Transcript Topic 11_2__Doppler effect

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 phenomena
11.2 Doppler effect
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
Topic 11: Wave phenomena
11.2 Doppler effect
Describe what is meant by the Doppler effect.
PRACTICE:
Who claims he is wearing a Doppler effect
costume?
SOLUTION:
Sheldon makes this claim.
Sheldon
Leonard
Raj
Topic 11: Wave phenomena
11.2 Doppler effect
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.
The crosssection is
symmetric.
Planar
crosssection
us = 0
us = 0
Since the ice cream truck is not yet moving, the
wave fronts are spherically symmetric.
Topic 11: Wave phenomena
11.2 Doppler effect
Explain the Doppler effect by reference to
wavefront diagrams for moving-detector and
moving-source situations.
Whether Dobson is in front of the truck, or
behind it, he hears the same frequency.
Us = 0
FYI
Even though the sound waves are drawn as
transverse, they are in actuality longitudinal.
Topic 11: Wave phenomena
11.2 Doppler effect
Explain the Doppler effect by reference to
wavefront diagrams for moving-detector and
moving-source situations.
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
us
a little bit during each
successive spherical wave
emission.
FYI
Don’t forget, the actual speed of the wavefronts
through the stationary medium of the air is the
speed of sound v (bunched or otherwise).
Topic 11: Wave phenomena
11.2 Doppler effect
Explain the Doppler effect by reference to
wavefront diagrams for moving-detector and
moving-source situations.
Dobson will hear a different frequency now,
depending on his position relative to the truck.
f’ < f
f’ > f
us
If Dobson is in front of the moving truck, f’ >
f.
If he is behind, he will hear f’ < f.
Topic 11: Wave phenomena
11.2 Doppler effect
Explain the Doppler effect by reference to
wavefront diagrams for moving-detector and
moving-source situations.
Now we want to look at TWO successive pulses
emitted by a source that is MOVING at speed us:
 = vT
Pulse 1
t=0
Pulse 1
t=T
Pulse 2
t=T
’
uS
d = us T
 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.
Topic 11: Wave phenomena
11.2 Doppler effect
Explain the Doppler effect by reference to
wavefront diagrams for moving-detector and
moving-source situations.
 = vT
Pulse 1
t=0
Pulse 1
t=T
Pulse 2
t=T
’
uS
d = us T
’ 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
Topic 11: Wave phenomena
11.2 Doppler effect
Explain the Doppler effect by reference to
wavefront diagrams for moving-detector and
moving-source situations.
 = vT
Pulse 1
t=0
Pulse 1
t=T
Pulse 2
t=T
’
uS
d = us T
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 ) ]. [ f = 1/T ]
FYI
If the source is receding, replace us with –us.
Topic 11: Wave phenomena
11.2 Doppler effect
Apply the Doppler effect equations for sound.
f’ = f[ v / ( v  us ) ]
Doppler
effect moving
v is the speed of sound.
source
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.
Topic 11: Wave phenomena
11.2 Doppler effect
Apply the Doppler effect equations for sound.
f’ = f[ v / ( v  us ) ]
Doppler
effect moving
v is the speed of sound.
source
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.
(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.
Topic 11: Wave phenomena
11.2 Doppler effect
Apply the Doppler effect equations for sound.
f’ = f[ v / ( v  us ) ]
Doppler
effect moving
v is the speed of sound.
source
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.
(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.
Topic 11: Wave phenomena
11.2 Doppler effect
Explain the Doppler effect by reference to
wavefront diagrams for moving-detector and
moving-source situations.
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
o
frequencies are the
same. 5 wavefronts pass observer in 12 seconds.
Thus f = f’ = 5 cycles / 12 s = 0.42 Hz.
Topic 11: Wave phenomena
11.2 Doppler effect
Explain the Doppler effect by reference to
wavefront diagrams for moving-detector and
moving-source situations.
The diagram shows which
wavefronts pass the
o
observer in the time
interval ∆t (red):
v∆t
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/.
FYI
Recalling the definition of f’ we have
f’ = #cycles detected /∆t = (v∆t/)/∆t = v/.
But v/ = f so that f’ = f. (EXPECTED)
Topic 11: Wave phenomena
11.2 Doppler effect
Explain the Doppler effect by reference to
wavefront diagrams for moving-detector and
moving-source situations.
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).
o
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.
Topic 11: Wave phenomena
11.2 Doppler effect
Explain the Doppler effect by reference to
wavefront diagrams for moving-detector and
moving-source situations.
The diagram shows which
wavefronts pass the moving
o
observer in the time
interval ∆t:
v∆t uo∆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/.
Topic 11: Wave phenomena
11.2 Doppler effect
Explain the Doppler effect by reference to
wavefront diagrams for moving-detector and
moving-source situations.
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.
f’ = f[ ( v  uo ) / v ].
Doppler
effect moving
v is the speed of sound.
observer
Approach (+uo), recede (-uo).
FYI
Pick the sign that makes f’ do what is expected.
Topic 11: Wave phenomena
11.2 Doppler effect
Explain the Doppler effect by reference to
wavefront diagrams for moving-detector and
moving-source situations.
f’ = f[ ( v  us ) / v ].
Doppler
effect moving
v is the speed of sound.
observer
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.
Topic 11: Wave phenomena
11.2 Doppler effect
Explain the Doppler effect by reference to
wavefront diagrams for moving-detector and
moving-source situations.
f’ = f[ ( v  us ) / v ].
Doppler
effect moving
v is the speed of sound.
observer
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.
Topic 11: Wave phenomena
11.2 Doppler effect
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.
∆f = (v/c)f
v is the relative speed between
the source and the observer.
Doppler
effect for
light
Topic 11: Wave phenomena
11.2 Doppler effect
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 5.6106 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/ = 3108 / 52010-9 = 5.81014 Hz.
The star is moving away from us so the effect is
to decrease the frequency. Thus v = -5.6106 ms-1.
f’ = f + (v/c)f = f[ 1 + (v/c)]
f’ = 5.81014 ( 1 - 5.6106/ 3108 ) = 5.71014 Hz.
’ = c/f’ = 3108 / 5.71014 = 5.310-7 = 530 nm.
Topic 11: Wave phenomena
11.2 Doppler effect
Outline an example in which the Doppler effect is
used to measure speed.
PRACTICE:
A
The absorption spectra of stars
of varying distances from Earth
B
are shown here. The sun is the
bottom spectrum (D). Which star
has the highest relative velocity C
to Earth? Is it moving towards us,
D
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.
This is the so-called
“redshift” and it is
Topic 11: Wave phenomena used as evidence for an
expanding universe.
11.2 Doppler effect
Outline an example in which the Doppler effect is
used to measure speed.
PRACTICE:
A
The absorption spectra of stars
of varying distances from Earth
B
are shown here. The sun is the
bottom spectrum (D). What is the
approximate speed at which star D C
is receding from Earth?
SOLUTION: Use c = f ( f = c/ ): D
From A, f = 3108/39010-9
= 7.71014 Hz.
From D, f’ = 3108/49010-9 = 6.11014 Hz.
Then ∆f = f’ – f = (6.1–7.7)1014 = -1.11014 Hz.
Finally from ∆f = (v/c)f we get
v = c∆f/f
= (3108)(-1.11014) /7.71014 = -4.31107 ms-1.
Receding…
Topic 11: Wave phenomena
11.2 Doppler effect
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-andreturned one.
Topic 11: Wave phenomena
11.2 Doppler effect
Outline an example in which the Doppler effect is
used to measure speed.
EXAMPLES: The following are all examples of the
Doppler effect.
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.
Topic 11: Wave phenomena
11.2 Doppler effect
Outline an example in which the Doppler effect is
used to measure speed.
EXAMPLES: The following are all examples of the
Doppler effect.
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.
Topic 11: Wave phenomena
11.2 Doppler effect
Outline an example in which the Doppler effect is
used to measure speed.
EXAMPLES: The following are all examples of the
Doppler effect.
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).
Topic 11: Wave phenomena
11.2 Doppler effect
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!
Topic 11: Wave phenomena
11.2 Doppler effect
Solve problems on the Doppler effect for sound.
Consider both the
stationary and moving
source S.
By placing a scale
from the first diagram
into the second…
Topic 11: Wave phenomena
11.2 Doppler effect
Solve problems on the Doppler effect for sound.
A is wrong because
f goes negative.
Beware!
Topic 11: Wave phenomena
11.2 Doppler effect
Solve problems on the Doppler effect for sound.
Topic 11: Wave phenomena
11.2 Doppler effect
Solve problems on the Doppler effect for sound.
Because the speed of is a constant c regardless
of the speed of the source of the light, its
frequency cannot change.
For Doppler light, the v in the formula ∆f =
(v/c)f is the relative velocity.
Topic 11: Wave phenomena
11.2 Doppler effect
Solve problems on the Doppler effect for sound.
For moving away the wavelength INCREASES.
FYI
Regardless of who is observing, the speed of the
wave is determined by the air, and nothing else.
Thus the speed is V.
Topic 11: Wave phenomena
11.2 Doppler effect
Solve problems on the Doppler effect for sound.
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.