull 1 minus119907119904
119907=
119891
119891prime
bull 1 minus119891
119891prime=
119907119904
119907
bull 119907 = 1199071199041
1minus119891
119891prime
= 30 times1
1minus440
484
=30
0091= 330119898119904minus1
copycgrahamphysicscom 2016
Example bull A radio signal from a galaxy is measured to be f =
139x109Hz The same signal from a source in the laboratory has f = 142x109Hz Suggest why the galaxy is moving away from earth and calculate its recession speed away from Earth
Solution bull Frsquo of moving source lt than f source is moving away less f longer λ
bull ∆119891 = minus119907
119888119891
bull 119907 =minus119888∆119891
119891=
minus30times108times 139minus142 times109
142x109
bull = 634 times 106119898119904minus1
copycgrahamphysicscom 2016
Using the Doppler effect bull Can be used to measure speed of galaxies
bull To measure speed in general
bull Source = transmitter this can be sound or EM waves
bull Frequency is constant
bull Waves are incident on reflector which is moving towards the source at speed v
bull Reflected waves are detected by receiver placed next to transmitter
bull 119907 ≪lt 119888 where c is the speed of wave from transmitter
copycgrahamphysicscom 2016
Continued
Reflector is moving observer
bull Observer is moving towards stationary source
bull ∆119891 = 119891prime minus 119891 =119907
119888119891
Receiver is stationary observer
bull Stationary observer watches source moving toward him
bull ∆119891 = 119891primeprime minus 119891prime =119907
119888119891rsquo
bull Mathematical manipulation
bull 119891prime minus 119891 =119907
119888119891
119891primeprime minus 119891prime =119907
119888119891rsquo
119891prime minus 119891 + 119891primeprime minus 119891prime =119907
119888119891 +
119907
119888119891rsquo
119891primeprime minus 119891 =119907
119888119891 + 119891prime
bull 119891prime is observer moving tw source
+
copycgrahamphysicscom 2016
continued bull 119891prime is observer moving toward source
119891prime minus 119891 =119907
119888119891
119891prime = 119891 +119907
119888119891 = 119891 1 +
119907
119888
bull 119891primeprime minus 119891 =119907
119888119891 + 119891prime =
119907
119888119891 + 119891 1 +
119907
119888
bull 119891primeprime minus 119891 =119907
119888119891 1 + 1 +
119907
119888=
119907
119888119891 2 +
119907
119888
= 2119891119907
119888+
1199072
1198882119891
bull But since vltltltc 1199072
1198882 can be ignored
bull 119891primeprime minus 119891 =2119907
119888119891 or short ∆119891 =
2119907
119888119891
bull If v = c we must use the full Doppler equation bull For EM radiation we always consider situations were vltltc
copycgrahamphysicscom 2016
Example bull The speed of sound in blood is 1500 times 103 m 119904minus1
Ultrasound of frequency 100 MHz is reflected from blood flowing in an artery The frequency of the reflected waves received back at the transmitter is 105 MHz Estimate the speed of the blood flow in the artery
Solution
bull Since vltltc we can use ∆119891 =2119907
119888119891
bull 119891primeprime minus 119891 = 105 minus 100 119872119867119911 =2119907
1500 times 103 times 10 times 106
bull 005 times 106 = 133334119907 119907 = 375m 119904minus1
bull We have assumed that ultra sound is incident at perpendicular direction to the blood flow
copycgrahamphysicscom 2016
Example bull Judy is standing on the platform of a station A high speed train is
approaching the station in a straight line at constant speed and is sounding its whistle As the train passes by Judy the frequency of sound emitted by the whistle as heard by Judy changes from 640Hz to 430Hz Find a) the speed of the train Solution
bull A)
bull (speed of sound = 330m119904minus1)
bull 119891prime(1 minus119907119904
119907) = 119891prime 1 +
119907119904
119907
bull 640 minus 640119907119904
119907= 430 + 430
119907119904
119907
bull 210 = 1070 119907119904
119907 119907119904 =
210
1070times 330 = 6477119898119904minus1
Towards Judy Away from Judy
119891prime = 640119867119911
119891prime = 1198911
1 minus119907119904119907
119891 =119891prime
1
1 minus119907119904119907
119891 = 119891prime 1 minus119907119904
119907
119891prime = 430119867119911
119891prime = 1198911
1 +119907119904119907
119891 =119891prime
1
1 +119907119904119907
119891 = 119891prime 1 +119907119904
119907
copycgrahamphysicscom 2016
hellipcontinued
bull b) the frequency of the sound emitted by the whistle as heard by a person on the train (speed of sound = 330m119904minus1)
bull Solution
bull 119891 = 119891prime 1 minus119907119904
119907= 640 times 1 minus
6477
330= 514119867119911
copycgrahamphysicscom 2016
Example bull A source emits sound of f=440Hz It moves in a straight line towards
a stationary observer with a speed of 30m119904minus1
bull An observer hears sound of frequency f = 484Hz
bull Find the speed of sound in air Solution
bull S moves towards observer 119891prime =119907
119907minus119907119904119891
bull 119891 = 440119867119911 119891prime = 484119867119911 119907119904 = 30119898119904minus1
bull 119891prime =119907
119907minus119907119904119891 = 119891
1
1minus119907119904119907
bull 1 minus119907119904
119907=
119891
119891prime
bull 1 minus119891
119891prime=
119907119904
119907
bull 119907 = 1199071199041
1minus119891
119891prime
= 30 times1
1minus440
484
=30
0091= 330119898119904minus1
copycgrahamphysicscom 2016
Example bull A radio signal from a galaxy is measured to be f =
139x109Hz The same signal from a source in the laboratory has f = 142x109Hz Suggest why the galaxy is moving away from earth and calculate its recession speed away from Earth
Solution bull Frsquo of moving source lt than f source is moving away less f longer λ
bull ∆119891 = minus119907
119888119891
bull 119907 =minus119888∆119891
119891=
minus30times108times 139minus142 times109
142x109
bull = 634 times 106119898119904minus1
copycgrahamphysicscom 2016
Using the Doppler effect bull Can be used to measure speed of galaxies
bull To measure speed in general
bull Source = transmitter this can be sound or EM waves
bull Frequency is constant
bull Waves are incident on reflector which is moving towards the source at speed v
bull Reflected waves are detected by receiver placed next to transmitter
bull 119907 ≪lt 119888 where c is the speed of wave from transmitter
copycgrahamphysicscom 2016
Continued
Reflector is moving observer
bull Observer is moving towards stationary source
bull ∆119891 = 119891prime minus 119891 =119907
119888119891
Receiver is stationary observer
bull Stationary observer watches source moving toward him
bull ∆119891 = 119891primeprime minus 119891prime =119907
119888119891rsquo
bull Mathematical manipulation
bull 119891prime minus 119891 =119907
119888119891
119891primeprime minus 119891prime =119907
119888119891rsquo
119891prime minus 119891 + 119891primeprime minus 119891prime =119907
119888119891 +
119907
119888119891rsquo
119891primeprime minus 119891 =119907
119888119891 + 119891prime
bull 119891prime is observer moving tw source
+
copycgrahamphysicscom 2016
continued bull 119891prime is observer moving toward source
119891prime minus 119891 =119907
119888119891
119891prime = 119891 +119907
119888119891 = 119891 1 +
119907
119888
bull 119891primeprime minus 119891 =119907
119888119891 + 119891prime =
119907
119888119891 + 119891 1 +
119907
119888
bull 119891primeprime minus 119891 =119907
119888119891 1 + 1 +
119907
119888=
119907
119888119891 2 +
119907
119888
= 2119891119907
119888+
1199072
1198882119891
bull But since vltltltc 1199072
1198882 can be ignored
bull 119891primeprime minus 119891 =2119907
119888119891 or short ∆119891 =
2119907
119888119891
bull If v = c we must use the full Doppler equation bull For EM radiation we always consider situations were vltltc
copycgrahamphysicscom 2016
Example bull The speed of sound in blood is 1500 times 103 m 119904minus1
Ultrasound of frequency 100 MHz is reflected from blood flowing in an artery The frequency of the reflected waves received back at the transmitter is 105 MHz Estimate the speed of the blood flow in the artery
Solution
bull Since vltltc we can use ∆119891 =2119907
119888119891
bull 119891primeprime minus 119891 = 105 minus 100 119872119867119911 =2119907
1500 times 103 times 10 times 106
bull 005 times 106 = 133334119907 119907 = 375m 119904minus1
bull We have assumed that ultra sound is incident at perpendicular direction to the blood flow
copycgrahamphysicscom 2016
Example bull Judy is standing on the platform of a station A high speed train is
approaching the station in a straight line at constant speed and is sounding its whistle As the train passes by Judy the frequency of sound emitted by the whistle as heard by Judy changes from 640Hz to 430Hz Find a) the speed of the train Solution
bull A)
bull (speed of sound = 330m119904minus1)
bull 119891prime(1 minus119907119904
119907) = 119891prime 1 +
119907119904
119907
bull 640 minus 640119907119904
119907= 430 + 430
119907119904
119907
bull 210 = 1070 119907119904
119907 119907119904 =
210
1070times 330 = 6477119898119904minus1
Towards Judy Away from Judy
119891prime = 640119867119911
119891prime = 1198911
1 minus119907119904119907
119891 =119891prime
1
1 minus119907119904119907
119891 = 119891prime 1 minus119907119904
119907
119891prime = 430119867119911
119891prime = 1198911
1 +119907119904119907
119891 =119891prime
1
1 +119907119904119907
119891 = 119891prime 1 +119907119904
119907
copycgrahamphysicscom 2016
hellipcontinued
bull b) the frequency of the sound emitted by the whistle as heard by a person on the train (speed of sound = 330m119904minus1)
bull Solution
bull 119891 = 119891prime 1 minus119907119904
119907= 640 times 1 minus
6477
330= 514119867119911
copycgrahamphysicscom 2016
Example bull A radio signal from a galaxy is measured to be f =
139x109Hz The same signal from a source in the laboratory has f = 142x109Hz Suggest why the galaxy is moving away from earth and calculate its recession speed away from Earth
Solution bull Frsquo of moving source lt than f source is moving away less f longer λ
bull ∆119891 = minus119907
119888119891
bull 119907 =minus119888∆119891
119891=
minus30times108times 139minus142 times109
142x109
bull = 634 times 106119898119904minus1
copycgrahamphysicscom 2016
Using the Doppler effect bull Can be used to measure speed of galaxies
bull To measure speed in general
bull Source = transmitter this can be sound or EM waves
bull Frequency is constant
bull Waves are incident on reflector which is moving towards the source at speed v
bull Reflected waves are detected by receiver placed next to transmitter
bull 119907 ≪lt 119888 where c is the speed of wave from transmitter
copycgrahamphysicscom 2016
Continued
Reflector is moving observer
bull Observer is moving towards stationary source
bull ∆119891 = 119891prime minus 119891 =119907
119888119891
Receiver is stationary observer
bull Stationary observer watches source moving toward him
bull ∆119891 = 119891primeprime minus 119891prime =119907
119888119891rsquo
bull Mathematical manipulation
bull 119891prime minus 119891 =119907
119888119891
119891primeprime minus 119891prime =119907
119888119891rsquo
119891prime minus 119891 + 119891primeprime minus 119891prime =119907
119888119891 +
119907
119888119891rsquo
119891primeprime minus 119891 =119907
119888119891 + 119891prime
bull 119891prime is observer moving tw source
+
copycgrahamphysicscom 2016
continued bull 119891prime is observer moving toward source
119891prime minus 119891 =119907
119888119891
119891prime = 119891 +119907
119888119891 = 119891 1 +
119907
119888
bull 119891primeprime minus 119891 =119907
119888119891 + 119891prime =
119907
119888119891 + 119891 1 +
119907
119888
bull 119891primeprime minus 119891 =119907
119888119891 1 + 1 +
119907
119888=
119907
119888119891 2 +
119907
119888
= 2119891119907
119888+
1199072
1198882119891
bull But since vltltltc 1199072
1198882 can be ignored
bull 119891primeprime minus 119891 =2119907
119888119891 or short ∆119891 =
2119907
119888119891
bull If v = c we must use the full Doppler equation bull For EM radiation we always consider situations were vltltc
copycgrahamphysicscom 2016
Example bull The speed of sound in blood is 1500 times 103 m 119904minus1
Ultrasound of frequency 100 MHz is reflected from blood flowing in an artery The frequency of the reflected waves received back at the transmitter is 105 MHz Estimate the speed of the blood flow in the artery
Solution
bull Since vltltc we can use ∆119891 =2119907
119888119891
bull 119891primeprime minus 119891 = 105 minus 100 119872119867119911 =2119907
1500 times 103 times 10 times 106
bull 005 times 106 = 133334119907 119907 = 375m 119904minus1
bull We have assumed that ultra sound is incident at perpendicular direction to the blood flow
copycgrahamphysicscom 2016
Example bull Judy is standing on the platform of a station A high speed train is
approaching the station in a straight line at constant speed and is sounding its whistle As the train passes by Judy the frequency of sound emitted by the whistle as heard by Judy changes from 640Hz to 430Hz Find a) the speed of the train Solution
bull A)
bull (speed of sound = 330m119904minus1)
bull 119891prime(1 minus119907119904
119907) = 119891prime 1 +
119907119904
119907
bull 640 minus 640119907119904
119907= 430 + 430
119907119904
119907
bull 210 = 1070 119907119904
119907 119907119904 =
210
1070times 330 = 6477119898119904minus1
Towards Judy Away from Judy
119891prime = 640119867119911
119891prime = 1198911
1 minus119907119904119907
119891 =119891prime
1
1 minus119907119904119907
119891 = 119891prime 1 minus119907119904
119907
119891prime = 430119867119911
119891prime = 1198911
1 +119907119904119907
119891 =119891prime
1
1 +119907119904119907
119891 = 119891prime 1 +119907119904
119907
copycgrahamphysicscom 2016
hellipcontinued
bull b) the frequency of the sound emitted by the whistle as heard by a person on the train (speed of sound = 330m119904minus1)
bull Solution
bull 119891 = 119891prime 1 minus119907119904
119907= 640 times 1 minus
6477
330= 514119867119911
copycgrahamphysicscom 2016
Using the Doppler effect bull Can be used to measure speed of galaxies
bull To measure speed in general
bull Source = transmitter this can be sound or EM waves
bull Frequency is constant
bull Waves are incident on reflector which is moving towards the source at speed v
bull Reflected waves are detected by receiver placed next to transmitter
bull 119907 ≪lt 119888 where c is the speed of wave from transmitter
copycgrahamphysicscom 2016
Continued
Reflector is moving observer
bull Observer is moving towards stationary source
bull ∆119891 = 119891prime minus 119891 =119907
119888119891
Receiver is stationary observer
bull Stationary observer watches source moving toward him
bull ∆119891 = 119891primeprime minus 119891prime =119907
119888119891rsquo
bull Mathematical manipulation
bull 119891prime minus 119891 =119907
119888119891
119891primeprime minus 119891prime =119907
119888119891rsquo
119891prime minus 119891 + 119891primeprime minus 119891prime =119907
119888119891 +
119907
119888119891rsquo
119891primeprime minus 119891 =119907
119888119891 + 119891prime
bull 119891prime is observer moving tw source
+
copycgrahamphysicscom 2016
continued bull 119891prime is observer moving toward source
119891prime minus 119891 =119907
119888119891
119891prime = 119891 +119907
119888119891 = 119891 1 +
119907
119888
bull 119891primeprime minus 119891 =119907
119888119891 + 119891prime =
119907
119888119891 + 119891 1 +
119907
119888
bull 119891primeprime minus 119891 =119907
119888119891 1 + 1 +
119907
119888=
119907
119888119891 2 +
119907
119888
= 2119891119907
119888+
1199072
1198882119891
bull But since vltltltc 1199072
1198882 can be ignored
bull 119891primeprime minus 119891 =2119907
119888119891 or short ∆119891 =
2119907
119888119891
bull If v = c we must use the full Doppler equation bull For EM radiation we always consider situations were vltltc
copycgrahamphysicscom 2016
Example bull The speed of sound in blood is 1500 times 103 m 119904minus1
Ultrasound of frequency 100 MHz is reflected from blood flowing in an artery The frequency of the reflected waves received back at the transmitter is 105 MHz Estimate the speed of the blood flow in the artery
Solution
bull Since vltltc we can use ∆119891 =2119907
119888119891
bull 119891primeprime minus 119891 = 105 minus 100 119872119867119911 =2119907
1500 times 103 times 10 times 106
bull 005 times 106 = 133334119907 119907 = 375m 119904minus1
bull We have assumed that ultra sound is incident at perpendicular direction to the blood flow
copycgrahamphysicscom 2016
Example bull Judy is standing on the platform of a station A high speed train is
approaching the station in a straight line at constant speed and is sounding its whistle As the train passes by Judy the frequency of sound emitted by the whistle as heard by Judy changes from 640Hz to 430Hz Find a) the speed of the train Solution
bull A)
bull (speed of sound = 330m119904minus1)
bull 119891prime(1 minus119907119904
119907) = 119891prime 1 +
119907119904
119907
bull 640 minus 640119907119904
119907= 430 + 430
119907119904
119907
bull 210 = 1070 119907119904
119907 119907119904 =
210
1070times 330 = 6477119898119904minus1
Towards Judy Away from Judy
119891prime = 640119867119911
119891prime = 1198911
1 minus119907119904119907
119891 =119891prime
1
1 minus119907119904119907
119891 = 119891prime 1 minus119907119904
119907
119891prime = 430119867119911
119891prime = 1198911
1 +119907119904119907
119891 =119891prime
1
1 +119907119904119907
119891 = 119891prime 1 +119907119904
119907
copycgrahamphysicscom 2016
hellipcontinued
bull b) the frequency of the sound emitted by the whistle as heard by a person on the train (speed of sound = 330m119904minus1)
bull Solution
bull 119891 = 119891prime 1 minus119907119904
119907= 640 times 1 minus
6477
330= 514119867119911
copycgrahamphysicscom 2016
Continued
Reflector is moving observer
bull Observer is moving towards stationary source
bull ∆119891 = 119891prime minus 119891 =119907
119888119891
Receiver is stationary observer
bull Stationary observer watches source moving toward him
bull ∆119891 = 119891primeprime minus 119891prime =119907
119888119891rsquo
bull Mathematical manipulation
bull 119891prime minus 119891 =119907
119888119891
119891primeprime minus 119891prime =119907
119888119891rsquo
119891prime minus 119891 + 119891primeprime minus 119891prime =119907
119888119891 +
119907
119888119891rsquo
119891primeprime minus 119891 =119907
119888119891 + 119891prime
bull 119891prime is observer moving tw source
+
copycgrahamphysicscom 2016
continued bull 119891prime is observer moving toward source
119891prime minus 119891 =119907
119888119891
119891prime = 119891 +119907
119888119891 = 119891 1 +
119907
119888
bull 119891primeprime minus 119891 =119907
119888119891 + 119891prime =
119907
119888119891 + 119891 1 +
119907
119888
bull 119891primeprime minus 119891 =119907
119888119891 1 + 1 +
119907
119888=
119907
119888119891 2 +
119907
119888
= 2119891119907
119888+
1199072
1198882119891
bull But since vltltltc 1199072
1198882 can be ignored
bull 119891primeprime minus 119891 =2119907
119888119891 or short ∆119891 =
2119907
119888119891
bull If v = c we must use the full Doppler equation bull For EM radiation we always consider situations were vltltc
copycgrahamphysicscom 2016
Example bull The speed of sound in blood is 1500 times 103 m 119904minus1
Ultrasound of frequency 100 MHz is reflected from blood flowing in an artery The frequency of the reflected waves received back at the transmitter is 105 MHz Estimate the speed of the blood flow in the artery
Solution
bull Since vltltc we can use ∆119891 =2119907
119888119891
bull 119891primeprime minus 119891 = 105 minus 100 119872119867119911 =2119907
1500 times 103 times 10 times 106
bull 005 times 106 = 133334119907 119907 = 375m 119904minus1
bull We have assumed that ultra sound is incident at perpendicular direction to the blood flow
copycgrahamphysicscom 2016
Example bull Judy is standing on the platform of a station A high speed train is
approaching the station in a straight line at constant speed and is sounding its whistle As the train passes by Judy the frequency of sound emitted by the whistle as heard by Judy changes from 640Hz to 430Hz Find a) the speed of the train Solution
bull A)
bull (speed of sound = 330m119904minus1)
bull 119891prime(1 minus119907119904
119907) = 119891prime 1 +
119907119904
119907
bull 640 minus 640119907119904
119907= 430 + 430
119907119904
119907
bull 210 = 1070 119907119904
119907 119907119904 =
210
1070times 330 = 6477119898119904minus1
Towards Judy Away from Judy
119891prime = 640119867119911
119891prime = 1198911
1 minus119907119904119907
119891 =119891prime
1
1 minus119907119904119907
119891 = 119891prime 1 minus119907119904
119907
119891prime = 430119867119911
119891prime = 1198911
1 +119907119904119907
119891 =119891prime
1
1 +119907119904119907
119891 = 119891prime 1 +119907119904
119907
copycgrahamphysicscom 2016
hellipcontinued
bull b) the frequency of the sound emitted by the whistle as heard by a person on the train (speed of sound = 330m119904minus1)
bull Solution
bull 119891 = 119891prime 1 minus119907119904
119907= 640 times 1 minus
6477
330= 514119867119911
copycgrahamphysicscom 2016
continued bull 119891prime is observer moving toward source
119891prime minus 119891 =119907
119888119891
119891prime = 119891 +119907
119888119891 = 119891 1 +
119907
119888
bull 119891primeprime minus 119891 =119907
119888119891 + 119891prime =
119907
119888119891 + 119891 1 +
119907
119888
bull 119891primeprime minus 119891 =119907
119888119891 1 + 1 +
119907
119888=
119907
119888119891 2 +
119907
119888
= 2119891119907
119888+
1199072
1198882119891
bull But since vltltltc 1199072
1198882 can be ignored
bull 119891primeprime minus 119891 =2119907
119888119891 or short ∆119891 =
2119907
119888119891
bull If v = c we must use the full Doppler equation bull For EM radiation we always consider situations were vltltc
copycgrahamphysicscom 2016
Example bull The speed of sound in blood is 1500 times 103 m 119904minus1
Ultrasound of frequency 100 MHz is reflected from blood flowing in an artery The frequency of the reflected waves received back at the transmitter is 105 MHz Estimate the speed of the blood flow in the artery
Solution
bull Since vltltc we can use ∆119891 =2119907
119888119891
bull 119891primeprime minus 119891 = 105 minus 100 119872119867119911 =2119907
1500 times 103 times 10 times 106
bull 005 times 106 = 133334119907 119907 = 375m 119904minus1
bull We have assumed that ultra sound is incident at perpendicular direction to the blood flow
copycgrahamphysicscom 2016
Example bull Judy is standing on the platform of a station A high speed train is
approaching the station in a straight line at constant speed and is sounding its whistle As the train passes by Judy the frequency of sound emitted by the whistle as heard by Judy changes from 640Hz to 430Hz Find a) the speed of the train Solution
bull A)
bull (speed of sound = 330m119904minus1)
bull 119891prime(1 minus119907119904
119907) = 119891prime 1 +
119907119904
119907
bull 640 minus 640119907119904
119907= 430 + 430
119907119904
119907
bull 210 = 1070 119907119904
119907 119907119904 =
210
1070times 330 = 6477119898119904minus1
Towards Judy Away from Judy
119891prime = 640119867119911
119891prime = 1198911
1 minus119907119904119907
119891 =119891prime
1
1 minus119907119904119907
119891 = 119891prime 1 minus119907119904
119907
119891prime = 430119867119911
119891prime = 1198911
1 +119907119904119907
119891 =119891prime
1
1 +119907119904119907
119891 = 119891prime 1 +119907119904
119907
copycgrahamphysicscom 2016
hellipcontinued
bull b) the frequency of the sound emitted by the whistle as heard by a person on the train (speed of sound = 330m119904minus1)
bull Solution
bull 119891 = 119891prime 1 minus119907119904
119907= 640 times 1 minus
6477
330= 514119867119911
copycgrahamphysicscom 2016
Example bull The speed of sound in blood is 1500 times 103 m 119904minus1
Ultrasound of frequency 100 MHz is reflected from blood flowing in an artery The frequency of the reflected waves received back at the transmitter is 105 MHz Estimate the speed of the blood flow in the artery
Solution
bull Since vltltc we can use ∆119891 =2119907
119888119891
bull 119891primeprime minus 119891 = 105 minus 100 119872119867119911 =2119907
1500 times 103 times 10 times 106
bull 005 times 106 = 133334119907 119907 = 375m 119904minus1
bull We have assumed that ultra sound is incident at perpendicular direction to the blood flow
copycgrahamphysicscom 2016
Example bull Judy is standing on the platform of a station A high speed train is
approaching the station in a straight line at constant speed and is sounding its whistle As the train passes by Judy the frequency of sound emitted by the whistle as heard by Judy changes from 640Hz to 430Hz Find a) the speed of the train Solution
bull A)
bull (speed of sound = 330m119904minus1)
bull 119891prime(1 minus119907119904
119907) = 119891prime 1 +
119907119904
119907
bull 640 minus 640119907119904
119907= 430 + 430
119907119904
119907
bull 210 = 1070 119907119904
119907 119907119904 =
210
1070times 330 = 6477119898119904minus1
Towards Judy Away from Judy
119891prime = 640119867119911
119891prime = 1198911
1 minus119907119904119907
119891 =119891prime
1
1 minus119907119904119907
119891 = 119891prime 1 minus119907119904
119907
119891prime = 430119867119911
119891prime = 1198911
1 +119907119904119907
119891 =119891prime
1
1 +119907119904119907
119891 = 119891prime 1 +119907119904
119907
copycgrahamphysicscom 2016
hellipcontinued
bull b) the frequency of the sound emitted by the whistle as heard by a person on the train (speed of sound = 330m119904minus1)
bull Solution
bull 119891 = 119891prime 1 minus119907119904
119907= 640 times 1 minus
6477
330= 514119867119911
copycgrahamphysicscom 2016
Example bull Judy is standing on the platform of a station A high speed train is
approaching the station in a straight line at constant speed and is sounding its whistle As the train passes by Judy the frequency of sound emitted by the whistle as heard by Judy changes from 640Hz to 430Hz Find a) the speed of the train Solution
bull A)
bull (speed of sound = 330m119904minus1)
bull 119891prime(1 minus119907119904
119907) = 119891prime 1 +
119907119904
119907
bull 640 minus 640119907119904
119907= 430 + 430
119907119904
119907
bull 210 = 1070 119907119904
119907 119907119904 =
210
1070times 330 = 6477119898119904minus1
Towards Judy Away from Judy
119891prime = 640119867119911
119891prime = 1198911
1 minus119907119904119907
119891 =119891prime
1
1 minus119907119904119907
119891 = 119891prime 1 minus119907119904
119907
119891prime = 430119867119911
119891prime = 1198911
1 +119907119904119907
119891 =119891prime
1
1 +119907119904119907
119891 = 119891prime 1 +119907119904
119907
copycgrahamphysicscom 2016
hellipcontinued
bull b) the frequency of the sound emitted by the whistle as heard by a person on the train (speed of sound = 330m119904minus1)
bull Solution
bull 119891 = 119891prime 1 minus119907119904
119907= 640 times 1 minus
6477
330= 514119867119911
copycgrahamphysicscom 2016