 It is fine. We understand that. So you can take the chemon test. If suppose there is a gap to like, you know, many days gaps are there between two exams, then you can take that time of the chemon test. But then we are not able to find any other appropriate time, because if we prepone the exam, then probably syllabus will not get over and we can't postpone because your school exams are there. So that's the reason why we have to start from 31st of January and 31st of January, many of you are writing KVPY also, right? So I think almost all of you are writing KVPY, right? So all the best for that also. So probably you will not be able to take the 31st chemon test as well. That also you can take later. We are flexible this time round. Don't worry much about the scheduled time of the chemon test and day. You can take it as per whenever you get time. But do take it, okay? This is the full class 11 syllabus test. Alright guys, so we can start now. We, you know, initially I thought that I'll be doing problem solving on rigid body dynamics, but then later I realized that there is a small chunk of the chapter which is, which we have to cover that is beats and Doppler's effect. So that will take around two hours to complete itself. So I thought that remaining one hour rather than starting something new entire class. Let us do waves only. We'll be completing the leftover portion and we'll do some problem practice from the same chapter waves. Okay. And then whatever classes are there, we can do some problem practice on let's say like chapters like rigid body or fluids. Okay. Alright, so without wasting any more time. Let me start today. We'll start with a topic called beats. Okay. Write down beat. So when you hear beats, what comes in your mind? Musical beats, drums. Okay. Okay, let me tell you, this is not that beats, which you think. Alright, this is the outcome. I'll show you, I'll show you exactly how you can hear the beats, which I'm talking about. First, all of you write down beats is the outcome of superposition, superposition of two waves, two waves having different frequencies. Okay. Till now, we were superposing waves. When they were traveling the same direction. They were giving another traveling wave. When two waves are traveling in opposite direction, they were creating standing wave. Right. And both the cases, the frequencies of both the waves were same. Okay. But this is the first time when frequency we are varying. And this will result in beats. So let me just show you exactly how it sounds like when you hear beats. See, all of you listen to this. This phenomenon is used. Can you hear this? Are you able to hear it? Okay. Great. So all of you listen to it carefully. Now, let's get to tune musical instruments. In an orchestra, for example, the oboe will give an A, and then all the other instruments in the orchestra will tune their instrument to the oboe. It's effective if the beats, if you can actually hear the beats. And to get beats you need the sound of the reference instruments, say the oboe, and the sound of the instrument being tuned to be roughly the same volume. And you get a loud, soft pattern from the combined effects of the two frequencies being played simultaneously, the oboe and the violin or the oboe and the French horn or whatever. So I'm going to demonstrate that with these tuning forks. They're supposed to be identical, but they haven't been quite constructed identically. And in order to tune them, actually previous to the demo here, I found the location of this movable piece of weight so that the two tuning forks come out with about the same frequency. So when they're tuned, you can hear a nice steady tone that doesn't vary in loudness. Right now, when he is beating both the tuning forks, the frequency from both are same. So same frequencies sound waves were superposing. Now he is now changing the frequency of one of the tuning forks. So now see what will happen. See, he just moved one part of structure upward. So structures, this order structure changed. So it's natural frequency also changed. So automatically the frequency that will come out from tuning fork also changes a little bit. Then you get the following phenomena. Do it again. The farther these are from being tuned, the faster the wah wah pattern is. So these are even further out of tune. What I'm doing by moving this, this weight is lowering the frequency of this tuning fork. And the other one is staying the same. You should be able to hear what really fast and so you can see that now you understand what is beats. Is it clear. Okay, so what is the use of that. What is the use of that in. have you seen when let's say someone comes to perform in the stage initially they try to tune their musical instruments let's say a drummer will come and along with it there will be some other musical instruments will be there so they will spend five ten minutes just tuning their musical instruments have you seen that have you observed that right so when they do that they're trying to eliminate the beats otherwise you'll not hear a very nice sound the different frequencies will come from different musical instruments they will superpose and you will not get a comforting music so they will try to eliminate the beats initially for first five ten minutes that is what they do when they say they are tuning the musical instruments okay fine so this is the use of it that it is something which is unwanted you want to eliminate it okay there is no use of it you want to eliminate it actually all right so now this is what it is now let us see mathematically how we can analyze it right in our textbook in our ncrt textbook they have taken two waves which are cosine waves for some reason okay they should have taken sine wave itself but since they have taken cos wave i'm also taking cos and cos and sine both are same only just pi by two phase differences there so this is one of the wave and second wave is let's say a cos kx minus omega 2t these are the two waves having same amplitude but their frequencies are different okay now i am superposing these two waves all right now it is my coordinate system that we can pick we can choose origin all right so what i'm doing is that i am superposing i am superposing both waves at origin so what i'll get i'll get y1 plus y2 so y is equal to y1 plus y2 when i superpose at origin x is equal to 0 so it'll be a cos of omega 1t plus a cos of omega 2t cos of minus theta is cos theta all of you able to understand this right this is a superposition now when you add them up you'll get 2a cos of omega 1 plus omega 2 by 2 into t multiplied by cos of omega 1 minus omega 2 by 2 into t okay now all of you can see that you'll get a function like this which has two time functions what i mean is that it has multiplication of this with this okay both are periodic functions right both are cosine functions all right now which one is oscillating faster can you tell me which one is having higher frequency the red one or the orange one all of you which one is oscillating faster the red one or the orange one the red one of course it has higher frequency right omega 1 plus omega 2 by 2 into t okay now if frequencies are high all right if frequencies are high then we'll not be able to hear the fluctuations fine for example when i'm talking to you all right i have certain frequency of my voice right but then you're not able to hear that fluctuation the actual frequency let's say i'm talking to you in a frequency of i'm not sure but maybe let's say 100 hertz so you're not able to hear my voice going up and down 100 times in one second right so if frequency is high you'll not be able to hear the fluctuation okay so basically all the fluctuation is happening because of the red one also but we as humans humans won't be able to hear the fluctuation of write down fluctuations of cos of omega 1 plus omega 2 by 2 into t simply because it is too rapid it's very very fast okay but that second term is fluctuating very slowly because even if let's say omega 1 is very high omega 2 is also very high and if omega 1 and omega 2 are little closer to each other when you subtract it it becomes very less okay so this one we can hear okay so this one we can hear as the frequency is supposedly less frequency is much lesser okay right and the function which you have here is a multiplication of these two with 2a okay so let me just show you graphically how does it look like so y is equal to 2a cos of omega 1 plus omega 2 by 2 into t and cos of this okay now since it is multiplication of these two functions if any of them becomes zero the y will become zero all right so it is difficult to plot this mathematically so I'll just show you how does it look like then it will make more sense to you see this is how it is this don't copy right now drawing an envelope and then this is what this is the plot for the above function okay so the rapidly you can see there are two frequencies one frequency is that rapid fluctuation it comes down and comes up quickly and second fluctuation is the fluctuation of the amplitude you can see overall there is an envelope this is the fluctuation which I am talking about you will be able to hear the fluctuation that is captured on the dotted line okay this fluctuation is because of this and that blue one which is rapid fluctuation is because of this term okay we as humans will not be able to hear this blue fluctuation we will feel as if the fluctuation is this only the white one okay we'll be able to hear only this okay so our ears will filter out now we are talking about something we hear only so we will ignore this term so beat frequency is related to what we as humans can hear so if I talk about beat frequency which we can hear this one right cos of omega one minus omega two by two into t so what is the frequency of this term everyone what is the omega for that term it is omega one minus omega two by two all of you fine right this is the omega for this term which we are able to hear okay so now one more thing you need to understand is that when fluctuation happens like this this is one complete wavelength okay in one complete wavelength once it goes to this extreme and then again it goes to that extreme okay this is plus a this is minus a if I am talking about what I can hear I will be able to hear plus a and minus a exactly the same way because I am not hearing amplitude I am hearing intensity and intensity is proportional to amplitude square do you all understand that in one cycle I will be able to hear two loudness plus a and minus a type in quickly see I am telling you things in detail so that you better understand otherwise in book it is not in so much detail let me know if it is clear type in quickly fine so if the frequency of this function is omega one minus omega two by two the beat frequency the frequency with which I will hear the loudness will be two times of that it will be two times of omega one minus omega two by two okay I didn't understand hearing the intensity part see what we hear is energy okay some sort of energy goes in our ears and that will convert into electrical signal and that goes in our brain okay so the energy is proportional to amplitude square okay so whether the fluctuation is above or below for energy it doesn't matter when let's say for example the some spring is there half kx square the energy remains half kx square whether x is plus two or minus two okay it will be same similarly when the oscillation is happening the energy when you are at plus a or minus a will be same that is what I am saying you will hear two loudness in one cycle clear so beat frequency is this so in our textbook they have written directly all right in our textbook they have written directly that is the beat frequency okay why not others you will be able to hear this also why not but beat frequency is defined it is defined as how frequently you hear the loudness loudness will be distinct getting it for loudness will come like that and then go down and then again will come so you will see that okay what is the difference between two loudness okay so this is the angular beat frequency the actual beat frequency will be equal to f1 minus f2 so if you are interested in only problem solving then beat frequency of two sources having frequency f1 and f2 is f1 minus f2 pretty simple okay to solve problem you just need to remember that okay now this is clear I suppose beat frequency so now tell me one thing that if frequency of one the frequency of first source is 10 hertz okay and you are getting a beat frequency you are getting a beat frequency of second source of four hertz then f2 should be what all of you what should be the value of f2 to get a beat frequency of four hertz okay good hurry around it can be 14 right f2 can be 14 as well as six all right the difference should be equal to four pretty simple all right so this is how the problems from this chap from this topic would be in fact we'll see there are certain varieties wherein you club with the other topics to with this concept okay but then standalone concept if there is a question it will look like this only just find a difference and you'll get a beat frequency okay fine guys so we are now moving towards the last topic of this chapter and probably the most important one because I have seen that if there is a question from this chapter there is a very high chance that it will come from this topic that we are going to discuss now and the the reason for it probably would be that you can mix mechanics with it so the variety become a bit larger okay the name of the topic is Doppler's effect do you know anything about Doppler's effect already ever heard of it no one no one has ever heard of it red shift okay good what else anybody else frequency is not absolute very good all right so have you ever wondered how radar works how sonar works it works under this principle so first I will write down what it is correct shashwat I mean not a hundred percent correct but then yes that is what it is write down it is the upper end it is upper end it appears it's not actual but it appears okay upper end change change of frequency because of the movement of the source or observer observer could be a detector it need not be a living being it can be just a machine component which is detecting something okay source is the source of sound or source of frequency all right so Doppler's effect tells us that you know if there is a movement of something which is generating sound or something which is capturing the sound then the frequency will change all right and there is a mathematical relation to how much frequency will change for how much movement or for what velocity how much frequency will change and that is one of the very very interesting discovery because you know write down there is there is a there is an equation or relation between velocities of source and observer with change in frequency okay so that is how radar works radar will just what radar will do it will try to capture the sound coming from the aircraft or radar will throw away its own vibration in the air it will hit the aircraft and then come towards it all right and then it will calculate how much frequency has been changed and because of that frequency change it will estimate where the aircraft is first and second with what velocity the aircraft is moving in its direction fine dolphin also do that to detect I'm not under present sure about dolphins but then yes I have heard that they they use the ultrasonic sound okay so this is what it is guys that it's I mean it's the it is the first practical thing in the chapter okay rest all we're just theory part of it rest all as in what is superposition what is beats what is sound what is the velocity of sound all those were introduction they were not let's say something which is directly useful it is like abcd of what is waves now we are getting into that okay fine now you know the basics of the waves let us see how we can use it and topless effect is one such example wherein you can use the property of the waves which property the upper end change in the frequency all right so we are going to derive a relation between how much how fast the source or observer should move to create a given amount of change in frequency all right and then we'll see some numerical steps out of it right and so there will be two cases case number one when source is moving observer is at rest case number two source is at rest observer is moving then you'll say case number three both are moving if both are moving will club case one and case two together and directly write the expression okay so we'll split the thing into these two sir but why does frequency change yeah we are trying to derive it so you will know why it change how it change because we are going to see entire derivation only okay what is frequency frequency is how many times you see the crests in a second or how many times you see the troughs in a second that changes okay suppose wave is coming towards you like that and you are also moving towards a wave suppose you are stationary then waves crests will hit you after let's say two seconds two crests will hit you in two seconds but if you are you are also moving against the wave then wave crests will hit you in less than two seconds don't you think so so you'll feel as if frequency is less understood okay case number one is the source is moving hold on aditya we we are going to derive it source is moving and observer is stationary all right since asking again and again I'll just quickly talk about it once suppose this is a wave okay this is coming towards you fine this is coming towards you you are stationary all right these two crests will come and hit your ears in let's say in some time okay now you are stationary right now then you also starts moving if you also start moving in this direction if you when you are stationary if these two crests one and two they're hitting you in two seconds then you are moving towards the wave so now these two crests will hit you sooner or later will they hit you in less than two seconds or more than two seconds what do you think if you also starts moving towards the wave did you get my question less than two seconds so you are observing the two peaks of the wave in less than two seconds but its time period is two second you are observing lesser time period so frequency is more understood but actually frequency is not more because of your movement it appears as if frequency is more all right but who cares whether it is appearing or it is the actual one all right what appears is more important than what actually it is because your detector will detect what appears to it fine so let's draw this one line and here is another line all of you be very this thing alert here because this is the derivation from your textbook nct textbook and this derivation I have seen that people find it tricky okay so this is let's say the source this is observer okay observer is stationary this this line represents observer it is stationary it is not moving but the source has a velocity of vs okay and speed of sound is c all right speed of sound is c this distance distance between source and the observer is l this distance is l c is the velocity of sound okay so let us say that after every time period t0 t0 is a time period with which I am throwing pulses from here okay now you will argue that here we should be throwing a wave okay whether I throw a wave or a pulse doesn't matter if time period is t0 after every time period t0 if I am throwing a pulse it is a wave only with time period of t0 okay just for the simplicity I am taking that way okay so what I am doing is I am throwing the pulses after every time t0 I am trying to find out after what what is the time period between which I get the two pulses here do you understand this all of you type in this understanding this thing is important before I start the derivation t0 let's say I am throwing after every five second I am throwing pulses from the source after every five second I am throwing pulses from the source so my time period is five second suppose I am receiving pulses after every four seconds so as an observer my time period is four seconds okay so I am trying to relate what is a time period with which I am throwing the pulses with whatever is the time period I am receiving the pulses getting it everyone type in I don't want to start the derivation before you guys understand this basic first type in quick Imancho, Karthik, Nikhil, Nishan, Pooja, Rehan, Vaibhav, Swaraj, Shreyas is it all clear what is the length here length is the distance between source and observer okay again I am repeating myself I am throwing disturbance from the source after every two seconds let's say after every two seconds I am throwing something from this side I want to see after what time interval and catching that disturbance from the source side okay so with whatever time interval I am catching it that is my time period whatever time period source is throwing the pulses that is a source's time period and frequency is one by time period only all right so it is better to understand the time period frequency will just be inverse of that okay fine enough of Gann now so one pulse is thrown from here can you tell me at it is thrown right now at let's say time right now is is zero at t equal to zero seconds a pulse is thrown from here when it reaches here what will be the time everyone what is the time when the pulse reached there with what velocity pulse is traveling with what velocity pulse is traveling everyone it's traveling with speed of sound this c c is given to you so something is traveling with c velocity distance is l so how much time it will take how much time don't you think it is l by c l by c only but the source is also moving with vs will that affect will the velocity of source affect the velocity of sound what do you think everyone source is moving and throwing the pulses so will the velocity of pulse correct velocity will depend only on the media it doesn't depend on it okay fine now i am throwing the next pulse t1 is what is t1 t0 after every t0 time period i am throwing pulses so at what time i will receive the second pulse here find out at what time i will receive the second pulse as an observer everyone when i'm okay when i'm throwing the second pulse don't you think that observer will move little bit sorry the source will move little bit this distance will be what this this distance will be vs into t0 okay so now tell me what time it will take what is the distance between the source and observer now source and observer the distance is l minus vt l minus vs t0 divided by see is that the answer is that the answer correct plus t0 no l by c plus t0 it's not that when pulse the first pulse the when first pulse reaches here second pulse is right behind it it's not that the second pulse has started when first pulse has reached are you getting it all of you okay so this is t1 now let's do one more exercise the the third pulse third pulse what is the time here tell me what is the time there 2 t0 right after every t0 i'm throwing pulses from there so why there is a confusion there are you getting it everyone keep it straight simple are you getting see the first pulse at equal to 0 second pulse at equal to t0 third pulse at equal to 2 t0 okay i am throwing pulses after every t seconds all right so now tell me what is the time here when the third pulse reaches all of you what's the time there it's more of kinematics question right do it yourself it is very easy for me to tell you directly but if you do it yourself you will remember okay here and got it others okay so now what is the distance between source and observer after 2 t0 the distance is l minus 2 vs t0 it will further move forward by 2 vs t0 c plus 2 t0 type in is everybody able to understand this the derivation is almost over now just go through it once ask me the doubts from source see source is continuously throwing out the pulses one after the other at equal to 0 at equal to t0 at equal to 2 t0 observer is receiving this at l by c this one at l minus vs t0 by c plus t0 and this one at that time that is what we did time period have to be same as what do you mean no no no time period is difference between the two time intervals t1 and t0 what is the difference that is a time period t2 and t1 what is the difference that is a time period getting it time difference between two pulses is the time period all of you understand type in quick so can you tell me what is the let me copy this on the other page so now that we have here so now time period from the source side is t0 that is very clear time period the new time period is t1 minus t0 or t2 minus t1 and you can see that both are same only both are same so t1 minus t0 is how much t1 minus t0 is not oh wait wait this I should put it as dash this is t0 dash 2 dash t1 this is different from that okay all right so basically this time minus that time is what I am doing okay so l minus vs t0 by c plus t0 minus l by c okay this is equal to this is equal to t0 minus vs t0 by c or it will be c minus vs by c times t0 this is the new time period this is what the this is what we will observe as an observer okay so frequency will be what the modified frequency will be 1 by t1 dash this is your modified frequency it will be inverse of this red bracket c divided by c minus vs into 1 by t0 which is n0 okay so you can see that the frequency has been changed the frequency which was n0 has become c divided by c minus vs into n0 so frequency has increased or decreased all of you it has increased or decreased decreased how it is decreased c divided by c minus vs into n0 is greater than n0 or less than n0 it is greater than n0 denominator is lesser than numerator okay right it is increased so basically if write down if the source moves towards observer frequency increases and if source moves away from the observer frequency decreases so what we have derived we had assumed that the source is moving towards the observer okay now if it is suppose moving away from the observer then how the formula will change n is equal to right now this is the formula wherein the source is coming towards the observer if source is going away from the observer the formula will change to what it will become c divided by c plus vs times n0 okay so all you have to do is to remember these two basic things you will not confuse with when to take plus when to take minus because plus minus will come again when observer moves source is stationary all right so if you remember the thumb rule that when source moves towards the observer frequency should increase and when observer moves towards the source then also frequency should increase so that way it will become easier for you to remember things otherwise frequency you can write the formula as c plus minus velocity of the source into n0 okay so this is the condition wherein the source is moving but observer is at rest now let us see the derivation of source is at rest but observer is moving okay so you have seen the first derivation in which the source is moving observer is at rest exactly the same way we are going to derive the second one so most of it i want you to do it i'll just give you hints okay so this line represent the source so let me first write down the heading case number two source at rest observer moving what happened here we release sound to be release sound relative speed of the sound is to be no no no speed of the sound relative to what relative to what observer no speed of the sound is the absolute speed okay observer was at rest initially isn't it source was moving observer was not moving now we are considering source is at rest and observer is moving second case why are you talking about second case i haven't even started when i asked you didn't you tell let us start you you derive it completely but let me first tell you the scenario okay this is the source this is the observer very similar scenario the distance between them is l0 okay speed of the sound is c and observer is moving with vo vo okay the sound is c okay now very very similarly we'll be doing it i'm throwing away first pulse from here at t0 equal to let's keep it at t1 t2 that'll be better and my first pulse goes at t1 equal to zero time at what time i'll receive here t1 dash will be what when i receive it what is the time on my watch all of you observer is moving yes observer is moving with v0 the pulse is moving with speed c pulse is trying to come with a speed of c and observer is moving with v0 okay so what is the velocity of approach for the sound towards the observer what is the velocity of approach c minus v0 okay distance is l0 so time when it reaches here is l0 divided by c minus v0 clear now i'm throwing second pulse from here my time is t0 time period so when i receive the second pulse what will be the time on my watch don't be in a hurry do it accurately that is more important remember observer is moving okay by the time the second pulse has been thrown what will be the distance between source and observer when the second pulse is about to be thrown distance between source and observer will become l0 plus v0 t0 by when if you wait till t0 observer will travel a distance of v0 t0 right it is moving so now the sound has to travel a distance of l0 plus v0 t0 that divided by c minus v0 okay plus t0 time which is already gone is it clear to everyone all of you l0 plus v0 t0 is the distance okay great now the third pulse let's do one more time third pulse is thrown at 2 t0 remember they are not thrown after first pulse is reached second pulse is not thrown after first pulse is reached they all three pulses are in the air only okay so this is thrown at equal to zero this is thrown at t0 this is thrown at 2 t0 like that it is okay so when the third pulse is received what is the time on my watch it's okay if you don't get it correct even when when i was a student i also did not get it when i was at your age but try it you may have got it wrong distance will become l0 plus 2 v0 t0 now in time 2 t0 the observer will travel v0 into 2 t0 okay that divided by c minus v0 plus 2 t0 okay so derivation is over now let's just accumulate whatever we have so new time period t dash will be equal to t2 dash minus t1 dash or t3 dash minus t2 dash both will be equal only so better to subtract t2 dash minus t1 dash easier that way so t2 dash minus t1 dash l0 plus v0 t0 divided by c minus v0 plus t0 minus this it will become what it will okay this thing is gone so it'll be v0 t0 divided by c minus v0 plus t0 okay so this will become c t0 divided by c minus v0 new time period c t0 divided by c minus v0 understood so the new frequency which is 1 by t new time period is equal to c minus velocity of observer divided by c into 1 by t0 1 by t0 is the original frequency so c minus v0 divided by c times and not okay so frequency has increased or decreased it has increased or decreased it has decreased why because the observer is going away all right so when observer goes away from source frequency decreases and when observer comes near to the source frequency increases so when it comes near it will become c plus v0 okay so the frequency is c plus minus v0 divided by c times and not got it so these are the two things when source is moving observer stationary when observer is moving source stationary now case number three when both are moving will not do any derivation will directly write it okay case number three both source and observer moves then the frequency I'll club both of them it'll be c plus minus vo divided by c minus plus vs into n again I'm telling you if source comes near the observer it'll be c minus vs because frequency has to increase and it is in the denominator okay when observer comes close to the source then it will be c plus vo because observer velocity is the numerator and it has to increase all right so do all of you understand this all of you is it clear to everyone this is the most important part of the chapter okay you can afford to still be light on the other parts but this part if you ignore it will become very very difficult this chapter as in 70 percent of numerical will be on this topic itself anyone has any doubts quickly type in no doubts fine so these are the few straightforward things which are written in the books okay now I'm going to tell you a few things that I have seen in the numericals how they make the numericals what are the some important things which you should keep in your mind while solving numericals otherwise it will become very tricky all right so here are a few points to remember very very important these points usually people make a lot of mistakes in numerical and then they learn these points but I'm giving you these things before you even solve numericals okay the first one is the velocity of source and observer are taken along the line joining source and observer it could be a point only source and observer what does it mean we need to take components we may need to take component of velocity along the line joining you have to take components of velocity that will go in the formula if source is moving like this and observer is here this velocity will not have any impact in the Doppler's effect okay the velocity should be along the line joining the source and the observer if it is not take a component along the line joining second if reflection happens then then we have to then we have to solve the situation the solve is not correct word we need to analyze the situation in two stages okay stage number one what is the stage number one the wall is wall is observer okay so as an observer wall will get a frequency that may not be equal to the original frequency because wall might be moving it might happen that wall is sliding or the source itself is moving okay so wall will get a frequency as an observer then stage number two wall will send back same frequency as source okay same frequency which frequency stage one's frequency whatever frequency the wall has observed as an observer the wall will reflect the same frequency but now wall is a source okay if wall is a source you know velocity of source comes where if wall is observer you know velocity of observer comes in the numerator okay so this thing is very important I have seen many mistakes here third okay if if anyone have any doubt you can type in you don't need to wait for me to complete it and then you type if wind is blowing let's say wind is blowing with velocity of wind VW okay it sends back the same frequency it apparently receives yes that's what I mean whatever wall receives as a source as a source suppose wall is a source forget about that it is a wall it is a source not not source sorry sorry sorry forget the fact that it is a wall assume it is an observer wall is an observer let's say so the original frequency is 400 okay let's say wall as an observer feel it to be 410 so observe as an observer wall is observed 410 but real frequency is 400 so which frequency wall will reflect as an source as a source what frequency wall reflect 400 or 410 everyone original frequency is 400 this is original wall as a source feels as a 412 410 is coming as an observer so which one wall will reflect 400 or 410 that is what my question is type in again I'm repeating stage number one wall is an observer okay as an observer wall will feel that 410 is coming towards itself but original frequency is 400 so when you go to stage number two wall has to act like a source so when it acts like a source it will reflect back 400 or 410 all of you type in don't type in two times okay or you want me to take your name then only you'll type in Pranam, Raghuram, Raihan, Rohit, Swaraj, Vibhav have taken six names type in Rishabh Patel can you hear me okay so the answer is 410 look at what is written stage number one wall is an observer wall is an observer so as an observer wall will not get the original frequency wall will get from the formula what you get c plus v not divided by c plus minus v s into n not okay so original frequency might be 400 but as an observer wall is getting 410 so wall will give out 410 only but now wall will become source in stage number two okay so be very clear about it all right stage number step number three point number three is if wind is blowing with v w then what will happen what will be the change what do you think what change will happen the speed of sound will become c plus v w or c minus v w depending on if it supports opposes okay speed of a sound is c only okay but it is like you know it it's like suppose there is there is a structure here let's say in my hand there's an insect which is moving on this okay with a small velocity I take this structure and I move it so the speed of the insect has increased or not insect might be moving with some velocity two meter per second but I have been moving this with five meter per second so velocity of insect becomes two plus five seven because it is also moving similarly inside the wind sorry inside the air the speed is c only but air is also moving along with sound air is moving so the air's the velocity gets added or subtracted understood is it clear to everyone the fourth point is just a point of caution that sometimes source and observer source and observer are same it's quite natural I am hearing my own sound so I'm source as well as observer and more so ever in the online classes I hear myself only okay so one example could be that I am let's say on a car and I'm moving towards a wall and I blow the horn okay when I blow the horn the sound goes reflects from the wall come towards me and I hear it so that is what the car along with me is a source of the sound also and after it get reflected on the wall and reaches my ears I become the observer also okay so you have to be very flexible in the thinking okay don't be in a rigid mindset that okay this is a formula I'll substitute this this this and that I'll get the answer no it will never happen like that in computer exams okay you have to be open minded when you think about the equations all right so let's take based on all these points to note let's take some numericals okay if a solid medium is moving we can apply the same rule yeah yeah anything okay so here is a small question you have a plane over here there is a wall okay this is you you are creating a sound of frequency and not all right speed of the sound is C and you are moving with velocity V okay you're moving with velocity V so you have to tell me when the sound get reflected from the wall it's a stationary wall when the sound get reflected from it what frequency of sound you will feel everyone what is the frequency of sound you will feel after reflection what is the frequency of sound I'll just write down the formula you might have noticed you might have written yourself though as a source we have to do it in two stages wall is first an observer and then it is a source so as an observer what is the frequency the wall will hear can you type in as an observer frequency heard by the wall everyone type in don't worry about being right or wrong if you're right not a big deal if you're wrong then also not a big deal but if you don't say anything then there is something terribly wrong as an observer the wall is observer this person is source everyone type in your answer what is the frequency heard by the wall velocity of these sources V velocity of observer is what what is the velocity of observer zero wall is at rest so frequency should increase or decrease the source is moving towards the observer frequency should increase or decrease increase and it is in the denominator so it will be C minus okay so it will be C divided by C minus V s into n naught this is what it will hear as an observer so what frequency wall will reflect back what is the frequency wall will reflect back n1 or n naught n1 because this is what it is hearing as an observer okay it will reflect it back so now the frequency heard by the person as an observer wall become source what it will be you can answer in terms of n1 tell me okay Hariran got it others Aditya got it as a way where the velocity V will come now there is no V s there is only V so now the person is observer so observer velocity observer velocity will come in the numerator only right it will come in the numerator because that is now observer and because of the observer's velocity frequency should increase or decrease it is coming towards the source sources the wall frequency should increase right it will it is coming towards so it will be C plus okay I have written V s there there is no V s this V only C plus V divided by C into n1 this is the answer that's it you need to substitute the value of n1 there so you can write down in terms of n naught it will be C plus V divided by C minus V into n naught just go through this question once and type in is it clear this is the first question on Doppler's effect there are many varieties but if you don't understand the first one itself it become difficult to proceed all of you go through it once anyone has any doubts type in I'll wait for 30 seconds we had a question on this and Sunday KVPY mock yeah Doppler's effect is there in KVPY so now you are done with that exact same question will never come Pranav there will be some variety all right do you guys like when you write the KVPY and when you write the mains came on test are you guys seeing there is a difference in the type or the way the questions are asked in KVPY big difference what is the difference what is the main difference you see in KVPY all questions have negative marking yeah of course in mains also KVPY correct do KVPY is more of imagination have you seen that have you observed that they want you to think they want you to visualize more than the mains exam okay mains is more mathematical you can say the physics paper anyways next question should I give you a little slightly different or difficult one wherein all the concept will be tested at once and then we'll take the easy ones later on all the concept will be tested here make sure you attempt this getting the right or wrong answer does not matter again I am telling you same situation there are some changes speed of sound is C you are moving with V the wind is blowing in this direction the wind is blowing with VW the wall is coming towards you with the velocity of U okay anything else and original frequency is N not with which you are whistling let's say you need to tell me the same thing what frequency I will receive after reflection from the wall okay first answer me this one whatever I have written here okay so speed of the sound will be speed of sound what you will take what you will take as speed of sound instead of C you'll be writing C plus VW that is the only change okay so source is moving or observer is moving source is moving or observer is moving source both are moving correct wall is the observer and the person is the source both are moving and because of whose motion frequency will increase and because of whose frequency will decrease both will increase all of you understand that both will increase both are coming towards each other both source and observer will try to increase the frequency contribution from both side will be to increase so C plus VW plus velocity of the observer which is U Y plus because it will increase divided by C plus VW velocity of sound minus V my Y minus it will also increase but it isn't the denominator so that is a minus it has to be minus to increase it that into N not anyone has any doubts here Charan is it clear okay now comes the final thing as as a observer person gets the frequency you can answer in terms of N1 N2 will be equal to what yeah yeah don't type it in terms of N1 you can type in terms of N1 what is the value of N2 now velocity of sound you'll take C plus VW or C minus VW sound is traveling like this it is going from after reflection it goes from wall toward the person so it will be C minus VW you're going against the wind okay now what C minus VW what should I write here plus V or minus V plus V this is observer now person is observer right and the wall is source for the second stage okay so now the person's velocity will be in the numerator and the wall's velocity in the denominator okay and both will try to increase the frequency they are coming towards each other C minus VW minus U times N1 where N1 is this so it becomes a huge expression actually so from one and two you'll get N2 to be equal to let me write it down meanwhile please ask any doubts if you have no doubts okay have you observed the Doppler's effect on the railway platform like when the train is coming with a siren you will see you will hear the siren with a different note or frequency and suddenly the quality of sound or the way you are hearing the sound suddenly changes as soon as the train crosses you you know a different type of sound comes if we ask to find standing wavelength which frequency do we use standing wavelength what do you mean standing wave in Doppler's effect where is standing wave talking about Doppler's effect or something else Doppler's effect is not super position you're confusing no superposition it is the same wave which after reflection comes towards you whatever you observed that is what your frequency is all right frequency will be speed of the sound divided by wavelength all the time that wave equation will be valid what you observe is your frequency okay all right let me see if I have more questions on Doppler's these are all simple questions okay you'll get you have done much more difficult question than these just now you should be getting the answer okay you know it helps to draw a diagram like this to represent the scenario otherwise you have to read the question again and again once you draw the diagram you don't need to read the question detector is your observer if you don't know it already anyone close to the answer okay Aditya got something the original frequency will be less than 1 9 5 0 or more than 1 9 5 0 what do you think it should be more right because both source and detector they are going away so they will try to decrease the frequency and this is the decreased frequency 1 9 5 0 so actual frequency should be more than this N should be equal to C minus 10 velocity of the observer divided by C plus 10 into N naught this is equal to 1 9 5 0 right so N naught is equal to C plus 10 which is 350 divided by C minus 10 that is 330 into 1 9 5 0 how many of you got this this so this is what the answer is okay so anybody solved it further why are you getting different different types of answers 2 2 5 0 2 0 7 0 2 0 6 9 is it exact 2 1 0 0 Pranam okay let me tell you how much it is rounded off 1 9 5 0 2 0 6 8 Arithra using calculator nowadays the exams are more calculation intensive they then knowingly give you difficult calculations okay then knowingly give you difficult calculation if you have if you see the latest exams so if you are not habitual of solving or dealing with numbers quickly then probably you will struggle a lot in the exam and that is very frustrating even though you know how to solve a question but because of calculations if you get stuck it becomes very frustrating all right so be good with numbers and practice all right next question this is also straightforward one there is no hurry solve it slowly get it correctly this is the third time we are doing something like this okay Pranam got something others all right Ajith there also fine many of you got the answer so you might have already known that c minus v I am directly writing the expression this is what we calculated for this scenario right this into n naught is equal to n okay so 330 minus v divided by 330 plus v no wait wait wait this plus here minus there into n naught which is 440 is equal to n which is 480 okay so now I leave the calculation to you 330 plus v divided by 330 minus v is equal to 12 by 11 okay and from here you get the answer how you solve this actually you can do component or dividend oh this divided by uh you add here itself it will become 660 this is equal to 12 divided by 23 okay now you can do it cross multiply I think some of you are getting v to be equal to how much 14 around 14 meter per second all right so sorry see so these are the questions on the topless effect let's say one more actually I did not get that kind in the book I'll just create that thing there is an aircraft which is moving like this with a velocity of v this aircraft the arrow is aircraft here is you you are running like that this is your velocity your height is very small compared to the level the aircraft is moving above from the ground the distance of aircraft from the ground vertical distance is let's take something else let's take it as a and the horizontal distance from you this distance is b okay the aircraft when it is traveling it is emitting sound of frequency and not all right speed of the sound is c fine you need to tell me what frequency you will hear find out what frequency you will hear this is the last question on topless effect I'm not adding the wind here unnecessarily but won't we hear multiple frequency yes you will hear multiple frequency I am interested right now what is the frequency you are hearing right now when aircraft is at a distance a vertically and a distance b horizontally angle will change by the time sound reaches only that angle matters when the sound has started from there when it has reaches doesn't matter where the aircraft is aircraft might have changed its direction also it does not matter while it is emitting how the aircraft is moving that matters only yeah you don't need to type just let me know you're done see I have just made it up you know so answer may not look very nice okay Pranav is done others what is the first thing you have to do connect the source and observer so velocities along this line matters not the absolute velocities getting it so if this is theta then this angle is also theta okay so aircraft is coming towards the observer with what velocity not v v cos theta the person is going away from the aircraft not with u blood but u cos theta okay so what is the frequency it will be c what should I write here who is observer this is observer this is source whose velocity comes in the numerator source or observer observer now because of observer frequency should increase or decrease in this case decrease it is going away so c minus of u cos theta divided by c now because of the aircraft frequency to increase or decrease frequency should increase or decrease everyone everyone increase it is coming towards along the line joining right so c minus of v cos theta into and not this is the answer and cos theta you can write separately cos theta cos of this angle right is b divided by root over a square plus d square let's say this is the answer all right all of you understood Doppler's effect please type in then we'll take a break this is the Doppler's effect