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Doppler effect formula when source is moving away

Let's look at the case where the source is moving away from the listener. Created by Sal Khan.

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  • orange juice squid orange style avatar for user Geoffry Gifari
    so, what will the formula be if the observer is in an angle from the source's direction of motion ( not exactly in front of or behind the source)?
    (16 votes)
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  • blobby green style avatar for user Vinayak
    For both of the equations, why must we divide by Vw ?
    To = Ts ((Vw - Vs) / Vw)
    (7 votes)
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    • aqualine ultimate style avatar for user Alex Li
      In this case, Ts(Vw-Vs) is the distance between the source and the observer after the source moved. Mathematically speaking, Time=Distance/Speed; which corresponds with this situation because To is the time, Ts(Vw-Vs), and Vw is the speed of the sound wave.
      (3 votes)
  • purple pi purple style avatar for user Mario Bobasti
    What happens if the observer is traveling at the exact same speed as the wave from the source or the source itself? What does the observer experience?
    (4 votes)
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  • purple pi purple style avatar for user Mario Bobasti
    Can the observer sitting in the source emitting sound, if traveling faster than the speed of sound, hear the source sound normally?
    (6 votes)
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    • piceratops seed style avatar for user Bereket Zelealem
      No, if the observer is moving faster than the speed of sound, then the wave being emitted will never reach him, assuming his speed is constant. lets say the wave moves at a speed of 8m/s for example but if the source moves at 10m/s the source would be 2 meters in front of the wave within the time interval of 1 second.
      (2 votes)
  • mr pink red style avatar for user Daniel Wiczew
    Why speed of the source doesn't affect speed of the wave ?
    And this is a law or an assumption ?
    (2 votes)
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    • leaf green style avatar for user stephen
      The speed of a wave depends on the medium it is travelling in. This is a law.
      It is a good question though, because the speed of the source is important for other things.
      For instance, since the speed of the wave depends on the medium, it is constant if the medium does not change, but the source can speed up. Some aircraft are capable of going faster than the speed of sound. When they do this the actually catch up to the sound waves they emitted earlier, and emit more sound waves at the same time, eventually they can break through its own sound waves, and when this happens a very loud sonic boom can be heard. If the airplane is not strong enough then this sonic boom can even break the plane.
      (4 votes)
  • spunky sam blue style avatar for user KushagraK7
    Can this change in frequency also be observed in electromagnetic waves?
    (2 votes)
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  • mr pants teal style avatar for user Udbhav Dalal
    Why does the frequency of an approaching object seem to be increasing gradually, and then decrease gradually? From the diagram above it seems that frequency suddenly drops as the source passes you by.
    (2 votes)
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  • purple pi purple style avatar for user nidhi bharadwaj
    Will the same formula hold true if the observer is not exactly in the path of source but sideways?
    (2 votes)
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    • starky ultimate style avatar for user Swarna
      If you mean the source is moving straight and the observer is stationed at an angle from it, this was an answer given by Andrew M for a similar question:
      The same concept holds true but now you just have to use trigonometry to get the velocity component that is parallel to the movement of the source given the angle between the source's path and the listener's location.
      (3 votes)
  • leafers ultimate style avatar for user Ghost
    at 8.00 I don't understand why this expression (10ms / 10ms-5ms) isn't evaluated in a way that the 10 m/s over 10 m/s doesn't jcancel out leaving you with 1 / 1-5.

    Instead he seems to do the denominator first 10ms-5ms giving him 10ms / 5 ms = 2 cycles. Is their a rule I've missed out when evaluating in fractions? because I always assumed that multiplication and division took priority over addition and subtraction.
    (1 vote)
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  • orange juice squid orange style avatar for user Siddharth Singh
    How can we observe the DOPPLER EFFECT in objects moving faster than the speed of sound??
    (2 votes)
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    • leaf green style avatar for user Geoffrey Sun
      I think you may be confusing the speed of sound with the speed of light. Things moving faster than the speed of sound are still quite observable. Moving faster than the speed of sound relates to topics like supersonic speed

      http://en.wikipedia.org/wiki/Supersonic_speed

      If you're still confused about anything else please, post your questions.
      Otherwise an example of the doppler effect when not faster than the speed of sound, could be sirens sounding of higher or lower frequency depending where they are as they pass you, I don't have an example with faster than the speed of sound of the top of my head at the moment.
      (3 votes)

Video transcript

In the last video, we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source. So the source is moving towards the observer. So this is the example where the train is moving towards you, and you perceive the train's horn as having a higher pitch or a higher frequency. And we were able to do that by doing a thought experiment. Saying, OK, my object starts here. After one period-- a period is just a measure of time, but it's the measure of time over which the source emits a cycle, so it emits a cycle every period. But after one period, we said, OK, where is that first wave front, or that first pulse, or that first crest? And where is the source? Because exactly one period has passed by, and the source will be ready to emit another crest or another cycle. So the distance between where the source is and that front of the crest, or that first crest, that is going to be the wavelength. Because this next thing that emits is going to be traveling at the exact same velocity, and it's going to be separated by that distance, which we saw is this expression. We said how long will it take it to travel that distance? Well, it's traveling at a speed of v sub w. That'll tell you what the observed period would be for this dude over here. We calculated it right here, and then the observed frequency is just the inverse of that. Now let's think about the situation where the observer is over here. So these equations, or these formulas that we came up with right here, this is observer-- or let me say source traveling in direction of observer. Now let's think about the opposite case, where the source is traveling away from the observer, and in this case, the observer is that guy over there. Maybe I'll do it in a different color. He'll be blue. So this is the observer. So when we started off, our source was right here. After exactly one period from the source's point of view, that first crest emitted has traveled radially outward that far. This is the distance. The velocity of the wave times the amount of time that passed-- velocity times time is going to give you distance-- and where the source of the wave will have traveled to the right exactly this distance. It's velocity times the amount of time that's gone by. Now, in the last video, we said, OK, that wave is just passing this guy. How long will it take for that pulse that's being emitted right then to also reach him? And then that tells us the period between two pulses or between two crests. Now let's think about that exact same situation here. That first crest is just passing this guy. And a period, or the t sub s, which is a period of the emitted wave has just passed by. So this guy is just about to emit another wave, So? That other wave is going to be right here. So what is the distance between the crest, or the cycle, or however you want to think about it, the pulse that is passing him by right now, what is the distance between that and the pulse, the front edge of that pulse that is being emitted right at that moment? Right at that moment. Well, it's going to be this radius, which is this value. It'll be v sub w times the period. That is that distance plus the amount of distance that our source has traveled away from this guy. So plus v sub s times the period. So that's what this distance is, and this is how far apart this wave pulse is going to be from that one or this crest is going to be from that one. So if he's seeing this first crest right now, right at this moment, how long will it take him to see the crest that's being emitted right now, that's this far away from him? Well, it's that far away from him. So let me write this down. So the amount of time it takes for him to see the next crest or the same point in the next cycle, that's the period. That's the observed period. That's going to be equal to this distance. The velocity of the wave times the period from the perspective of the source plus the velocity of the source, because the source has gotten that much further away from him, velocity of the source times the period of the source, so that's how far the next crest is. And then you divide it by the speed of the wave, by the speed of each of the crests, which is just the velocity of the wave. And we can just factor out the t sub s's here and say this is t sub s times v sub w, the velocity of the wave plus the velocity of the source divided by the velocity the wave. So this will be a larger observed period than if this guy was stationary and especially if the observer was in the path of the guy. That make sense, because every time this guy issues a cycle, he is moving a little bit further away. So every crest the same point in the cycle is going to be further and further apart, so you're going to have longer wavelengths, longer periods. And if you want the observed frequency for that guy over there-- I'll do it in the same color. The observed frequency for a guy where the source is traveling away from him is just the inverse of that. So one over the period, one over the period, same argument we did there, one over the period from the point of view of the source is the frequency of the source. Let me Color code it. So 1 over t sub s is equal to the frequency of the source. This is the inverse of that. So I'm just taking one over everything here. So 1 over t sub s is the frequency of the source, and then we take the inverse of this over here: the velocity of the wave divided by the velocity of the wave plus the velocity of the source. So we're done, at least for the simple cases. Obviously, it becomes a little more interesting when someone isn't exactly in the direction of the source or exactly being moved away from, but these are kind of the two extreme cases. So this is the situation when it is moving away from you. Now, just to check our math and maybe make it a little bit concrete in relation to the video we did where we introduced to the idea of the Doppler effect, let's actually apply those numbers. So on that video, two videos ago, we had a situation where the velocity of our source was 5 meters per second to the right, and the velocity of the wave was 10 meters per second radially outward, and the period of our wave-- I'll call it the period of our-- let me do it in another color-- from the point of view of the source was 1 second per cycle, and the frequency was just the inverse of that. So 1 cycle per second, or 1 Hertz, which is a cycle per second. So using those numbers, let's see if we get to the exact same answer we got in that first video where we first learned about the Doppler effect. So let's look at the frequency from the point of view of this gentleman right here. So the frequency of the source is going to be 1 cycle per second, 1 Hertz. The velocity of the wave is 10 meters per second. Let me write this. 1 cycle per second. The velocity of the wave is 10 meters per second. The velocity of the wave is 10 meters per second minus the velocity of the source is 5 meters per second. So what's this going to be equal to? The observed frequency for this guy right there, is going to be 1 cycle per second times 10 over 10 minus 5, and the meters per second cancel out. Meters per second in the numerator, meters per second in the denominator. So 10 divided by 10 minus 5, or 10 divided by 5, is going to be 2. So it's going to be 2 cycles per second. And if you want the observed period for this guy, its going to be the inverse of that, or it's going to be 1/2 second per cycle. And this is exactly what we got in the previous video, or actually, it was two videos ago. Now what about the guy who this guy's running away from? Well, we'll do the exact same thing. You have 1 cycle per second, or 1 Hertz. That's the emitted frequency from the point of view of the source times the velocity of the wave divided by the velocity of the wave plus the velocity of the source, because it's moving away from him. So it's 10 over 10 plus 5. That's 10 over 15. That is 2/3, so this is equal to 2/3. The units over here all cancel out. This was cycles per second. So 2/3 cycle per second, which confirms the numbers we got in that first video, so that should make us feel good and it also makes a lot of sense. This guy is going to see the wave crest more frequently. He's going to observe a higher frequency. If this is a sound, a higher pitch. This guy, since each crest or the cycles are getting spread out, he's going to see them less frequently, and if this is sound, he's going to observe a lower pitch.