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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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

+

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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

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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

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

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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

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

+

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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

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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

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

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