Transcript for:
Understanding the Doppler Effect in Physics

good day everyone uh we meet here again for another episode of physics um and today i want us to cover quickly uh the doppler effect now um one of the things that i want us to do when it comes to this section is really to make it as easy and simple as possible uh as the doppler effect actually is so please uh don't forget to subscribe and when you need to subscribe if you haven't done so already and again you can reach me on my email lung ac dot m.tenkosi at gmail.com that's u n g i s i dot m for mary or for mama dot n go c n k o s i at gmail.com all right let's get into it very quickly so when we talk about the doppler effect okay first of all let's talk about the definition we say that the doppler effect is the apparent change in frequency or or the pitch of a sound okay that's detected by listener when the source and this and the listener have different velocities relative to the medium of sound propagation now that's a definition and it sounds very formal but what i want us to do is to quickly just look at it first of all um sound um actually is formed by vibrations okay so as you're hearing my voice right now it's because of the vibrations that are forming in my voice box okay and those vibrations can actually be quantified in terms of amplitude now sound can be um can be seen as a wave right now if you look at this so we talk about the amplitude which is um this height here okay depending on where you want to uh take it from your reference point from so we talk about the amplitude of sound so you can either have a high volume of sound you know we usually use that word a high volume so uh i can in in the way that i talk i can even talk i can talk louder or i can talk softly so that has to do with what with the amplitude of sound but then um you find people that have a baritone like myself hollow okay or some people who have a uh um you know a s you know a lower pitch that is a soprano hello okay so what's the difference there that's not the loudness of the sound but that has to do with uh now the frequency of the sound so for me you know you'd find that this is the wave that i form hello right and then for someone with a a tinier voice all right you find that these are the waveforms that you find now what is the difference between the two waves all right so first of all the difference is that this one has got a longer wavelength okay so we call that a wavelength the wavelength of sound okay all right or off a wave and then we've got in this case for this one it has a shorter wavelength okay so if you compare wavelength one and wavelength two that's a shorter wavelength but now if you notice the two lengths what you will notice is that um this one with a longer wavelength there are fewer peaks that will be formed can you see that so we call that the frequency now we'll get into that in just a little while okay so there are fewer peaks that are formed and as a result um in this case the frequency is low so this one that has um a shorter wavelength can you see that i've been able to make far more uh or many peaks within the same period okay i've been able to form many peaks within the same time so as a result we say it has a high frequency now i want you to notice something so the longer the wave length right the smaller the frequency so there are fewer peaks that i make okay so the shorter the wavelength in this case the more peaks that i'm able to make so the higher the frequency so obviously that means that there is an inverse proportion to the relationship between wavelength and and frequency now there's a formula by the way and that we normally use for waves okay i won't go too much into it because i really want us to get into the doppler effect so we usually say v is equals to f times lambda now if you think about it sound is able to travel okay right uh in this case if you want your sound to travel between two points okay so it means that sound does have a a speed it does uh it it is able we can actually have the speed of sound now the speed of sound depends really on the medium that you have so uh the speed of sound in air would be different to the speed of sound in water and so on and so forth and as a result because of that we know that because the speed is different in different mediums okay what will then tend to happen is that we will then need to investigate or find out what is the speed of sound in a particular medium we'll talk about that as we go on now please remember when we use this formula i just give i want to give us a warning all right so when we talk about v that has to do with the speed of sound okay um and obviously this would be this is measured in meters per second right and then this is frequency all right this is frequency uh and this is measured by the way in hertz heads just simply means per second all right and then we've got um this one here which is wavelength the one that i've i've shown this is a greek symbol called lambda okay so in this case uh wavelength is usually measured in meters okay uh i didn't uh so this is wave length okay right now let's get into uh some of the nitty gritties what i want to rush to uh when it comes to the doppler effect is just simply how to apply it because i know many people have complained about ah i don't know when to use the plus or the minus and so on i want to make it i i promise you by the time that you finish watching this video you will understand every single aspect of it i promise you that much okay so um obviously we're going to use the the formula just uh something that i wanted to uh you to note is that this v that we are talking about here just a pre-warning remember that it is neither the sound of the speed rather of the source or the listener this is just the the speed of the propagation of sound in a particular medium so normally that is either given or or in some cases you know i've seen questions where they require you to calculate that just remember when you use this it's neither the sound of the or the speed of the source or the listener but it is rather the speed of sound okay beautiful now i want you to imagine okay uh you know i think sometimes because we cannot see the waves of sound maybe let's use something that's a little bit more visual right i want you to imagine something imagine uh we've got a b okay uh you can see i'm very good at drawing okay this is the stinging part of the b all right let's put some eyes on it okay so imagine you've got a b right and of course um i'm just going to be a little bit cruel okay let's imagine that we uh we catch that b and we hold it by its wings so you can imagine it's wiggling right so it's making a buzz okay and what we're going to do is we're going to put that b in water as it's buzzing so can you imagine what's going to happen is that that water now because of the buzzing of the bee okay hagen's principle simply says that the waves will move in all directions at the same time right so this is kind of a wave form that will be formed because of that b okay agree with me right uh now um what what's happening here waves are forming right so if you think about the lines uh if you imagine the lines to be the peak of that wave so it's this part here and then these parts where there's nothing just imagine it to be the crests of the wave right uh these deeper parts right so what will happen is that we'll see those waves forming and they'll be moving in all directions at the same time okay now um so this is when we keep the b stationary all right so i want you to imagine in this case whatever it is that we observe right would be in a sense what it would be the same everywhere all around okay now if i take the very same b and i start moving i'm moving it i'm going to just imagine that we're moving it in that direction okay now all of a sudden this pattern changes and what we'll then begin to see all right so here's our b here okay so we're moving it in that direction so a pattern will now begin to form and we'll see something that looks like this okay now remember the the buzzing of the b has not changed okay what has changed the fact that we are moving the b all right now all of a sudden what are we able to observe now look at this on the side that the b is moving towards the wave forms remember the buzzing of the b has not changed right the waveform that is now formed seems as though and i i use that word uh very advancedly right it seems as though that the waves now are actually much closer together the peaks are more closer together so if you remember our pattern right wavelength becomes shorter so what happens it seems as though that the frequency the buzzing on the side seems a bit more frequent than on that side there so what we are saying is that as it moves if it moves towards a particular object okay so let's imagine there was someone who was observing this from here right you can very you can see very well that i'm very good at drawing hey right so if it was moving towards someone what are they able to see they're seeing ripples that are forming okay and these ripples are actually if you think about it right they're much closer to each other but is that what is happening no what's actually happening is this one here right it's still buzzing in the same way so in this case what appears to be you see that's why we say it is the apparent change okay in the frequency or the pitch of sound right that is detected by the listener so now on this side where it's moving towards they're much closer together the wave length is shorter so as a result we say ah but this one observes a higher frequency than the frequency that is emitted by the b please i want you to remember this picture here okay right so this one observes a higher frequency than what is emitted by the b right now think about a person that would be standing or i'd need to squeeze in a person somewhere there okay that's standing on this side okay and observing the same thing right to them what are they beginning to observe they are seeing the very same ripples but to them it looks as though their wavelength is much longer and a longer wavelength would mean a smaller frequency you're quite right right so a smaller wavelength so the frequency that is observed on this side of when the b is moving away from us right so it seems like the frequency is much shorter okay so now that's essentially the the the the doppler effect what we are observing here happens when it comes to sound okay right so um i want you to imagine i don't know if you've er ever heard you know when you are listening to an ambulance passing right okay so if it's coming towards you you'd hear that sound but as it moves away from you it's as though uh you know there's there's there's a a difference or a change in that sound you know um or sometimes it happens when you're listening to uh when you hear uh you know say a car that has music playing inside right as it passes you it's as though the beat of the music is getting slower right so all of that is an indication of the doppler effect okay right now let's keep it short man you know we don't wanna bog you down with all of these details now there's a formula that we are going to use and i want you to please remember what i'm saying to you right now okay so the formula that we use we say okay the frequency of the listener is equals to v plus minus v l okay right that's v plus minus v s times fs now let's uh let's just indicate quickly what does this actually tell us about so this is the frequency of the listener right the observer okay whoever it is that's observing so if you think about this picture here we've got an observer here we've got another observer there right so the frequency the apparent frequency what appears to this person okay right is equals to v now this v is the speed of sound and i do want to say this again this v and that v over there are exactly the same thing i want you to please remember that okay right um perhaps if i can add something else please also just remember that frequency is also can also be designated as one over period so what would be the period in this case is the time that it takes to complete one full wave right so frequency is one over period so in this case if you're given the time that it takes to complete one full wave or one full cycle right you can get uh the frequency in that way right um so as i was still saying so this is the formula that we're going to use this is the speed of the listener okay so in this case if the listener is moving right um that would actually be the speed of the listener right and in this case again that's the speed of sound and this is the speed of the source so in this case it was our b that was moving our listener or our observer was stationary and then this is the frequency of the source okay so in this case uh this is the formula that we are going to use right now ladies and gents i want to make the doppler effect as i said to you i really want us to make this as easy and simple as possible okay right now what did we say we said all right um when the source is moving towards a listener we know what are we going to expect we expect a higher frequency we expect a frequency that would be greater than the frequency that is emitted now can i just go back a little bit all right suppose we had an observer here okay when the b was stationary okay so when the b is stationary the frequency that's emitted by the source and the frequency that is observed by the listener would be exactly the same there would be no change remember the change only or of course and condition that the the the listener is also stationary the change only begins to take place when either one of them starts moving i want you to please remember that when both of them are stationary no change right what the listener what the source emits the listener hears when one of them begins to move right either the source moving towards the listener or the listener moving towards the source right the apparent frequency now would now seemingly become higher right and then if you know it's moving away obviously it would seem lower now because we know what to preempt right so what i want us to do is to um you know i i i i don't like um you know making people remember things like saying okay when it's moving away you use a plus when it's moving towards you use a minor i don't like all of that because what it does is that the moment you get under pressure and you you see you you you tend to forget those kind of things so what i'm going to do is i'm going to use a method right and in this method i want you to please listen carefully okay and what i'm going to say now note you're going to have two scenarios when it comes to the doppler effect and i promise you that much okay you're going to have uh two different scenarios scenario number one it's when you've got a stationary source okay so suppose our source is an ambulance okay right so a stationary source and a moving listener okay so v is zero right this is our source of sound okay but then you find that the source is stationary but it's the listener that's either moving away or moving towards the source okay that is scenario number one okay or in scenario number two you're going to have a moving source okay right and a stationary listener so v is equals to so la it was uh the velocity of the source was zero here the velocity of the listener is zero okay and our source can either be moving towards the listener or moving away from the listener i want you to please remember that those are the two scenarios that we will have when it comes to um the doppler effect now i want us to quickly do something for ourselves right let's go back to maths and you know something that you learned either when you were still in primary school or you know when we when we did fractions when we when we're introduced to fractions now i want you to please note something there are two types of fractions that you have all right and i promise you this will help you in terms of answering questions when it comes to the doppler effect there are two types of fractions that you have there's what we call a proper fraction okay now what is a proper fraction it's a fraction that has a smaller denominator right okay i mean a smaller numerator rather and a greater denominator so if you take 1 over 2 that's an example of a proper fraction right so if i take a proper fraction and i multiply by any number okay let's take um let's take four for argument's sake right so anything that i multiply right by a proper fraction the product of it or the result of it is always smaller than in this case uh what i multiplied the number that i multiplied with right so in this case if i take that fraction proper fraction right and i multiply it by number i the result is is is is something that is smaller right on the other hand i've got an improper fraction now what is an improper fraction okay it's a fraction whereby the numerator we meaning the number at the top let's take three over two the numerator is bigger than the denominator right so when i multiply an improper fraction okay by a number let's take four again right so if i take that an improper fraction and i multiply by any number the result of it is always bigger right so notice this if i take that over 2 times 4 that will give me 6 okay so if i note this result is bigger than the the number that i multiplied with okay right you see if you remember this i promise you half our work is done why am i saying that now this is our result all the time okay so this is our result and now this is our fraction i'm going to show you how simple this is okay right so this is our fraction and so this is the number that we multiply with now if the source is moving let's say towards the listener what am i expecting i'm expecting a higher frequency isn't it so the frequency of the listener should be higher than the frequency of the source right so in that case what type of fraction should this be if i want a higher result i should have what an improper fraction it means this top part must be bigger than the bottom part okay i'm going to show you this practically all right so um but if it's moving away okay for argument's sake then i know the frequency of the listener must be smaller than the frequency of the sources in it so all that's simply going to happen very simple it means i must multiply with a proper fraction so if that's going to be a proper fraction so it means that the numerator must be smaller than the denominator isn't it right so um what i want us to do now if you remember this if you remember all of this i want us to take just a typical question i don't want to put too much into the question right i just want us to take a very simple scenario and what i want us to do is to look at it in terms of what we've just done here and thereafter i want us to start tackling past exam typical past exam questions is that okay all right let's get into it all right so i want us to look at this uh simple scenario quickly right so suppose we've got an ambulance all right that's our source that's able to emit a hundred hertz right so frequency of a hundred heads all right and say our ambulance is moving at 20 meters per second right now suppose we've got a stationary listener over there okay so let's show that our listener is stationary okay and uh we want to know okay let's show that it's a listener uh let's put in the big ears there okay so there's our listener over there right so we want to know what will be the frequency that will be observed by this guy over here all right in relation to the frequency that's emitted by that source so here's our ambulance okay moving at 20 meters per second and we want to apply the top layer effect of course the thing that we must always get right ladies and gents please write down the doppler effect equation right as it appears in your answer sheet okay so here's our doppler effect equation okay so we're going to say fl that's going to be v plus minus vl over v plus minus vs multiplied by fs now let's just make an assumption let's say the speed of sound in a medium in this medium let's say the speed of uh sound in this medium is a the medium brother is air right so let's say the speed of sound just for argument's sake is 340. so v remember this is not the speed of the uh source or listener this is just simply the speed of sound propagation right uh in a so let's say the speed of sound is 340. now ladies and gents i want you to please stay with me all right now what do we anticipate the source is moving towards the listener you remember when i told you about the two scenarios i promise you um in our syllabus you'll only get those two scenarios stationary source or stationary listener with the moving source right or the other way around so in this case we are simply saying all right so our source is moving our listener is stationary okay right now typically what do we expect in terms of the frequency of the listener of course the frequency of the listener should be higher than the frequency of the source do you agree with me right so the frequency of the listener definitely will be higher than the frequency of the source so in this case i'm going to say all right we want the frequency of the listener right now i want you to please note the listener is stationary so what are we left with this is going to be zero so we just have a v at the top isn't it okay because v l is already zero okay the velocity of the listener right and then here i've got v and v s now i'm not going to put a plus or minus you're going to tell me which one are we going to put right multiplied by the frequency of the source now we are going to multiply these two things and i've got a fraction you remember that no right so in this case i know i want my answer to be bigger than what i multiply with isn't it so i want my answer to be bigger so what type of fraction should this be i think you get it aha this should be an improper fraction so if it's an improper fraction it means that the numerator must be bigger than the denominator or we can put it this way the denominator must be smaller than the numerator so now in order to make this smaller right do i use a plus or a minus i think you you can see that very clearly right obviously in order to make this smaller right so i'm going to put what a minus remember this and this are exactly the same thing right so in this case i'm going to have a minus y because i want this to be a and improper fraction so now we know what v there is that's 340 divided by 340 remember i want this bottom part to be smaller minus what is the speed of the source the speed of the source is 20 right multiplied by what's the frequency of the source we said the frequency of the source is 100. so in this case i've got 340 right and divided by okay let me just quickly grab my calculator over there all right um so okay there we go i have all right that's 340 divided by uh 340 minus 20 and i'm going to multiply all of that by 100 okay and guess what definitely i get a higher value can you see that value it's going to be 106.25 heads so the frequency that is perceived by the listener is actually higher as we anticipated than the frequency that is emitted by the source i hope that makes sense ladies and gents eh right okay right um you know what maybe just to uh to push this a little bit uh so that we really get um we get this in full now let's say obviously the ambulance is going to move okay towards the listener and then ultimately it's going to move past the listener and start moving away so let's say um in our next question suppose same listener who are stationary okay to show that the person is listening and now the source is now moving away okay still with the same velocity so say it's still moving at 20 meters per second right and still our listener is stationary right we want to know what would be the passive frequency okay so this would be f sorry fs the frequency of the source we said it was a hundred hertz right so what would be the perceived frequency okay so once again we're going to use our formula so fl is equals to v plus minus vl nothing changes you still use the same formula right so v plus minus vs multiplied by fs now all right now in this case what do we expect we anticipate that when the source is moving away from the listener right what will happen you remember the story of the bee right we anticipate that when the source is moving away the frequency that will be perceived by the listener will now be less than the frequency that is emitted by the source agree no right so typically this is what's going to happen so we know the listener is still stationary right so this is going to be 340 divided by once again uh okay i just wanted to say v first okay sorry about that so let's say fl is equals to v over v and v s times f s now we want to have a result that is smaller okay in this case so what type of fraction should this be i think you've got it of course it's going to be a proper fraction so it means that the numerator should be less than the denominator or maybe let's put it in case of the denominator so the denominator should be bigger than the numerator so how do we make it bigger i i know you've got it of course you put a plus i think once we look at it that way it always makes things a little bit easier right so and let's substitute so we're going to have 340 so this is 340 plus what is our velocity speed of the source that's 20 multiplied by okay the frequency of the source we said it's a hundred okay and then what should be the speed uh the frequency rather of the listener so once again we're going to do the same thing so that's 340. uh yeah that's 340 uh divided by 340 plus 20 right multiplied by 100 and what do we get ah can you see so we get 94.44 hertz so the frequency that is perceived in this case remember the source is emitting 100 hertz but the listener is perceiving it appears to him as though that frequency is actually 94.44 hertz i i hope that you can um you you understand this analogy of uh using the uh the the the fractions you know proper and improper fractions okay now with that said i think um what i wanted to illustrate to you is quite clear so i think without any further ado ladies and gents let's do this i want us to take past exam questions right i've taken these from you know booklets and then um but they are actually from past exam questions so as a result what we're going to do is we're going to look at them and try to solve them together is that okay right and in the process what i'm simply going to do is i'm going to be explaining some of the things that we will require okay to solve the problems or some of the questions um particularly when it comes to issues about uh you know the expanding universe the red shift and and all of that if you don't mind i'm not gonna explain that now but i'll explain it as i as i go on ahead okay right all right so uh here's a question right it's from uh past exam question paper okay so um what we want to do is try to solve this question uh and it says the siren of a stationary police car sound waves of wavelength 0.55 now notice they are not talking about frequency this time they are talking about the wavelength okay right so they've given us the wavelength and then they say with its siren on the police car now approaches a stationary listener you remember i said to you there are two types of scenarios right so we've got a moving source here and a stationary listener so they say um [Music] now approach is a stationary listener at a constant velocity on a straight road assume that the speed of sound you remember the speed of sound our v value right assume that the speed of sound in air is 3 45 meters per second now all right now let's try and look at our uh scenario right so we said first of all okay they said a siren of a stationary police car emits a sound wave of 0.55 now so initially um they told us that the car was stationary but then later on they say with its siren on the police car now approaches a stationary listener so here's our police car you know my fancy cars okay all right so here's our police car and now it's moving towards this stationary listener so i already know vl is zero right because they told me that the listener is stationary right so we know the chi is approaching okay right and then remember they also told us that the speed of sound okay in a is 345 okay now they ask us this question will the wavelength of the sound observed by the listener be greater than smaller than or equal to 0.55 now ladies and gents please i don't want you to make a mistake of this they are not asking about the frequency now if they were asking about frequency we know that the frequency that will be perceived because the ambulance is moving towards this person what would happen to the um to the waves that become more compressed that be actually much closer together than on this side where the car is actually moving away from right so in this case what do we know we simply know that the frequency would be higher but remember what about the wavelength the wavelength is the distance between two crests of a wave right so in this case the higher the frequency the smaller the wavelength i hope that makes sense to you right so they are asking uh towards the the listener will the wavelength of the sound observed by the listener be greater than smaller than or equal to 0.55 well this guy's emitting 0.55 wavelength so definitely the wavelength here towards the listener would actually be would actually be smaller than okay so the answer for 2.1 would be smaller than i hope that that makes sense uh to us okay then they say name the phenomenon observed in question 2.1 of course i think you already know that that's the doppler effect right okay now they say to us calculate the frequency of the sound waves observed by the listener if the car approaches him at a speed of 120 so we know them the car would be moving at 120 kilometers per hour all right please just be careful of that we want typically our speed to be in meters per second isn't it so if i were you um the first thing i would start doing is that i would convert that speed uh to meters per second okay so how do you convert that now i'm gonna show you the longer way and then i'm going to show you the the shorter way of doing it so kilometers right so in this case what would be the speed it means that my speed would be 120 but i want to change those kilo kilo is a thousand right so a thousand meters so it's multiplied by 1000 right per hour now how many seconds remember i want it to be meters per second how many seconds are there in an hour well there are 60 seconds in a minute okay and there are 60 minutes in an hour so that's 60 times 60 isn't it right so in this case and it's divided by hour remember it's per hour meaning it's divided by hours right so in this case it's divided by 3600 where does this 3600 come from from 60 times 60 60 seconds all right in a minute okay and 60 minutes in an hour so this would be 3600 so um let's see what you get there so this would be 120 multiplied by 1000 divided by 3600 and you get 33.33 okay now remember we've now converted it to meters per second okay right um i said i wanted to show you the easier way of doing this okay right to convert from kilometers per hour uh to meters per second right now think about this if i said um a thousand going it goes into itself once okay um and it goes into six 3600 how many times three comma six times so you can just simply say it's 120 divided by three comma six so whenever you want to convert from kilometers per hour to meters per second you just divide by three comma six okay and i promise you you still get to the very same answer which is 120 divided by 3.6 and we get okay that 33.33 okay right just a a nice little shortcut okay hopefully it helps someone right so now we know that the car the police car was moving at 33.33 meters per second so we don't need to use that 120 kilometers per hour anymore right and what do we want to get we want to find out um what is the frequency of the listener now here's another issue we did not have frequency we're not given um the the the frequency of the source but remember what they gave us they gave us the wavelength you remember that they told us that the wavelength is what 0.55 okay so are we able to get frequency from wavelength remember we had that formula about um v being equals to f multiplied by lambda right and if you remember what did i say this v stands for this is the speed of sound in a medium okay so remember this v here that we calculated this was the speed of the source so let's maybe let's just write it as vs if you don't mind so this v here right this v here that we use in this formula we said this should be the speed of sound in a medium right and what did they say to us the speed of sound is 345 okay so this would be 345 okay we want the frequency and we were given the wavelength as 0.55 okay right so what would i do divide both sides by 0.55 okay okay so i would say this is uh 345 divided by 0.55 and typically i get an answer of the frequency is 627.27 hertz now what is this this is the frequency of the source okay remember our police car they told us that it emits uh a sound waves with a wavelength of 0.55 so essentially what did we just calculate here we just calculated the frequency of the source okay right now let's apply the doppler effect okay now they wanted us to calculate the frequency of sound observed so we want to find out what will be the frequency that will be observed by this guy who is our listener right so what will they will be the frequency that is observed by that guy there right now um let's try and tackle that question okay so now we're going to say now we want the frequency of the listeners in it so we're going to say the frequency of the listener is equal to okay right so now we're saying that's v plus minus v l v plus minus vs times fs okay right i'm just going to continue over this side here okay so we are not out of the shot of our camera right so now what are we going to say vl what do we know we know that our listener is stationary so that's going to be v so vl is zero okay so we just left with v there and then at the bottom we've got v and v s now please i want us to note again multiplied by f s right so now i know that the frequency of the listener should be what should be higher than the frequency of the source isn't it okay so the frequency of the listener should be higher than the frequency of the source so now what am i going to say ah so what type of a a a fraction should this be if i want this one to be higher you are quite right it's going to be an improper fraction right so it means that the denominator must be smaller than the numerator so to make it smaller i'll just put a a minus there okay so this is going to be 345 you remember the speed of sound in this particular question they said 345 okay divided by this is 345 minus what is the speed of the source you're quite right this is going to be 33.33 okay multiplied by what is the frequency of our source okay the frequency of our source we said that 620 7.27 okay and all that's left for us to do okay let me just say multiplied by okay that's 345 divided by 345 minus 33.33 okay and i get an answer of 694.35 okay of course um if you had rounded this answer off um you might get just a slightly different answer but i mean it would be more or less the the same thing okay right i hope that makes sense to everyone okay so there were two things that you needed to take heat off in this question first of all it was that calculation the conversion of kilometers per second of kilometers per hour rather to meters per second and secondly it was that calculation of uh the frequency okay using this equation here all right so please don't think that it was um an atypical question right uh all of those by the way all of those formula you are you are actually given um besides the one that uh has to do with the frequency uh uh rather the the conversion of kilometers per hour to meters per second you just need to know that okay right and then the very last question they say how will the answer in question 2.3 change okay how will our answer in 2.3 change if the police car moves away from the listener at 120 kilometers per hour all right so what will happen to our frequency if the uh source moves away of course it should decrease isn't it right so uh obviously the answer there should be decrease okay super um i hope that answer makes perfect sense or the way that we tackle that question made sense all right so what we're going to do is let's take just another question all right um so that we just explore a little bit further and how the doppler effect is applied all right just to save a little bit of time um yeah i'm just going to do this question on the same sheet um so they say to us and here's the thing that i sometimes don't like about physics it's that you know it's it's embroidered with so many words and you know if you you're not a first language speaker of english sometimes that's what may make it seem as though it is a difficult subject but nonetheless i want us to always be mindful that it's not a difficult subject at all i think i've tried to prove that in my videos that you can actually do any question on physics now and i hope it's helping right now let's look at this it says the cyrena first stationary ambulance emits a note of frequency okay now note it's the siren of uh the ambulance is stationary right emits a note of frequency one one three zero heads when the ambulance moves at a constant speed now what happens the ambulance starts moving right so uh when it moves at a constant speed a stationary observer do you see so it's a moving source yet again and a stationary observer right so a stationary observer detects a frequency that is 70 hertz higher than can you see that higher than the emitted that emitted by the siren okay so typically um so here again we've got uh here's our ambulance it's going to emit a frequency okay and then um they say when a stationary observer detects okay and the siren it's 70 hertz higher than now when is the frequency higher than the one that is emitted of course this is when the source okay because the source is the one that's moving this time when the source is actually moving towards the listener so in this case here it is once again right so you've got a moving source you've got a stationary listener and the frequency that is detected okay that this guy here so they're telling us the frequency of the listener is 70 hertz higher all right so they had given us the frequency of the source okay so the frequency of the source they had said remember it's one one three zero heads right okay and then they telling us that this guy who's listening he has a higher frequency than the one of the source and his frequency is one one three zero plus it's higher than okay one one three zero plus seventy right obviously uh that should give us what a thousand two hundred hertz okay so that's what he hears okay um and then they say state the doppler effect in words obviously we've spoken about that earlier on and we said it is the apparent change in the frequency as perceived by the source right uh if the source and the listener are moving relative to the speed of sound propagation right so you must be able to to state that um and then they say is the ambulance moving towards or away from the observer okay right so so so we've already concluded that obviously if the perceived frequency is higher than the one that is emitted so it must be that the ambulance is actually moving towards the observer they say give a reason for your answer okay right um what we've just stated right the perceived frequency is higher than the um the the frequency of the source so it means that the the wavelength of the emitted or of the passive sound waves right uh is actually much shorter okay right so and then they say to us calculate the speed at which the ambulance is traveling so what are they looking for they want the speed of the source okay right and i want you to please remember they did say that our stage our listener is stationary okay right so we want to find out the speed of the ambulance so again we're going to use our equation right fl is equals to v plus minus vl okay v plus minus vs times fs okay right so let's substitute we know fl okay before we substitute everything right we know vl is zero so that's going to be v over v and v s m n times f s now ladies and gents we know the answer is higher than so what type of uh fraction should this be and you're quite right this is going to be an improper fraction to make it improper it means that the denominator must be less so in this case it's going to be minus okay right now let's substitute all that we know so this is going to be okay i'm just going to lift this a little bit higher uh if you don't mind okay uh yeah so that it's within range so in this case it's going to be fl right that's a thousand two hundred okay all right they told us that the speed of sound take the speed of sound is three four three right so this is going to be three four three over three four three minus vs which is what we're looking for isn't it and multiplied by the frequency of of the source okay that was one one three zero okay right now i know many of you tend to panic when you get to this i promise you ladies and gents you see when you've done what i've done here you already have got in your maximum max all that's left now is just you know your mathematical gymnastics you know some of you are laughing at me when i call them mathematical gymnastics um so all that is left in in in all honesty is just the mathematics okay so in this case um let's make it as simple as possible for ourselves right i'm going to say um okay let's do this let's divide both sides by one one three zero okay so that i get rid of that okay so this is one one three zero so what i did on the left on the right i'm also doing on the left and then if you want to you can cross multiply right so i'm going to say um okay now in fact let's just make it as easy as possible i'm going to say 1200 okay divide by 1130 okay 1 3 0 and that gives me 1.062 okay so in essence i've got 1.062 okay which is equal to 3 4 3 all over [Music] 3 4 3 minus v s now please remember all i just simply did was just uh to cancel that out so you can take that as um divide by one right so now let's cross multiply this multiplied by one okay i'm just going to continue over to that side okay so three four three times one i'm cross multiplying now okay three four three times one that would give me three four three okay which is equal to okay one point zero six two okay into three four three minus v s okay now note this is just uh sorry i tried to squeeze that in there okay uh i hope you can still see it right maybe let me just rewrite it 343 is equals to 1.062 into 343 minus vs okay still within our brackets there so if you want to you can just multiply those in so that's three four three okay 1.062 fortunately i didn't erase that of my calculator i can just multiply that by 343 okay i get three six four point two five okay right minus okay remember it multiplies in here it multiplies in there so that's minus 1.06262 times vs okay please don't forget that okay right if i take that to the other side what do i have 343 minus 364. so that's minus uh 343 there okay you should be able to get minus 21.25 okay which is equal to right remember you still had minus 1.0 vs okay right i really went for you know these questions that um i know would give you guys a little bit of a headache right so what do i do divide both sides by 1.062 okay but what i do on the left i do on the right or the other way around so that cancels with that and your vs value right you have 21.25 divided by one point uh sorry um one point six okay right and i seem to get an answer around 20 meters per second okay right you can try it out of course um you know i played around with uh mathematics a little bit uh and what it did is that um it gave me 20.0 something uh some of you probably may get something a little bit less than that 19.9 something okay nothing wrong with that it would be more or less in the same ballpark okay right okay that's out of five marks and please note in all honesty in all honesty if i can just quickly show you you know this is how of course we'd mark the correct use of the equation would mark your correct substitution there your correct substitution there and all that we've done actually is just worth one mark okay because typically this is not a maths exam okay so we're not really going to um you know be fussy on the mathematics side of things okay right now let's talk very quickly the last question they say the study of the spectral lines obtained from various stars can provide valuable information about the movement of the stars the two diagrams below represent different spectral lines of an element diagram one represents the spectrum of an element in a laboratory on earth so they're taking an element on earth and then diagram two represents the spectrum of the same element from a distant star now ladies and gents without necessarily wanting to go too much into the details right now what happens is that when you take a certain element on earth it gives you a specific absorption spectrum okay um so these lines that you are actually observing here would be absorption spectras right as i said i don't want to go too much into the details unless i make this lesson what it's not about okay so now there are two things that happen now i want you to note when we talk about the visible spectrum we talk about this roiji beef red orange yellow green blue indigo and violet now when you talk about the visible spectrum when we go in the direction from red red has got the shortest uh or rather the the longest wavelength meaning that the frequency of red light is actually um the frequency is the smallest okay so frequency of raid is the smallest and as you go towards violet it increases going towards violet so all i'm simply saying is that frequency here okay so if we think of it in terms of frequency so it's smaller on the red side okay and it increases as you go towards the violet side okay right so it increases as you go towards violet so now what we're simply saying is that if you notice spectral lines okay which is how basically they would determine whether a star is moving away or or towards the end so they observe the spectral lines of a particular element on earth and they observe the spectral lines of the same element uh from that star now if you notice here here's this the element on earth here's the spectral lines that you observe there now when you observe it from the distant star look at what has happened the lines can you see this line that was appearing here has shifted but where has it shifted to it has shifted towards the blue okay so it's shifted more towards the blue so they put blue on this side they put red on that side okay right so now it's shifted more towards the blue and now if you remember what i said i said when we go towards the violet the blue side right so we know that the frequency is increasing in that direction okay right now if you think about the uh the doppler effect when do we observe a higher frequency when something is moving towards us right so this is simply telling us when you notice it's moving towards the higher frequencies which is the blue frequencies right so therefore this is telling us that particular star must be moving towards earth all right so in this case why how do i know that uh because it's uh the spectral lines are actually moving to the blue side on the other hand ladies and gents obviously there are instances and uh more prevalent rather is that we'll find instances where when we observe the spectral lines of an element from another star uh it would actually be showing us that it's moving more towards the red okay they call that red shift by the way this one they we call it a blue shift when it's moving towards the blue now if it's moving towards red remember the frequency of red is a smaller frequency right so what is that telling us it's telling us that that particular star is actually moving away from the earth all right and by the way when they study and most of these stars they realize that um you know the the the spectral lines are actually more shifting towards the red and when they observe um their spectral lines so as a result this is what has given rise to their theory that actually the universe is expanding all right so when you think about redshift always i want you to think about an expanding universe okay right so um i wanted to to to explain that okay so they say is the star moving towards or away from the earth in this case it's blue shift so it must be moving towards the edge and they say explain by referring to the shift in the spectral lines in the two diagrams okay and then you'll explain that it's moving towards uh the blue which has a higher frequency right and that's showing us that um the the the star is actually moving towards earth okay right um i hope that kind of makes sense now what i want us to do very quickly is just to tackle the last question and then i think i'm going to leave it there okay in terms of the doppler effect all right um quickly so they say the data below was obtained during an investigation into the relationship between different velocities of a moving sound source and the frequencies detected by a stationary listener once again ladies and gents sources moving stationary and the listener is stationary for each velocity uh the effect of wind was ignored in this investigation right so here's our experiment so in experiment one they had a velocity of zero right and for the source so and typically it means both the source and the listener were both stationary can you see that now if you remember what i said at the beginning so if both the source and the listener are stationary right so here it is so here's the source stationary okay if you remember b and the listener is also stationary so whatever the source emits all right the listener perceives so this would be the only time when both of them are stationary it means the frequency of the source would be equal to the frequency of the listener are you with me now right now notice when that was the case the frequency was 900 the frequency of the listener was 900. now i want you to think about that okay so if this guy hears 900 when this guy is stationary then what is this guy emitting it means he's emitting a source of sound at 900 hertz okay right i hope that makes sense right so now look at what happens all of a sudden when the uh when the source starts moving in experiment two the source starts moving at ten meters per second look at the frequency now all of a sudden it begins to decrease okay so now the frequency of the detected frequency is smaller than the frequency that is emitted what does that already tell you the frequency of the listener is smaller than the frequency of the source definitely for me it's telling me that the source must be moving away remember listener is stationary the source is moving away from the listener okay right now um they say write down the dependent variable for this investigation okay so remember what depends on what it's the frequency of the source that depends on the speed of the uh so it's the frequency of the listener that depends on the speed of the source the more i change the speed the more the frequency changes okay so our independent variable what we keep changing in the experiment what is it that we kept changing it's the speed of the source right the velocity of the source but what changed as a result that's the dependent variable what changed as a result is our is the speed of the i mean it's the frequency of the um of the listener okay right i hope that makes sense right they say write down okay we've answered that they say state the doppler effect in words okay i know that you already can do that okay so i want to dwell on that now look at this was the sound source moving towards or away from the listener all right we know definitely now that the source was moving away why is that because the detected frequency is less than the frequency that is emitted by the source okay and they say use the information in the table to calculate the speed of sound during the investigation okay right oh my goodness i i really hope if there are any teachers that are watching this video please just send a shout out because i really want this to not only help our students but i'm hoping that it will also help us to teach this subject better okay to find better ways of explaining because sometimes we can make physics difficult unnecessarily so so please just send a shout out you can send me an email and tell me how you finding this lesson okay right they say use the information in the table uh to calculate the speed of the sound during the investigation okay now they want us to find out what the speed of the sound is now in this case you can use uh any of the data there now so we know the frequency of the source right so frequency of the source is 900 okay um let's take experiment two if you don't mind right let's say so the frequency of my listener when is 874 right but this is when the source is moving at 10 meters per second so the speed of the source would be 10 meters per second and remember it's moving away right so so if the sauce was moving in that direction then i know there's my listener there okay uh he's hearing something that is less than the sauce what the sauce is emitting okay right and then um right let's use that information there so this will be f l that's v over v plus minus v l okay just want to get s times f s and then uh what's the frequency of the listener okay so we know that's 874 right now our listener is stationary so this is going to be v we don't know what v is that's what they wanted us to investigate or to to find out so divided by v and v s times f s now i've already substituted there so that's 874 is equal to v now i want for the answer to be less than the the the you know the frequency of the source so what type of fraction should this be okay that should be a proper fraction isn't it so proper fraction the denominator is bigger than the numerator so definitely this will be a plus okay so this is v over v plus now in this case what is v s we said that's the velocity of the source we took the instance where it's 10 someone could be asking yeah but what if i took the one experiment three it really doesn't matter we'll get the same answer right right multiplied by the frequency of the source if you remember there we said it's 900 okay so all that we simply going to have here ladies and gents is just a mathematical gymnastics right so um okay let's do this so we can say this is equal to okay that's 900 times v that's 900 v over v plus 10 okay which is equal to 874. now please remember this is over one isn't it okay right i'm just going to lift that a little bit so that you can see it okay so uh we can cross multiply 900 v times one that's 900 v okay and 874 into v plus 10 this is going to be 874 v plus remember it's this times that this times that okay so this is going to be eight seven four zero okay take that to the other side 900 minus 874 v right and this will give us 26 v okay which is equals to 8 7 4 0 we can divide both sides by 26 okay and our answer is v okay let me write it here so our answer is v um okay so that's eight seven four zero divided by 26 and we get 336 so the speed of sound we get that to be 336.15 please if you can verify that for me okay i'm just doing this in a rush now because uh i want i didn't want this video to be this long and then they say the spectral lines of a distant star are shifted towards the longest wavelength of light so the longer wave length of light so it means that's towards the red isn't it they say is the star moving towards or away from the earth so definitely if it's a red shift this is telling us that um it is moving away from the earth and it's moving towards the longer wavelength okay great stuff um ladies and gents uh i know i promise that this is the last question but i got a question from one of the students um on on on something that they were that they were doing um so i want us to quickly look at that question i'm not going to go into the details of the question uh but i want us to just quickly look at the approach of it um one of my students had asked me and i i thought maybe it would be of benefit to you if you were to look at that question so i'm just going to go through that question with you now all right so the question had to do with um a scenario where there was an ambulance um which is a source that is moving but we didn't know what the frequency of the ambulance is nor do we know what the velocity the speed of the ambulance is the frequency of the source and then they told us that when the ambulance i'm just going to make an example when the ambulance is moving uh towards a stationary listener so we had a stationary listener the listener perceives let's say a frequency of 170 hertz right and then they say the same ambulance now passes the listener and then they find out still it's moving at this at a constant speed and then they find out that uh when it passes uh the frequency of the listener now changes to 130 hertz right so now the question was calculate the speed of the source as well as the frequency of the source i just wanted to show you how to tackle that question all right so now what you're going to have is a simultaneous equation okay so now what i'm simply going to do is i'm going to have an equation okay just a separate equation on this side okay i'm going to say all right for scenario number one remember when the frequency of the listener or rather when the source is moving towards the list i know the frequency should be higher than okay right uh so that's our frequency day so in this case i'm going to say f l it's v plus minus v l over v plus minus v s times f s okay so now i'm going to have f l is equal to i know my listener is stationary right that's vvs times fs once again i know it's moving towards so in this case i know that um the answer that i'm expecting here should be higher so the perceived frequency should be higher than the emitted frequency so what type of a fraction should this be right once again we know that's an improper fraction right so it means that the denominator must be smaller than the numerator so we know our frequency of the listener that's 170 okay and then our v value oh sorry um sorry v was given as 340 meters per second that's the speed of sound in a medium right so this is going to be 340 divided by 340 minus vs remember what we didn't know was vs the speed of the source as well as the frequency of the source okay so what i said is that um i don't want to actually bother myself much with the mathematics here so what i did was 340 times fs this will give me 340 fs right divided by 340 minus vs and this is equal to 170 okay right now in this case we can cross multiply okay this is over one so this is 340 fs is equal to okay sorry about that so that's 340 fs which is equals to 170 over 3 times 340 minus vs okay so i left it as this equation here and i call this equation one and then i went to the second scenario right where i said okay i need to do the exact same thing but now the frequency of the listener it's going to be v over v plus v s remember now i'm going to have a an improper rather proper fraction here right times fs right because it's moving away i know i want that frequency to be smaller than so in this case it's going to be a proper fraction right so now i've got 130 which is equals to v that's 340 over 340 plus vs we didn't know what vs is remember it's the same one as the side multiplied by fs okay so again i'm going to do the very same thing i'm going to say okay this is over 1 so that's 340 fs times 1 so now i'll have 340fs okay which is equal to 130 into 340 plus vs okay so i'm gonna call that equation 2. now if you notice on both equations i've got 340fs on the left hand side so if their left hand sides are the same then obviously it means the right hand sides should be the same so what does that say it means that equation one is equals to equation two right so 170 into 340 minus vs should be equal to 130 right which is the right hand side here into 340 plus vs okay right and then what you can simply do is just uh say divide by 130 just want to make the mats easier for myself here okay so i've got 170 divided by 130 okay you can put that as 17 over 13 uh actually in fact let me do exactly that okay into 340 minus vs okay which is equal to now what you have on the side that's 340 plus vs okay right and then um ladies and gents i think uh all that's left for us to do here is just the mathematics so that's 17 over 13 times 340 times 340 uh and i get um this is four four four point six two minus seventeen over thirteen vs which is equals to 340 plus vs okay right i'm taking it in detail so that um you'll see what i did there right so i group all the like terms together so triple four six two minus 340 when i bring it over to this side minus 340 um so i've got 104.62 okay which is equals to now if i take this to the other side you see it becomes positive right and i've got vs there so i've got 17 over 13 right plus a vs so i've got 17 over 13 vs plus vs now remember there's a one here right so uh i'm going to say 17 over 13 17 over 13 okay plus one okay um i get 30 over 13 vs if you wanted to keep this as a fraction uh there's nothing wrong so that's 10 104 points and then to get rid of this fraction what do you do you just multiply by the inverse of it okay so that that cancels that but what you do the side plus 13 over 30 okay so vs um i'm just going to write it on this side so vs would be equals to 1 18 over 30 so you've got 104.62 multiplied by 13 over 30 okay i got a value of 45.334 sorry that's in meters per second sorry i'm out of space there okay and then how do you get the frequency all you simply do is that you're going to substitute that answer into any of the two um uh so you can either take this one here okay uh so i'm just simply going to say 340 fs is equal to [Music] 170 into 340 minus 45 point okay right and then all you simply do okay so that's 170 into sorry 340 minus 45.34 okay and then you divide all of that by 340 ladies and gents you can do that for yourself at home i seem to get a frequency of 147.33 heads and now please i want you to note um this frequency here if we compare it to what we had uh in that question there i knew that the frequency of the source the perceived frequency should be higher than um the frequency of the source when it's moving towards and look at this definitely 170 is greater than 147 and i know that the frequency of the listener should be smaller than or we should be lower than the frequency of the source when it's moving away and um definitely 147 is greater than 130 so the frequency that the listener hears is much smaller than that of the of of the of of the source uh so it this is how you would tackle this type of a question and i hope that has helped um even to those who had questions so ladies and gents this is all there is to the doppler effect i hope this video helped please you are more than welcome to send those questions through and um i still do advice if you need to get in touch with me my email but otherwise you can throw a comment [Music] as i said if you're a teacher and you find this material helpful please just also let me know how we can collaborate together to try to make our nation a better nation but otherwise i'd like to leave it here and thank you so much for um and listening thank you remember to subscribe and i'll see you when we do another lesson next shop