 So, in the last class, we had started talking about natural modes of rooms. We had started with one-dimensional room space, very very long ducts. Then, we had talked about 2 D rooms and I think we also covered 3 D rooms. We did not cover 3 D rooms. So, we will recap a little bit of that and we will extend the same analogy to 3 D rooms today. And we will also introduce slightly new concepts also. So, for 1 D rooms, we had said that the natural mode and the associated natural frequency, let us say it is f n is basically n times c over 2 l, where l is the length of the room, c is velocity of sound and n is the as an integer number. And I can represent this in a vector form like this. So, this is my frequency, this is my origin 0. So, the existence of a node is after every period, distance or unit distance, where a unit distance is c over 2 l long. So, every c over 2 l, I have a node f 3 for instance will represent this is f 3. So, the third natural frequency of this room will be f 3, where f 3 could be is defined as frequency associated with 3 c over 2 l. In terms of wavelength, each of these units is lambda over 2 in the direction of the length. So, that is for 1 dimensional. Now, we will extend similar thought process for 2 D rooms. So, in case of 2 D rooms, the frequency is associated with an x direction parameter and a y direction parameter. So, f n x n y equals c over 2 n x over l x square plus n y over l y square. And we had also shown that the angle of incidence of n x n y th mode or natural frequency of the room has to be tan inverse of n y times l x over n x times l y. And based on this understanding, we had shown some natural modes of rooms. Let us say I have a room, it is again it is a 2 dimensional room, its dimension in y direction is l y, dimension in x direction is l x. Then there could be a crisscross pattern of nodes and crisscross pattern of nodal lines in the room. In the boundary condition, the number of nodes in y direction is essentially n y, the number of nodes in x direction along the length of the room is n x. So, that is the physical interpretation. Also on each line, let us say I consider this particular line, I am just highlighting it and making it thick. So, along this line, my normal velocity v n is 0 and that is basically because the boundary condition at the point where this line hits the wall. If I have one node line going in this direction and another node line going in this direction, then these two lines, the velocity condition will be v n 1 equals 0. Let us say this is line 1 with the normal n 1 and v n 2 equals 0 for this line which has a normal n 2. These lines do not have to be necessarily at right angles to each other because essentially these are reflections. So, at point of intersection, this is by velocity condition. v n 1 is same as v n 2 is equal to 0. If the angle of intersection of these two lines is 90 degrees identically, then the total velocity is 0 at the point of intersection. We had developed this kind of a concept for a 1D room. We will represent similarly in 2D space the frequency of a 2 dimensional room and we do it in this way. So, I have an x and a y coordinate system. On the horizontal axis, I am plotting f of cosine theta and in the vertical direction, I am plotting f y sine of theta where theta as we had explained in earlier class is the direction of the incident wave. This is my x axis in a room. This is my y axis in a room. Theta is the angle of incidence. So, just as in this case, we had put a unit length as c over 2 L. If I extend this analogy in 2 dimensional space, what will be the unit along the length direction, x direction, c over 2 L x and in the vertical direction, c over 2 L y, 3 or 4 lines and you will have to bear with my not such a good drawing. So, this is c over 2 L x where L x is the length in next direction for the room c over 2 L y times 2. This is 4 c over 2 L x. This should be and this is 3 c over 2 L x. Similarly, in vertical direction, I have c over 2 L y, 2 times c over 2 L y, 3 times c over 2 L y and so on and so forth. So, the natural frequency of a room will be any intersection point of these two of the grids. So, this could be a natural frequency. This could be a natural frequency. This could be a natural frequency. This is not a natural frequency at this point. It is not a natural frequency because this is my intersection. So, for instance, if I consider this point e, then this is f 1, 2, 3, 4 in the x direction, 4 comma 3. Similarly, another natural frequency could be f 4 comma 2. So, for 4 comma 2, I can write the relation as 4 times c over 2 L. Basically, the length of this vector is the magnitude of the frequency. So, it is 4 times c over 2 L x whole square plus 2 times c over 2 L y whole square, the square root of the whole thing. And the angle of incidence associated with 4 comma 2, which has to be maintained to create this mode in the room is theta. And that is essentially tan inverse of this component divided by this component. So, it is 2 c, 2 c over 2 L y times 2 L x over 4 c. I take the inverse tangent of it. So, once again, this is my f 4 2. This is f, oh, I am sorry, it is 4 3 and this is f 4 2. So, essentially, we have extended this kind of a representation to two dimensional space for 2 D rules. So, now, we take the next jump and we go to three dimensional, 3 D rules. So, you can have a room. It could have a height, width and length. Again, these are rooms with parallel walls. All this analysis is assuming that the walls are parallel. So, this could be my x axis, y axis, z axis. The dimensions along x, y and z are L x, L y, L z. So, if I draw a three dimensional coordinate system and put a point in it, I have a point p. This is my z axis, y, x. I make an angle phi with respect to the z axis of this vector. And in the x y plane, the projection of this vector p, vector o p is o p prime. And o p prime makes an angle theta with respect to x axis. So, this is standard 3 D Cartesian geometry. Such a system, what will be the natural frequency for a room? If I use similar approach, it is basically you add one more term here, n z over L z whole square. So, f n x n y n z equals n x over L x square n y over L y square n z over L z square. The whole thing, we take a square root of times c over. And the way we express this frequency in two dimensional space for 2 D rooms, we will use a similar approach to figure out the value of theta and phi in three dimensional space for 3 D actual rooms. So, this room is more or less like a three dimensional room. It has a little bit of projection non rectangular structure there, but otherwise more or less a 3 D room. So, what I do is, now I represent this f n x n y in 3 D space. So, this could be a point P and if this point P is located at an intersection point of c over to L x c over to L y c over to L z lines, then it is a node and it is a valid frequency. Similarly, like we developed for 2 D. So, it has to reside at the node points at the grid point of these parallel these lines. It has to lie at the intersection of these lines. So, in this case I have to extend it a little bit. So, my vector O P, O P equals f, what will it be? This is my y axis, x axis, z axis. So, this is c over to L x L y twice thrice 4 and it could be 4 4 and in the z direction I have 3 0 1. I have to draw a better picture to show it in three dimensional space. It appears, but the point is that it is sitting somewhere which is 4 times c over to L x along the x axis, 4 times c over to L y along the y axis and 3 times c over to L z in the z direction. That is the location. So, maybe you can draw a better picture and, but that is what I am trying to show and some this kind of a representation I can figure out tan theta. So, tan phi is, man can you help with what is the value of tan phi in this case? Well, I mean tan phi, ok, I will, in the z direction it is 3 times c over to L z and in the projection of it in the x y plane will be. So, first we should take projection in the y direction first you have defined phi with z axis. So, sir then first term should be projection with y axis. In the x y plane. Sir, in the diagram you have defined tan phi phi is the angle with the z axis. Yes. So, then the y term should be taken this the z term should come should come in the denominator. Yes. So, it will be in the denominator it will be c c over to L z right and in the numerator it will be 4 c over to L x plus 4 c over to L y this entire thing right and tan theta is basically 4 c over to L y times 2 L x over 4 c. So, you have a process and an approach through which you can locate the physical nodes of different frequencies for a given 3 D rectangular room. You find out its coordinates and for each frequency in a very methodical way you can say that oh this is a node this is a node this is a node. So, at the nodal lines my pressure is minimum or maximum at points of where these things intersections are happening it is maximized right and my velocity is minimum. These node locations vary for with respect to each frequency. So, the challenge is that if there is a no room where there are lots of nodes big band of frequencies what does that mean in terms of perception of sound that as person moves from one place to other he will see big changes in the volume related to specific frequency. So, his understanding of sound or his perception of sound will change from place to place as he moves from one place to other place. So, that is not a good room to have right. There are some place he will see a higher level of SPL then he moves to an anti node location where he will have a lower level of sound then he moves to another one and so on and so forth. So, we have to figure out how to construct a room which does not have this challenge. So, that actually is a segue into the next concept which is all related to all these approaches we have developed to figure out what is the what is the location of a node in a room. So, now what we will try to do is we will try to quantify how many nodes are there in a room. So, we will develop that for a 2D room and then we will extend it for a 3D room. So, in 2D rooms I will draw the same x y system my x is same as f cosine theta and this is f sin theta and then I am drawing a. So, each of this is c over 2 L x 2 c over 2 L x 3 c over 2 L x and here I have c over 2 L y this is 3 c over 2 L y and so on and so forth. I want to find the number of nodes in this 2D room associated with the frequency f naught then what I do is I draw a quarter of circle with the radius f naught. So, whatever number of nodes are there inside this circle or one fourth of a circle that is the number of nodes this room is going to have right area of circle actually one fourth area of circle is pi f naught square over 4 and area of one of these guys these rectangles is what. So, number of node is how much I just take the ratio of these 2. So, pi f naught square over 4 times 4 L x L y over c square what do I get pi f naught square over c square times L x L y and this is pi f naught square over c square times area of room. I can make this generalization to area of room only if it is a rectangular room actually this is a more fundamental relation L x L y. So, what this says is that n is dependent on the square of frequency. How does this represent number of nodes? How does this ratio represent number of nodes? It represents how many rectangles are there in this c circle. So, but a frequency of this circle. There is an area left like in which it will contribute to this. So, this is a very broad estimate because you will see that at higher frequencies this number of nodes runs into millions it runs into millions. So, you can omit 2, 3, 10 you know, but it is a fairly good representation. It is a discretization of a continuous function. So, what this says is that number of nodes is dependent on the square of frequency as frequency doubles up number of node becomes 4 times. That is one thing and obviously, if you have a larger room you will have larger number of nodes. If you have smaller room you will have smaller number of nodes. Also, the larger room will have very closely spaced nodes because L x and L y will be larger. This c over 2 L x is almost a node at all the frequencies like if the room is larger. No, the tightness of one node to other closeness of one node to other will depend on the wavelength. The L x and L y are independent. L x and L y are independent, but what you have here is f naught over c which is wavelength. If you look at wavelength then f naught over c is the wavelength. So, the spacing of the nodes will not change from one large room to a small room. So, what is spacing is c over 2 L x h? This is a representation. This is the actual relation. You have to look at this relation in its totality. What this says is if you have big area of the room you have large number of nodes. What this says is f naught over c is basically lambda. If you have to relate it visualize it in terms of spacing of the room you can do an experiment yourself numerical experiment that what is the spacing of room whether if this if you have a room this big and then you make the half half as big as it and at the end of the day everything ends up in terms of functions of lambda because you have a node at the room away from that as I move inwards in a 1 d tube after a lamp distance lambda over 2 I have another node then after another lambda over 2 I have another node and so on and so forth. So, the spacing is not changing. What is changing is how many nodes you have in 1 d room. So, for 3 d rooms what will we do? We use the same approach for a 3 d room. So, instead of area of a circle what do we compute? 1 8th area is what will it be? 4 by 3 times pi f naught cube over 8. 4 naught 4 by 3. 4 by 3 right volume and volume of a brick I can call it of unit size in that x y z coordinate system where each dimension is C over 2 L x C over 2 L y C over 2 L z. This is C over 2 L x times C over 2 L y times C over 2 L z. So, number of nodes n is basically I take the ratio of these two guys what that gives me is pi over 6 f naught cube into 8 L x L y L z over C cube which gives me 4 pi over 3 f naught over C cube times volume of room. So, in a 3 d room the number of nodes explodes as frequency goes up. If I increase my frequency by a factor of 10 I have 100 1000 as many nodes in a 1 d room it just is linearly dependent in 2 d rooms it goes up in a squareish way in 3 d rooms there is a cubic dependent. So, we will do a very small example let us say my f naught is 20 kilohertz that is the highest frequency which a human being can hear and let us say L x times L y times L z is 25 feet times 15 feet times 8 feet. So, this gives me 3000 cubic feet and C is I am making a very crude approximation 1000 feet per second. So, I can either convert it to S i or this is cubic and when I do a C cube I get feet cube per second cube. So, it will so n is 4 over 3 pi times 20 times 10 to the power of 3 cube over 1000 cube times V room V room is 3000. So, this gives me 10 to the power of 8 there are 100 million nodes in a room at 20,000 hertz. If I bring this down to 2000 this number will become approximately equal to 10 to the power of 5. If I bring this down to 200 then this goes to 100 nodes. So, for low frequencies you will hardly hear change in the sound as you move from one location to another, but at higher frequencies at very high frequencies you will hardly hear the change in frequencies also because the spacing will be very tiny. Yeah wherever your ear can perceive you have typically you not typically always we hear from 2 ears. So, if the frequency the spacing is very tight this may hear something this may hear something and the brain will kind of average it and give you a kind of a mean or an RMS value, but if this spacing from this ear to ear is less than the spacing of the node then you may hear changes in pressure. Very big auditoriums let us say 55 as our auditorium and there would be some nodes for base as well. For very big auditoriums there will be nodes for base as well. There are bigger issues and very big auditoriums which is we will talk about it may be in next lectures reverberation and echoes, but yes for large auditoriums you will have change in perception of sound for low frequencies, but for most of these rooms of this size that is not an issue at all. So, very simple math, but it shows you a lot if I do dn over df then I get f naught square times 4 pi v room over c cube. So, again I will just do a very quick example for this example for these data my dn over df at f equals f naught comes to be 15 I calculate 15000. So, dn equals 15000 df df. So, if I make a 1 percent change in frequency as change in n where n is a delta n is a small change is basically 15000 times delta f naught or actually f at f equals 20 kilohertz. So, what that means is that at 20 kilohertz if I increase again just for a appreciation if I increase my frequency from 20 kilohertz to 20000 1 I get 15000 extra nodes in the room of the dimension. If I increase my frequency by 1 percent increase in frequency then extra number of nodes which I get is about 3 million you can use this relation that higher frequency is very sensitive to changes in frequency number of node changes very rapidly. So, what we will do now is a very simple experiment. So, I have a very basic speaker and what I will do is from the computer I will generate two two frequencies the first frequency will be 125 hertz and what you can do is that when the sound is coming you can close one of your ears and you can move your ear like this that whatever and see if you can sense the change in sound pressure level then I will stop that and I will play another frequency at 3 kilohertz or 2500 hertz and then try to figure out whether you can sense changes in sound pressure level. The other thing is that if I play the sound directly from my computer it is a very clean sound, but looks like there is some electronic issue. So, even if I am not playing it and when I run it there is some background noise static noise you can ignore that volume from the laptop volume from laptop volume from laptop is, but for purpose of this experiment I have checked it this noise will not lead you to wrong conclusions. So, what I am going to do is I am going to maximize this and play the ok. So, that was 125 and hold your thoughts. So, now, I will change the frequency and this is now 25 and 2500 hertz. So, what did you sense? Did you see is what was the difference when you play between 125 and 2005? Sir we can sense in both cases, but for the 125, the distance was large and I had to move my head from here to here. Then also the variation was not much significant. So, what we will do is for this room this is like a rectangular 3D room kind of there is some extra non-rectangularity here what we will do is compute the number of modes and see. Because of reflections and everything at the end of the day you will get all the modes which will be developed. What this room has is l x equals 7.06 meters l y equals 7.06 meters l z equals 3.12 meters. So, my v room is 155.5 meter cube. So, if I compute the number of nodes in this room is 4 pi f cube over c cube 3 times v room and if I do the math I got to calculate it today after long number of years I get 1.586 times 10 to the power of minus 5 f cube. So, for a 125 hertz frequency I get 1.586 times 10 to the power of minus 5 into 125 to 31 nodes and for n 2500 basically I can just multiply this by 8000 and I will get the same number right. So, it is about 2.48 10 to the power 5 equals 2.48 lakh nodes or 248000 lambda 125 hertz c over 125 is 2.76 meters. So, lambda at 125 hertz divided by 2 this is the spacing of nodes in an approximate sense because it also has to depend on theta which we are not talking about here is 1.4 and lambda for 2500 is 0.138 meters. So, lambda over lambda 2500 over 2 is 6.9 cm. So, the first thing which you may have sensed which Abhishek mentioned was that for 2500 hertz tone he had to move his head very little and it is directly related to the fact that lambda over 2 is just 7 cm about this. Sir, one question which may not be given but how do we can did particular frequency sound from that? I was browsing the net and there is some website where they have a lot of frequency tones mp3 files. So, I just downloaded two frequency and Shatij found some tools which I have not used where you can actually now input frequency and it will give you that particular tone. So, again to Shatij's point what you were saying earlier for low frequencies 100 hertz 60 hertz 40 hertz actually I tried playing 40 hertz on this but it will not it was not producing sufficient I could not listen 40 hertz sound on it because it is too small. You have seen that for 40 hertz to have a significant SPL level first is resonance point has to be low which we saw earlier probably its resonance point is may be 90, 80, 100 hertz and second thing is it has to move a lot or its pistonic area has to be large it could not generate also I tried playing 125 hertz on this computer and I could not hear anything just by computer. So, you can generate the frequency but you may not be able to hear it unless you have the right speaker. The perception of low frequencies does not change significantly in rooms of this size or even in larger rooms because you have to physically move a long distance from one place to another to measure that which brings me. So, to the next point that if I am going to plot the pressure at my ear right I am sitting in a chair in a room someone is speaking or some music is coming out or some noise is getting generated. So, my ear is hearing something and that is P ear and the source is generating some noise or sound content that is P source. So, if I plot P ear over P source and in the horizontal axis I am just varying frequency because of some of these effects which you just now experienced my plot will look something like this. As I move from one node to other node there will be variation in the pressure and in a lot of cases this is about 20 decibels let us say this is 20 dB. So, the maxima and minima are off by about 20 dB let us say and that is basically because if the person is not moving then for different frequencies he will have here different SP levels and there will be some play frequencies where he will hear the minimum it will be some frequencies where he will hear the maximum and that spacing will get tighter and tighter as frequency goes. You have to draw an exact curve but basically what I am trying to show here it is not sinusoidal it will be because it is a continuous function it is a continuous function it would not be jumpy think about this. This is P ear you can also replace it by a microphone and a P source could be a radio or a tape recorder or CD player. So, you have a CD player with speakers and everything and you have a microphone and then you take from the mic and then you record it that sound. This CD player could be producing a flat frequency response and what the mic is going to record will be something like this then you play this sound again and you hear it then the overall threshold. So, the first thing is in the first step this will be a flat frequency response this will be a wiggly frequency response. So, the even though this is producing sound faithfully mic is not recording it faithfully because there are echoes there are reflections there are this interference and because of all that the mic is recording something totally different when you are measuring sound this is very important. You may conclude based on this data that your source is producing this kind of a sound but that may not necessarily be the case that is why in reality there are lot of times people make measurements in anechoic chambers where there are no reflections but anyway going back. So, you are recording this. So, in next step your CD is producing this sound because I am feeding that recording back to the CD. So, it is producing this and then the mic will record from here it will this threshold will go down. The top will remain same but this threshold will go down. So, it will become like this do you understand what I am saying. So, each time you record an original soundtrack the threshold at certain frequencies will keep on going down and down and down and further that is the reason whenever you are trying to duplicate sounds. A lot of times it is not directly from what is being heard through a sound system rather than rather what is preferred is that you just directly take the electronic output in terms of voltages and recorded. If you want to replicate but if you reproduce sound it will also capture you will also be capturing the impact of reflections and echoes and that will destroy the original soundtrack totally. If you do this iteration several times and the more you do the more corrupted your overall sound recording becomes and it could be for engineering purposes it could be for entertainment and all that. So, that is something I wanted to share with you. So, that is all for today. In the next lecture we will try to capture we will try to explain a little bit about mufflers which is a little bit of a jump from what we are doing today. In several lectures back in time we had talked about Helmholtz resonators right. We are using a volume and a tube we could eliminate specific frequencies or not specific frequency a specific frequency or a narrow band located around a specific frequency. So, that is Helmholtz resonator what we will talk about is little bit about a mufflers which are used in a lot of automobiles or all the automobiles scooters motorcycles cars trucks how do they work and then we will also start introducing the impact of echoes what is a reverberant field in rooms how can you reduce the impact of reflections in rooms. So, that is what we will cover in next couple of lectures. That is all I wanted to talk about. Thanks.