 So, let's take up numerical now. We'll take up the numerical from your textbook. A pi 30 centimeter long, let go. See, if you have to consider the edge effect, diameter will be given and then only you should consider edge effect. If diameter of the pipe is not given, ignore edge effects. In your school level exams, edge effect will not be there. In J kind of numerical, edge effect will be there once in a while. They'll be mentioning it that ignore edge effect or consider edge effect something like that. So, the length of the pipe is 30 centimeter. It is open at both ends. You need to find out which harmonic mode the pipe resonates with 1.1 kilohertz source. The velocity of sound is given as 330 meter per second. Which harmonic mode? Will even harmonic exist here? Yes, it will. It's open, sorry. Second harmonic, what do you think? Do the second harmonic. How about it? Do the second harmonic. Exactly. This is the second harmonic. What are you getting? Second harmonic? Yes, second harmonic. And this will do 1, right? No, and this will do 1. What is the formula for the frequency? N by 2L times velocity. Velocity is root p by rho. This should be equal to what? 1,100. So, N will come out to be equal to 1,100 into 2 into L, 30 centimeter divided by V, 330. This comes out to be 2. Second harmonic. How simple is it? Right? Now, one end is closed. Find out which harmonic now. If you close one of the ends, then which fit? Harmonic. It won't exist. It won't exist. Will the wave know that the one end is closed now? The wave was there, it suddenly closed one of the ends. So, basically, this is what was happening. N equal to 2, right? So, second harmonic, this is what was happening. This is what was happening? Yes or no? Now, you are closing this end. This end is closed. How can the closed end be the anti-node? It is not possible. You need to tell me if the cross section area is... Cross section area is 100 centimeter square. You need to find out how much water should you pour on this? How much water you should pour on this, so that it starts resonating? Ignore edge effects. Pouring water means what? If I pour water exactly, what is happening? The length is decreasing. How much water I have to pour? I have to pour water at least until here. This much water I should pour, then only the closed end will be node and now 1.1. The frequency is unchanged. Just one part of the wave is removed and it is resonating now. Are you getting this? 750, 750. How much is this distance? Height of the water column is how much? Lambda by 4? Lambda by 4. So the volume of the water is lambda by 4 into 100 centimeter square. That is much meter cube. So the mass of the water is density which is 1000 into 100 into 10 square minus 4 into wavelength divided by 4. This came sometime in J, sometime back. That time there was no concept of advance. What we will do is that, we will finish the chapter first and then one full class will do only problems of this chapter. By the way, any doubts? Yes, what is the doubt? Sir, in this question are we considering the wave travelling in water? Wave travelling inside water, you are saying. We will go. It will go, it will get transmitted, but we are ignoring that. How do we know when to ignore that? It will be mentioned, right? That is what you didn't say. Right, right, right. That is why we use water. Which experiment? Same experiment. Same experiment. Oh, you pour water and see where the sound of the, where the sound is maximum. So when sound is maximum, it starts resonating actually. Same thing. Okay, so let's finish the chapter first. They go beats. Have you heard of beats? What is beats? What is that? It's those... So what are the beats in songs? No? It's not that. Look at the beats. I can make you hear beats. Let me find out if I can. Okay, hear this out. This single frequency. Only one frequency is there right now. I am changing the frequency of one of the waves. They are super posing. Listen carefully. Or... Ding, ding, ding, like that. It has like... No, I'll open up another video. On YouTube, I search beats. Beat physics. What does that mean? This one. Watch this at home. Write down. Beats is an interesting phenomenon arising from interference of waves. So basically this is all the superposition. But the two waves that are meeting, they don't have same frequencies. Their frequencies are different. Okay, so first time we are super posing two waves having different frequencies. And let's see what is happening here. So in your textbook, they have taken cosine wave for some reason. So let's take cosine only. First sound wave is... Sound wave is longitudinal. So we'll take S1 is equal to... Don't talk. A cos kx minus omega 1t. Frequency is a difference. Everything else is same. A cos kx minus omega 2t. Here are the two sound waves that are meeting each other. They are super posing. Okay, now the trick is I am putting my coordinate axis in such a manner that wherever I am super posing, I am taking that as origin. If x will become 0, right? So I am assuming that I am standing at the origin and I am hearing these two sounds. A cos omega 1t and A cos omega 2t. So add these two and see what you will get. Then it will be 2A cos of omega 1. This is what you will get. Angular frequency. Omega is angular frequency. And this is what you get. The interesting thing here is that you get two time terms. Till now you are getting one time term and one x term. Now you have two time terms multiplied with each other. So visualizing this becomes tricky. So we will take case for the beats. Take a case if omega 1 and omega 2 are high frequencies but close to each other. But they are not so close that omega 1 minus omega 2 becomes 0. One could be like for example 300 other could be 294. Something like this. If time now tell me what will happen? This will change with time. This bracket term. With time. This will also change with time. Who is changing faster? The frequency of second one is higher? Very high. Frequency of the first term is? Very low. Are you getting it? So we humans can't hear a frequency which is very high. So we will not be able to hear this fluctuation. But we will be able to hear that fluctuation. So if omega 1 and omega 2 are close to each other we are going to hear this fluctuation only. Our ears will automatically filter out that. So let's see how the graph will look like for this. So let's say I am drawing on this. This fluctuation will rapidly fluctuate. So it will fluctuate something like this. So this is the kind of fluctuation you will see. On top of it there are two frequencies. One frequency is frequency of fluctuation. The other frequency is the frequency with which the amplitude is changing. Are you getting it? So we are going to hear this fluctuation. That fluctuation our ears can hear. But this rapid fluctuation our ears can't hear. So this is the phenomena of beat. When two frequencies which are close to each other when two frequencies close to each other superpose, two frequencies close to each other superpose beats will get generated. The beats will get generated. Now let's talk about beat frequency. With what frequency I hear the sound? Whatever I am hearing the frequency of that will be equal to the frequency of the wave or not. What I hear? I hear the wave or the intensity of the wave. Intensity of the wave. Intensity depends on amplitude square. Whether it goes to plus A or minus A. The effect of two mile ears will be same. So in one time period starting from here to here there will be two ups and two downs. Are you getting it? So basically if you just focus on one wave part of it let's say this is what I am focusing on. So there will be one portion and two portion. Two things will be there which will have the same effect. So what I am hearing is two times of that frequency. So the frequency with which I am hearing is omega 1 minus omega 2. This is called the beat angular frequency. This angular frequency, so beat frequency will be what? Frequency of beat will be equal to f1 minus f2. First write it down. Frequency of beat will be f1 minus f2. Now tell me can f2 be more than f1? Then beat frequency will be negative? It will be more than that, isn't it? So basically 300 and 296 will also give you 4 beats. 300 and 304 will also give you 4 beats. So it could be 4 less or 4 more also for the beat frequency. Getting it? Any doubts? So why is your beat wave the standing wave? It's not a standing wave actually. I should not have drawn this portion. It's like this. It looks like standing wave for the orange one, but that is not a standing wave. It's a rapid fluctuation. It is not a standing wave. So that one entire thing is one beat, right? Yellow part. This is one wave. In this one wave you will hear two loud sounds. You will hear loud sound corresponding to this situation and corresponding to that situation. So you hear two loud sounds per wave? Per wave. Per one wavelength you hear two loud sounds. So the frequency of beat is two times the frequency of that thing. So that is two times of that is the beat frequency. So this frequency of beat is not the frequency of the difference? No. How many times do you hear the difference? The entire beat phenomena is what we hear. In reality this is also happening, but we are ignoring this. Why? Because we are not able to hear it. So beat is with respect to what humans can hear. In this case both the beats will sound the same way. Intensity is amplitude square. Any other doubt? No doubts.