Class 11 Physics (India)
- 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
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.
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- 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
= 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?
- 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)
- 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)
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.