 So, the next equation we are going to develop is abba is from the material constitutive behavior. So, how and that will explain or help us understand how is volume or change in volume related to change in pressure and so on and so forth. This piece of fluid which we are considering it could behave in a range of ways and we have to figure out which is the most appropriate model which predicts its behavior. So, this could be behaving in an adiabatic way that could be one model or the fluid here could be behaving in an isothermal way or it could be behaving in an isobaric way and so on and so forth. So, the question is how is gas behaving in this piece of material volume? Is the behavior adiabatic? Is the behavior isothermal? Is the behavior isobaric or is there some other thing and how do we know which that what is the exact behavior of this piece of fluid? So, this question was also address by Newton himself and he made an assumption that as sound propagates in air the fluctuations in pressure and fluctuations in volume associated with this propagation of sound they are related in a way which mimics an isothermal behavior. So, in that case he said that p times v is constant and if I use this relation then I can figure out how the wave is propagating. But it turns out that Newton was not correct in this particular assumption that propagation of sound is an isothermal phenomena. Later Lord Rayleigh he also tried to understand this question and answer this question and he made an assumption that the propagation of sound is essentially an isentropic happens in isentropic conditions that is my p times v to the power of gamma is constant or in other words it is an adiabatic behavior. And subsequent measurements showed that he was indeed correct and Newton was in error. So, the question is that why is that the case what was erroneous about Newton's assumption of isothermal behavior and what was correct about Rayleigh's assumption of adiabatic or isentropic behavior. So, to understand this we have to see how heat propagates in fluid medium and how sound travels and then we see how are these two phenomena interlinked and how do they influence each other at a physical level. So, if I compress a piece of gas it will generate heat now that heat could leak out from the gas and go away get dissipated or it could remain contained in it in the gas itself. So, the question is it leaking out in such a way that the temperature of the gas remains constant then that would be an isothermal behavior or is it contained in the fluid volume in some sense and if that is the case then that would be saying that the behavior of gas is something close to an isentropic process or adiabatic process. So, to understand this let us look at some again let us look at some observed data. So, let us assume that I have a sound wave and it is propagating at 10000 hertz. So, my auditory range is from 20 hertz to 20000 hertz. So, I am assuming that there is a particular tone which is propagating in air at 10000 hertz. Now, as it is propagating in air in some parts air is getting compressed and in some parts air is getting expanded where it is getting compressed heating is getting generated where the air is experiencing rare fraction heat is being absorbed or cooling is happening. So, there is some thermal process also associated with this acoustic process and the speed of travel of this heat wave which is also called thermal diffusion wave. So, thermal diffusion wave speed at 10000 hertz is 1.5 meters per second. This is based on some observed data as available in engineering literature which means that the wavelength of this heat wave wavelength of this heat wave at 10000 hertz is velocity of heat wave which is 1.5 over frequency. So, that is 10000 and that comes to 1.5 times 10 to the power of minus 4 meters. 1.5 of this wavelength is essentially this number divided by 2. So, that is 7.5 times 10 to the power of minus 5 meters. When sound travels the speed of sound as measured at room temperature conditions is about 345 meters per second. So, lambda for sound is lambda for sound at 10000 hertz is velocity of sound which is 345 divided by frequency. So, that comes to 0.0345 meters and half of this wavelength is 0.01725 meters. So, now consider medium and what I will do is I will plot the sound wave. So, let us say this is as this is a pure tone it may be it will be having some sinusoidal shape. So, this is my x. So, wave is travelling along the x direction and this is its amplitude of pressure. This number is 0.01725 meters and here x is 0.0345 meters. So, in the first phase of this sound wave first part of this sound wave pressure is rising above atmospheric pressure and what that means is heat is getting generated heating is happening and in this phase cooling is happening. Now, when heat is getting generated and that happens in this 0.01725 meter of a range and then in the next 0.01725 you are having cooling. The first thing as heating is getting generated heat travels at an extremely slow speed which is 1.5 meters. So, the travel of heat as it is getting generated it does not travel far enough to enter into the cooling range and by that time you have the next cycle of the sound coming up. So, what is happening is that heat is getting generated in a small place a fluid it moves a little bit because of its propagation at this speed 1.5 meters per second and then once that happens sound starts experiencing in that area you have rare fraction of the pressure and once that happens you start having cooling. So, heat is getting generated moves a little bit and it has not moved sufficiently enough to leak out of the system and then you have cooling happening because of rare fraction because the sound velocity is extremely high 345 meters per second. So, heat is not able to leak out from the system in an appreciable sense and essentially what that means is that we have an adiabatic process adiabatic process is something which captures this kind of a behavior fairly accurately. So, it is for this reason Rayleigh when he made the adiabatic assumption he was correct and Newton when he assumed that the gas is behaving in an isothermal way he was in error in context of sound waves. So, using this understanding we will develop the gas law for sound. So, we know for adiabatic process adiabatic process P T V T to the power of gamma equals constant where gamma equals 1.4 it is a gas constant and the value is 1.4 for gases which are diatomic in nature whose molecules have 2 atoms. So, I now differentiate this so essentially what I get is d P T over d T times P T to the power of gamma plus P T V T to the power of gamma minus 1 times d V T over d T equals 0. So, I am differentiating this equation in time and then if I rearrange this essentially what I get is d P T over d T equals so I move this on the other side of the equation. So, what I get is minus P T over V T and I am sorry there should be gamma here times gamma times d V T over d T. Now, we know that P T equals P naught plus P. So, essentially what that means is the differentials of P T are same as differentials of P naught which is 0 plus d P over d T. So, using this understanding I can replace d P T over d T by d P over d T. So, I get d P over d T equals minus P T gamma over V T times all also I know that V T equals V naught plus tau. So, differential of volume V T with respect to time is differential of V naught which is 0 because it is a constant plus differential of tau with respect to time. So, this thing I can rewrite it as d tau over d T. Further I know that P T is approximately equal to P naught. So, I make that change here and my final relation is so or my next relation is d P over d T equals minus gamma P naught over V T times d tau over d T. Making one final simplification we know that d P over d T is partial derivative of pressure with respect to time plus partial derivative pressure with respect to x times del x over del T and that is u times partial derivative with respect to x. And we had seen earlier that this term is extremely small this non-linear term compared to this one. So, I can replace d P over d T by just its partial derivative. So, my final equation in differential form for the adiabatic process is partial derivative of pressure with respect to time equals negative of atmospheric pressure times gamma over V T times d tau over d T. So, this is the second equation and this is essentially the equation for adiabatic process, but in a differential form with the assumptions embedded into this equation as they relate to propagation of sound. So, I have developed an equation for momentum and now I have developed an equation for adiabatic gas process. So, the third and final equation which I am going to develop and then after that I will start synthesizing all these three equations. The third and final equation is it relates to conservation of mass and it is also called continuity equation. So, the basis of continuity equation is that this there may be some inflow from this side of the surface in this fluid and there is some outflow happening from this and if I add these two numbers up that the total amount of variation in mass should remain 0 because these are constant mass particles. My outflow minus inflow equals change in volume. Now, once again I will draw the same picture this is inflow here position is x and u is a function of x and time I have an outflow here u is a function of x plus delta x and time. So, my outflow is u which is a function of x plus delta x times time and time minus inflow velocity outflow velocity minus inflow velocity and then that I have to multiply this area. So, that area is delta a and as this is velocity I have to also multiply it by d t and this equals change in volume because I am assuming that changes in density are negligible. So, this is the equation I am getting. So, I can rewrite this equation as u x plus delta x time minus u being a function of x and t and then here I am dividing this by delta x and multiplying it by delta x. So, delta x times delta a. So, I have multiplied by delta x and then I am dividing it by delta x times d t equals change in volume. This term delta x time delta a is v t. So, I can replace that by v t. So, I can rewrite this as u x plus delta x t minus u is a function of x and time and in this case what I am going to do is I am going to take time and move it on the right hand side of the equation. So, I get delta tau over d t. So, actually these are all finite quantities small, but finite quantities. So, I am going to replace d by delta. Now, if I assume that the size of this fluid volume is extremely small such that it is approaching 0 that is one assumption and I am also assuming that the amount of time which is getting elapsed which is delta t is also extremely small. Then in that case I take the limits for this expression and the limit for this expression and what I get is here I get partial derivative of u with respect to x times v t equals derivative of volume with respect to time. So, this is my third equation and this is continuity equation. This is the third equation and the continuity equation. So, next I synthesize the continuity equation, gas equation and Newton's law and try to develop one single equation for pressure. So, let us rewrite these equations. So, continuity equation is v t times del u over del x equals d tau over d t. I call this a, this is my continuity. Then the gas law is del p over del t equals minus p naught gamma over v t times d tau over d t. This is gas law, adiabatic gas law. So, this is equation b and the Newton's law or the momentum equation is negative of rate of change of pressure with respect to x equals rho naught times rate of change of velocity in time and this is equation c and this is momentum. This is the momentum equation. So, what we are trying to do here is that we have three equations and we are trying to eliminate other variables such that we get one final equation for pressure or we can do something same for velocity also. So, in this case all what we are trying to do is develop an equation for pressure as to how pressure is changing in x and in time. So, if I use a and b, these two equations, they have this d tau over d t term and if I use these two equations to eliminate the tau variable, then what I get is, so del p over del t equals minus p naught gamma over v t and here I have d tau over d t which equals this term. So, I get v t times del u over del x. So, I get partial derivative of pressure with respect to time equals minus p naught gamma times partial derivative of u with respect to x and now if I differentiate this whole equation. So, again my aim is now to eliminate u from this equation and equation c. If I eliminate then I get an equation in pressure and see I can figure out how it is changing in x and time. So, that is what I am trying to do. So, the way I am going to do is that I will differentiate this equation with respect to time and I am going to differentiate my momentum equation with respect to x and then eliminate u in that way. So, if I differentiate this equation with respect to time what I get is second derivative of pressure with respect to time equals minus p naught gamma times second derivative of u with respect to del t del x. So, I call this equation d and then from c I get if I differentiate this whole equation in x what I get is second derivative of pressure with respect to x equals rho naught second derivative of u with respect to x and t. So, here if we assume that del 2 u with respect to second derivative of u with respect to x and t is same as cross derivative of u with respect to t and x because if we are assuming here that u and its subsequent derivatives in x and t at least up to the second level they are continuous and if that is the case then this equation this relationship will hold to. In that case I can eliminate the terms encircled in green color from equations d and e from equations d and e and I can get the final equation. So, that is what I do and finally, what I get is by eliminating u from d and e what I get is negative of 1 over p naught gamma times second derivative of pressure with respect to time equals negative second derivative pressure with respect to x into 1 over rho naught or I get del 2 p over del t square equals p naught gamma over rho naught times second derivative of pressure with respect to x. So, this is 1 d wave equation for pressure this is 1 d wave equation for pressure. Now, if I replace this term in parenthesis by a constant and this constant is a positive number because p naught is positive rho naught is positive and gamma is positive it is a positive number. So, if I replace this by a constant called c square and we will understand the meaning of c square later then the final equation which I get is this where c square is p naught gamma over rho naught. The value of c if I compute through this relation is c equals 344.2 meters per second the value of c comes to 344.2 meters per second. If I measure the speed of sound in air then what measured data tells me is that speed of sound equals 344.8 meters per second. So, this again from published literature. So, what appears here is that c is awfully close to the measured value of speed of sound and what we will try to find out is the how speed of sound and c are related and we will find out that in few minutes from now that they are actually exactly the same c is actually the speed of sound. But before we get there let us try to solve this equation and develop some understanding of this equation. So, what we are going to do is first we will solve this equation and after that we will try to understand the implications of this equation and how they relate to propagation of sound in one dimension. So, what we will try to do is we will first solve this equation in a general sense and then try to understand what is the meaning of this equation in context of wave propagation in one dimension. So, you have this equation where you have on one side differentials in x and on other side differentials with respect to time only and what that immediately tells you is that if we use a variable separable approach where pressure is dependent on x and time in a variable separable form of sort of way then that kind of a format may work out in terms of solution for this equation. So, we assume that let p x t be a product of p 1 which is dependent only on time actually let us say x times p 2 which is dependent only on time and now we plug this form of pressure into this and what we get is c square del second derivative of pressure not pressure p 1 with respect to x equals p 1 times del 2 p 2 over del t square or we can also write the same thing in shorter format as c square p 1 prime prime times p 2 equals p 1 times p 2 double dot. So, now I separate the variables I bring everything dependent on x on one side and everything dependent on time to the other side of the equation. So, what I get is c square p 1 over p 1 p 1 double prime over p 1 equals p 2 double dot over p 2 and this equation can be only true this can be only true if the terms this term in parenthesis and this term in parenthesis they are both constants they do not dependent on time or in x because this equation has to be valid for all values of time and all values of x. So, we assume so based on that we say that p 1 prime prime over p 1 and the term on the right side is equal to a constant and this constant we call it as k square and this k can be complex this k can be complex. So, then we have two columns. So, so what that gives me is c square p 1 prime prime over p 1 equals minus k square this is the equation in space and the other one is p 2 double dot over p 2 equals k square this is equation in time. So, now we solve both these two equations separately for p 1 and p 2 and once we get the relationship for p 1 and p 2 we multiply those two using this and we get the actual value of p. So, we have a column for equation in space and then we have another one for equation in time. So, for the equation in space I get p 1 second derivative plus k square over c square times p 1 equals 0 and on the time side I have p 2 double dot plus k square p 2 equals 0. So, now we will guess. So, let us guess that p 1 equals upper case p 1 times e to the power of g x and here we guess p 2 equals upper case p 2 e to the power of h times t because it is dependent on time on that side. So, we plug this in the respective ordinary differential equations and what we get is g square plus k square over c square equals 0 and here we get h square plus k square equals 0. So, here g equals plus minus k over c j and here h equals plus minus k times j and again p and g and h all these things can be complex p 1 p 2 h and j all of them they can be complex. So, my p 1 is upper case p 1 e to the power of minus k over c x times j and p 2 is upper case p 2 e to the power of plus minus k j t. So, finally I now synthesize these multiply these two and I get the pressure. So, pressure which is a function of x and time is p 1 times p 2. So, I get p 1 times p 2 e to the power of minus k x over c times j and actually this should be plus minus and again it is plus minus k j t or if I simplify. So, I can write p 1 times p 2 is p then this is essentially p e to the power of plus minus k x over c plus minus k t j. So, this is the solution for the wave this is the solution for the wave. So, this is one approach, but in a more general sense I can also say I can also say that p x t is nothing, but a function f 1 which depends on t minus x over c or excuse me or it could also be a function of another function f 2 which is dependent on t plus x over c and this relationship also we see holds true of in this case because if I take k out from the parenthesis essentially I get x plus c plus minus t and so on and so forth. So, we will finish today's lecture at this point and in the next lecture what we will do is we will try to understand the implication of this form of a solution and it will help us understand some very fundamental concepts as they relate to propagation of wave in one dimension what is a forward travelling wave, what is a backward going wave, why is c the same as speed of sound which is the speed at which sound wave propagates and so on and so forth. So, what we have done in today's class is we have developed the wave equation using momentum equation, continuity equation and adiabatic gas law, we synthesize them and we developed the wave equation for pressure in one dimension and we have developed a solution for this wave equation and in the next class we will elaborate on this particular format of the solution and see what it tells us in terms of characteristics of one d wave. So, with that we conclude today's lecture. Thank you very much.