Topic 11: Wave phenomena 11.2 Doppler effect

11.2.1 Describe what is meant by the Doppler effect.

11.2.2 Explain the Doppler effect by reference to wavefront diagrams for moving-detector and moving-source situations.

11.2.3 Apply the Doppler effect equations for sound.

11.2.4 Solve problems on the Doppler effect for sound.

11.2.5 Solve problems on the Doppler effect for electromagnetic waves using the approximation f = vf/c.

11.2.6 Outline an example in which the Doppler effect is used to measure speed.

Topic 11: Wave phenomena11.2 Doppler effect

The Doppler Effect

When a car passes you on the street, the frequency of its engine note appears to change. This is due to the ‘Doppler effect’:

It can occur due to any relative motion; i.e. Either due to the source or the observer moving.

The Doppler effect is the change in observed frequency of a wave due to relative motion between the wave source and an observer.

The Doppler effect

The diagram below represents waves emitted by a source of sound, S, which is stationary relative to the air. The velocity of the sound waves relative to the air is v.

Johann Christian

Doppler1803-1853

Describe what is meant by the Doppler effect.

The Doppler effect has to do with moving sound sources (or moving observers).For example if a sound source of frequency f approaches you at a speed us its wave fronts will bunch together and you will hear a frequency f’ which is higher than f. ( f’ > f ).On the other hand if a sound source recedes from you at a speed us its wave fronts will stretch out and you will hear a frequency f’ which is lower than f. ( f’ < f ).


us

The next diagram represents waves emitted by a source which is moving (with velocity vs) relative to the air.

Motion of the source produces a change in the wavelength, longer wavelength behind, shorter wavelength in front of source.

λ

λ

Speed same, frequency and wavelength change

Doppler effect for soundHyperlink

http://www.colorado.edu/physics/2000/applets/doppler.html

Doppler effect for sound with moving observer

Explain the Doppler effect by reference to wavefront diagrams for moving-detector and moving-source situations.

Consider the ice-cream truck that parks in the hood playing that interminable music.

Since the ice cream truck is not yet moving, the wave fronts are spherically symmetric.


Planar cross-section

The cross-section is symmetric.

us = 0 us = 0


Whether Dobson is in front of the truck, or behind it, he hears the same frequency.


Us = 0FYIEven though the sound waves are drawn as transverse, they are in actuality longitudinal.


Suppose the ice-cream truck is now moving, but ringing the bell at the same rate as before.

Note how the wave fronts bunch up in the front, and separate in the back.

The reason this happens is that the truck moves forward a little bit during each successive spherical wave emission.FYIDon’t forget, the actual speed of the wavefronts through the stationary medium of the air is the speed of sound v (bunched or otherwise).


us


Dobson will hear a different frequency now, depending on his position relative to the truck.

If Dobson is in front of the moving truck, f’ > f.

If he is behind, he will hear f’ < f.


us

f’ > ff’ < f


Now we want to look at TWO successive pulses emitted by a source that is MOVING at speed us:

is the “real” wavelength of the sound.

Since the speed of sound in air is v, = vT where T is the period of the sound source. Note that Pulse 1 has traveled = vT in the time t = T.


uS

Pulse 1t = 0

Pulse 1t = T

Pulse 2t = T

= vT

d = usT

’

Which gives you the most chocolates?


’ is the wavelength detected by the observer.

The distance d between the emission of the two pulses is simply the velocity of the source us times the time between emissions T. Thus d = usT.

Since = d + ’ then ’ = – d = vT – usT. So

’ = (v – us)T


uS

Pulse 1t = 0

Pulse 1t = T

Pulse 2t = T

= vT

d = usT

’


From v = ’f’ and ’ = (v – us)T we see that

f’ = v/’ = v/[ ( v – us )T ]

f’ = ( 1/T )v/( v – us )

f’ = f[ v / ( v – us ) ].


uS

Pulse 1t = 0

Pulse 1t = T

Pulse 2t = T

= vT

d = usT

’

FYIIf the source is receding, replace us with –us.

[ f = 1/T ]

Apply the Doppler effect equations for sound.


Doppler effect moving

source

f’ = f[ v / ( v us ) ]v is the speed of sound.Approach (-us), recede (+us).

EXAMPLE: A car horn has a frequency of 520 Hz. The car is traveling to the right at 25 ms-1 and the speed of sound is 340 ms-1.

(a) What frequency is heard by the driver?

SOLUTION:

(a) The driver has a relative speed of us = 0. Thus f’ = 520 Hz.




source



(b) What frequency is heard by Observer 1?

SOLUTION:

(b) For approach use -us.

f’ = 520[ 340/( 340 - 25 ) ] = 560 Hz.

Be sure f’ is higher since car is approaching.




source



(c) What frequency is heard by Observer 2?

SOLUTION:

(c) For approach use +us.

f’ = 520[ 340/( 340 + 25 ) ] = 480 Hz.

Be sure f’ is lower since the car is receding.


Suppose the source is stationary.If the observer is not moving, he will hear the true frequency of the source.EXAMPLE: A sound source with wavefronts is shown. A stationary observer (the blue circle) is immersed in the sound. Observing the clock and the wavefronts, determine the frequency f of the source (and the frequency f’ detected by the observer).

SOLUTION: Both frequencies are the same.

Thus f = f’ = 5 cycles / 12 s = 0.42 Hz.


o

5 wavefronts pass observer in 12 seconds.


The diagram shows which wavefronts pass the observer in the time interval ∆t (red):

Because the wave speed is v, the length of the region in question is v∆t.

If we divide the length v∆t by the wavelength we get the number of cycles (or wavefronts). Thus

#cycles detected = v∆t/.


o

v∆t

FYIRecalling the definition of f’ we have f’ = #cycles detected /∆t = (v∆t/)/∆t = v/.But v/ = f so that f’ = f. (EXPECTED)


Suppose the source is stationary.If the observer is moving TOWARD the source at uo, he will hear a higher frequency.EXAMPLE: A moving observer (the blue circle) is now immersed in the sound. Observing the clock and the wavefronts, determine the frequency f of the source (and the frequency f’ detected by the observer).

SOLUTION:

f = 5 cycles / 12 s

= 0.42 Hz.

Now 8 wavefronts pass the observer in 12 s.Thus f’ = 8 cycles / 12 s = 0.66 Hz.


o


The diagram shows which wavefronts pass the moving observer in the time interval ∆t:

The red region v∆t is still due to the wave speed itself (still 5 cycles in our example).

The blue region uo∆t is due to the oberver’s speed uo (3 cycles in our example).

Now for the observer #cycles detected = (v∆t + uo∆t)/.

Thus f’ = #cycles detected /∆t

= [(v∆t + uo∆t) / ]/∆t = v/ + uo/.


o

v∆t uo∆t


From f’ = v/ + uo/ and the relation v = f we can write

f’ = v/ + uo/

f’ = f + uo/(v/f)

f’ = fv/v + fuo/v

f’ = f[ ( v + uo ) / v ].

If the observer is moving AWAY from the source, substitute –uo for uo.



observer

f’ = f[ ( v uo ) / v ].v is the speed of sound.Approach (+uo), recede (-uo).

FYIPick the sign that makes f’ do what is expected.




observer

f’ = f[ ( v us ) / v ].v is the speed of sound.Approach (+us), recede (-us).

EXAMPLE: A car horn has a frequency of 520 Hz. The car is stationary and the speed of sound is 340 ms-1.

(a) An observer approaches the car at 25 ms-1. What frequency is heard by the observer?

SOLUTION:

(a) For approach use +us.

f’ = 520[ ( 340 + 25 ) / 340 ] = 560 Hz.

Be sure f’ is higher since the observer is approaching the car.




observer

f’ = f[ ( v us ) / v ].v is the speed of sound.Approach (+us), recede (-us).

EXAMPLE: A car horn has a frequency of 520 Hz. The car is stationary and the speed of sound is 340 ms-1.

(b) An observer recedes from the car at 25 ms-1. What frequency is heard by the observer?

SOLUTION:

(b) For receding use -us.

f’ = 520[ ( 340 - 25 ) / 340 ] = 480 Hz.

Be sure f’ is lower since the observer is receding from the car.

Solve problems on the Doppler effect for electromagnetic waves using the approximation f = vf/c.

We have discussed the Doppler effect for sound waves, but we can also discuss it for electromagnetic waves, with this major difference: The speeds of source us and observer uo are very much less than the speed of light c.

We define v to be the relative velocity between source and observer. Thus v = us – uo.

Then f’, the frequency detected by the observer, is related to f, the frequency of the light source, by the formula f’ = f + (v/c)f.


Doppler effect for

light

∆f = (v/c)fv is the relative speed between the source and the observer.

Solve problems on the Doppler effect for electromagnetic waves using the approximation f = vf/c.


EXAMPLE: A star in another galaxy is traveling away from us at a speed of 5.6106 ms-1. It has a known absorption spectrum line that should be located at 520 nm on an identical stationary star. Where is this line located on the moving star?

SOLUTION: Use c = f to find f = c/: f = c/ = 3108 / 52010-9 = 5.81014 Hz.

The star is moving away from us so the effect is to decrease the frequency. Thus v = -5.6106 ms-1.

f’ = f + (v/c)f = f[ 1 + (v/c)]

f’ = 5.81014 ( 1 - 5.6106/ 3108 ) = 5.71014 Hz.

’ = c/f’ = 3108 / 5.71014 = 5.310-7 = 530 nm.

Moving observer

Wavelength fixed, speed and frequency change

Doppler shift for SOUND source

Be careful to apply the + or – the correct way.

Doppler sound problems

Answers

Questions

Tsokos page 248 Q’s 1-7

Hamper page 136 Q’s 17-19.

Solve problems on the Doppler effect

for electromagnetic waves using the

approximation

Students should appreciate that the approximationmay be used only when v << c.

Caught by the fuzz.

Take the case of the car and the radar speed trap. The emitted frequency of the police radar gun is f, the car acts like a moving observer receiving it at f'. It reflects it back at f', acting now as a moving source and the police receive it back shifted once again as f''.The total change in frequency (f'' - f) is therefore doubled.

It can be shown that the relative Doppler shift for electro-magnetic radiations like light, radio waves etc is given (approximately) by

where c is the speed of light and v is the relative speed of source and observer. The speed of a car is being measured by a police-person using a "radar speed-measuring gun". The frequency of the transmitted signal is 5GHz. When "mixed" the transmitted and received signals beat with a frequency of 750Hz. If the speed limit for the road is 110kmh-1, should the driver be fined or not?

Question

Adding colour

http://en.wikibooks.org/wiki/File:DopplerBloodFlowDiag.gif

http://images.google.com/imgres?imgurl=http://94.136.37.172:81/fmf/fmc/Umbilical%20cord%20Color%20Doppler.jpg&imgrefurl=http://www.fetalmedicine.com/fmc/ultrasound/wellbeing-scan/&usg=__E6jNo8OHBqkT7dRHcJz20SNYXLE=&h=466&w=355&sz=120&hl=en&start=5&um=1&tbnid=NZiqxqgVnd2rlM:&tbnh=128&tbnw=98&prev=/images?q=colour+doppler+ultrasound&hl=en&rls=com.microsoft:en-us&um=1

Analysing blood flow

Questions

Hamper page 138 Q’s 20-22.

Outline an example in which the Doppler effect is used to measure speed. PRACTICE: The absorption spectra of stars of varying distances from Earth are shown here. The sun is the bottom spectrum (D). Which star has the highest relative velocity to Earth? Is it moving towards us, or away from us?SOLUTION:From ∆f = (v/c)f we see that the higher the relative velocity v the greater the shift ∆f. Thus A is our candidate.As a reference look at the heavy black line in D.It is shifting to the red region. Thus its frequency is less, and thus A is receding.


A

B

C

D

Outline an example in which the Doppler effect is used to measure speed. PRACTICE: The absorption spectra of stars of varying distances from Earth are shown here. The sun is the bottom spectrum (D). What is the approximate speed at which star D is receding from Earth?SOLUTION: Use c = f ( f = c/ ):From A, f = 3108/39010-9 = 7.71014 Hz.From D, f’ = 3108/49010-9 = 6.11014 Hz.Then ∆f = f’ – f = (6.1–7.7)1014 = -1.11014 Hz.Finally from ∆f = (v/c)f we get v = c∆f/f = (3108)(-1.11014) /7.71014 = -4.31107 ms-1.


A

B

C

D

This is the so-called “redshift” and it is used as evidence for an expanding universe.

Receding…

Outline an example in which the Doppler effect is used to measure speed.


EXAMPLES: The following are all examples of the Doppler effect.

1. The radar gun used by police. It uses the form v = c∆f/f to find the velocity of the car. It actually measures the difference in frequency between the emitted radar beam, and the reflected-and- returned one.




2. A Doppler ultrasound test uses reflected sound waves to see how blood flows through a blood vessel. It helps doctors evaluate blood flow through major arteries and veins, such as those of the arms, legs, and neck. It can show blocked or reduced blood flow through narrowing in the major arteries of the neck that could cause a stroke. It also can reveal blood clots in leg veins (deep vein thrombosis, or DVT) that could break loose and block blood flow to the lungs (pulmonary embolism). During pregnancy, Doppler ultrasound may be used to look at blood flow in an unborn baby to check the health of the fetus.




3. A Doppler radar is a specialized radar that makes use of the Doppler effect to produce velocity data about objects at a distance. It does this by beaming a microwave signal towards a desired target and listening for its reflection, then analyzing how the frequency of the returned signal has been altered by the storm’s motion. This variation gives accurate measurements of the radial component of a storm’s velocity.




4. If you look at a rotating luminous object like the sun, you see that one side is moving away from the observer while the other side is approaching the observer. Once again, the formula v = c∆f/f comes into play. The sun’s right side will be red shifted, whereas the left will be blue shifted. We can then find the speed of rotation of the sun (or any star, even if it is quite distant).

Solve problems on the Doppler effect for sound.


Since the source is receding ’ had better increase.

It will increase by how far d = VT the source has traveled.

Note that the source is NOT traveling at v, the speed of sound!



Consider both the stationary and moving source S.

By placing a scale from the first diagram into the second…



A is wrong because f goes negative. Beware!



For Doppler light, the v in the formula ∆f = (v/c)f is the relative velocity.

Because the speed of is a constant c regardless of the speed of the source of the light, its frequency cannot change.



For moving away the wavelength INCREASES.FYIRegardless of who is observing, the speed of the wave is determined by the air, and nothing else. Thus the speed is V.



For moving observer we use f’ = f(v+uo)/v.

In terms of the given symbols f = f0(v + 0.1v)/v.

Thus f = 1.1f0 IF the observer O and the source S were perfectly IN-LINE. They are not. Thus B is the answer.

E.g.

Calculate the frequency heard by the driver of a (silent!) car moving at 30.0ms-1 away from a sound of frequency 750Hz.

A f2 = 682 Hz

EM Radiation and the Doppler Effect

As the speed of light cannot change, it can be shown that for EM radiation such as light, the change in frequency (or ‘Doppler shift’) is given by...

Δf = v f0

c

Where...Δf = change in frequency (Hz)v = relative speed of the source and observer (ms-1)c = speed of lightf0 = emitted frequency (Hz)

Note:This formula is an approximation and can only be used if v << c

E.g. (You need to be able to remember this example)

A ‘Gatso’ speed camera emits microwaves of wavelength 2.6cm. These are reflected off a car moving at 160kmh-1.

a.Calculate the change in the frequency (Doppler shift) that would be detected in the car.

b.If the wave is then reflected, calculate the total Doppler shift in the frequency detected back at the Gatso camera.

a. Δf= 1700Hz

b. Δf= 3400Hz

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Topic 11: Wave phenomena 11.2 Doppler effect

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