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Doppler effect formula for observed frequency

The Doppler effect is the change in frequency of a wave as the source moves relative to an observer, and explains why the pitch of a sound sometimes changes as it moves closer or further to or from an observer. Learn how to derive the formula for perceived frequency of a sound using an equation that accounts for a sound and observer’s velocities relative to one another. Created by Sal Khan.

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• But, isn't the velocity of the wave for moving particle the sum Vw and Vs, because the speed of the wave would also include the speed of the object emitting the light?
(25 votes)
• The speed of the sound wave is not the speed of the particles moving, but the speed at which the peaks (compressed air) propagate through the medium that the sound is travelling in. It is not related to the speed of the emitting source. If you think this is unclear, you should watch the earlier videos in this series.
(5 votes)
• im still really confused... can someone explain the Doppler effect in simple terms??
(18 votes)
• When the sound source is approaching the observer or when the observer is approaching the sound source, the apparent frequency of the sound source increases. You notice this when a train whistling is approaching you. The apparent frequency of the whistling train is larger compared to the frequency of the whistling when both you and the train are stationary. However if the train is approaching you, then you'll notice an apparent increase in frequency, which is perceived by the higher pitch of the train's whistling sound, keeping in mind that higher pitch is associated with higher frequency sound.
(8 votes)
• What happens if the wave source is moving faster than the waves themselves?
(11 votes)
• Since the source is moving faster than the sound waves it creates, it actually leads the advancing wavefront. In the case of Sound, The sound source will pass by a stationary observer before the observer actually hears the sound it creates. " SONIC BOOM" is heard and seen when source or object breaks the speed of sound
(15 votes)
• What would happen if the source travels faster than the wave?
(6 votes)
• this can only happen with sound waves in which one experiences a sonic boom by the pressure difference at the front and rear of a aircraft. since other electromagnetic waves travel at the speed of light,no source can travel faster thus making such a situation impossible
(7 votes)
• What happens when things approach the speed of light
(5 votes)
• When an object approaches the speed of light very closely (Very little difference), it's mass increases and goes up, and so does the energy it is carrying (Kinetic, Mechanical, and Potential), at one point both reach infinity when it hits the exact same speed of light. This arises a lot of complications, such as dealing with the Laws of Conservation of Energy and Matter, Time Dilation, and much more, which is why he wanted to avoid it. Traveling at such a speed (or very close to it) means that waves can't travel as fast enough, including sound and light, causing them to stay in front of the object, and they would be unable to leave the object, and instead pile up on the front. (Fun fact, we don't actually know the true speed of Light in one direction, only an mean of light travelling in two directions. It is also impossible to travel at the speed of light.)
(1 vote)
• Can't we write To = Ts [ ( Vw - Vs ) / Vw ] as
To = Ts [ ( 1 - Vs / Vw ) ] ?
(5 votes)
• In fact, this is the best way to express this equation
(2 votes)
• I didn't exactly understand how Sal represented the crest of the waves in the form of circles. Please help..
(3 votes)
• It's like when you throw a stone in a pond. It makes circular ripples that spread out from the stone.
(6 votes)
• Sal says that in the last video he showed the formula for doppler effect when the object is moving towards you. Where is that video
(3 votes)
• Is Sal saying that the time taken by source and time taken by wave both are equal? because in calculating its distance he has used the same time(T sub S). And if the time is different how is the calculation going to differ?
(4 votes)
• He just calculated the distance after a fixed time to both the source and the wave.
(1 vote)
• It seems like Mr. khan related frequency to "loudness" in this video and the last. However, if the amplitude isn't changing (which I assume is true given that word has not been said in the last two videos), why do we perceive the loudness to be changing?

I assume the effect (or affect) is a combination of attentuation + changes in frequency, but I just want to make sure I have the concept correct - loudness vs. pitch (frequency).
(2 votes)
• Frequency is not related to loudness, it's related to pitch
(3 votes)

Video transcript

I've got this source of a wave right here that's moving to the right at some velocity. So let's just say that the velocity of the source-- let's call it v sub s to the right-- so we're really going to do what we do in the last video, but we're going to do it in more abstract terms so we can come up with a generalized formula for the observed frequency. So that's how fast he's moving to the right, and he's emitting a wave. Let's say the wave that he's emitting-- so the velocity of wave-- let's call that v sub w radially outward. We've got to give a magnitude and a direction. So radially outward. That's the velocity of the wave, and that wave is going to have a period and a frequency, but it's going to have a period and a frequency associated from the point of view of the source. And we're going to do everything. This is all classical mechanics. We're not going to be talking about relativistic speed, so we don't have to worry about all of the strange things that happen as things approach the speed of light. So let's just say it has some period of-- let me write it this way. The source period, which is the period of the wave from the perspective of the source, so the source period, we'll call it t sub source, And the source frequency, which would just be-- we've learned, hopefully it's intuitive now-- would be the inverse of this. So the source frequency would be-- we'll call it f sub s. And these two things are the inverse of each other. The inverse of the period of a wave is its frequency, vice versa. So let's think about what's going to happen. Let's say at time equal zero, he emits that first crest, that first pulse, so he's just emitted it. You can't even see it because it just got emitted. And now let's fast forward t seconds. Let's say that this is in seconds, so every t seconds, it emits a new pulse. First of all, where is that first pulse after t sub s seconds? Well, you multiply the velocity of that first pulse times the time. Velocity times time is going to give you a distance. If you don't believe me, I'll show you an example. If I tell you the velocity is 5 meters per second, and let's say that this period is 2 seconds, that's going to give you 10 meters. The seconds cancel out. So to figure out how far that wave will have gone after t sub s seconds, you just multiply t sub s times the velocity of the wave. And let's say it's gotten over here. It's radially outward. So, I'll draw it radially outward. That's my best attempt at a circle. And this distance right here, this radius right there, that is equal to velocity times time. The velocity of that first pulse, v sub w, that's actually the speed. I'm saying it's v sub w radially outward. This isn't a vector quantity. This is just a number you can imagine. v sub w times the period, times t of s. I know it's abstract, but just think, this is just the distance times the time. If this was moving at 10 meters per second and if the period is 2 seconds, this is how far. It will have gone 10 meters after 2 seconds. Now, this thing we said at the beginning of the video is moving. So although this is radially outward from the point at which it was emitted, this thing isn't standing still. We saw this in the last video. This thing has also moved. How far? Well, we do the same thing. We multiply its velocity times the same number of time. Remember, we're saying what does this look like after t sub s seconds, or some period of time t sub s. Well, this thing is moving to the right. Let's say it's here. Let's say it's moved right over here. In this video, we're assuming that the velocity of our source is strictly less than the velocity of the wave. Some pretty interesting things happen right when they're equal, and, obviously, when it goes the other way. But we're going to assume that it's strictly less than. The source is traveling slower than the actual wave. But what is this distance? Remember, we're talking about-- let me do it in orange as well. This orange reality is what's happened after t sub s seconds, you can say. So this distance right here. That distance right there-- I'll do it in a different color-- is going to be the velocity of the source. It's going to be v sub s times the amount of time that's gone by. And I said at the beginning, that amount of time is the period of the wave. That's the time in question. So period of the wave t sub s. So after one period of the wave, if that's 5 seconds, then we'll say, after 5 seconds, the source has moved this far, v sub s times t sub s, and that first crest of our wave has moved that far, V sub w times t sub s. Now, the time that we're talking about, that's the period of the wave being emitted. So exactly after that amount of time, this guy is ready to emit the next crest. He has gone through exactly one cycle. So he is going to emit something right now. So it's just getting emitted right at that point. So what is the distance between the crest that he emitted t sub s seconds ago or hours ago or microseconds ago, we don't know. What's the distance between this crest and the one that he's just emitting? Well, they're going to move at the same velocity, but this guy is already out here, while this guy is starting off from the source's position. So the difference in their distance, at least when you look at it this way, is the distance between the source here and this crest. So what is this distance right here? What is that distance right there? Well, this whole radial distance, we already said, this whole radial distance is v sub w, the velocity of the wave, times the period of the wave from the perspective of the source, and we're going to subtract out how far the source itself has moved. The source has moved in the direction, in this case, if we're looking at it from this point of view, of that wave front. So it's going to be minus v sub s, the velocity of the source, times the period of the wave from the perspective of the source. So let me ask you a question. If you're sitting right here, if you're the observer, you're this guy right here, you're sitting right over there, and you've just had that first crest, at that exact moment that first crest has passed you by, how long are you going to have to wait for the next crest? How long until this one that this guy's emitting right now is going to pass you by? Well, it's going to have to cover this distance. It's going to have to cover that distance. Let me write this down. So the question I'm asking is what is the period from the point of view of this observer that's right in the direction of the movement of the source? So the period from the point of view of the observer is going to be equal to the distance that the next pulse has to travel, which is that business up there. So let me copy and paste that. So it's going to be that. Let me get rid of that. It shouldn't look like an equal sign, so I can delete that right over there. Or a negative sign. So it's going to be this distance that the next pulse is going to travel, that one that's going to be emitted right at that moment, divided by the speed of that pulse, or the speed of the wave, or the velocity the wave, and we know what that is. That is v sub w. Now this gives us the period of the observation. Now. If we wanted the frequency-- and we can manipulate this a little bit. Let's do that a little bit. So we can also write this. We could factor out the period of the source. So t sub s we could factor out. So it becomes t sub s times the velocity of the wave minus the velocity of the source, all of that over the velocity of the wave. And so just like that, we've gotten our formula for the observed period for this observer who's sitting right in the path of this moving object as a function of the actual period of this wave source, the wave's velocity and the velocity of the source. Now, if we wanted the frequency, we just take the inverse of this. So let's do that. So the frequency of the observer-- so this is how many seconds it takes for him to see the next cycle. If you want cycles per second, you take the inverse. So the frequency of the observer is just going to be the inverse of this. So if we take the inverse of this whole expression, we're going to get 1 over t sub s times v sub w over the velocity of the wave minus the velocity the source. And of course, 1 over the period from the point of view of the source, this is the same thing. This right here is the same thing as the frequency of the source. So there you have it. We have our two relations. At least if you are in the path, if the velocity of the source is going in your direction, then we have our formulas. And I'll rewrite them, just because the observed period of the observer is going to be the period from the point of view of the source times the velocity of the wave minus the velocity of the source-- that's the velocity of the source-- divided by the velocity of the wave itself. The frequency, from the point of view of this observer, is just the inverse of that, which is the frequency. The inverse of the period is the frequency from the point of view of the source times the velocity of the wave divided by the velocity of the wave minus the velocity of the source. In the next video, I'll do the exact same exercise, but I'll just think about what happens to the observer that's sitting right there.