Grade 12 Physics Class | Doppler Effect | Members Only

Added: 2026-05-07

[00:05:02]All right. Good afternoon. Good afternoon everybody. Once again we are back together. Sona Lutando and Phillip and Valerie and Lumi.

[00:05:17][laughter] Good afternoon guys. Are you well?

[00:05:22]>> [snorts] >> Awesome. Awesome. All right. Uh good afternoon to those of you that are joining us from our YouTube channel. Uh welcome to all our members on the channel and uh it is really such a great pleasure uh to be with all of you. Well, firstly I want to remind you uh for those of you that are intending to apply at university, please in fact if you haven't applied, what have you been up to? Um for those of you that are intending particularly to apply for engineering and the likes uh and medicine, please note that you might need to write NBTs.

[00:06:02]Okay? and NBTs are really a requirement if you're going to go into UCT into VETS and maybe some other universities as well require them. Uh please um uh just note that we do have a course that is running an NBT course guys I have included everything on that course and I've made sure that you can it can prepare you so that you do well on your NPTs. All you need to do is just click on the link that is on the description of this video but also uh you can also get in touch with us. Um this will also be on your uh on the comment section. It will be running as the stream is running. So please uh just click on that and you can use the code malum 11 and that will get you a 15% discount for that. Okay. Right. And the second thing that I just wanted to let everybody know about and this is to all of you. Uh please do note that we've got workshops that are running at the moment. But now I want you to note for those of you that are upgrading. These are meant to prepare you for your May June exams. Right. Uh if you are in metric, uh please do note that there will be workshops that are meant specifically for you. Um for now these are for our upgraders. However, if you are in metric and it happens to align align with your time for writing exams, you are more than welcome to join those workshops because ultimately the content is exactly the same. So, um please do look out. We will have uh those for our metriculants specifically.

[00:07:50]Well, the last thing that I want to invite everyone into is memberships.

[00:07:55]Please note you can be a member of the channel and it will continue to give you these videos. Uh they will come in once a week. But if you want to uh join all the people that are in my class at the moment and you want to become um uh part of the group uh please uh just get in touch with us and our number 06454719.

[00:08:22]Hey guys, I need to do all of that before we can get started.

[00:08:28]So this afternoon I'm going to be touching on a topic that is the Doppler effect. Right? And if you are doing IE uh you are excused for now. All right? Because uh the IEB curriculum does not have this. Oh my Phillip I I wasn't aware IB. Um so for now uh Phillip is uh you can excuse us but uh we will definitely cater for you in our next session. I'll only touch this uh on this particular session. All right. And and I will ensure that uh the next one is accommodating all of us.

[00:09:14]Okay. Right. Now, I want to talk about the uh the Doppler effect, but because this is only going to be one a time event, okay, it's a one-time lesson. I'm going to be very very short and brief and I will try and include everything that we need to know in terms of the Doppler effect. Now, firstly I want us to talk about what exactly is the Doppler effect. So the definition and I'm going to only mention this once, right? So we say this is the apparent change in the frequency. Now note the word apparent, right? It's the apparent change in the frequency or the pitch of a sound, right? That is detected by a listener. Now because the sound source and the listener are moving at different velocities in respect to the uh velocity of sound propagation right so in essence what we are saying is suppose that you've got a source of sound okay so here's a person here right and they are listening in on uh something that has sound. Now let's assume that here's our ambulance that is moving towards the sky right and we know that it has this wonderful sound that is it is emitting now I want you to please note if let's remove this person so the sound waves that are made by the ambulance if it was stationary we know that we will have sound waves that are propagating according to Hagen's principle we know that they are going to propagate in all directions at the same time. Right? So what happens is that the sound waves have got a consistent frequency have got a consistent wavelength. Okay. Now what happens is that once the uh the uh let's say the ambulance or the sound source starts moving right it now creates a false impression that the sound waves now begin to have a wavelength that is compressed. Now you can imagine I am moving in the direction of sound propagation right. So as a result this as I'm emitting I'm also moving in the same direction. So it creates an illusion that now the wavelength between each different crest of sound is now closer together. They are much closer together and as a result it will now have an elevated pitch. So to the person that is listening. Okay.

[00:12:08]Now here's our listener right here.

[00:12:10]Okay. You can see they are hearing everything right to the person that is listening it almost seems like the pitch is now higher than the one that is emitted. So the frequency of the listener and I want you to note in this case will be greater than the frequency of the source. So what the ambulance or what the sound source is emitting will be higher than what is perceived. And that's why you'll see in our definition we said it is the apparent what does that mean? It appears as though the frequency of sound is higher. It's the apparent uh change in the frequency of sound because the source and the listener have different velocities in respect to the velocity of sound propagation. So I want us to please note the following right and I'm I'm going to show you the uh the you know just the diagram of it. Now of course in this case once the ambulance or once the sound source has moved away what begins to happen? So in this case I am moving in the opposite direction from the propagation of those sound waves. So what happens as I am emitting I'm moving away. So meaning that it now begins to appear as though my wavelength is much longer. So if the person that's hearing is here, right? So what they are hearing seems to reach at a different pitch and that frequency is lower.

[00:14:02]Okay? Uh the frequency is lower than the frequency of the source. Why? because as I'm emitting the sound waves, I'm moving away from the direction of propagation of the sound waves. Now guys, I don't want to spend as much time explaining here. I want us to look at the mathematical relationship uh of proper and improper fractions. Now, if I may ask, I know that this is uh not a maths lesson, right? Um right, Philillip, do you remember what's the difference between proper and improper fractions?

[00:14:47]>> So, I can tell you improper.

[00:14:49]>> Okay, tell me about improper. So improper fractions are um let's say for example the the numerator is greater than the denominator.

[00:15:03]>> Oh there we go. So the numerator is greater than the denominator. So let's say 3 over two that would be an improper fraction isn't it?

[00:15:13]>> Yes.

[00:15:13]>> Right. So if I multiply anything by an improper fraction, right? In this case, this is our multiplicant. Okay? So in this case, if I multiply something with an improper fraction, note, if I take 3 over2 and I multiply by six, the answer is nine. Okay? you'll realize that the product comes out higher than the multiplicant.

[00:15:45]Okay? So in this case it means that anything that I multiply by an improper fraction will give me a much higher product. So if I take a proper fraction now what is a proper fraction? Well, Philip told us about an improper h. Now, a proper fraction is when we've got the numerator less than the denominator. So 1 / two, that's a a a proper fraction.

[00:16:13]So in this case, if I take 1 / two and I multiply by six, six being our multiplicant in this case. So what do I get? I get a smaller value, right, than what the multiplicant is.

[00:16:28]So what am I trying to say?

[00:16:31]I'm trying to say when we multiply for this guy right so when we take our equation or formula for the Doppler effect this is what it looks like okay now often times you will realize we will get rid of one of the var of the variables right so sometimes we will talk about a stationary source or we will talk about a stationary listener, you'll have one of those two scenarios.

[00:17:03]Okay, for now we will not have a a situation where both the source and the listener are moving. So let's assume in this case that it was the listener that was stationary. So what does that mean?

[00:17:18]It means that the speed of sound sorry the velocity of the listener rather is h is the one that's going to be zero. So meaning this guy is going to be zero. So what do I have? I end up with V / V + - V S multiplied by FS. Now here's the part that I want you to appreciate.

[00:17:44]So now you've got a fraction over here.

[00:17:50]Now if I've got a scenario where I say to you our source is moving towards the listener.

[00:17:58]Okay.

[00:18:00]So what do I expect in that case? I expect that whatever pitch is given by the source in this case the listener will perceive a higher pitch. [snorts] I know that. Does that make sense guys?

[00:18:19]Right? I know I'm expecting a higher pitch in this case. So the frequency of the listener must be greater than the frequency of the source.

[00:18:31]So let me ask a question. Lieutenant frequency of the listener to be higher, right? And this is my multiplicand, right? Here's my fraction. What type of fraction should that be? So that the product remember this is this should be our answer is greater than the multiplicant.

[00:19:02]What type of fraction should this be?

[00:19:08]Lieutenant uh Lutando.

[00:19:15]Okay. I've noted your hand, Philip. I just want I'm checking with the Lutando.

[00:19:22]Is your mic on?

[00:19:26]Okay, today. All right, Zanda.

[00:19:33]Good evening.

[00:19:35]>> Good evening, sir.

[00:19:37]>> Are you well?

[00:19:38]>> Yes, sir. I'm good.

[00:19:40]>> All right. So I was still asking what type of fraction should this be if I want the product the answer to be higher than the multiplicant >> uh improper >> should be an improper fraction. Now I want you to think about it. So in both instances we've got v being there right?

[00:20:05]So now we need to make an a a a determination. Are we going to use plus or minus? Remember, I want my denominator to be less than the numerator. So, am I going to use plus or minus in that case to make it smaller? So, remember I've got V and V, which is the same thing.

[00:20:29]That's the speed of sound, right? So, in order for me to make this smaller, am I going to use plus or minus?

[00:20:40]Well, and I want you to think about it Zena, right? So, when I use minus in this case, it makes my denominator smaller. So, I end up with a proper fraction. Uh, sorry, not a proper fraction, but an improper fraction.

[00:21:00]Okay. Does that make sense? Uh, Z.

[00:21:04]>> Yes, sir. Sorry. I didn't know you asking me.

[00:21:07]>> Oh, okay. No, it's okay. It's okay.

[00:21:09]>> It's okay.

[00:21:10]>> It makes sense.

[00:21:11]>> Right. So, in this case, we've got an improper an improper fraction. So, I want us as we look at this to appreciate the fact that we've got an improper fraction and so we end up with an answer that is um uh that is higher. In this case, the frequency of the listener ends up being higher. Right? I want to be very quick about this. And of course, if we are going uh if the frequency or rather if the source is moving away from our listener, we know that the wavelength is stretched, right? So the apparent frequency in this case will be lower, right? than the one that is um that is emitted by the source. So as a result if that were the case then I know that I would need a an improper I mean a proper fraction. Okay. So in that case I'm going to use a plus. Now I want us to apply this. Um I've got two questions that I have. Uh one was from our exam.

[00:22:23]It's a very interesting question and I want to kind of do that with you right now. Let's go for it quickly. They say to us in the diagram we've got A, B, and C below. Um, the wave patterns are shown for the sound, sorry, for the sound of a siren from an ambulance. I was talking about an ambulance just now. And a stationary observer. So they've already told us our observer is stationary.

[00:22:54]Okay, let's keep that in mind. They say uh the observer remains stationary in each case.

[00:23:02]Right now I want you first to realize in picture A both the source as well as the as the observer both of them are stationary. So what happens what the source is emitting and what the observer is hearing are going to be exactly the same. So what we hear on the source side will be what we hear from the observer side. But when does that happen? When both of them are stationary. Okay. Right. But now look at image B. What happens at image B? You can see that the sound waves become more compressed. So what does that tell me?

[00:23:50]It tells me that my ambulance if the observer is stationary, it tells me that my ambulance must be moving towards the listener. So what the the listener will hear, right? So the frequency that the listener perceives here will be greater.

[00:24:08]Notice that they are compressed. So the pitch is higher meaning that the wavelength is shorter. Now remember that the the relationship between frequency and wavelength are inverses of each other.

[00:24:24]Right? So F is equal to V divided by lambda. Right? So what we're saying is that our ambulance was moving to the left in this case towards the listener and so the apparent frequency will be much higher in C.

[00:24:42]I can see that uh the compression is more towards this side and my apparent frequency over that side is stretched.

[00:24:52]What does this tell me? it tells me that my ambulance must be moving away from my listener. Okay, [snorts] so it must be moving in that direction uh in image C.

[00:25:05]I hope that makes sense guys and um as as we did say I I will continue to explain this as we continue right now.

[00:25:14]Um okay, my [clears throat] laptop is about to die. Let me just make sure that I'm charged up. Okay. Right, there we go. So now they say to us the the question 6.2.1 and 6.2.2 either. Okay, they said define the Doppler effect. Okay, I've already defined it. H I'm not going to go through that again. So they say um so they they say state either A, B or C.

[00:25:47]Right now they say firstly the sound source is stationary. Where is that? We said that's at a right. The propagation of the sound waves is the same in all directions in line with what we call uh we said Hagen's principle, right? And then they say to us the sound source is moving away from the observer. Well, we said where is it moving away? That must be at C. Okay. And they say give a reason for the answer in 6.2. 2.2 right when we look at 6.2.2 two we can see that the wavelength right is longer. So the the wavelength of the sound waves is longer right. So the wavelength is longer.

[00:26:43]Okay. So therefore it means that uh it is moving away. You can also state that the frequency okay is lower right than the emitted frequency of the sound waves the original frequency and as a result it means that it must be moving away right I want us to get into the calculations in this case right and if there are any questions you you're more than welcome uh to ask all right let's go for it so they say to us if the ambulance is moving away from the observer at a speed of 25 m/s.

[00:27:25]Right now I am going to draw the image for myself uh so that we're able to do the calculation.

[00:27:35]All right.

[00:27:37]So here's the ambulance.

[00:27:44]All right. I'm very good at drawing. Uh, in my next lifetime I'm going to be Picasso.

[00:27:53]Right. So, here's our observer.

[00:27:57]The ambulance is moving away. They said to us it's moving at a speed of 25 m/s.

[00:28:05]So, what are they giving to us? They're giving us the speed of the source. So, this is going to be the velocity of our source. Remember the ambulance is the one that is responsible for emitting the soundwave, right? And [snorts] they say to us and this and the frequency of the siren, okay? Um and the frequency of it siren is heard by the driver uh rather as heard by the driver is 900 htz. Now please I want you to note when there's a driver hear inside the car remember that what the driver hears will be exactly as what is emitted. Why?

[00:28:53]Because the driver as well as the ambulance are moving at the same velocity.

[00:29:01]Okay. there is no change in velocity that takes place between the driver and the ambulance as opposed to in this case when you look at the ambulance and the listener. Okay, so remember that the change in the pitch is caused by the relative velocity between the source and the listener. So meaning that what they are trying to tell me here is that the frequency of the source is also 900 htz and I want you guys to please appreciate that. So they're not giving us information like they are not just handing it on a silver platter. They are telling us so that they also um you know kind of invoke our understanding of the section. Right? Uh are are we together guys? Does this make sense?

[00:30:01]All right. Awesome stuff. H Tando, you haven't said anything so far. No. Men, right?

[00:30:13]>> We're learning. Sir, >> you are learning. Is I like that in what a response. What a response. All right.

[00:30:22]So what I was saying is that the frequency of the source and the frequency of the driver they are the same. So meaning that what when they're telling me about the driver it means they're telling me about the frequency of the source. Now uh the part that I want you guys to notice now they say to us calculate the frequency observed by the observer take the speed of sound in air as 340.

[00:30:50]Now this is very very important. Now so firstly I want us to note we are looking for the frequency as heard by this listener but they have told us that V now please note the speed of sound V in a medium. Which medium are we using this time? We're using air as a medium.

[00:31:16]Right? They're saying that's going to be 340 mters per second. Now note because the ambulance is is moving away I know that the frequency of the listener will be less than the frequency of the source. So let's start and calculate there. So our Doppler effect equation firstly we'll write it as it is uh sorry and that's V + minus V L V + minus V S multiplied by FS. Now what am I doing? I'm trying to avoid you guys having to memorize. you know they'll say to you if it's moving away it's it's going towards then you have to memorize whether it's plus or minus or whatever the case is because it changes based on whether it's the source that's moving or whether it's the listener that that's moving but if you understand what I'm doing with you right now it will actually help quite a lot right [snorts] uh so in this case I'm going to now do this with the equation remember what do we know we know that our listener is stationary. So which means V L is zero.

[00:32:35]So that will be V divided by V plus - V S time FS. Now once again I said to us okay I'm going to just switch this thing off. I said to us we already know that the frequency of the listener should be lower. So, what type of fraction should this be?

[00:33:02]H Let me ask Marian. Marian, good evening.

[00:33:09]Um, Marian, is your mic on?

[00:33:13]Not you didn't respond the first time.

[00:33:19]Okay, I think uh Marian is experiencing technical issues. All right, Philillip, what type of fraction should this be?

[00:33:31]I want my product to be less than my multiplicant.

[00:33:37]>> Would it be a um a proper fraction?

[00:33:47]>> A proper fraction. Remember we said if we want a lower number right as a product we need to use a proper fraction right? So meaning if I want this to be a proper fraction it means my denominator must be bigger than my numerator. Does that make sense?

[00:34:09]Does that make sense?

[00:34:11]>> Yes sir.

[00:34:12]>> Right. So, do I use a plus or a minus in that case to make the denominator bigger?

[00:34:21]>> You would use a plus. There we go.

[00:34:24]Right. So, there's my equation. That's my formula. So, let's let's substitute.

[00:34:30]So, I'm going to have 340 divided by 340 plus the speed of the source. And the source is 25, right?

[00:34:43]multiplied by the frequency of the listener. They said this is 900.

[00:34:49]And all we need to do now is to use our calculator and say that's 340 / 340 + 25.

[00:35:06]Oops. 340 + 25.

[00:35:10]uh this is multiplied by 900.

[00:35:14]Okay. And I get the pitch or the frequency to be 8 uh I forgot that number 838.36.36 hertz. Please remember we are looking for the frequency of sound and in this case that frequency is less than what was our emitted frequency or the frequency of the source that was 900 hertz. All right. I hope that makes sense ladies and gents. Right. Now um I am going to skip this question for now because um I will explain this at some point uh much later um where they are they are talking about red shift and blue shift right I will take some time to do that a little bit later um yeah for for now I'll leave this uh uh for now now I want you guys to please note sometimes they might want us to calculate the wavelength of sound And here's the part. Now I've got the wavelength I mean sorry the frequency of the sound of the listener.

[00:36:28]But what would be the wavelength? And please I want you to note it means that I'm going to use f is equals to v over lambda. Now the part I want to caution you about is that hey we know a frequency that's uh 838.36.

[00:36:47]Now the speed that you use here is not the speed of of of the source or the listener. This is the speed of sound in a medium.

[00:37:03]So if our media sorry is air, we were told right now that our speed is 340.

[00:37:12]And so this would be 340 divided by lambda. And we're looking for the wavelength. And of course what we will do in that case is that we'll find the wavelength. H if we cross multiply then it means that wavelength will be 340 divided by 838. I won't go through the calculation. What I just wanted to make you aware of is that anytime that they're looking for wavelength, please just note that we will be able to use uh or rather we should use the speed of sound in that particular medium. Okay.

[00:37:49]Right. I want us to uh take a somewhat well more challenging example and this I'm going to take from how I mean sorry uh from the national exam in 2023.

[00:38:03]Right. I thought it was a a particularly interesting question that they gave.

[00:38:09]Right. And I want to go through this question with you. So firstly they say to us the device emits uh or rather let's start right at the beginning. They say to us, "A car moves at a constant velocity of 22 m/s on a straight horizontal road towards all right so they've unlocked it for us.

[00:38:36]It is towards a stationary device. So if this device is stationary and the device is the one that is picking up the sound then it means that the velocity of the listener is stationary. Okay. Right. So they say which which can both emit and detect.

[00:39:00]Now note the device is able to to to emit sound waves, but it's also able to detect sound waves. Right? Now, stay with me on this one. They say the device emits sound waves at a frequency of 24,000 hertz, right? Uh these sound waves are are reflected off the car. All right?

[00:39:29]are reflected off the car and and the reflected sound waves are then detected by the device. Now what's happening the device in fact let's do this sends a signal okay I don't know if you guys have uh seen the police that would hide and have this device where they are uh you know detecting the speed okay of of the car and if you go above a certain speed you know obviously they would it would flash and it means that you'd get a ticket or you would get uh you know a a a a speeding fine, right? Um and if you're within the speed limit, of course, they they just uh you know, it would just be as it is, right? Now, in this case, how does that device work? It's exactly this. So, it sends out a signal. So, it would send out a signal and that signal would go into the car, right? Well, not into the car, but it would bounce off from the car. And in this case, then it would be able to then get a reflected signal, and it will now compare that signal to what it actually emitted and the difference will tell the device what type of uh speed the car was actually moving at. All right. All right. So, I want to uh to go through this very quickly.

[00:41:00]So they say to us um these sound waves are reflected and then you know then the device detects uh they are detected by the device. Now they say to us I'm going to skip the first question. If the speed of sound is 340 so again we're back to 340. Calculate the frequency of the reflected sound waves detected by the device. Now guys, I want you to listen very carefully here, right? Because there are different roles that are being played by each of the scenarios.

[00:41:39]Now, scenario number one, this is where the device is the source, right? It means the frequency of the source. This is scenario number one.

[00:41:54]The frequency of the source is 24,000 htz.

[00:42:01]Are you guys with me so far?

[00:42:05]The velocity of the source.

[00:42:09]Now remember they told us the device is the one that emits sound waves.

[00:42:16]So the velocity of the source is 0 m/s because remember the device is stationary. So meaning initially the car is the one that is acting as the listener. All right. So I want to find out what is the frequency that will be picked up by the car. Does this make sense guys?

[00:42:42]Right. That's not the end of the story.

[00:42:46]So that's scenario number one. So let's deal with uh scenario number one where the car is the one that is the listener and it's picking up a frequency from this uh from this device. So again L that's V + minus V L plus minus V S time FS right now again I want us to note for this formula our listener is the one that is moving this time around our source is the one that is stationary. So, V S is zero. So, I'm going to have V + minus V L divided by V. Why? Because this guy here is stationary.

[00:43:46]Does that make sense, guys?

[00:43:48]Are we good so far?

[00:43:52]Beautiful stuff. Okay. Right now I know that when I am or rather when the source is moving towards the listener now I I want you to to think about it right or rather our listener is the one that is moving towards the source. So what's happening? It's colliding against these frequencies or these peaks of sound and so it looks as though because it's moving in the direction where the sound is actually being emitted from.

[00:44:29]Right? So it collides more often with the sound waves of the car. So the apparent pitch because the source is moving or rather the listener is moving towards the source then the apparent frequency will be higher than the emitted frequency.

[00:44:48]Does this make sense to you guys? Okay, are we all good in the hood?

[00:44:58]All right.

[00:44:59]So in this case, I know my frequency that will be picked up by the car is greater than the frequency that is emitted by the source. Okay. So I want the frequency to be higher. So what type of um what type of uh fraction should that be? Tando. Uh good evening once again. So I was about to ask you to repeat what you said because >> Okay, this is you running away from answering my question.

[00:45:35]>> No, I was.

[00:45:37]>> Okay. So what I was saying is so here we are tandun we are emitting sound waves right >> now. I want you to imagine all right and I'm I'm going to try and make a sound here. Imagine if these sound waves are kind of crashing against the car, right?

[00:46:02]If both the car and the device are stationary, right? Let's imagine we hear a sound, right? We're making that sound.

[00:46:14]What? Now if the car is moving towards the sound wave, >> do you agree with me that that sound will actually appear more frequently?

[00:46:29]Right? So when the car is moving in that direction, it will be da da da da da.

[00:46:35]Right? Why? Because I am colliding in a in a sense against the sound waves.

[00:46:44]Would that make sense to you?

[00:46:48]somewhat. So, somewhat >> somewhat.

[00:46:53]How do I make this make sense? Um, okay. So, I I I want you to think of this as, you know, if it were cones, right? You are you are going to collide against these cones. Would you agree with me that if I collide against these cones, if the car is stationary, right, of course it's not going to collide against the cones, but the higher the speed of the car, the more it will be able to go through those cones. Right.

[00:47:27]Ha da da da da. Right.

[00:47:30]>> Mhm.

[00:47:31]>> Right. So in that direction. But if I was moving at a higher speed, what kind of sound would I hear?

[00:47:39]it would be da da da da da da >> right which means it's not so much that I've moved the cones the distance away from the or or rather increased the distance or they decreased the distance from the cones do you agree with me >> yes >> but what's happening is that the car is now moving at a higher speed and as a result it feeling as though right >> the sound is much more frequent than it was before.

[00:48:11]>> So it's there's no change in the wavelength.

[00:48:14]>> There's no change in the wavelength, right?

[00:48:17]>> Okay.

[00:48:18]>> To the one who's listening, it appears as though >> there's a sudden change, >> right?

[00:48:26]>> Okay.

[00:48:26]>> I don't know if you listen to Formula 1, you know, when you listen to a Formula 1 car, you can tell between when the car is moving towards and when it's moving away, right? you hear that zoom sound, right?

[00:48:41]>> So in that case, when you hear the sound beginning to be of a lower pitch, you know it's moved away, right?

[00:48:50]>> But when it's moving towards you, you know the car is moving towards. Okay?

[00:48:55]And that's exactly what is happening. So you you you do understand what what I was trying to present there.

[00:49:02]>> Yes, sir.

[00:49:03]>> All right. So I'm saying because now the car is moving in the direction right against the propagation of sound it will appear as though the frequency is higher >> higher.

[00:49:19]>> Does that make sense?

[00:49:21]>> Yes sir.

[00:49:22]>> Now the part I wanted you to help me with what type of fraction should this be? Should this be a proper or improper fraction? Remember I want this to be higher.

[00:49:36]>> Higher.

[00:49:38]>> All right.

[00:49:39]>> Um we want our so can I say improper.

[00:49:49]>> Improper. You are absolutely right. So which means my numerator must be greater than my >> denominator.

[00:49:59]>> Beautiful stuff. Now, if that is the case, Tando, tell me, should this be a plus or a minus at the top? Remember, I want the the numerator to be higher.

[00:50:11]>> It should be a plus.

[00:50:12]>> There we go. So, get rid of this guy, guys. Am I making sense?

[00:50:20]Right. Okay. So, we know we want the frequency to be higher. And so the detected frequency going to say 340 plus the speed of the car is moving at 22 divided by 340 and this is multiplied by the frequency of the source and uh we said it's uh 24,000.

[00:50:48]Right? Please help me out quickly. Um let's calculate that.

[00:51:00]So this is 340 + 22 divided by 340 and this is multiplied by 24,000.

[00:51:09]Okay. Right. I get a frequency a detected frequency by the car of 25,52.

[00:51:25]52. Um, so this is the perceived frequency. Okay, so that's 94.

[00:51:35]Okay, hertz. Right, I'm going to leave that answer there. Right now, remember, what am I looking for? That wasn't what the question is asking me for because at the moment I am ask I I I am um taking the emitted frequency to the car but the question is asking me about the reflected sound waves meaning I want to know what is the device picking up. Am I making sense guys?

[00:52:10]Right. So what is the device picking up now?

[00:52:16]Which means we are switching roles.

[00:52:20]The device is the one that will now become the listener.

[00:52:30]Right.

[00:52:33]And it means the velocity of my listener is zero.

[00:52:39]And the car is now going to be the source. What frequency is it emitting?

[00:52:46]It is this 25,550.

[00:52:56]So the frequency of the source uh point we said 94 Okay, so the frequency of the source is the 25,000, but I also know the velocity of the source is 22 m/s.

[00:53:16]Right now, let's get to the last portion of this.

[00:53:24]Again our formula says L that's V + - V L that's V + - V S time FS. Now once again right remember who is detecting the sound waves it is this device now and the source of the soundwave is moving in the same direction and it is moving towards the device. Now what is the device picking up? it is picking up a frequency in this case that would be higher. Now if if you remember that image that I was I I I gave to you guys if something is stationary then the sound waves are you know uh moving in all directions at the same rate right but if the car is moving all of a sudden this is what happens because it's moving in the same direction as the sound waves it means that they will appear to be more compressed in this case.

[00:54:36]the frequency will be much higher than the frequency that it is emitted at. So again frequency of the listener should be higher.

[00:54:48]Okay.

[00:54:57]Our frequency now should be higher once again.

[00:55:02]>> Yes sir.

[00:55:03]>> Right. I want to know. We know our listener is stationary. So we know our equation will be V V plus minus V S time FS. Now may I ask what type of a fraction does this need to be proper or improper? Remember we want the frequency the detected frequency again to be higher.

[00:55:28]>> Improper. So >> should be an improper. Now if I want it to be improper in the denominator, do I use a plus or a minus?

[00:55:39]>> Minus.

[00:55:40]>> Right. There we go. So this is going to be 340 / 340 minus 22 and this is multiplied by now please note the frequency that we're going to use now is the one that we calculated earlier which is the 22 552.94 I'm going to Leave it as it is. Now uh without rounding off.

[00:56:17]So I'm going to take my previous answer that we got and I'm going to multiply it by a fraction 340 divided by 340 minus 22.

[00:56:37]All right. And ultimately I am I get 27,320.

[00:56:50]This is 775.

[00:56:56]Okay.

[00:56:58]75 and this now becomes the frequency that is detected by the device. So I want you to note the device played two roles in the initial part. The device was the source and the car is the one that was detecting.

[00:57:22]But the same soundwave is then bounced off. Right? They said it is reflected.

[00:57:28]So it reflects from the car and it is now going to be picked up by the device.

[00:57:34]Okay? And so in this case the reflected soundwave that is picked up by the device is going to be 27,320.75 hertz. Guys, does this make sense to you?

[00:57:55]All right, awesome stuff. Um, I just want to quickly as I close uh you know I I just want to I I will talk about spectral lines but uh they will require me to kind of I'll explain this uh to you guys well on on Saturday. All right, which won't take a long time. Um but I I will do that uh with you. That is the people that are in my class. Um but the final part that I just want to talk about before I close this session you know there's a question that was given um I don't remember which question paper it is all right and I I want you guys to just appreciate um you know this this question so they gave us time frequency against time and they said to us we have a detector that picked up frequency initially and then this happened. So let's just say for argument sake our frequency was 900 htz right and then it then changed to let's say 720 htz right so in this case uh we've got a detector a detector rather was that was detecting a car initially it had a reading of 900 htz and then afterwards the pitch then lowered to 720. Now, what is this trying to tell us? It's trying to tell us that the car that it was detecting initially must have been moving. Let's say this is our person. Let's say they've got a detector there, whatever that detector looks like. And initially, it means that the source was moving towards the listener. How do I know that the frequency was higher, right? And afterwards the source was moving.

[01:00:11]Okay, the source was moving away from the listener.

[01:00:19]How do I know this? The frequency suddenly changed to a lower frequency.

[01:00:26]So this tells me it means that the source was now moving away from the detector.

[01:00:33]Okay. Now in that question [clears throat] what they wanted you to get is what is the speed of the source and what is the frequency of the source.

[01:00:47]It requires you to do simultaneous equations. So in the initial case you are going to have to find an equation use an equation right where you've got uh let's just say that the speed of sound in air again is 340 m/s. Now you're going to have to set up an equation of the detected frequency sorry when the car was moving towards and another equation of the detected frequency when the car is moving away.

[01:01:20]And you're going to have to take those equations and solve simultaneously and be able to find both the velocity of the source as well as the frequency of the source. All right, I don't want to h exhaust that. You can find that question and and and possibly uh for you guys I will try and look at that question first when we meet on Saturday. But uh we will be actually also looking at chemical equilibrium as well. All right. So, I want to leave it here for today. All right. Um, please do remember our our workshops for those of you that are upgrading. Uh, please do remember that we've got those workshops that are coming up. Uh we do want to also alert you that if you do want to join uh the people that are in my class, you can uh get in touch with us which is 06454 uh 719 and that will help you a lot. You know, unfortunately a lot of people wait until it's late and they are already not doing well. U but I want you guys to be shining stars all the time.

[01:02:34]You're going to be a shining star.

[01:02:40]>> All right. And uh just the last thing here that I also want to say.

[01:02:44]>> Yes, sir.

[01:02:45]>> Yes. Yes. Yes. uh if if you want to come part of those workshops, please do note that the link uh is on the chat uh section. But if you will be watching this later, it will be on the description of the video and you're more than welcome to join us on those workshops. As we are saying goodbye, uh please have a good evening guys and um we'll see you again next time.

[01:03:13]Shop.

[01:03:15]Enjoy your evening.

[01:03:27]>> Bye, sir.