 Welcome again. So, in the previous lecture, we had introduced this concept of volume velocity and we had developed the relation for pressure field attributable to a monopole and the relation looked something like this. So, pressure field due to a monopole is the real component of volume velocity divided by 4 pi times j omega rho naught times E minus j omega r over c times E j omega t. And where this volume velocity was defined as complex amplitude times r naught square times 4 pi, where r naught square times 4 pi equal the surface area of sound source emitting radially symmetric sound, sound pressure. So, if I have a sound source and if it is let us say radius is r naught and it is generating this sound which is radially symmetric, then that is the surface area which is emitting sound, I multiply that by velocity and that gives me volume velocity. The other thing we had done as we were developing this relation was that we had assumed that r naught which is the size of the sound source is extremely small compared to lambda over 2 pi and as long as this assumption held valid, this relation would be reasonably accurate. If this particular relation is not valid, then we have to go back to the first principles and modify this particular relation, so that we also include the effect of size. At this stage, I also wanted to introduce two more terms, so one is that if the device is small and I am doing measurements which are very close to the device, so such that my measurement are near the device such that is which is the position where I am taking measurement is 0 to 2 times lambda over 2 pi, then these type of measurements are called near field measurements and if measurement is far away such that r is very large compared to lambda over 2 pi, then I have these measurements are known as far field measurements. So, I have near field measurements when I am extremely close to device which is emitting sound in the sense that the distance between the device and my observation point is less than 0 to 2 times of lambda over 2 pi, then I will be doing near field measurements and if my location is extremely far such that the radius or the distance between observation point and the device is very large compared to lambda over 2 pi, then what I am doing is they are known as far field measurements. So, that is about volume velocity near field measurements and far field measurements. Now, we move one step further and we start exploring what happens if we have more than one monopoles and then how does the sound field get affected and at this stage we will start talking about the concept of interference. The concept of interference in case of sound waves is not in any fundamental way different from light ways when they interfere with each other or for that say any type of ways when they interfere with each other. So, we will use similar mathematical tools to predict the sound pressure field when there are more than one monopoles in the neighborhood of each other and essentially what we will do for purposes of predicting the sound field is that we will compute the sound field attributable to one particular sound source. The same thing will be done for a sound field attributable to another sound source and then if we have two sound sources we just add these two components and then we get the total final consequential sound pressure field at the place which we are interested in. So, we will start talking about interference and specifically in this lecture we will just talk about two sound sources. We will just talk about two sound sources to make the discussion simple and relatively easy to understand and then we will see how these two sound sources interact with each other to produce an interference sound pattern at some distance away in the pressure field. So, let us say I have two sound sources and this sound source is S 1 this is another sound source S 2 and this is a geometric line which connects these two. So, these two sound sources are separated by a distance D such that this is distance D over 2 and this is another distance D over 2. Now, I am interested in finding the sound pressure level at this point and let us say this point is P and the pressure here is lower case P and that is what I am interested in finding. So, this is my origin this point. So, these are two sound sources and this is my origin. So, from with respect to the origin this point P is located and the distance from the origin to point P is R and this angle is theta. So, I will just make it a little larger this angle is theta. So, the pressure at point P is lower case P and that depends on three variables R which is the distance between P and origin theta and of course, time. The distance of sound source S 1 is R 1 and similarly the distance of sound source S 2 is R 2. Finally, I draw some construction lines. So, I drop a normal from S 1 on the vector R and if that is the case then this angle is again theta this angle is again theta and if I extend this line. So, this is 90 degrees and from the origin which is point O to this point let us say this point is B then O B is basically D over 2 times cosine of theta because the angle between line O S 1 and the normal is again theta and O S 1 is the hypotenuse. So, the base of this triangle is nothing which is O B is D over 2 times cosine of theta. Now as so again I have two sound sources S 1 and S 2 I have a point P which is distance R away from origin. I am interested in finding the sound pressure field at point P and see how this pressure varies with respect to R theta and time and that is the basic problem which I am trying to address here. Now for purposes of this discussion we will make a couple of assumptions. So, assumptions so the first assumption is that S 1 and S 2 are monopoles. So, ideally they are point sources of sound, but in reality you cannot generate sound from point sources, but if they are if they still have to be like monopoles or approximate to monopoles then the size of these devices sound these sources has to be extremely small compared to the wavelengths they are emitting. So, S 1 and S 2 are monopoles. The second one is that the volume velocity of S 1 or I will call it V V 1 the magnitude of this volume velocity is same as magnitude of V V volume velocity of the second one. So, this is another assumption. So, essentially I have two monopoles they emitting sound and the strength of these two monopoles is same and their volume velocity is V is equal to V V this is the second assumption and the third assumption is that even though they have similar strengths the phase difference between the sounds being emitted by them is not exactly zero and there is a phase difference between the volume velocities of these two monopoles. So, phase of V V 1 minus phase of V V 2 equals phi and that is phase difference. So, these are the three assumptions. So, once again I will formulate the problem or I will restate the problem I have two sound sources S 1 and S 2 they are in terms of size they approximate monopoles they generate sound waves and because of these sound waves an interference pattern is getting generated at point P which is located at a distance r from the origin and which is also at an angle r theta with respect to the origin and I am interested in finding the sound pressure field in the neighborhood of in this entire area the strengths of these two monopoles are same even though they are off in terms of phase by an angle phi. So, with that statement I now proceed to start solving the problem. So, the first thing is that I will write down expression of V V 2. So, V V 2 is essentially it will have some magnitude and that magnitude we know is V V and it will also have some phase E j phi over 2 and similarly V V 1 is equal to some magnitude component and some phase and we know that because the phase difference between S 1 and S 2 is phi. So, it is magnitude it is phase will be minus j times phi divided by 2. Now, P of r theta t equals pressure due to V 1 plus due to V V 2. So, I will write first complex pressure and complex pressure is complex pressure due to V 1 and plus complex pressure due to V V 2 and then I will add these two up take the real part and then I can get the actual pressure. So, this is essentially V V divided by 4 pi r 1 times E minus j phi over 2 times j omega r 1 rho naught E minus j omega r 1 over C times E j omega t. So, this is my complex pressure due to first source plus now I have complex pressure due to second source. So, instead of r 1 now have r 2 now I am going to simplify this whole thing. So, I have several things common. So, it is V V over 4 pi and then I will also make j omega rho naught is common and so is E j omega t and then I have two terms in parenthesis minus j phi over 2 times E j omega r 1 over C and the entire thing is divided by r 1 plus E j phi over 2 times E minus j omega r 2 over C divided by r 2. Now, if r is extremely large compared to D. So, now, we are making an assumption here that if r which is this distance between the point of observation and origin if this distance is extremely large compared to the distance between these two sources then if r is extremely large compared to D then r 1 is approximately equal to r minus D sin theta over 2 and r 2 is approximately equal to r plus D sin theta over 2. So, now, I introduce these approximate definitions of r 1 and r 2 in my expression for complex pressure and from that I get complex pressure which depends on r theta and t is V V over 4 pi j omega rho naught E j omega t E minus j phi over 2 E minus j omega over C and then I am replacing r 1 by r minus and then again on the denominator I am replacing r 1 by r minus D sin theta over 2 and then I have another term E j phi over 2 E minus j omega over C and r plus D sin theta over 2 divided by r plus D sin theta divided by 2. So, I have this somewhat complex expression and now I will at this stage I will start simplifying it. So, one simplification is that the denominator this is approximately equal to r similarly this denominator is approximately equal to r. So, what I get is P r theta t equals V V over 4 pi r j omega rho naught E j omega t times E minus j phi over 2 E minus j omega over C r minus D sin theta over 2 plus E j phi over 2 E minus j omega over C times r plus D sin theta over 2. So, this is a somewhat simplified expression of the original version. Now, I can make this version further simple by taking E j omega r out of the bracket. So, what I get is I have taken r from outside from here. So, in the parenthesis what I am left with is E minus j phi over 2 times E j omega D sin theta over 2 plus E j phi over 2 times E minus j omega D sin theta over 2. So, this is my further simplified expression where I have taken r outside the parenthesis and I have clubbed it with E j omega t. So, now, what I see in the parenthesis which is this term that the first term is E minus j phi over 2 times E j omega D sin theta divided by 2 and the second term is just the inverse of it E j phi over 2 times E minus j D omega D sin theta over 2. So, this is essentially excuse me. So, we know from complex algebra rules that these two terms they add up if I expand them in terms of cosines and sines then the sin terms will get cancelled because the sin terms will get cancelled and we will be left with only cosine terms. So, that is what I do. So, I make it further simple. So, I get V V over 4 pi r times j omega rho naught E j omega t minus r over c times twice of cosine of omega D sin theta over 2 minus phi over 2. So, this is the simplified expression here. Now, we know that omega I think I am missing a c here. So, there should have been a c here and a c here. So, there is a c here as well. So, now, omega over 2 c equals 2 pi f over 2 c equals pi over lambda. So, I can replace omega over 2 c by pi over lambda. So, my complex pressure relation is further simplified and also I cancel this 2 by 4. So, I am left with only 2 pi r here. So, V V over 2 pi r times j omega rho naught E j omega t minus r over c times what I am left with is cosine of pi D over lambda times sin theta minus phi over 2. So, this is my final expression. So, what this relation shows is that if I know the volume velocity of these two sources and I am aware of the phase difference between the two sources, then I can calculate complex pressure using this relation and once I have the complex pressure, then I can take its real component and that will give me the actual pressure. So, I will rewrite this equation again. So, P and what I will also do is that I will drop the t term because that is just E j omega t. So, I will just take the drop that dependency you know for purposes of convenience. So, P is dependent the complex pressure magnitude is dependent on r and theta and that is V V over 2 pi r j omega rho naught E minus j omega r over c times cosine of pi D over lambda sin theta minus phi over 2. So, what we are seeing here is that the first term depends on r and then this term depends on theta. The second term depends on theta. So, even if the value of r is fixed and if I go around a circle, the value of the pressure at different values of theta will be different. So, what we are going to do is we will assume some value of lambda and theta and then see some value of lambda and some value of D and some value of V and then see how does for a particular set of variables the dependence of theta manifest itself as. So, what we will do is we will do an example and example is such that the phase difference is pi over 2 and D equals lambda by 4 D equals lambda by 4. So, for these two parameters phi equals pi over 2 and D equals lambda over 4, we are going to see what kind of dependence of on theta is there. So, we will develop a table. So, the first column in table is theta, second column in the table will be pi D over lambda times sin theta minus phi over 2 and the third column will be cosine of pi D sin theta over lambda minus phi over 2. So, when theta is 0 degrees or 0 0 radians, when theta is 0 sin theta is 0. So, first term is 0 phi is pi over 2. So, this becomes minus pi over 4 and cosine of minus pi over 4 is 1 over root 2. When theta is pi over 2 then what I get here is pi over 4 minus pi over 4 equals 0 and here I get is 1. When theta is pi then again sin theta is 0. So, the first term is 0 and what I get is minus pi over 4 and here once again I get 1 over root 2 and when this theta is 3 pi over 2 then sin theta is negative 1. So, I get here minus pi over 4 minus pi over 4 is equal to minus pi over 2 and then I get here is 0. So, if I do a polar plot of cosine with respect to theta this is how it is going to look like. So, at 0 it is I am just making some marks. So, at 0 degrees the value is going to be 1 over root 2 which is 0.707. So, this is 1. So, it is going to be little less than 1. So, this is 1 point at pi over 2 when theta is pi over 2 this value is 1. So, this is another point when theta is pi then again I get 1 over root 2. So, this is. So, this is half this is 1. So, at theta equals 0 it is 0.707. So, it is going to be this is 0.75. So, it is going to be somewhere here. So, this is here and then similarly at pi again it is going to be 1 over root 2. So, this is 0.725. So, again it is going to be somewhere here and then at 3 pi over 2 it is going to be at 0. So, these are the 4 points. So, now I develop a curve and I hope I do a an job in developing this curve. So, it is going to look something like this. So, this is the polar plot for term T 2. So, what this polar plot shows is that when theta is 90 degrees I have the maximum amplitude, when theta is minus 90 degrees the amplitude is minimized and I have very interesting polar pattern. So, even if I keep my radius fixed the plot changes the amplitude changes as with respect to theta and this is purely because of interference of sound waves and at different places they add up or they interact in a positive or constructively or destructively and we get variation of sound pressure levels as I move in the theta direction. So, this is the essence of interference of two sound waves. Similarly, we can use similar theory. We can extend this theory very easily to have three sound sources interacting with each other four sound sources and so on and so forth. One there are several significant application areas of when sound waves are interfering and how do we predict these sound waves. For instance, we use this approach to figure out the performance of radars. We also use similar theory to understand how sound propagates when it is being broadcasted over a large area. Also, there are directional microphones and also directional sound sources where the sound pressure field is predicted based on this particular theory. So, I hope this approach of using interference helps you in understanding what happens when more than one monopoles start producing sound in a 3D field and how do they interact with each other to produce constructive and destructive interference pattern in the 3D sound field. So, that closes this particular lecture and thank you very much. Bye.