 Hello again, welcome to today's lecture on acoustics. In the last class, we had started developing mathematical relation for wave propagation or sound propagation in a radially symmetric fashion. And specifically what we had developed were two things. One was a wave equation for a monopole. A monopole I had explained earlier is ideally a point source of sound which emits sound in all the directions in a radially symmetric way. So for a monopole, a sound field generated due to monopole and if there are no reflecting surfaces then the variation of sound field in the theta direction and also in the psi direction. So they are both these variations, they are exactly zero. The only variation which happens for the sound field in a spherically symmetric system is in the radial direction. So with this kind of a framework, we had developed the pressure wave equation for a monopole which looked something like this. And once we had developed this equation, we also went ahead and developed a general solution for this pressure wave equation and we found that a general solution for this pressure wave equation looks something like this. Unlike 1D wave propagation in a Cartesian frame where we had forward traveling wave and also we had accounted for reflections because of the fact that here we are not talking about any reflections till so far. So because of that we have only a function f 1 which depends on t minus r over c. There is no equivalent function for a reflected wave which would have looked something like f 2 of t plus r over c because there are no reflections in this system and there is no provision for reflections in this kind of a system and that is the constraint which we have imposed upon ourselves as we are developing the theory for propagation of sound which is getting emitted through a monopole. Finally we had said that a particular, so this is a general solution and then we had stated that a particular solution which is consistent with this general solution is something like this function where p plus of r omega is nothing but a complex number p plus divided by r times e minus j omega r over c. So these are the three things we had talked about in the last class and today what we will do is that we will extend our understanding of a sound field attributable to a monopole further and by addressing two or three important questions. The first question is how does velocity relate to pressure in this kind of a radially symmetric field. Second question would be what is the impedance of what is the impedance in such type of a sound field and the third would be we will be introducing a new concept called volume velocity. So we will start talking about velocity and its relationship with pressure. To address this question that how does velocity in a spherical, a spherically symmetric sound field relate to pressure we will use the Newton's law which we had developed earlier and what we had seen was and we are just drawing analogy from what we had developed earlier for a 1 d wave propagation equation for a Cartesian frame of for a Cartesian frame. So what we had seen was that p is equal to del p over del r is nothing but minus rho naught del u over del t. So this is essentially directly coming out of Newton's equation or the momentum equation and what this equation says is that the partial derivative of pressure with respect to r equals partial derivative of velocity with respect to time multiplied by density. So this is one equation and then we also simultaneously write the wave equation for velocity and the derivation of this wave equation for velocity is very similar to the derivation of pressure wave equation for a monopole. So I am not going to develop this particular velocity wave equation explicitly but this looks like partial derivative of u times r with respect to r equals 1 over c square 1 over c square times partial derivative of u times r with respect to time. So this is the velocity wave equation. Now from this wave equation just as we had developed a solution for pressure in this form similarly it is very easy to see that for the velocity wave equation for velocity wave equation a possible solution could be of this form where u plus is a function of r and omega which is equal to a complex number divided by r times exponent of minus j omega r over c. So this is equation 3. So the solution for a velocity wave equation we have drawn analogy from the solution for a pressure wave equation which is here and then we have just mapped that understanding while we wrote down the solution for a velocity wave equation through equation 3. So let us look at these three equations, equation 1, equation 2 and equation 3 and if I put this equation 3 in the RHS of equation 2. So this is my RHS. So if I put in this equation this thing and on the LHS side if I put this on the LHS side then what do I get? Let us look at it. So what we get is so we are combining equation 3, 2 and 1. So what we are going to get is so I am just doing on the pressure side. So I know the expression for pressure, I have put it here and I am going to differentiate it with respect to r. So this is how I have picked up my equation for the left side of Newton's equation and then that equals minus rho naught times partial derivative of u with respect to t. So I am going to plug value of u in Newton's equation. So it is minus rho naught and then I am going to partially differentiate u plus over r e j omega t minus r over c. So once again what I have done here is I looked at the Newton's equation. On the left side I have partial derivative of pressure with respect to r and here I replaced p with the expression for p which I had developed earlier and on the right side I have inserted the expression for u and now I am going to do further mathematical operations. So this becomes, so we know that p plus now is a pure number and so I will take it out of the parenthesis and now I am going to differentiate the entire thing with respect to r. So what I get is minus e j omega t minus r over c divided by r square minus j omega e omega t minus r over c divided by r c and this equals minus rho naught and now I am going to differentiate on the right side. So what I get is u plus over r times j omega t minus r over c divided by r c and this equals minus e j omega t minus r over c. So I will process this now further and what I get here is p plus and I have negative here, I have negative, negative, negative so they all can go away right away. So what I am left with is 1 over r square and also the fact that I have e j omega t minus r over c in every single term. So I can also get rid of this it. So what I get is p plus times 1 over r square plus j omega over r c equals rho naught u plus over r times j omega or if I raise it to be 0. So rearrange this equation I get u plus over p plus equals 1 over j omega rho naught r plus 1 over rho naught c and this is same as I can multiply the numerator and denominator by the same term. So this is same as u plus e minus j omega r over c p plus e minus j omega r over c times r over r. So this term is u plus r omega and this term is p plus r omega. So I can write this relation as u plus of r omega as divided by p plus of r omega equals 1 over j omega rho naught r plus 1 over rho naught c and this term is p plus r omega. So I can write this relation as u plus of r omega as divided by p plus of r omega equals 1 over j omega rho naught r plus 1 over rho naught c. Now I know that z or the impedance depends on omega and r is essentially p plus of r omega divided by u plus of r omega. So from these two equations what I get is that z of r omega z impedance depends on omega as well as radius is equal to 1 over rho naught c plus 1 over j omega rho naught r and this whole thing is inverse and if I simplify this further what I get is 1 over rho naught c plus 1 over j omega rho naught r and this is nothing but rho naught c times j omega r divided by j omega r plus c. So impedance for a radially propagating wave as emitted through a monopole is 1 over rho naught but so this is the relation for impedance of a monopole and please be aware that this relation is valid to the extent that there are no reflecting surfaces. There are no reflections happening in the system. Now there are several observations which we can make as we look at this particular relation. One is that for a 1 d plane wave which would be travelling through a wave guide or tube for a 1 d plane wave the impedance was z naught and this was equal to rho naught c. Here z is for monopole and it changes with r it changes with radius. For a 1 d plane wave the impedance was not changing with radius it was or position it was constant it was constant and its value was rho naught c. So this is one very significant thing. The second thing is if I am far away from a monopole then the numerator which is j omega r and the denominator which is j omega r plus c the gradually approach 1 because the value of c becomes increasingly small with respect to j omega r. So if omega r plus c is approximately equal to omega r which means omega r is extremely large compared to c then z of omega r approaches rho naught c. So if I have a monopole and if I am extremely far from the monopole then the impedance which I will observe for the spherical wave which is moving out will be same as that of a planar wave. So essentially what it means is that if I am far away from a monopole then even though in a strictly mathematical sense the wave propagation is spherical in nature but it approximates to a monopole. So if I am far away from a monopole then even though the strictly mathematical sense the wave propagation is spherical in nature but it approximates to planar wave traveling out in one dimension. So that is the second implication. Now this is valid only when omega r is extremely large compared to c and we will qualify that now. So this is the second implication. So this is valid only if omega r is extremely large compared to c which means that r is extremely large compared to c over omega which means r is extremely large compared to c divided by 2 pi f which means r is extremely large compared to c over omega r. So this is lambda over 2 pi. So whether I am far away from a monopole or not it depends on the wavelength of the wave which is being emitted. If compared to one sixth of the wavelength I am far away from a monopole then my observations of that sound source will be very similar to that of a planar wave, 1D plane wave. So that is another important observation which we see here. So once again in a 1D plane wave the impedance is Z naught which is same as rho naught c and it does not have any dependency on r or it does not have any dependency on omega. In a spherical wave which is being emitted by a monopole the situation is different and in such a case the impedance depends on omega and it also depends on the position of observation but if I am far away from the monopole in the sense that the distance between the observation point and the monopole is significantly large compared to one sixth of wavelength then in that case the impedance will be approximately same as that of a 1D plane wave which is rho naught c. So this is how the velocity and pressure are related for spherical waves through this term called impedance. Now we will move on to a new concept called volume velocity. So we know for a monopole we know that pressure is real of velocity is nothing but real of p plus over r times e minus j omega r over c times e j omega t. So I have just rewritten the expression of pressure but now I am going to divide it by impedance and that will give me my velocity. So I have a relation for pressure and I have a relation for velocity and if I know the impedance and if I also know this p plus then I can find how pressure and velocity are varying with respect to r and also with respect to time but in a lot of situations it is difficult to find this value of pressure using physical or actual experimental measurements. So the way this challenge is addressed is by finding a different quantity called volume velocity and which we will develop which we will explain in a while and once we have figured out what the volume velocity is then from that particular measurement then we can calculate the pressure field and velocity field for a monopole. So it is for ease and for convenience that we have developed the concept of volume velocity and we use this concept to estimate and measure pressures estimate pressure field and velocity field for monopoles. So we will continue to expand the relation for velocity and what we will do is we know that complex velocity is this entire thing in parenthesis and this is nothing but p plus over r times e minus j omega r over c times 1 over z and this is nothing but p plus over r e times minus j omega r over c times 1 over rho 0 c plus 1 over j omega rho 0 r. Now suppose I have a sound source and I am at a distance r 0 away from this. So let us assume that I am r 0 away from this source then u plus r 0 omega equals 1 over rho 0 c plus 1 over j omega rho 0 r. Now suppose I have a sound source and I am at a distance r 0 away from this. So let us assume that I am r 0 away from this source then u plus r 0 omega equals so what I have done here is I have rewritten the expression for complex pressure and amplitude complex pressure amplitude and I have replaced r by r 0 then I am going to rearrange this particular relation and so that I can express it develop a relation for p plus. So p plus is nothing but u plus r 0 omega times r 0 and then I have to move this exponential term on this side as well. So exponent of j omega r 0 over c and then I also bring the term z on the other side and what I get is j omega rho 0 r 0 c divided by j omega r 0 plus c and this is same as u plus r 0 omega times r 0 excuse me times r 0 square. Times j omega rho 0 exponent of j omega r 0 over c divided by and I am dividing the numerator and multiplying the numerator by c. So what I get is I have eliminated c from the numerator and what I get is 1 plus j omega r 0 over c. So at this time we introduce this term called volume velocity and we define it. So suppose I have a diaphragm suppose I have a diaphragm and it is moving back and forth and let us say it is velocity is v then the normal area of the diaphragm normal in the sense suppose the velocity is like this and the diaphragm is like this then the normal area of the diaphragm times the velocity will be the total amount of fluid which will be displaced by the diaphragm as it moves back and forth. So that is what is known as volume velocity. So v v s I call as volume velocity of source and this is nothing but dot product of velocity of source and area of source. So it could be a diaphragm it could be a pulsating sphere if there is a pulsating sphere again it will be total surface area times the velocity at which the sphere is expanding and contracting expanding and contracting. So I have to take the dot product of the velocity and the area which is moving out back and forth. This is nothing but velocity of source and then there is a dot product times area. Now in case I have a diaphragm in case I have a diaphragm and it is a circular diaphragm then and if it is a circular diaphragm and let us say its radius happens to be r naught or for that matter it could even be a sphere and if its radius is r naught then a equals 4 pi r naught square. So you have a sphere this radius is r naught and its pulsating sphere so that is the area. So with that I get I rewrite this particular equation and what I have here is u plus times r naught square is basically v v s divided by 4 pi because u times r naught square times 4 pi is the volume velocity. So it is v v s divided by 4 pi times j omega rho naught e j omega r naught over c divided by 1 plus j omega r naught over c. Now once again because u plus of r naught omega could be a complex entity so v v s could be complex entity and it may change with respect to omega and it will also depend on the value of r naught. So I have this volume velocity so pressure magnitude complex magnitude pressure is same as v v s divided by 4 times times this complex function. Now if omega r naught over c and is extremely small compared to 1 that is 2 pi f r naught over c is extremely small compared to 1 which means r naught is extremely small compared to lambda over 2 pi then so if this is the case that the size of this sound source is extremely small compared to 1 sixth of wavelength if that is the case then the denominator term just collapses to 1 and in this case p plus is equal to v v s divided by 4 pi times j omega rho naught e j omega r naught over c and that is it the denominator is 1 instead of 1 plus j omega r naught over c. So if that is the case then pressure field I can write it as real of v v s times 4 pi r naught over c and that is it the denominator is 1 instead of 1 plus j omega r naught over c. So if that is the case then pressure field I can write it as real of v v s times 4 pi times j omega rho naught times e j omega r naught over c times e minus j omega r over c times e j omega t. So I am using my original expression for pressure and I realize that because of this condition r naught is extremely small compared to lambda over 2 pi this term also is approximately equal to 1. So if that is the case then I can simplify this as so this is why simplified expression for pressure. Essentially what this relation tells us is that if I have a monopole and I know its approximate size and let us say it is emitting sound of a frequency f and monopole is extremely small compared to the wavelength then in that case it is very close to an ideal monopole which has of a 0 size that is 1. Second thing is then the total area of the this sound source if I multiply that in a dot product sense with the velocity of the oscillating or pulsating surface and then I then from that I can get volume velocity and then I can use that volume velocity to compute the pressure field generated by the monopole. That is essentially the crux of this entire development which we have talked about till so far. So we have volume velocity of a system it is with dot product of velocity and area surface area from that I can compute volume velocity and then using this volume velocity I can calculate the pressure field using this particular relation with the condition that the size of this source has to be extremely small compared to the wavelength of the sound which is being emitted by the source. Thank you very much.