So recording is on now. I heard that you have some inter-school competition going on. Is it correct? Is there such event? Satyam, there was in NPSHSR. Okay. Anyways, so all right, so now we should start. The others will join a little late and will miss something. So in the last session, we started this chapter ways. Okay. And in the last session, we talked about the kinematics of the waves. All right. And we also discuss about the velocity of the waves. Okay. So in the last session, it was mostly, you know, theoretical stuff. And we did not get a lot of chance to do the numericals. Fine. And this is the last chapter also. So I want you to end the physics 11th in a good note. This chapter, let us solve it more and let us solve more and more questions from this chapter so that we are very comfortable with it. Fine. So today's session is mainly problem practice. Okay. So I'll do some problems related to what we did in the last session. And then I will teach a little bit. And then again, we will get into the problem practice. Fine. So that is the agenda for today. What is going on in your school? Can anyone tell me what is happening in Rajesh in a girl and in. What is that naffle? In the S H M is going on naffle waves is going on. Okay. I oscillation just finished. So I just got over. Right. So now you know that you in your school oscillation just got over and one more chapter to be finished. And immediately after chapter finishes, you have and semester exams or end of the year exam, whatever you want to call it. Right. So you can see how tight the entire grade 11th curriculum is. And even we have to finish the curriculum. You remember when we started, we started, I think in the month of March. Right. And we are sitting on mid of January. So till end of the January grade 11th will go on. So it is definitely not like how it was in grade 10th. Right. So that's where you complete the curriculum and revise five, six times. Right. So that is the reason why I was telling you every class that do the assignments, whenever you're doing it for the first time, do it properly that time itself as if you'll not get time for the revision. Okay. So whatever it is, maybe some of you have listened initially some of you have listened in the middle of the session and change your style. You are probably realizing now. Okay. But still you have one more year. Fine. In grade 12th, at least whatever mistakes you have done, do not carry that to the next year also. Fine. So keep that in mind. And anyways, we'll discuss about the grade 12th also. But after finishing the grade 11th curriculum. So grade 12 is a lot simpler than how grade 11 physics is. Okay. Anyways, so like what is the agenda, let us start the problem practice. And before I project questions in front of a screen, I'll quickly write down all the equations, formulas, whatever we discussed in the last session, so that it's a ready reference for all of us. Okay. We started the last session with wave equation. Okay. Wave question is what y is equal to a sign of kx minus omega t plus five. I'm looking at this equation. Which direction the wave is traveling? Can you tell me anyone? Everyone. Which direction the wave is traveling? All of you. Right. It is in positive direction. Good. Positive x direction. And which direction the particles of the mediums are oscillating. What is that direction in which the particles of the medium, they are oscillating y axis along the y axis. Right. So which wave it is? Which wave it is? Transverse or longitudinal? It's a transverse wave. Okay. If I have to write down the longitudinal wave equation. Okay. There, the oscillation is happening along the x axis only. The particles of the medium, they oscillate on the x axis itself. Wave is also moving along the x axis. So there is nothing in the y direction to track. So what do you track? You track how much away the particle is from the mean position. And that you're tracking in the transverse wave also. But since they're traveling along the y coordinate. So whatever movement they have, the particle that is a y coordinate. But along the x axis, their mean positions are at different, different locations. And here particle is oscillating, here it is oscillating everywhere. The particles are oscillating along the x axis. So we introduce a variable s. S is the distance from the mean position along the x axis only. And even the longitudinal wave equation looks similar. This is a longitudinal. So since wave equations are very similar to each other, we don't study kinematics of transverse and kinematics of longitudinal separately. It is same for both. So the wavelength, how will you write if these are the wave equation, how will you write wavelength as? Lambda is equal to everyone. Correct. 2 pi by k it is. Time period is what? Right. 2 pi by omega. And what is A? A is referred as amplitude. And velocity of the wave is what? V is equal to what? Good. Omega by k. So these things you can make out from the wave equation itself. And wave equation need not be sine or cosine waves. Fine. k is referred as wave number. If you are looking for the name, what is the name of k? k is called wave number. But that is just a name. Now sine and cosine waves are the special kind of waves in which what happens, you already discussed that right down sine cosine, if the sine or cosine wave is there particles perform SHM. They perform SHM. All right. Now, if you differentiate y with respect to time, what do you get, everyone? At a particular location of x, if you take a derivative of y with respect to time. What it is? What does it represent? First tell me that. Good. Now, how does slope of the wave? It's not slope, right? Slope is dy by dx. All right. This is velocity of the particle. Velocity of the particle. So this is minus of A omega cos of kx minus omega t plus 5. Fine. This can be written as minus of omega under root of A square minus A square into this. Now, what is the second term in the square root? This is what? This is y only. So you can see that the velocity of the particle is minus of omega root over A square minus y square. So it is like the way, like the SHM equation only, right? Omega under root A square minus x square was the SHM equation. The same way you get the particles equation also. So this proves that the particles are performing SHM. All right. Don't worry about the minus sign. If it would have been kx plus omega t, it would have been plus sign. It just tells you that which direction it moves. Nothing other than that. Okay. Now these are the kinematics related things. Suppose I ask you what is the acceleration of the particle, then what you will say? Acceleration of particle is what? Particle doing oscillation. So equation is omega square y. Okay. Always keep in your mind that the particles are doing SHM. All the SHM equations are valid for the particle, but the wave is moving with velocity along the x axis. So whatever wave does and whatever particles are doing are two different things. Okay. So this is the kinematics. Then in the last session, we spoke about the transverse wave speed on the stream. What is the velocity? Power t by mu. Correct. Okay. Then we spoke about longitudinal wave inside a fluid. What will be the velocity? Correct. Under root b by rho. Bulk modulus, if you remember, is minus of v dp by dv. Okay. In a fluid like air, bulk modulus is gamma times p for the sound waves. Now there is also one equation related to longitudinal wave inside a solid. Okay. So sound can travel to the solid also. So the velocity of the sound on the solid would be under root, Young's modulus by rho. That is the only change. Okay. Here the bulk modulus is of the air. There the Young's modulus is for the solid. So clearly as expected, Young's modulus of solid is very large compared to bulk modulus of the air. So that's the reason why when the sound travels, it travels faster in the solid medium and slower in the air. And that is the reason why I think I've given you this example already. That's the reason why when in the earlier times, if you have to understand the enemies approaching you towards you or not, if you put your ears on the ground, you'll hear the horses, you know, they are approaching to your side. That is how you can determine much earlier. Maybe they are 15, 20 kilometers away. But when you put your ears on the ground, you'll be able to hear the tap of the horse foot. Because the sound travels faster on the solid medium, which is the ground. So in the air, if you try to hear it, it may not be very easy to hear, first of all, because of the wind blowing and everything. And also in the air, the speed of sound is a lot lesser. Okay, so these are the things that we discussed already. Let us solve questions now. Anyone has any doubts? Quickly type in. Anyone wants to ask any doubts or anything you want to highlight? Nothing, no doubts. Okay, so here are the questions. Okay, somebody is asking something. Sir, what is the difference between phase difference and path difference? So did you guys learn about the path difference in the school? If yes, then I can tell that. Did you guys learn that? Okay, it was there in the assignment. Okay, so we will talk about it. I will discuss about it. Maybe after we are done with these three, four numericals, if I forget, you please remind me again. I will talk about it. Okay, solve this question right now. A part, what is the answer? It's very straightforward, I think. A is what? All of you. 3 centimeter is A, B, wavelength. Okay, 6.28. Are you guys able to recognize 6.28 is two times of pi? 6.28 is 2.2 times of pi. So the wave equation can be written like this. 3 sine of 2 pi into 0.5 is pi, pi x minus 314 t. And 414 is 100 pi. So the answer for B is 2 pi by K. K is pi coefficient of x. So this is 2 centimeter. B. Now what is the answer for the C? C is 1 by time period is the frequency. Omega divided by 2 pi. Omega is 100 pi divided by 2 pi. So this is 50 hertz. That's B, sorry, C. D, what is the answer? Everyone. Okay, I can see some of you already got it. Others. Velocity is omega by K, right? Omega is 100 pi. K is pi. So this is 100 centimeter per second. One meter per second. Okay, so this is how you do this. This is the school level kind of question. Let's go to the next one. Do this. In the previous question, our focus was what wave is doing. In this question, the focus is what particles are doing. Okay, particles perform SHM. A part is what? Part A. Vishal, Harshita got two different answers. September A part. Now, whenever a particle is doing SHM, the maximum speed in any SHM is what? Do you remember this? What is the formula? A omega, right? And you anyway know the formula is velocity is omega root over A square minus Y square. So when Y is zero, then only V is maximum. Omega A, that is at the mean position. Omega is 314, which is 100 times pi into 3. So that is 300 pi centimeter per second. So you have to keep a tab on the units also, okay? This is A. What is B? Satyam got something, others? Vishal, okay. So you can see that if you differentiate Y, keeping X constant, you get velocity. If you again differentiate, you get acceleration. And if you already know that it is doing SHM, then ultimately you'll get an SHM equation that A is equal to minus of omega square Y. SHM equation, right? Do you remember this? What is it doing SHM? So it boils down to finding what is the value of Y at X equal to 6 centimeter and at equal to 0.11 second. So that will be 3 times sine of at X equal to 6 centimeter. So this is pi. So it will be 6 pi minus 314 is 100 pi. 100 pi into 0.11 is 11 pi. So you'll get minus of 3 sine 5 pi. What is the value of this? What's the value of this? Everyone, what is the value of this? 0, right? This is equal to 0. So Y is 0, the acceleration is also 0. Is it clear to all of you? Type in. Okay. Let's move on to the next question. Do this. Should I give hints? Satyam got something. That tension is due to the stress. And stress by strain is Young's modulus. Do it now. Others. All right. Okay. So all of these things are given to us. This is the string. It is clamped. There's a tension in this. Okay. If tension is three, then the stress is T divided by S area of cross section. And if extension is delta L. Strain is delta L by L. Fine. Now you will see that when you write velocity, velocity is equal to root over T by mu. Okay. Mu is mass per unit length. We are considering it as a uniform rod. Uniform wire. Under root T divided by M by L. Okay. Now T is not given to us. Rather they asked us to find delta L. So I can use that formula. Sigma divided by epsilon is Young's modulus. So Sigma is strain. Multiplied by the Young's modulus. Sigma is T by S. So T is delta L by L. Y into S. This is T. So the speed of the wave is T L. Sorry. T you have to write delta L by L. YS divided by M by L. L is gone. Okay. So delta L will come out to be U square M divided by YS. Is this clear? Fine. These are all straight forward questions. Fine. They are very simple ones. Okay. Won't the string be extended initially? What do you mean by that? Somebody asking that won't the string be extended initially? I didn't get what you mean. You should speak up. I'll unmute you. Speak. Sir, in the formula for a strain, we have written delta L by L, but L must be natural. But L is not natural. Yes. Given to us in the question. Yes. We are assuming delta L is very small and L plus delta L is roughly alone. Okay, sir. Thank you.  The hint is frequency is same throughout the strength as it depends on the source. Frequency below is equal to frequency top. You can roughly tell me the answer or the expression also is fine. No one. No one till now. Is there anyone about to get the answer? Okay. The first step is drawing a freeboard. A rope of 12 meters. And mass of 2 kg. Hanks from a rigid support. A mass of 6 kg. Transverse wave of some wavelength is produced here. And it is moving upwards. You need to find out what will be the wavelength when it reaches the top. That is a question. Let's say this point A, that is point B. Between point A and B, frequency is same. Is velocity same? Tension is different at different points. So the velocity is not same. Okay. Tension at point A is how much? Tension at point A is 6g only. If you draw the freeboard diagram of 6 kg, there is TA, there is 6g. So TA minus 6g is 0. Fine. So TA is 6g. If you take GS10, TA is 16 meters. Fine. At the top, what will the tension? Let's say tension over there is TB. What is the value of TB? Okay. Others, right? It will be 18 Newton, right? It will hold now 6 kg as well as the ropes mass also. Till now, many times we have been ignoring the mass of the rope. But here the mass of the rope is comparable to the mass of the block. We should not. That is 18 Newton. Now tell me one thing. If this entire thing is in the lift and it is accelerating with addition A, will the tension remain same or they will be different? Will it remain same or different? Tension changes, right? So you need to draw the freeway diagram every time. Don't assume that tension will be mg. Because it is all at rest, that is why tension is equal to mg. All right. So the velocity at point number A is root over TA by mu. And velocity at point number B is root over TB by mu. So velocity of, you know, the frequency at A, let me recognize FA. Frequency at A is velocity of A divided by the wavelength at A. Frequency at B is velocity of B divided by wavelength at B. The frequencies are same. It is the same system. Same system cannot have two different frequencies. So VA by lambda A should be equal to VB by lambda B. So lambda B is equal to VB by VA times lambda A. When you do VB by VA, you will get under root of TB by TA. Which is root over 80 by 60. So 4 by 3, that multiplied by 0.06 meters. Clear to all of you? All right. So these are the few questions I wanted to highlight. Now let us move forward with a new set of theory. And once the theory gets over, we'll again come back to the numericals. Okay. In case any one of you have any doubts, you can highlight. Okay. So the next thing that we're going to learn is the case 3 of superposition. Do remember, what are the first two cases? First two cases that we have learned previously. What was it? Case number one, amplitude is same. And frequency and wavelength is also same. Case number two, amplitudes are different. Frequency and wavelength, they are same. Okay. In both the cases, there was one thing that was common other than frequency and wavelength. What was one thing that is same throughout? Both are sine waves. Okay. What else? Correct. Both are traveling in the same direction. Yes or no? In case number one, and in case number two, both the waves were traveling in the same direction and they meet and that's how things are happening. Okay. Now we are talking about a case in which the amplitude is same. Frequency and wavelength is also same. But one is traveling in positive x direction. Other is traveling in negative x direction and they meet. That is the case that we're going to discuss here. Okay. So let us do that. If this is the first wave, y1 is equal to a sine kx minus omega t. This wave is traveling in positive x direction. What is the second wave which is traveling in negative x direction? What should I write here? A times sine of kx plus omega t. This is traveling in the negative x direction. Okay. So I want you to simplify what will happen when both of them meet. Okay. So y will become y1 plus y2. Find out, let me know once again. Correct. When you add, you can take a common sine of kx plus omega t plus sine of kx minus omega t. So this becomes a case of sine a plus sine b.  So it will become 2 a sine of kx cos of omega t. This is what it becomes. Now, does it look like a wave equation to you? Everyone, does it look like a wave equation? It's not a wave equation. It is not a wave equation. But suppose you call this, this bracket term, suppose you call it as amplitude. So it will look like a cos of omega t. What does it look like? A cos omega t is which type of equation? Correct. Others, which equation comes into mind when, I mean, what is that name of the situation for which? Y is equal to a cos omega t is the equation. Correct. Others, it is the equation of the SHM. Okay. But here what is happening is that the amplitude a is equal to 2 a sine of kx. It depends on x. Amplitude depends on x. As in, all the particles are doing SHM only. Okay. But then not all the particles will be able to move up the same distance. So one particle will only go up to here, other one can go up to there. So they will together go up, together comes down. So while they're oscillating, they have different different amplitude at different different x at x equal to one centimeter, one amplitude at x equal to two centimeters, some other amplitude. Okay. Will there be some positions x for which amplitude will become zero? Yes or no? If kx becomes 180 degree, kx is pi. Sine of kx will become zero or not? Okay. So kx is pi. So x is pi by k. So at x equal to pi by k, amplitude is zero. So at that location, particle doesn't move at all. Because amplitude is zero. Then at x equal to two pi by k, amplitude is again zero. Fine. So this is something like this. This is x equal to zero. This is x equal to pi by k. X equal to two pi by k. See what is happening when these two waves meet, this particle can go up and down like this. This can go up and down like that. The particles are oscillating at different different amplitudes. Okay. Now, how it is different from a traveling wave. In a traveling wave, when the wave is moving forward, what happens is that they're all the particles, they're oscillating in the same amplitude. All the particles can go down to a distance of A and can go up to a distance of A. But they are moving at different different times like this. But here, what happens is different locations, the amount of distance the particle moves up is different. But all the particles together go up, all the particles together go down. So that is what is happening in this case. In traveling wave, when it is an actual wave equation, all the particles move, but they're not moving together. They're going like this. When this particle is down, that particle may be up and vice versa. All right. But in this case, all the particles together go up, together go down. All right. So this kind of situation, this kind of situation, we call it a standing wave. All of you clear what exactly is happening here? Anyone has any doubts? Type in, is this clear? Do you want to see the standing wave? I'll just show you so that there is no confusion. Before showing it to you, have you ever seen, let's say, a guitar string or something like this, which is tightly wound like that? Okay. You just pluck at the center. And have you ever seen such kind of vibration in the string? It shows as if an eye has been formed. Okay. You're seeing it. Let me now show you the, even rubber band. Yes. Yes. Good. Here's a frequency generator. It's connected to what's called wave driver. It's just a speaker. It's going to pump up and down. And whatever frequency I set this guy to, when this guy pumps up and down, well, he sends away, sends away, down to this end, the way you fly. The first one will be half this number. So 8.5. I'll just show you. Look at this. This is a standing wave. All the particles, they are together moving up and down. Okay. This is what is happening. So you may be wondering that you need two waves. So where is the second wave that is traveling in the opposite direction? One wave, what happens? This machine is sending from here. The, this wave goes, hits an obstacle here. There is a reflection that happens. And second wave gets produced from here. This is incoming wave. And that's the reason why this is getting created. Fine. I will talk about that later in detail. But this is the standing wave. All right. And standing wave is a very, very common phenomena. At least in all the musical instruments, you'll have standing wave on the flutes, guitar, tabla, Santur, all those instruments that, that people use, they work on the standing wave only. The working principle is standing waves. So that's why we are studying it. Let me now take this equation, standing wave equation. And if, if amplitude, which is to a sign of KX at a location becomes zero, then that point doesn't move at all. Okay. So what, what are those points to a sign of KX should become zero. So from here, sign of KX becomes zero. And if it is zero, KX should become equal to n pi or the X value would be n pi by K and can be zero, one, two, like that. Fine. So at, let's say X equal to zero, where n equal to zero and X equal to zero. So at X equal to zero, you will have amplitude zero. That point is not moving at all. Then at X equal to pi by K, when you put n equal to one, then you put n equal to two, two pi by K also. Like that you can keep on getting different, different values. And you'll get like infinite values. But practically speaking, your length of the string of the finite length. Okay. So you all, you need to also see how much long the string you are using. It should not be that mathematically, you're getting at 100 kilometer. Also there is amplitude equal to zero, but that 100 kilometer is not on the string. Right. If string is let's say one meter, so you need to stop substituting the value of n as soon as X becomes more than one meters. Like that. Okay. These points, there's a special name to these points. They are called nodes. Okay. They are called nodes. Fine. Anyone has any doubts? Okay. Now the maximum possible amplitude is what? This amplitude to a sin of K X. How much maximum possible amplitude this can have? Plus minus two M. Right. What does minus means? Minus only means that particle is going down. Okay. Minus is like a sin convention. Minus a and plus a are same. Okay. So wherever to a sin K X becomes equal to plus minus two a at that location, maximum possible amplitude happens. So for that, sign of K X should be equal to plus minus one. So what is the value of X for that? K X for that. What K X should be equal to everyone K should be able to generate solution to tell me like this and pie in terms of N. What it should be. Okay. Plus minus N pi by 2 K. What is N equal to two can N be even number. Can N be even number if you say N pi by 2 K. It cannot be an even number. It has to be odd. Right. So if it has to be odd, odd digit like this to N plus one. Pi by two. Like this. Okay. So at this location where X is equal to two N. Plus one. Pi by two K. Maximum possible. You know, amplitude happens. These are called anti-nodes. Now look at this. X equal to all of you again. N pi by K. They are notes. And it's equal to two N plus one. Pi by two K. Anti-nodes. Okay. So for nodes, the value of X. Write down few values. Zero. Put N equal to one. Pi by K. Three pi by K. And so on. Okay. For anti-nodes. Value of X. Put N equal to zero. You'll get. Pi by two K. Then I'm going to do one. Three pi by two K. Then five pi by two K. And so on. Now, when you look at the locations of nodes and anti-nodes, what can you make out one clear observation? What do you think it is? Everyone. Good. Look at this. I mean, whatever you have said, that is also correct. But I was looking at a simpler observation. So look at this. Between zero and Pi K. What is the point? Pi by two K or not. That comes in the middle of this. Three pi by two K comes in the middle of these two. Five pi by two K comes in the middle of these two. Right. Similarly. Pi by two K comes middle of these two. Two pi by K comes middle of these two. So it implies what? It implies that between, between two nodes, one anti-node is present. And between two anti-nodes, one node is present. Clear all of you? Anyone has any doubts? Anyone has any doubts? Any doubts? Type in. So now, a little bit of hint. I have already given you how the standing wave is created. You need what? You need one wave traveling in one direction. And the other wave should travel in the opposite direction. Okay. What is the best way to create a wave that travels opposite to your direction of the wave? What is the best way, best way to do that? I've already told you. What is that way? You need two waves, right? One wave you can create. How will you create the other wave having the same frequency which travels in the opposite direction? Reflection. Reflection is the way. Whenever I teach reflection, one scenario comes in my mind. Like when I was at your age, I used to sit in the balcony and study in the late evenings. So every time, you know, our building was very near to the boundary wall of the society. So every late evening, you know, it is a routine of, there was a dog, a stray dog, who came from near the wall and he started barking. And his own voice gets reflected from the wall and he hears his own voice. So that dog used to think that there is another dog on the other side of the wall. So he keeps on barking for hours every day. So one fine day I have to put water on him so that he doesn't come back. But then that is how it happens. Okay, so reflection is very common phenomena.  Theaters, when they are created, special attention is given to acoustics so that the sound doesn't get reflected again and again multiple times and you hear echo. So all those things are very important. Anyways, so write down reflection of the waves can generate standing waves. Okay, can generate the standing waves. Now, we will see the different cases of the reflection because it's not that same kind of reflection happens everywhere and everything is similar. So let's say what all, how the reflection happens. For that, we have to take examples. Okay, so we will take example of a string, the wave on a string. Write down case number one. One end of the string is fixed. Let's say this. This end is clamped. From here, you are sending a wave. Okay. This wave moves forward. All right. When this wave moves little bit forward, let's say it goes there. Okay. So what happened to this point when it was here, later when move wave move forward, what happened to that point? When this wave became a dotted line as it moves forward, what happens to that? Correct. It moves up. Okay. So if wave has to move forward, the point has to move up wherever it goes. Okay. So wave is trying to lift the rope in upward direction everywhere. So let's say the wave reaches here at the end. The wave has reached over there. This is at T equal to zero. T equal to T one. The wave has reached here. Okay. This is let's say point P. Okay. So what this wave is trying to do with point P? Tell me. What does this wave is trying to do with point P? Point P is on the wall. The wave is trying to do what? It wants to lift it. Right. It wants to lift it. Right. So the wave is trying to lift the point P on the wall upwards. So that it can move forward. It is simple. That is what everywhere the wave is trying to do. Right. This is what the wave is trying to do. Now point P can't move. It is, it is as it is a part of wall. So wave cannot move the wall. So the wave is applying force on the point P. Or you can say the wave. Or indirectly, let's say you're saying that the rope is applying force on the wall in upward direction. Right. Wave is writing on the rope. Rope is fixed to the wall. So the rope is trying to pull the wall in upward direction. So what wall will do? What wall will do? What wall will do? Good. So according to Newton's third law. According to Newton's third law, it will apply equal and opposite force. So the wall will apply equal and opposite force. Basically downwards. This is what it tries to do. Now when rope is applying force on the wall, it can't move. But when wall is applying force on the rope, will rope move or not? Rope will move. And which direction is applying downward force? So what happens is ultimately this kind of wave get generated. The wave get inverted like this. And start moving this way. Clear to all of you how this wave has generated. You're pulling the rope down. So that wave gets created. The downward wave is getting created. And it is moving in the opposite direction. Clear. Type in type in case it's here. Okay. So the incoming wave equation, let's say is a sin of kx minus omega t. They're incoming waves. All right. The reflected wave equation. That will be, tell me. Will it be a sin kx plus omega t? Does this sound correct? Correct. Then what is the difference between this wave traveling in this way or that wave traveling in this way? Is there any difference or not? There's a difference. There is a difference. Okay. So what is that difference? The difference is of the phase actually. There's a phase difference between these two ways. How much is the phase difference? Look at this. If you connect it like this. It's a sine wave, right? So this angle, let's say is zero. This is pi. There's two pi. Okay. So that point. This point is pi angle away. Of this point. So this is zero and suddenly it has become pi. So all the corresponding points are ahead by pi. So you can say that there is a phase. That get introduced. Plus five. Okay. This is also referred as hard reflection. Hard reflection. So you'll study about it next year also in wave optics. Hard reflection. Introduces a phase of, not only the wave is traveling in opposite direction, it gets inverted also during the hard reflection. Okay. Now this example that we have taken by the way is the transverse ways. Let us see if something similar happens with the longitudinal wave, like for example, sound when it gets reflected. So for longitudinal waves, let's say this obstacle, longitudinal wave is nothing but compression and refraction. Somewhere the air will get compressed more. Some places the air will get compressed lesser. Like this. Compression refraction like that. I've shown it to you, right? So wherever it hits the obstacle, at that point, the air will get compressed or will get elongated. Sorry. The compression or refraction. What will happen when it hits the wall at the point where touches the wall. Compression, right? It is compression that happens. Right? So now when it gets reflected off after reflection, what will happen? The compression will remain compression or will become refraction? What do you think? After reflection, will it remain compression or it will be refraction? Just imagine. Just imagine the spring is getting compressed over here and when it push it back, it is anyway getting compressed from this side. You are putting from that side also after reflection. Will it get compressed or not? I mean, why are you saying refraction? Some of you are saying refraction. It will get compressed. Okay? So refraction won't happen. All right? That is unlike how it happened in the transverse wave. In transverse wave, the wave gets inverted. Inversion means whatever the compression suddenly becomes refraction. But now, even after reflection, compression remains compression at that point. Okay? So with this, if incident wave, incoming wave equation is a sin kx minus omega t, the reflected one will be simply a sin kx plus omega t. No phase change happens. Fine? So in a longitudinal wave, reflection creates no phase change. Okay? So after reflection, I'll also talk about the path difference or just like phase difference, there is something called path difference also. But right now, let me complete the reflection part of it. Anyone has any doubts here? Even to visualize the compression, even after it gets reflected, there will be compression near the wall. Type in. You have to visualize it. All right. All right. Okay. Now let us see the case two. These longitudinal wave and transverse wave, they were encountering the hard reflection. We have seen what happens after the hard reflection with the transverse and with the longitudinal wave. So let's see what will happen to the transverse wave in this soft reflection. First of all, we should know what is soft reflection, right? So case number two, soft reflection. What it is, we will take an example of a string to explain that. This is the scenario, the end of the rope. There's a ring, a massless ring that can move up and down. A massless ring can move up and down. Okay. And from here, you are, let's say sending a wave like this, this wave, this wave move forward. Okay. Now when this wave goes over there, will it try to move the ring in upward direction? Will the ring move or not? Will the ring move? It will, it will move. So it will be like this. When the wave reaches over here, the ring will be like this. When the wave reaches here, the ring is like that. And slowly and slowly wave disappears and the ring comes here, right? Initially the ring will move up and then ring will come down as the wave is passing through the wall. But when the ring is doing this, it is as if somebody stood on the other side, took the ring and did like this. So that movement of ring, will it create a wave that way or not? It is as if somebody is standing on the other side, took the ring and did like that. Will it create a wave? It create a wave itself, right? Because the ring is moving up and down, all right? And hence reflection happens. But now when the reflection is happening, the wave, the reflected wave that get generated will be in upward direction only. It is like this. Facing doesn't happen in the soft reflection. Okay. So if the incoming wave is a sin of kx minus omega t, the reflected one is a sin of kx plus omega t. This is a soft reflection. Okay. Now tell me, is the light transverse wave or longitudinal wave? What it is? Light wave. Light is transverse wave. Okay. Light is a transverse wave. So does the reflection happen with the light? Of course, yes. Reflection does happen with the light. So whatever we have learned, same thing you can apply to the light also. All right. And when we talk about reflection from the light, have you observed that when you take a transparent piece of glass, you can not only see on the other side, but you can see your faint reflection also on that transparent glass. Have you observed that? Another example could be suppose you are inside the car. When you're inside the car, you can not only see outside the car, but also your own on that mirror. Okay. So what I'm telling you this is that whenever there is, let's say, light goes from one medium to the other medium. It is not that entire light passes through the other medium. Some part of it get reflected. And because of that, you are able to see yourself and some part of it gets transmitted. Right. So it is not 100% refraction all the time. So 80% will be reflected. 20% will be reflected. So like that, it happens. Fine. So keep that in mind. So coming back to the reflection of the light, the first of all reflection from the mirror is considered to be hard reflection. It's a hard reflection. Okay. Then if the light is in denser medium. Okay. And it is going to the rare medium reflection from rare to denser medium. Did I say opposite of it? I think. Anyway, since I've written reflection from rare to denser medium. Suppose you suppose the light is in the oil and the light enters the water. So oil is rare medium. So light is trying to go to the denser medium from the rare one. And if reflection is happening over there, that is a case of again hard reflection. I'm saying hard reflection because hard reflection means a phase change of pipe. And phase can play a big role when two ways superimposed. It is a phase difference that matters. Okay. Reflection from denser to rare. This is a soft reflection case. Okay. So for light, we are not going to study much. We just since we are talking about reflection, the best reflection which we are aware of in our day to day life is reflection of the light. So I thought I could just tell you little bit about the light reflection also. Okay. So these are the cases and you have a chapter called wave optics in which you are anyway going to discuss the superposition of the light waves later. Okay. Somebody is asking, sir, in the soft reflection, what is the phase difference? Yes, there won't be phase difference. The velocity of the light changes, but the direction of velocity of light changes, but the phase difference remains zero. Okay. All right. So now somebody was asking the path difference. Phase difference we have seen. Right. So I'll just talk about the path difference over here.  So two ways are there. Y1 is equal to a sin of kx minus omega t. Y2 is equal to a sin of kx minus omega t plus 5. Now what is the phase difference between them? Everyone. What is the phase difference? Phase difference is 5. Okay. It is 5. Path difference is what? Path difference is pretty simple. Path difference is difference. Path difference is equal to distance traveled by wave 1 minus distance traveled by wave 2. Path difference is the difference in the paths. Okay. Now why it is so important because the distance traveled can itself create the phase difference. For example, let me take an example on the other slide. I'm first telling you why it is important. What it is, I'm coming to it. Let's say this is the wave, one wave, and this is the other wave. Okay. Now at the start, when both started at t equal to 0, what are the phase difference? What does it look like? Is there any phase difference when both started 0? It is 0. At t equal to 0, no phase difference, t equal to 0. Both the waves are moving. Suppose the above wave has reached here, this is point P, this is point Q. Now tell me, is there a phase difference between these two points, P and Q? They may not have reached at the same time, but is there a phase difference or not? What does it look like? Looks like, right? P and Q are not same as A and B. A and B, they are very similar. There's no phase difference. P and Q, there's a phase difference. And why there's a phase difference? The reason is wave one has traveled more distance than wave two. Wave one traveled more distance compared to wave two. That is the reason. So how do you take into that? How do you account for that? Let's see. This is a phase of 0. All of you agree? This is pi. It's a sine wave. This is 2 pi, 3 pi, 4 pi. And distance is traveled along this line. So can you correlate distance in terms of phase? Think of anything. Can you correlate distance with the phase? Like if you travel this much distance, this much phase will be added up like that. Can you think of? Correct. How should I got it? So you can see that one wavelength distance corresponds to 2 pi phase. Yes or no? One wavelength from here to there. It is 2 pi. Every time you travel a distance of lambda, 2 pi phase difference will get added up. So if let's say above wave traveled the distance of L1, so the phase of the above wave will be 2 pi by lambda is phase per unit length multiplied by L1. Below wave, the phase will be distance traveled is let's say L2, 2 pi by lambda into L2. So the phase difference between P and Q, 5 1 minus 5 2 becomes 2 pi by lambda multiplied by L1 minus L2. So if you multiply the path difference with 2 pi by lambda, you will get a phase difference. And vice versa, if you multiply lambda by 2 pi to the phase difference, you'll get the path difference. Is it clear to all of you? Now it is clear path difference. Type in everyone. Is this clear? Is L1 L2 a distance move in the same time? Yes, you can say that. I can show you a situation, but is this clear whatever I have written here? Okay. I'll just show you a situation in which it happens. Suppose this is a screen, two openings. These two openings, light can enter. Okay. So the light enters from here. It travels here, there. From here when light enters, it travels there to meet. So suppose these two light have, these light is a wave, so it travels as a wave and meets at point P from these two points. It can meet here also. So the path difference here is zero, but path difference when these two meet over here is not zero. Is it clear to all of you? The wave two has traveled more distance compared to wave one when they meet. Satyam, is it clear? Okay. So there are a lot of such scenarios. We are going to learn all this in grade 12 actually. Part difference was not as important in 11th as it is in 12th, but since it was there in the assignment, we discussed it. Is there anything in the assignment that you have seen and we haven't discussed? Anything else? No. Let me first ask how many of you tried the assignment? Okay. Someone is asking if, if phase difference is zero, it's not compulsory that part difference is zero. If phase difference is zero, phase difference could be zero or two pi or four pi. It is similar because after every two pi phase, things get repeated. Okay. So if you are, if you say zero, it should be exact zero, then path difference should be exactly zero. But if you're like phase difference can be two pi, it is same as phase difference zero. Then the path difference is lambda because lambda corresponds to two pi. Four pi. Then path difference is two times lambda. Okay. All right. So let us move forward. Coming back to the standing waves. How do you find out if it is zero or two pi? Don't worry. It will be there in the question like that scenario. It will be there. You'll understand. Okay. Like, for example, here. In this case, if L2 minus L1 is lambda, the distance difference is lambda. The phase difference is two pi. If it is, it comes out to be two lambda phase difference is four pi, but we assume that while entering these two, do not have any phase difference. Got it. While the light enters from here and here, initially there is no phase difference. If initially there is a phase difference, you need to account for that also. Initial phase difference plus final phase difference. That is a total phase difference. So more and more doubts you ask, it becomes. Okay. So let's move forward. And what is the time right now? 15 minutes for the break. We will again come back to the standing base. Okay. So when standing wave gets created, when two waves travel in the opposite direction, they're superimposed. All right. And the best possible way of superimposing or best possible way to create a wave that travels opposite direction to the initial wave is by having reflection. Okay. Reflection. When it happens, there can be many scenarios in our curriculum. A couple of standard scenarios are there. Let us discuss them. Okay. So now things will be pretty simple. Less mathematical, more conceptual. Write down standing wave on string. That is what we are going to discuss. But when we discuss standing on a string, there are some thumb rules that we need to keep in our mind. Point number one is if a point is clamped, or you're holding that point, you're not allowing that point to move at all, then what that point should be? Then automatically that point becomes what? Then that point must be what? Node, anti-node, what it is? A node is a point having zero amplitude should be node. Okay. And if a point, this is like a constraint things. If a point is free to move, the end point, okay? Not just any point. The end point of the string is free to move. Then that must be anti-node, maximum possible amplitude of that point. If the end of the string is free to move. So there are two cases. Case number one, both ends are clamped. Case number two, one is clamped, other is free to move. We cannot have case three where both are free to move. Then how will you have a string? It will just fall down. So case number one is when both the strings, sorry, both ends are clamped. Like this. If both ends are clamped, can you draw largest possible wavelength here? What's the largest possible wavelength so that these two points are the nodes because A and B cannot move. Can you draw largest possible wavelength standing wave here? Okay. This is one full wavelength. This is one full wavelength. Okay. Let me know once you're done. Done. Okay. So all of you agree that this is the largest possible wavelength you can have because you have to make sure A and B are the nodes. Is it clear to all of you? Largest possible wavelengths in the standing wave. Type in, is this clear? So if wavelength is lambda, what is the relation between wavelength and length of the string? Lambda and L, what is the relation? Lambda by two is equal to L. Half of the wavelength is L, right? So the wavelength is two L. And if tension in the string is T, mass per length is mu. The velocity is under root of T by mu. The velocity of the wave. Velocity of the wave has nothing to do with what wave is doing. Velocity of the wave is always under root T by mu. Okay. It is a medium's property that drives the velocity of the wave. Which is under root T by mu. So the frequency is velocity divided by wavelength. This will be one by two L root over T by mu. Okay. So this is the largest possible wavelength. And this frequency is what? Smallest possible or this is also largest possible? Smallest possible? Smallest possible? Right. And anything which is smallest possible? In physics, we call that as fundamental unit. Okay. So since it is the smallest possible frequency that is possible, we call it fundamental frequency. Okay. This is the fundamental frequency. Now fundamental frequency, just write fundamental. Okay. Writing it like fundamental is, it is like the common name. You have learned the nomenclature in organic chemistry. Right. There you have two kinds of nomenclature. One is common name. Other is IOPSE name. Similarly here also you have two names for the same thing. You can call it fundamental or you can call it in a more formal manner. First harmonic. First harmonic. Now let's see the next possible wavelength. Can you all draw the next possible wavelength? We have drawn the largest possible wavelength. Can you think of next possible wavelength? Having A and B as nodes. Let me know once you're done. Done. Okay. So do you all agree that the next possible wavelength is this? Yes or no? It should be this only. There is no other way. So here the wavelength is half of the previous wavelength. Right. So between A and B it is one full wavelength. One full wavelength. So now the length is the wavelength only. Lambda is the length. The velocity of the wave is root over T by mu. So the frequency is velocity divided by wavelength. So one by L root over T by mu. Which is two by two L. Why am I writing two by two L? You'll understand. You can see it is the frequency is two times the previous frequency. All right. The name is first overtone. First overtone means after fundamental whatever comes is first overtone. And it is second harmonic. Why is called second harmonic? It is second harmonic not because it is the second possible frequency. It is second harmonic because it is two times the first frequency possible. Okay. That is why it is second harmonic. So can you tell me the formula for the nth harmonic frequency? What do you think that should be? Can you see some pattern here? Nth harmonic what would be the frequency? N by two L root over T by mu for the nth harmonic. All right. Now, what is the time? Should we take a break now or later? Let's complete one thing. Then we'll take a break. Case number two. Case number two is one and is free to move. The situation is like this. You can move this point up and down so you can create a wave. So can you draw the first possible standing wave here? A should be node. It is fixed. A is fixed. You're not doing A up and down. Sorry about that. A is fixed. B is free to move. So the first possible standing wave. How does it look like? Let me know once you're done. Done. Okay. So the other end should be anti-node. So this is the first possible standing wave. How many times is wave length it is? Length L. This is lambda by four. One fourth of the wavelength it is. So lambda is four L. Okay. Velocity is root over T by mu. So the frequency is one by wavelength which is one by four L. Multiply that by under root T by mu. This frequency is called the fundamental frequency. There's also the first harmonic. First one is always first harmonic. Now try drawing the next possible frequency and derive the expression yourself. Let me know once you're done. For the next possible frequency. Done. Good. It will be like this. Okay. This point mandatory have to be anti-node and that point mandatory have to be a node. Okay. How many times this wavelength it is? Lambda by four, lambda by four, lambda by four. Three times lambda by four the length is. So three lambda by four is length L. So lambda is four L by three. Velocity is root over T by mu. So the frequency is velocity divided by the wavelength. So three divided by four L. This is root over T by mu. Okay. So it is called first overtone. Which harmonic it is? Which harmonic it is everyone. No. This is not second harmonic. How many times this frequency is of the fundamental or the first harmonic? How many times it is? Three times. So this is third harmonic. So you can see that even harmonics don't exist in this case. Okay.  It is not that you can have any frequency possible. The minimum frequency for the standing wave is the fundamental. Next possible frequency for standing wave is this. First overtone. You cannot have any other frequency of standing wave between these two. So that is why these frequencies are important. All right. I'll just give you one situation here. Tell me, can you draw the next possible frequency? Second overtone. You have already drawn. Draw the second overtone and find out the frequency. Everyone, right? Then this point have to be anti-node. The rightmost point, leftmost point have to be a node. So this is what you will get. Typing, is this clear? Next possible frequency is something like this. Everyone, type in. So how many times the wavelength, the length is? Count lambda by four. One, two, three, four, five. Five times lambda by four. Okay. So five lambda by four is L. So lambda is four L by five. So again, when you see the frequency, it is five by two L. Even harmonic doesn't exist. Four by two L doesn't exist. This is the second overtone. And fifth harmonic. Okay. So if you're looking for a generic equation, two N minus one eight. Harmonic. The frequency of that is two N minus one divided by four L root T by mu. Yes, there is a four in the denominator. Okay. So let us take a break. Now at least. Wait. Sorry. Meet after the break. All right. Can you hear me? All right. So let us proceed. Next is the, you know, when we study the longitudinal wave, it is the outcome of two waves moving in opposite directions. Fine. Now these two waves, when they move, they can be transverse wave as well as longitudinal waves. But the example that we have taken is the example of a transverse wave because a wave on a string that example we took. Similar to what we did just now, there can be longitudinal standing wave also, but longitudinal standing wave. Like sound. So standing wave was sound. Okay. So that can be inside. Let's say a pipe. It can't be on a string. Fine. So we will take next example for the standing wave as. Okay. I have to show you a quick video, which I forgot. This video. Okay. I wanted to show it to you because we have learned enough to understand this one. So just all of you see this. Can you hear him? It's going to pump up and down. Whatever frequency I set this guy to. When this guy pumps up and down, well, he sends away. Is it audible? Whatever the sound is coming from the video. Okay. Now listen to him. Sends away. Down to this end. The way the flex comes back. If I set it to just any old frequency. So it's sending waves back and forth, but these waves are out of sync. If you like, they're they're not exhibiting consistently constructive or destructive interference at any one point. But if I set it here, for example, which harmonic it is. This one is which harmonic. Correct. I have what's called a standing way. That is the way that's that's sent down and reflects. Constructively interferes with the next wave. It comes down right at this point. Now the waves that reflect. That come down and reflect. Destructively interfere with the incoming waves right here. And that's called the note. I can touch it and it's still working. The note is not moving. Here it's moving up and down. And here it's moving up and down. And we'll see that in a second. You can see all the particles. They are together moving up together going down. Here we have what's called the second harmonic. It's the second possibility for a standing way. For a string that's held at both ends. Because notice this is a node. This is a node in the middle. We have three nodes for this way. Now, like I said, second harmonic, what's the first one? The first one will be half this number. So 8.5. This is the scenario of a jump. I can find any of the other harm. All of the other harmonics are multiples of this number. So we just saw that two times this number was the second harmonic. Second possibility. Well, if I go three times this number, what's the third harmonic? Well, three times 8.5. So that's the third harmonic. How many wavelengths is this? Well, it's one, one and one-half wavelengths. It's easier to see in high speeds. So check out this shot with a high speed camera and verify for yourself that it is one and one-half wavelengths. So again, third harmonic, one and a half wavelengths. Well, let's go to the sixth harmonic. Let's go to the sixth harmonic. The sixth harmonic will just be double this frequency. So we go up to 50. And there we go. That's the sixth. So you can see like that you can keep on increasing and get the various frequencies.  Now we are talking about longitudinal standing wave. Just a second, someone is calling. Sorry. Right now longitudinal standing waves. Now here also you can have different cases. And just like we had some thumb rules with the string that the clamp will be node and the end which is free to move will be the empty node. Similarly here, we'll follow these two rules. The first one is the open end of the pipe. Open end acts as anti node. Okay. Whereas the closed end acts as node. So node is nothing but you can say compression. Compression zone. So wherever it hits the wall, that's a compression, right? And this is rare fraction. Okay. So let's do that now. We'll have case number one. Both ends open right now. Both and open. Now, can you draw a wave having both ends open? Sorry, having the anti nodes on both the ends. Draw the largest possible wavelength. Having both ends as the anti nodes, because both ends are open. Let me know once you're done. So I hope all of you understand that this is the maximum possible wavelength. Okay. You can't have longer wavelength than this. Some of you can say that why not this frequency like this?  Straight line is not a wave. It has to be some sort of curve. Okay. So we have drawn it like this. My question to you is, does it look like a longitudinal wave? Does it look like a longitudinal wave? This. Longitudinal wave doesn't look like this. Longitudinal wave is the rare fraction, rare fraction, compression. It is like this. Okay. But then it is very difficult to visualize standing wave. If you draw a longitudinal wave like this, right? So to visualize it properly, we are drawing longitudinal wave as if it is transverse wave. Okay. Mathematically, both are same. That's why. So if length is L, this is lambda by four, lambda by four. So two times lambda by four. Lambda by two is L. So lambda is two L. Okay. Now what is the velocity of sound? Velocity of wave is what? In this pipe. Everyone. Velocity of the wave in this pipe. How will you write? Perfect. Under root of B by rho. This is a longitudinal wave. So the frequency velocity by wavelength one by 12 root over B by row. This is called fundamental frequency and first harmonic. First harmonic. Now I want you to draw the next possible wavelength. Let me know once you're done. So if two ends have to be anti-nodes, then right? So all of you understand that it will be like this. Type in next possible wavelength will be like this. Is it clear? Type in, right? So if this is the next possible wavelength, can you tell me what is the formula for the frequency of this? If length is L bulk modulus is B and density is 0. Tell me what is the formula? Type in. Wavelength is how much? You can see lambda by 4, lambda by 4, lambda by 4, lambda by 4 times lambda by 4 is L. So the wavelength is L only. One by wavelength, which is L root over B by row, which is 2 by 12. There's first overtone, second harmonic. Similarly, you can write down nth harmonic. It will be n by 12 under root B by row. This is nth harmonic. Which overtone this will be? Everyone? Which overtone this will be? n minus 1th overtone. Good. So this is the case in which both ends are open. Now, let's say one end is closed, other end is open. Theory is about to get over. Just two, three minutes. Theory is over. Then we will do numericals only. So pay utmost attention. Case number two. One end is closed. Other one is open. Let's say like this. If length is L. Okay. Bulk modulus is B, density is row. Can you tell me the fundamental frequency, what is the formula? Do it yourself. Largest possible wavelength. Draw it. This end must be a node. This end must be anti-node. Satyam got it. Others. All of you, I hope you agree that fundamental frequency. The largest possible wavelength is this, and this is one fourth of the wavelength. Right. So lambda by four is L. So lambda is four L. The frequency is one by four L under root B by row. Fundamental first harmonic. I want you to do the same thing for the next possible frequency. The same thing for the next positive frequency. Derivate your own. You'll see that if you do that, numericals will appear very, very simple. Okay. Satyam got it. Others can be got it. All right. So I hope all of you understand you'll get this. Smriti also got it. One, two, three, three times lambda by four is L. The lambda is four L by three. The frequency is one by lambda, which is three by four L under root B by row. This is first overtone, which harmonic it is. Third harmonic three times. Three times the first harmonic is third harmonic. So even harmonic doesn't exist here also even harmonics. Don't exist. Okay. So if you want to write the generic expression. Two and minus one divided by four L under root B by row. This is two and minus one at harmonic. Okay. I mean, these are the final expressions. You can choose to memorize it. The final expression. Or whenever numerical comes, you can quickly, you know, derive it hardly take let's say five, six seconds. All right. So if you try, if you have a habit of quickly analyzing and deriving it, you'll never go wrong. If you memorize it, probably memory can trick you sometimes. Fine. So there are certain situations. For pipes, last bit of theory I'm talking about. For pipes. Whenever we deal standing wave with the pipes. There are some, there are some and corrections. Okay. What is that and correction? The end correction is the anti-node gets formed roughly at a distance point three times diameter away from the pipe. It doesn't get formed anti-node when you draw it like this. We assume that anti-node gets formed over there. But in reality what happens is the anti-node gets formed slightly above this distance is point three D. This is, this is the correction, which many times you'll ignore it for school exams. They don't consider end correction at all. But I've seen in computer exams, they will give you the diameter of the pipe so that you consider the end corrections. If they are giving you diameter of the pipe, then they want you to consider the end correction. So that is the idea here. So does it mean that our formulas and equations everything gets changed? The answer is no. All you have to do is this. If the pipe is open from one side, one end, then you just write the effectively the length would be as if it is original length plus point three times diameter. In all the derivations, rather than considering it a length of L naught, you treat it as if it is L naught plus point three D length pipe. And if pipe is open from both ends, then you have to consider the end correction both sides. So consider it as L naught plus two times point three D. It is L naught plus point six D. Consider this as the effective length as if your length of the pipe is as if your length of the pipe is not L naught. It is L naught plus point six D. Okay, so whatever theory I have to do today. I am done with that. Anyone has any doubts on theory? Whatever we did since the start. Quickly type in. I got it. Any clarification? Any doubts? No doubts. Okay, so it looks like all of you will get all the questions correct. So let us solve some questions. Fine. We have about 40 minutes. We can do a lot of questions. Do this. Try solving it. Resonates with 420 hertz or 490 hertz means that frequency is 420 hertz. Resonate means the same frequency it has. Okay. Don't worry about that. Satyam got some answer. Satyam, check for the calculation mistake. Aja, can you can solve first for this? Find the find which harmonic 420 hertz is which harmonic. The second harmonic, third harmonic, fourth harmonic, what it is? No one. What else? Okay, I'll do it now. So we know that the frequency. If it is tied from both the sides, the frequency is N by 12 under root T by mu. Okay. So this is let's say 420. If this is 420, 490 would be what? Instead of N, what should I write? It is the next high frequency, right? Next high frequency. Instead of N, you'll have N plus one. Are these two equations clear to all of you? Type in. Okay. Now my question is what is N then? How do you find N? Divide these two. You'll get N divided by N plus one. Equal to 42 by 49. Which is 6 by 7. So what is N? It is an integer and has to be 6. Okay. Once you get N to be equal to 6, now you can substitute the value of N over here and 6 by 12 root over T by mu, right? T is 450. Mu is 5 into 10 is 4 minus 3. This you can equate to 420. This is what you did, Satyam. I can do whatever answer will be. Answer you should get is this 2.14 meters. Okay. This is how you solve this question. All right. We'll move to the next one now. This is a simple one. Everyone solve it. Okay. Satyam got something. Let me check the answer. Check for the silly error. What is that something which will be same for both the pipes? One thing that is same for both. Velocity of the sound. Good. Velocity of the sound will be same. So the fundamental frequency of the closed organ pipe. So it is closed from one side. Let's say length is L1. Fundamental frequency is 1 by 4L times root over D by rho. Okay. Now open from both the sides. Length is L2, let's say. First overtone as in second harmonic of an open organ pipe. Second harmonic is like this. Where wavelength is the length. That is, this is mu1. This is mu2. 4L1. 1 divided by L2. Under root B by rho. Okay. Both are equal. Right. Both are equal. So mu1 is equal to mu2. This L2 is equal to 4L1. Length of open pipe is 60. So L2 is 60. So L1 is 60 by 4. Which is 15 centimeter. All of you understand where you made silly error. Type in those who got the wrong answer. You should not have got the wrong answer because these are very, very simple questions. Fine. You just cannot afford to lose marks in this. Whenever you write any combat exams. Even came on test for that matter. Have you ever observed that in a came on test also at least 50 to 60% questions are very simple. The problem is we do not know where they are. So you need to find them and solve it and then go to the difficult ones. So easy questions are more important for you. Give them more respect and be more attentive. Never make silly error in an easy question. Okay. Next one is a question from your textbook. Solve this. I have taken from your NCRT. Guys, do you know tomorrow you have chemistry session? Do you know this? Tomorrow chemistry class will happen. So as told last week, right? Tomorrow is extra chemistry session. Okay, please come to the class. We want to finish the curriculum well before your exam. So that is the reason why tomorrow will be extra chemistry session. He will send a reminder anyway in the group. Okay. 200. We shall put the answer others. Okay. So some of you have answered already. Let us see. Let's see. If you have made silly error this time or not. A pipe of length 30 centimeter. So 0.3 meters is the length. Is open both ends. Frequency that is given to us is 1100. Which is 1.1 kilowatts. Which harmonic mode. It runs. It is open at both ends. So n by 12. Into speed. Speed is directly given to us. We don't have to find out this should be equal to. 1100. Okay. Length is. 0.3. So you'll see that n by two into 330 is equal to 330. So n becomes equal to two. So it is second harmonic. Okay. It's a second harmony with the resonance with the same source be observed if one end of the pipe is closed. Whatever is a frequency. That frequency must match with this formula. Two n minus one. By four. Into the velocity. V. Which could be under root b by row. But velocity directly given to us. This should be equal to 1100. So two and minus one divided by four. Multiplied by 330. Is equal to 1100. All right. If I solve this equation. And must belong to integer. If n doesn't come up to be integer. If n is not integer. When I solve this. Then it is not possible. Okay. Let us see that. 11. 10s. 11. 11. Okay. Okay. Let us see that. 11. 10s. 11. 3s. So two and minus one. By four. Is equal to. 10 by three. Minus one is. 40 by three. So two n is. 43 by three. That does it. How I make a silly error. I didn't took the length here. There should be length L. In the denominator. Lent L in the denominator. When. This into. Lent. Lent is what. Lent is point three. So two and minus one becomes equal to four. So two and is equal to five. And comes out to be. 2.5. Not integer. So not possible. Okay. All of you is this clear. This is from your textbook. All of you type in. Okay. It's a very simple question. Though it now. Actually it is not on standing wave actually. But. Very. You can say it is. Question from wave chapter but not exactly on the standing. Anyone got the answer. A part. Is very simple. We shall how. That is a wavelength. I thought you're telling point zero three meter is a depth of the sea. One zero three meter is like three centimeter. Okay. September got party others. Part A answer. What it is. Okay. So let's say this is the sea. All right. So the speed of the wave is 1500. The signal is generated here. It goes into the sea.  Comes back. How much distance it has traveled. If length is L. It has traveled a distance of two L. Two L distance is traveled. Right. It at this. This much time at this velocity. So velocity into time. So the length is. One five zero zero. Into point four. That is. 600 meters. That is. Length or you can say the depth of the sea. Be part. Wavelength of this signal. How will you find out. Frequency is velocity divided by wavelength. So wavelength is velocity divided by frequency. Velocity is. 1500. Divided by 50 kilohertz. And comes here. I'm here. So point three. How many zeros. Three. Four. One, two, three, four. One, zero, three. One, zero, three meters. Okay. A straightforward question. This is the school level stuff. Okay. Just pay a little bit more attention. All of you will be able to solve it. All right. Let us find out something else. Do this. On standing wave. Yeah. Point zero. Did I write point zero five? What did I write behind? Do this. And while focusing on calculation. Focus on the concept calculation. It's your. Shall we do it now? Okay. Anyone have to get the answer? It's a guitar string. Of length. Point nine meters. The fundamental frequency is. 124 hertz. Where should it be pressed to produce a fundamental frequency of 186 hertz. Okay. So what will be the same between the two scenarios. What will be the something that will be same. For the two scenarios. Let's say it is pressed over here. Effective length would be now this one. Velocity of the wave will be same. Okay. So this frequency. 124 is a fundamental frequency for this much length. This is equal to one divided by two L. Into V. Right. 186 should be equal to one divided by two L one. Into V. These are two equations. You can divide it. We will be gone. 124 divided by 186. Anybody did like this and get the answer. What do you get for this multiplication? This is equal to. 0.6 meters. Is it. Let me quickly. Check it. 124 divided by 186. Yeah. Okay. You guys are also checking with the calculator. Don't use calculators. Okay. I mean you will not get it in the exam. So you should have a habit of doing. Hand calculations. Do this. Okay. One answer is there. Let us do this. Let us do this. Okay. It's a sonometer wire. Length is one meter. Where should that two bridges be kept? Let's say this is one bridge. Other bridge. Okay. Is L one. This is L two. This is L three. So that their fundamental frequencies are. Of the ratio one issue to three. So there'll be a standing wave here. Standing wave there. And standing wave there also fundamental frequencies. Okay. Fundamental frequency for length L one is one by two. L one into velocity of the sound or rest of the wave. F two is one by two L two velocity of the wave. F three is one by two L three. And also the way. We know that F one is to F two is to F three. This is one by L one to one by L two to one by L three. This is one is to two is to three. Okay. So one by L one is let's say K. Then one by L two is two K. One by L three is three K because the ratio is such. So alone is one by K. Two is one by two K. L three is one by three K. And L one plus L two plus L three should be equal to one meters. So basically one by K plus one by two K plus one by three K is equal to one. What do you get the value of K? What is K? K comes out to be K comes out to be six by eleven. So L one is one by K eleven by six. L two is one by two K. So eleven by twelve. Then eleven by eighteen. So like that we got the answers. Clear to all of you type in. Actually this thing is slightly together. Mathematics part is slightly twisted. Other than that, it is all fine. Arjita is saying K is eleven by six. Is K eleven by six or six by eleven? September what do you get? You want me to do the calculation also here. K is in a denominator Arjita and Vishal. K is in the denominator. You haven't noticed it. This is six K. Six plus three plus two is one. What is K? Eleven by six. Oh, what's up to them? All right. Yeah, of course. When you add up, it can't be written one. So eleven by six is K. Six by eleven is L one. Okay. So like that you can get the value of. Now use these equations. This one and this one to solve it. All right. Let's take one last question for today. And here is the lucky question for you. What could be the fundamental frequency? What do you think? And what kind of tube it is one enclosed or both hands open? Can you see that 600 is two times 300. So can it happen for one enclosed? Even multiple times one frequency is the other one. Can it happen? No. One frequency can never be even multiple of the other frequency like this. Okay. All right. So we need to first find out the fundamental frequency. So if one of them is fundamental frequency, it can be 300 only fundamental is the least frequency. All right. So it is two times 300 well and good, but 750 is not integer times 300. So it cannot be 300. It is less than 300. Okay. It cannot be 100. Okay. Can it be 50? The answer is yes. But if it is 50, then 50 into six is 300. So there are, there are not six possible frequencies. There are not many possible frequencies then. So it cannot be 50. So is 150. Looks like, because 150 into one is 150. That is fundamental. 150 into two second harmonic 300. 150 into three, 450. 150 into four, 600. 150 into five, 750. And 150 into six is 900. These are the six harmonics. 150 is a fundamental one. And this is missing. And which one? 750, 900 is there. 150. 150 and 450 are missing. All right. So this is how you solve this particular question. Fine. We will again meet next week and continue this chapter. Next week will be the last session on the waves. Okay. And then we'll sit down and understand what we should do in couple of more classes to come after the next week. Fine. So bye for now.


Hello and welcome to CMC Markets on Tuesday the 28th of October and the weekly market update and this week we're going to be disseminating some of the weekend events with respect to the ECB stress test, the effects they've had on the equity markets and also looking a little bit further ahead to the potential ramifications for this week's FOMC meeting, the penultimate FOMC meeting of 2014. Now we've seen a significant bounce back in equity markets over the past couple of weeks. In so much as we've seen the DAX recover back above 9000, we've seen the Italy 40 actually rebound quite strongly as well and we've seen the S&P return back to the levels around about 1,970. So really it's about where can we go to next against a backdrop of an ECB that still remains fairly hamstrung with respect to what it can do to boost the recovery in Europe and the FOMC which still seems likely will end its tapering program at its monthly meeting in October tomorrow on Wednesday. I think it's very unlikely that we will see any change to the timetable outlined by the Fed policy makers earlier this year. I think what we will see is that policy makers will reiterate the guidance that they've been issuing for the past few months, i.e. they will keep rates low for a considerable amount of time. They want to anchor interest rate expectations over the course of the next few months and in that context I think that makes it very, very difficult going forward for equity markets to rally with any degree of what I would call conviction given the very weak backdrop that we're currently seeing play out in Europe. Now for today I'm going to be looking at the German DAX as a follow up to last week's update. I talked about a potential move back above 9000 could actually prompt a break higher. I'm also going to have a look at the Italian 40, the FTSE MIB, the Italy 40 index to see whether or not we can see further gains there. Also going to have a quick look at crude oil and look for the potential of whether or not we could actually see a break lower towards that $75 mark and also look at the potential breakout that I identified in Cable last week. We do appear to be making progress on that and we could actually potentially go as high as $163 or $164. So let's make a start with the German DAX and we're going to look at a daily chart here because I think for the purposes of this presentation it's probably the easiest. We've seen a significant bounce back from the October lows at 8,350. We've pushed back above 9000 but at the moment we're struggling to break back above the 9,120 area. Now that is a 50% retracement from the September highs just below 10,000 to the lows that we saw earlier this month. Just before that also we posted a potentially key reversal day which is essentially a higher high and a lower low. So there is potential as long as we stay below 9,120 for the DAX to rebuff this particular level and head back lower and also looking at the slow stochastic it is starting to embark onto slightly overbought area. It's not just about the DAX though. Let's look at some other European indices to see whether or not we get a similar confirmation on them as well and we do. We get one on the Spanish Ibex 35 and we've also got a key reversal day on the Italy 40 and that's going to be seen by this chart in front of you. Here a very strong negative candle on the Italy 40. We opened higher, we closed below the lows of the previous two days. Now we could get a rebound but as long as we stay below the highs for this week then the potential for a lower Italy 40 remains a very distinct possibility. We've heard an awful lot about the plunging oil price over the past few weeks and the prospect that maybe the OPEC producers could actually intervene and try and underpin prices. Well let's look at the WTI contract because this particular contract does seem to suggest that maybe we could actually push quite a bit lower. At the moment this four hour chart we're trading sideways in a triangular consolidation and the prospect of the sharp move up or down will increase if we break either to the top side or the downside but we need to break out of this triangular consolidation at the moment. Now triangles are generally continuation patterns but trying to preempt a breakout is always a very dangerous thing to do so we wait for the breakout to occur and trade in a direction of the breakout with a very tight stop loss. If we break lower then the potential is for a move towards $75 on a break below $79. So keep an eye on that, it's certainly worth keeping a watch out for. So we're going to finish up with a pound against the dollar. Now you may recall I showed you this chart last week, we've moved on a bit since then, we've been in a descending wedge on this daily candle chart for quite some days now. We do appear to be showing signs of breaking out of it but again it's not really that conclusive. We've broken above $161.50, that's important. We need to stay above $161 for this move to unwind but let's also look at another pattern. Yes we've got a descending wedge but we've potentially got something else. So let's quickly flick from this chart to another daily chart but with a different pattern. Now what does that look like to you? Looks like an inverse head and shoulders. So this inverse head and shoulders if it plays out and stays above the neckline breakout we could well see a significant 325 point move higher from the breakout level which is around about $161.50-60 to around about $164. So certainly we've seen some evidence that the pound could be starting to squeeze those short positions. If this move is sustained we could actually see a significant short position shake out. Okay so that's pretty much it for this week and this month. Thanks very much for listening. This is Michael Houston talking to you from CMC Markets until next week and next month in November.