Download Doppler Effect 11.1 Quiz

Physics 102
Conference 11
Doppler Effect
Conference 11
Physics 102
General Physics II
Monday, April 14th, 2014
11.1
Quiz
Problem 11.1
You are traveling towards a cliff at a constant speed w – at time t = 0, you
clap your hands, and at time T you hear the echo – how far away was the cliff
when you clapped your hands (call v the speed of sound in air)?
Suppose at t = 0 you are a distance d from the cliff. It takes tc = d/v for the
signal to reach the cliff face. At that point, you are a distance d − w tc from the
cliff. Now call your distance to the cliff: x(p) = d − w tc − w p, so that p = 0
represents the time at which the reflected signal left the cliff. The signal needs
to go a distance x(pr ) = v pr in a “time” pr , so that d − w tc − w pr = v pr
giving:
d − wv d
d − w tc
pr =
=
.
(11.1)
v+w
v+w
The total time here is then tc + pr and that is the measured T , so
T =
so that d =
1
2
1 − wv
d
2d
+d
=
,
v
v+w
v+w
T (v + w).
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(11.2)
11.1. QUIZ
Conference 11
Problem 11.2
Can a charge q moving at constant velocity v generate electromagnetic waves?
No – a charge q moving at constant velocity has no forces acting on it, so
will continue to travel at constant velocity. If it could generate waves, they
would carry energy away from the charge and it would have to slow down in
the absence of any force. Another way to see it – if the charge is at rest with
respect to you, there is no wave (the electric field is not time-varying and so
cannot act as a source for a magnetic field). Someone else moving at constant
velocity with respect to you (and the charge) is in another “inertial” frame (no
external forces) and their physical predictions must agree with yours – if you
see no waves, they must also see no waves.
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11.2. DOPPLER
11.2
Conference 11
Doppler
For an observer moving with vo away from a stationary source emitting f (vo > 0
moving away), the observed frequency is:
vo f0 = 1 −
f
(11.3)
v
where v ≈ 343 m/s is the speed of sound in air. If the source is moving and
the observer is stationary, we have (for vs > 0 moving towards the observer):
f0 =
1
f.
1 − vvs
(11.4)
These equations hold when the motion is occurring in one dimension.
Problem 11.3
A stationary bat emits frequency f that bounces off an object that is moving
with constant speed w – what frequency f 00 does the bat receive from the
object? How would the bat know if the object is moving towards or away from
it?
The frequency that is “observed” by the object is f 0 = 1 − wv f where V is
positive for the object moving away from the source. Now the object reflects
this f 0 back to the bat – the bat is a stationary observer, and the object is a
moving source that is going away from the bat for positive w – the observed
1− w
frequency at the bat is f 00 = 1+1 w f 0 = 1+ wv f . Suppose the object is moving
v
v
away from the bat – then w > 0 and
00
f =
1−
1+
|w|
v
|w|
v
f <f
(11.5)
so the frequency the bat hears upon reflection is less than the emitted one. If
the object is moving towards the bat, w = −|w|, and
f 00 =
1+
1−
|w|
v
|w|
v
f > f,
so the frequency is higher than the emitted one.
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(11.6)
11.2. DOPPLER
Conference 11
Problem 11.4
You are jogging toward the base of a cliff that is off in the distance – you’d
like to figure out how fast you are running, so you start screaming at f = 880
Hz. After a short delay, you hear the frequency F = 882 Hz. How fast are you
running towards the cliff?
You emit f = 880 Hz – the cliff receives f 0 =
towards cliff). The cliff then re-emits this
an observer moving towards it:
f0
f
1− vvs
with vs > 0 (source moving
as a stationary source, and you,
vs 0 1 +
F = 1+
f =
v
1−
vs
v
vs
v
f,
(11.7)
and we can solve this for vs :
vs =
(F − f ) v
≈ .4 m/s.
F +f
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(11.8)
11.2. DOPPLER
Conference 11
Problem 11.5
We developed the Doppler effect in the context of one dimension. In two
dimensions (as below), the Doppler effect, for a moving source, reads:
f0 =
f
,
1 − v̄vs
(11.9)
where v̄s is the component of the source’s velocity along the line connecting the
source to the observer (v̄s > 0 for the source moving towards the observer). For
a train moving with constant vs in the x̂ direction, and an observer a distance
s away from the tracks (at x = 0), find f 0 as a function of time (set the clock
so that t = 0 as the train horn passes x = 0, so t < 0 for x < 0) given a horn
frequency of f .
vs = vs x̂
x=0
f
x̂
|x|
s
✓
From the geometry, v̄s = vs sin θ and sin θ = √x|x|
– we want v̄s to be
2 +s2
positive for the train at x < 0 and negative for x > 0. The train’s location is
x = vs t with t < 0 for x < 0 so
vs t
v̄s = −vs p
.
(vs t)2 + s2
(11.10)
The received frequency is then:
f0 =
f
1+ √
v
vs2 t
(vs t)2 +s2
.
(11.11)
An example plot, for f = 400 Hz, with s = 10 m and vs = 20 m/s (v = 343
m/s) is shown below (for t = −2 to 2 s). The smaller s is, the steeper the
transition.
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11.2. DOPPLER
Conference 11
f in Hz
420
410
400
390
380
-2
-1
1
6 of 6
2
t