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## Class 11 Physics (India)

### Unit 19: Lesson 7

The Doppler effect- Doppler effect introduction
- Doppler effect: Wavefront diagrams and word problems
- Doppler effect review
- Doppler effect formula for observed frequency
- Doppler effect formula when source is moving away
- When the source and the wave move at the same velocity
- Doppler effect for a moving observer
- Doppler effect: reflection off a moving object

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# Doppler effect introduction

Learn about the Doppler effect and how it explains the change in frequency of a wave when its source and an observer are moving. Created by Sal Khan.

## Want to join the conversation?

- I don't get this. Why is Sal going backwards in time to explain this, and why is he adding the cycles of waves from the end point?(25 votes)
- The different colored circles (wavefronts) the left hand side are local, because the center of origin of the waves did not move. The outer circle is the oldest, since he originated first at the center and then propagated outwards with the wave front velocity.

The right side image is the same succession of wavefronts, but the origin point for the waves moves to the right.

Sal just started to draw the inner circles first, but the outer circles have been present earlier in time, since they already traveled outside and provided space for new wave fronts.(33 votes)

- At9:16I don't understand how the observer further away from the moving object has a wave every 1.5 seconds if the object is moving 5m/s and the wave is 10m/s.. Wouldn't it be 2 seconds between each wave? How did you get 1.5 seconds? Is there a calculation that was not shown?(6 votes)
- Hello, desrosierskloe. I read your question. At first it confused me to but later I came up with the solution. As you know the formula for time that is Distance/Speed. Here we know that the waves are travelling at the speed of 10m/s and distance between crest is 5m( towards which the train is coming).Here if you want to know the time period, you simply apply the fomula given above you get to the answer..

The wave is travelling with speed 10m/s and the distance between the crest is 5m. So Time Period = 5m/10m/s

= 5m * 1/10s/m

= 1/2 s

That is what that you get above the answer and hence the frequency 2 hertz.

If you apply this to the observer 2 (from which the train is moving away) you will get the Time Period = 15m/10m/s

= 15m*1/10s/m

= 1.5 s

That's how you got this answer. And in hertz if you reverse this you would get 10/15 that is equal to 2/3. Same as given in the answer.....

Hope you will find that useful..................(12 votes)

- I don't understand this, what is the Doppler Effect?(0 votes)
- The Doppler effect is the change in frequency that you hear when a source and an observer are moving with respect to each other. The classic case is when you have an ambulance racing by you: the pitch of the siren is higher when it is moving toward you and lower as it goes away. The Doppler effect is the mathematical description of this variation as a function of the velocities of the source and of the observer.(21 votes)

- Correct me if Im wrong sal messed up the direction of the wave sources saying it goes to the right but then saying later it's going left. At0:14and3:45(5 votes)
- Since he is going back in time, every time Sal goes
**back**in time, the source will have been more and more to the**left**, since it is moving to the**right**, as time progresses**positively the other way**.

When time progresses positively, the source moves right.

When time progresses**negatively**, the source moves left.

(this is kind of like slope of a line, if you go forward, it goes one way, but if you go the other way, the slope is opposite too)

The source is moving right, but in the past it was more to the left.(2 votes)

- I am confused- when a source emits waves at a certain frequency, will the observed frequency continually increase as the source (moving at a constant velocity) approaches the observer (stationary), or will the observed frequency be higher but constant?

Thanks!(2 votes)- Higher but constant if the source approaches the observer at a constant speed. Note that the source being at a constant speed does not mean that it approaches the observer at a constant speed. For example, imagine a car is moving on a straight road at 10 meters per second, and the observer is not on top of this road (so he doesnt get ran by), instead he is 10 meter by the side of the road. The car's approaching speed is not 10 m/s, because the car is not going towards the observer, the approaching speed changes according to the car's position on the road, therefore the observed frequency changes with the position as well.(4 votes)

- What exactly does 'pitch'mean ? I mean i understand frequency period or amplitude but how is pitch different ? ( ı am not a native english speakr ?) thanks(2 votes)
- Pitch is another word for the frequency of sound.(2 votes)

- What if the source moves faster than the wave?(3 votes)
- so how do we know the wavelength of the sound either stationary and moving object?(3 votes)
- Well, if the perceived frequency is higher or lower than the actual frequency, than the sound is moving and if the perceived frequency remains the same than its stationary.

Also, wavelength is inversely proportional to frequency.(1 vote)

- i dont understand it when sal draws circles and he says they are crests.... makes no sense to ME(1 vote)
- The circles represent the locations of the crests of a wave. The sound wave travels in all directions, so the crests are equidistant from the center.

Imagine that the sound is just beeps and don't worry about the waves.(4 votes)

- how come he is drawing the waves for the stationary point forward in time and the moving point backwards in time(2 votes)

## Video transcript

What we're going to do in
this video is think about two wave sources. But one of them is going to be
stationary and the other one is going to be moving. And just to have a concrete
number, it's moving at 5 meters per second
to the right. And what we're going to think
about is where are the crests of the wave that it's
been releasing for the last 3 or 4 seconds? So let's say that in both cases,
they are releasing a wave. So the velocity of the
wave is going to be 10 meters per second. You could visualize this, maybe,
as a sound wave, but sound and air move much, much,
much, much faster than 10 meters per second. But this'll make the
math work out easy. Especially relative to this
guy, who is moving to the right at 5 meters per second. And that's the whole point, to
give you the intuition and make the math really
simple here. And both of these guys are going
to be, the wave that they're emitting is at
10 meters per second. And the period of the wave
is going to be at 1 second per cycle. And if the period is 1 second
per cycle, you take the inverse of that-- the frequency
of the source, I guess you could call it, of the
wave, as being emitted is going to be the inverse
of this. The inverse of 1 is just 1. But 1 cycle per second. If it takes a second for a
cycle, it's also in 1 second, you're also going to see 1
cycle, so 1 cycle per second. So let's think about what's
happening here. So let's think about where--
let's say it emitted a crest of it's a wave exactly
1 second ago. Where is that crest
going to be now? Let's think about the stationary
character. Well, this guy emitted
a crest 1 second ago. It's moving outward. So this is outward. It's movie outward radially. We need to give a direction if
I'm giving a vector quantity. So it's moving outward at
10 meters per second. So if it emitted it 1 second
ago, it's going to be 10 meters radially outward
from the source. So maybe that is right
over there. Well, let me draw it a little
bit neater like that. So that's where that
crest will be. Now what about this guy? Where's the crest that this
guy emitted 1 second ago? You might want to just draw a
radius 10 meters around this guy as well. But he wasn't here
1 second ago. He was 5 meters to the left. Remember, he's moving 5 meters
per second to the right. 1 second ago, he was 5
meters to the left. So maybe that places him
right over there. So the crest that he emitted 1
second ago isn't going to be 10 meters from this guy. It's going to be 10
meters radially outward from that guy. So this is where he is now. That's where he was 1 second
ago, where he emitted this crest that has now traveled
10 meters away. This a little inexact,
so I could draw it a little bit like that. This is 5 meters. That's 10 meters away. But you get the general idea. Now let's keep going. Let's think about the crest
that both of these guys emitted 2 seconds ago. This guy's been stationary
the whole time. If he emitted it 2 seconds
ago, it's traveling at 10 meters per second. It's going to be 20 meters
radially outward from the center, from the source. So it will look something
like that. I'm just drawing the
crest of the waves. If you think of water, a pebble
being dropped into the pond, these are just the high
points on the wave that spreads radially outward from
where the pebble was dropped. Now this guy, once again, you
can't just draw a circle around this because he wasn't
here 2 seconds ago. He was right here. He was right here
2 seconds ago. 1 second ago, he was 5
meters to the left. A second before that, he was
5 meters more to the left. So that wave that he emitted
then is going to be 20 meters radially outward from
this point. So the center isn't going
to be that or that. It's going to be that point
where he was 2 seconds ago. So what about the crest that
either of these of sources emitted 3 seconds ago? Well, it would be 30 meters
radially outward, so another 10 meters from the last one. So it will be out here. It will be out there
just like that. This guy has been stationary
the whole time. But what about this guy? He wasn't here 3 seconds ago. He was here. Right? 1 second ago, here. 2 seconds ago, there. 3 seconds ago, there. So we're going to be
30 meters radially outward from this point. So once again, I can just copy
and paste this right here. And it should be centered
around that point. Now, let's think a little bit
about what the perceived frequency of this wave would be
for a couple of observers. So we could put an observer
here, really anywhere around this guy. We could put an observer
right here. And then we could put another
observer right here. Now, what's this guy
going to perceive? Well, every second he's getting
a pulse-- well, there's a couple of things
to think about. First of all, what is the
wavelength of, at least this wave right here. Well, every second, he's
emitting a pulse. So, a second ago, the pulses
out there would have traveled 10 meters. And then he emits
another pulse. So the pulses are going to be 1
second apart, but since they traveled 10 meters in that 1
second, they are also going to be 10 meters apart. So the wavelength,
in this case, is going to be 10 meters. The distance between these
crests are 10 meters. Now what about this situation
right here? It depends on what side, whether
the source is coming towards you or whether it's
going away from you. That's the situation
with this guy. When it's moving towards you,
it's emitting a pulse-- so let's say it emitted a pulse
right here-- and then, it moves 5 meters to
the right before emitting the next pulse. So instead of them being 10
meters apart, in this case, this guy has closed
the distance by 5 meters over here. So these pulses are only going
to be 5 meters apart. So over here, the wavelength
is only 5 meters. You can see it visually. This distance right here is
half of this distance. These are 5 meters apart. And on the left-hand side, if
you're on the side of the source that the source is moving
away from, it would be 10 meters, but every second the
source is also moving 5 more meters away from you. So this perceived wavelength
right here is going to be 15 meters. And we can see it visually. That's the whole reason why
I drew it this way. Now what's going to
be the perceived frequencies for this? Well, this guy has one crest
passing him right now. It's going to take exactly 1
second for the next crest to get to him because
it's traveling at 10 meters per second. So he is going to perceive 1
crest, or 1 cycle per second, or a frequency of 1 hertz,
which makes sense. This is stationary. They're both stationary relative
to each other. And we're also talking about
classical physics. We're not getting to relativity
and all that. But the observed frequency is
the exact frequency that was emitted by this guy
right there. Now what about this situation? Each of these crests are 5
meters apart for this guy. If you imagined that this was
some type of a train coming towards this guy, each of these
crests are only 5 meters apart, but they're traveling
at 10 meters per second. So how many crests are you
going to see in a second? Well, you're going to
see two of them. This one's going to take 1/2 a
second to return you, and then in the next 1/2 a second, this
one's going to get to you. Or you could say this one takes
1/2 second to reach you, and this is going to take
1 second to reach you. So you're going to see two. So there's two ways
to think of it. You could say your period in
this situation is 1/2 of a second per cycle. Or you could invert it and you
could say that the frequency-- I mean we could put the observed
frequency-- is going to be 2 cycles per second. And already notice, this guy's
experiencing a higher frequency then this guy over
here, because these wave fronts or these crests are
just passing by him more frequently. Because this guy is moving in
the same direction as this guy, they are closer together. Now this guy's going to
experience the opposite thing. Let's say that this crest
is just passing him by. How long will it take for
the next crest to cover that 15 meters? Well, they're going at
10 meters per second. It's going to be 1.5 seconds per
crest. That's going to be the observed period
for this guy. You have to take the inverse of
that-- that's 1.5 is 3/2, that's 2/3 of a --or you could
say crest or 2/3 of a cycle per second. So when the source is moving
away from this observer, the frequency, or the perceived
frequency, is the lower than the frequency of the actual
emitted wave. When the source is moving towards the
observer, the frequency is higher. This might seem some like some
type of bizarre thing but you've experienced it before. It's called the Doppler
effect, which you've probably heard of. And that's exactly what you
experience when you sit at, maybe, a train crossing. Be careful not to
sit too close. And as a train is approaching
you, you noticed-- say it has its horn going on. It'd be very high-pitched. And then right when it passes
you, and it starts moving away from you, it has a
much lower pitch. And that perceived pitch, that's
your brain and your ears' way of sensing
frequency. So when the train is coming
toward you, it's a high pitch, high frequency. When it's going away from you,
low pitch, low frequency. And hopefully, drawing it out
this way gives you a visual understanding of why that is;
why these points on the cycle, or these crests, are closer
together when it's moving in your direction then when they're
farther apart-- when it's moving away from you. In the next video, we'll do
these with more abstract numbers so we can actually
figure out generalized formulas for relating the
observed frequency with the emitted frequency.