 Hi, I'm Zor. Welcome to Nezor Education. I decided to add this particular lecture to a topic which I thought was actually completed, but apparently it's not completed. So we will talk about Doppler effect. And today we will talk only about Doppler effect for longitudinal waves like sound. So I will use sound as an example basically. Now this lecture is part of the course which is called... It's Physics for Teens actually, but you see the purpose of this is... I will use probably this in the relativity in the later topic. So it's written here relativity for all, but actually it's about the previous course. It's part of the Physics for Teens. It's part of the chapter called Waves. So if you will go to websiteunisor.com and go to Physics for Teens and then choose the chapter called Waves, you will see the topic which is called Waves in Medium. So this is about Doppler effect in sound because the sound actually is waves in the Medium and the Medium is air, just generally speaking. Okay, so let's get to the point. First of all, let me remind you a couple of characteristics of waves. First of all we have something which is called Frequency. I will use the letter F. So frequency is number of oscillations per second. And it's related to the period. Period is the time between two peaks of the waves. So obviously it's inverse to the frequency. If this is the time between the peaks and the one peak is actually one wave, so the number of waves per second would be one over G. Now lambda is the distance, physical distance in meters basically, between the peaks, so that's the wavelength. And there is also a concept of a speed. Well, now what is speed? Well, for example, you can say that speed is you have one length of one wave divided by the time this wave takes to pass. So that would be speed, right? Or in other words, it's lambda times F. The F is the frequency, it's reverse the T, right? Or if you wish, F is equal to U divided by lambda. And this is from this. And this is something which I will definitely use, basically this latest incarnation of all these very simple formulas. Okay, now what is Doppler effect? Now Doppler effect is basically a difference between the frequency of the sound produced by some source and the frequency you perceive, you hear. Because it depends on many different circumstances and in particular, as far as the Doppler effect is concerned, it's related to a movement, mutual relative movement between the source and the receiver. So I will consider only the simplest case when the source is standing still and the receiver is moving to or from this particular source. All other cases are actually very easily derivable from this one. For example, if the receiver stands still and the source is moving, well, that's basically the same thing because it's a relative distance between them and the speed of the distance is changing, that's what's actually important. If both of them are moving, and not along one line as I will consider, but something in space, like different trajectories, well, that's a little bit more calculations, but idea will be exactly the same, just more tedious calculations which will give you basically the same thing. Well, in general, if you are approaching the source of sound, your frequency would be higher, your perceived frequency would be higher than the one which source actually emits. If you go away from the source of sound, your perceived frequency would be less, so the tone will be lower than the one produced, and that's what I'm going to basically prove right now. Okay, so let's assume we have a source which has certain frequency, let's put that zero, certain, obviously, period, certain wavelengths. This is a speed of sound in this particular case, so that's the source. Source is fixed, let's just say that this is zero, and my first case is when the receiver, when observer is moving towards the source with speed v. One-dimensional case, I'm not going any more complicated calculations. So initially, at time is equal to zero, my observer is at point A, and it moves towards the source. From the source you have waves of sound, this guy is producing. So let's just fix some time period. Let's say this is a location after certain time t. What will be the coordinate in this case? It will be A, I'm using A basically as a distance. So the distance would be A minus v times t. If v is the speed, t is time during which it moves so that would be a location. Well, obviously I have to choose t in such a way that we are still on this side of the sound. I don't want to cross the sound. So t should be less than whatever, A divided by B. So it's still, in this particular case, it's still some kind of a positive distance. So this is distance A, this is A minus vt from zero, from the source actually. Okay, now what kind of, how many actually waves this guy will cross while he is moving? Well, if waves do not move, then obviously he will cross this distance which is v times t divided by the length, wavelengths, but waves are moving. So not only these waves will cross his way, but also some waves which are here. Now, how many? Well, by the time he moves here, all waves which are here will move towards his zone of perception, so to speak, right? So during the same time t, waves which are moving this way with the speed u. So if this is u times t, so these are all the waves which also will be able to come into his zone of perception during the same time t. Right? So all the waves from this point to this point will cross his way while he will be moving from this to this. Right? So what is this way? The length is A minus vt minus ut. So A minus vt which he himself, the observer, covers plus these will actually add to his perceived number of waves. So this is a distance basically. All the waves which are in this distance initially at time t is equal to zero will be crossing his way as he is moving during time t. Right? So this is a distance, right? And the time is t. So basically if I will divide this by lambda zero, I will have number of waves. So this is number of waves, and lambda zero is wavelengths, and this is the total distance A minus this and minus this. Well, I'm sorry. Actually I don't need A anymore. So it's just this plus this. Sorry. So this plus this, yes, that would be probably the whole length. This is all the waves which the guy will cross. So this is the wavelength. So this is number of waves. Yeah, that's it. Now, if I have number of waves and I have the time, if I will divide one by another, I will have frequency, right? Number of waves per certain amount of time. So my perceived frequency is equal to n divided by t, which is equal to, by t it would be u plus v divided by lambda zero, or u divided by lambda zero, one plus v over u. So this is the formula. And from this, look at this. This is my f zero. This is initial frequency of the sound emitted by the source. Now I have a factor which is greater than one. So the frequency perceived frequency is greater. Perceived frequency is greater than initial frequency by this factor. So basically, that's the formula, which means that the pitch is higher. So whenever you're moving towards a source of sound, you hear higher tone than produced by the source. Okay, so let's do the opposite thing. Let's, for instance, consider we are moving away from the source. The picture will be very similar. Okay, so this will be my source and this will be my initial location at the time equals to zero. And this would be my location at sometimes t. Okay, now this guy goes with a speed v and these waves are going with a speed u. Well, we should assume right now that the u speed of waves, speed of sound in this case, is greater than v because if v is greater, we will never hear anything at all. We are moving faster than the sound. So let's assume that u is greater than v. Okay, now u is greater than v. So at moment t equal to zero, there are some sounds. So now this guy starts moving Now the waves which are at this point and further are moving faster, this will never, so he will never cross these waves. He will only cross the waves which will overcome here because from this piece, which will overcome him while he is moving from here to here. They're moving faster and they will overcome him. So these waves, so which waves he will basically hear? Well, obviously he will start hearing those waves which are immediately here because they're just right behind him but they're moving faster. So he will definitely hear them. Now, which waves actually he will be able to hear until this moment? Well, this distance should be equal to u times t because by this moment, all the waves which are on this particular distance or less from him, he will be able to hear. They will overcome him. So by the time he will cover this distance, they will cover this distance. So he will basically hear only these waves from v, t. So this is vt and this is ut. So u times t minus v times t. This is the distance where the waves he will be able to hear are located. Now, their wavelength is this. So that's how many waves he will cross and therefore his frequency is equal to n divided by time which it takes which is u minus v divided by lambda 0 which is u lambda 0, u lambda 0 times 1 minus, in this case it's minus v over u, initial frequency times 1 minus v over u. Okay, so here is basically very close to the previous formula but it's minus instead of plus here which means that the frequency is lower which means that the whole tone is lower. So whenever we are going away from the sound we hear a lower pitch than the one which was produced by the source of sound. Now, obviously many experienced this. Whenever, I mean, I live in a big street, in a big city and whenever the ambulance is passing by I hear when it's coming to and then I hear when it's coming from and the sound is different, definitely. Well, actually it's not exactly like this particular case because in this particular case they are in the same line when you observe let's say an ambulance, the ambulance goes this way and you stand in here so you have a little bit more complicated calculations. So sound is not like definitely higher pitch before fixed and fixed higher pitch lower after when it's passing. It's actually kind of changing gradually and again it's very easy to do the calculations but they are kind of cumbersome and well, if you wish you can do it yourself that's really easy kind of thing. So that's it. That's all I wanted to talk about Doppler effect for in this particular case, we're talking about longitudinal waves. However, the same kind of Doppler effect should probably exist with transparsal waves but I will talk about light Doppler effect light in some other lecture. One more interesting topic, the Doppler effect this is some kind of a classical Doppler effect. The theory of relativity is bring some more color into this picture but again that would be a separate topic. So that's it for today, thank you very much and good luck.