 So, in the last class, we had established equivalence between the electrical side and the mechanical side. We had mapped velocity on to voltage, force on to current, compliance which is inverse of stiffness on to inductance. And the moving mass of an object on to capacitance and then we introduced a term which is inverse of damping called mechanical resistance and that we mapped to the electrical resistance. As we developed this analogy, we had seen that we had mapped u which is the velocity to voltage. One thing in this context, it may be important to understand this that if I draw a circuit, it could be any circuit. I have a voltage source and there is an impedance, there is another impedance and then I am closing the circuit. So, here the impedance is Z 1, here the impedance is Z 2. It is actually physically possible for me to measure the voltage across this impedance by using a voltmeter. And in doing so, I do not have to cut the cable here. If I have to measure the current, then I have to introduce an emitter here and then I have to actually physically cut the circuit. Now, there are newer current measuring devices which use magnetic fields, but the least some of the less expensive ways is that you have to cut the thing and then measure current going through that circuit. Similarly, when I have to measure this velocity, I do not have to break the mechanical circuit. If something is moving, I can use it, use a laser or some displacement meter to measure the velocity. If I have to measure force which is analogous to current, I have to again break the circuit because force is something which passes through the stream of things. So, it is important to understand that and that is another benefit of this analogy which we are using. So, this analogy where we had equated u with not equated, we had established analogy between u and voltage and force and current and so on and so forth is called mobility analogy. It is also called inverse analogy. There is another set of analogies which is called direct analogy or impedance analogy where these mappings are just the opposite. So, we will very quickly go over that. So, what we are going to talk about is direct or impedance analogy. So, we will start very quickly with a dash part. So, force equals c x dot which gives me force equals c times u which is velocity which gives me force equals r m times u. So, here it is in upper case. So, r m is same as the damping coefficient. So, from this I can say force equals r m times u and we will see that and if I say that force is similar to voltage u is similar to current and my element variable which is r m or damping coefficient r m is similar to electrical resistance. So, this is for dash part. I see that it is called direct probably because you have force which is mapping on to voltage where voltage itself like a force type of a quantity. It relates to electromagnetic of course, it is I have no clear answer for that question, but this is a terminology and we are just using it, but I do not know exactly the history behind the nomenclature. So, the other one is spring. So, force equals stiffness times displacement is equivalent to force equals stiffness times integral of u d t. So, is same as force equals 1 over c m times integral of u d t where c m is my capacitance compliance. So, again force maps to voltage velocity maps to current and my compliance maps to capacitance and then I go to mass and here force equals mass times acceleration that is force equals mass times first derivative of velocity over in time. So, I mean we had earlier used m m that is moving mass. So, I will just use the same notation and I know that in electrical area voltage equals l d i over d t. So, again my force looks similar to voltage, I looks similar to velocity looks similar to current and my moving mass m m looks similar to inductance. So, this is my other analogy. Now, if I do a four point transformer in this, you will see that the turn ratio instead of t is to 1 comes as 1 is to t. So, we will do couple of examples where we will just construct mechanical circuits not electro mechanical circuits just mechanical circuits and so that we develop some practice. Let us say I have a spring mass and a damper system like this. So, I have a spring and also it I have a dash part this is connected to a moving mass acted upon the force due to gravity. So, it is m m the force on it is m m times g it is an external force and let us say I am exciting at this end u s here it is mechanical resistances lower case r m and the compliance of this is c m. So, if I have to construct a equivalent electrical circuit then what I will do is I have a voltage source which is u s then also I have a moving mass m m and in some of the earlier lectures we had said that moving mass always refers to an inertial frame which in electrical engineering translates to a ground line which is this. So, I have I put a capacitance of value m m then I have to ground it which is my reference and then the mass is connected in parallel to a spring and a dash part. So, I have a resistance of value r m and I have a inductor which mimics the performance of a spring and the value is c m. So, I am missing one thing in this picture what is that this is I have also gravity I have to put and the gravity is acting on the mass and that current flows through the dash part and also the spring. So, I can put the current source also here and the value of that is m m and c like that. So, in this case what I can do is initially when the thing is set up and it is having no excitation of u s then I can solve using this current source of m m m m times g and basically that will give me a d c offset in the whole displacement and velocity. And then once I have that velocity will be 0 because it is a quasi static situation, but they will be the mass will move down a little bit then after sometime it will stabilize I can figure out what is the displacement of that then I apply an external voltage and then again solve it. So, whatever my total displacement will be initial displacement which is a consequence of gravity plus the displacement due to this excitation velocity. So, that is how I can solve it. So, what you are seeing here is that instead of writing long equation if I am able to develop a lumped you know system for lumped mass system for this entire mechanical network then fairly in a very straight forward way I can solve for complex systems this is a simple system, but I can also solve for complex systems. So, we will do couple of more examples let us look at another one where I am having two masses. So, I have a base I have a piston let us say a cylinder and a piston and inside this I have a piston the mass of piston is m m 1 the mass of the cylinder is m m 2 also I am exciting my piston by a velocity source u 1 it could be a this could be driven by a motor and the piston could go back and forth. Then I have a rigid frame in which this cylinder is on which this cylinder is placed and I have a spring I have a rigid frame and here is a spring of compliance c m the friction between the piston and the cylinder could be we designate it as r m 1 and the friction between the cylinder and the frame on this surface is r m 2 and this is ground this is ground plane. So, this is rigidly held. So, if I have to develop a circuit for this again I have a velocity source which is like a voltage source generating velocity of u 1 how many masses I have in this two masses m m 1 and m 2. So, I have two masses and I am using mobility analogy this is m m 1 m m 2 and I have to again ground the masses I have to provide an inertial frame to them. Then my first mass is linked to the second mass to this and the interface between that I have mechanical resistance r m 1 of course the first mass is also seen excitation through u 1 and the second mass and the ground are connected in parallel through two elements c m and the resistance. So, I have an inductance of value c m and then I have another resistance mechanical resistance value r m 1 because whatever force which is being exerted by the second mass will be distributed amongst the compliance member and the friction member. So, they are in parallel the force that will split into f 1 and f 2 to those two members. So, you have to like what may I ask you have to think a little bit carefully how I have to construct those elements, but once you have the circuit network correct industrially straight forward to solve these problems. We will do one more example and then we will move on and in this case what we will do is the example of an automobile suspension. So, typically you have a car four wheels on each wheel you have a spring and also a dash pot a damper to kill the vibration and then on that structure you have the mass of the whole car distributed. So, if I just take one wheel plus one dash pot plus one stiffness member then that system will see one fourth of the mass of the car. I have ground then I have the wheel, wheel could have a mass let us call it m m and of course, there is air inside the wheel also. So, it is acting as also spring. So, I have c m then the wheel is connected to a dash pot and also it is connected to a spring. So, let us call the spring stiffness to be c m s and the damper resistance to be r m s. So, I have two springs in this case one spring is inbuilt into the wheel itself thereby it is very structure. You can also argue that wheel also has some damping properties because it is made of rubber and all that you can also incorporate that, but typically the damping provided by the wheel itself is not significant. So, we can drop it and then on top of it what do I have the mass of the car how much mass one fourth. So, I put a mass here and I call it m m of car. What am I interested in when I am doing this that if I have any disturbance in the vertical direction in this direction coming from the road suppose the car is going like that and now you have a bump on the road. So, it will induce a velocity of u w right or u road you can call it and then what am I interested in that the person who is sitting here in the this m n c how much of a velocity he sees ideally you would like that that value is 0 right that is the ideal situation. So, what you are interested in knowing is what is u car. So, is everyone clear about these terminologies m m c is mass of the car one fourth u car is velocity of the car where the passengers sitting c m s is compliance of the suspension stiffness r m s is damping mechanical resistance of the damper m m and c m r the wheel of the mass and compliance of the mass. So, we will construct the mechanical circuit for this. So, again this is my excitation which is acting like a voltage source. So, I call it u r and then I have two masses the first mass is the mass of the wheel I make it m m w I will connect it to ground the second mass is mass of the car m m c and again again grounded there is a stiffness member between this mass and the excitation source which is c m w. You could ask why is this c m w not in parallel to this mass so any thoughts on that the compliance in the wheel why is it not in parallel to the mass of the wheel itself for that you have to see the physics of the problem. So, where is mass of the wheel most of the mass of the wheel is concentrated at the center where you have iron or steel and between that steel and the road you have that here getting compressed. So, it is not in parallel consistency. So, from here I have two mechanical elements in parallel c m s and then these guys are connected. So, this value is c m s and this value is r m s what I am interested to know is what is the velocity of this mass. So, if I can figure out the voltage difference here which is u r for given a value of u r if I can figure out what is u r then I have solved the problem. So, when you are looking at this circuit think about for very low frequencies the impedance offered by a by an inductor is for all sets of frequencies is omega l as frequency is very low omega is going towards 0 the total impedance offered by an inductor is going towards 0. So, for very low frequencies an inductor behaves like a short circuit as in frequency goes up omega goes up and the impedance offered by that inductor which are these compliance members this is a compliance member this it keeps on going up significantly and at very high frequencies the impedance offer it becomes a closer and closer to an open circuit or a capacitor is just the inverse because the impedance is 1 omega 1 over omega l. So, at low values of frequencies look it tries to become behave more and more like open circuit and at high values of the same thing it behaves more and more like short circuit. In the previous example we added a current source sir how do we determine the direction of the current source. You have to assign the direction the current source in your circuit itself like in this case you assign a positive or negative based on the physics of the problem you assign a positive voltage and a negative voltage. Similarly, you have to assign the direction of the current in the circuit itself. So, it is your understanding of the physics that will help you understand we will talk a little bit more about this problem and then we will talk about acoustic elements. So, this circuit here the current can go in this loop which we call loop 1 and it can also in go in this loop which we call loop 2 and then within this loop the current can go either through like this to this branch or it can go to this branch. So, we will very quickly without doing very elaborate calculations try to figure out what are the parameters at which these break points happen. What is the break points before which current goes through this branch and what is the break point after which current goes through this branch and so on and so forth. So, what are the break points if we are trying to develop a transfer function which is u car over u r suppose we are trying to develop this transfer function and then I develop a board plot for this. Then the board plot will have some break points what are those break points going to be today we will just talk about those break points figure out those break points and later class we will develop an exact board plot without doing solving the whole equation. So, my first break point is going to be. So, my first break point is going to be related to this part C m s and a mechanical resistance in parallel. So, this is my first break point C m s r n. So, for very low frequencies because omega is very low the impedance offered by this inductor will be low. There will be a point where this in the impedance due to C m will equal r m and after certain set of that frequency critical frequency r m will be larger than omega times C m. So, that is my first break point. So, the criteria for that will be omega times C m equals r m s or omega 1 which is my first break point is r m s over C m or f 1 equals 1 over 2 pi r m over C m. Now, we will assign some values to these properties. So, C m w equals 4 10 to the power of minus 6 meters per Newton. The mass of the wheel is 30 kilograms C m m s which is the compliance of the string is 1.7 times 10 to the power of minus 5 meters per Newton m m c which is one fourth the mass of the car including the weight of passengers is let us say 455 kgs. These are realistic numbers. So, that you get a feel. Did I miss anything r m s is 7 times 10 to the power of minus 4 meter over Newton second. So, I plug r m s and C m this is C m s. What I get is f 1 is 6.6 hertz. What does that physically mean? That physically what that means is that the influence of this mechanical damper becomes significant only after 6.6 hertz. Before that this is playing a bigger role stiffness number is playing a bigger role. So, that this is my first break point. So, this will give you some insight how you design or you know identify what is the right damping coefficient and the right stiffness for a system which is similar to this circuit. Now, the second break point would be the current is going in this loop which is loop 2 and before a certain threshold that is when C m s is significantly important. There will be a resonance point because C m s and m m c the impedance offered by these two will be out of phase because the impedance offered by C m s is j omega times C m s and the impedance offered by capacitor is 1 over j omega m m c. So, they are out of phase, but as omega develops grows there will be a point when they equal and cancel each other. So, my second break point is so, before below 6.6 hertz I can neglect this resistance and I have this circuit in an approximate sense. There will be some current flowing through the resistance, but in an approximate very approximate sense I have C m s and here I have m m c and the condition for resonance would be 2 pi f 2 times C m s. Equals 1 over 2 pi f 2 m m c. So, f 2 when I solve for it is 1.8 hertz. So, in my board plot there will be another break at 1.8 hertz as I am developing a board plot which will be happen later. I should be aware that I have to do there will be something special happening at 1.8 and similarly I have an inductor and a capacitor in series in loop 1. So, there will be a break point associated with loop 1 also. So, break point the third break point is for loop 1 circuit will be something like this m m w c m w c m w c m w c m w c m w and my break point will be f 3 just as we develop the second break point it comes to 14.45 hertz. So, this is all we will do today in this class at least in context of this problem, but we will develop this problem further in some of the subsequent lecture. But, we are in mind that what we have done here is we have developed electrically equivalent circuit just by looking at circuit we were able to figure out what are the different transition points based on visual understanding of the circuit and then later we will see what is significant about those transition points and may be in your once you go back to your class home you can try to develop it would be an informal homework. So, now what I will move on to is acoustic circuit. So, we have developed electrical networks for mechanical systems now we will develop similar network for acoustic systems in an acoustic circuit is an open tube. So, I have a tube which is open at x equals 0 and the tube is l long my axis for axis going away from the tube. So, my boundary condition at x equals 0 as we had talked about earlier also is the pressure at the exit point is 0 the fluctuation. So, my relation for p and u which are both functions of x and omega I can write them as p plus p minus p plus p minus p minus p plus over z naught p minus over z naught and then I do e minus j omega x over c e j omega x over c and once I impose the boundary condition I find that p plus equals minus p minus which we have also seen earlier in earlier classes also we have proven. So, using this equivalence between p plus and p minus what we get is p u again the functions of x and omega I get p plus minus p plus p plus over z naught and then I get p plus minus p plus p plus over z naught and then I get e minus j omega x over c e plus j omega x over c and if I simplify this then I get minus minus 2 j p plus sin omega x over c and I get plus 2 p plus cos omega x over c over z naught. So, what we are interested is trying to develop an expression of impedance of the tube at x equals minus l as we are looking at the looking into the tube. So, the impedance is z equals p over u x equals minus l and when I put x equals minus l in this entire relation what I get is minus tangent omega x over c times j times z naught and when I put x equals minus l I get plus z naught j tangent omega omega l over c taking minus sign out and eliminates the negative sign. I am putting x equals minus l here. So, when omega l over c is very significantly smaller than 1 basically what that means is l is significantly smaller than c over omega or l is significantly smaller than lambda over 2 pi. So, for this condition when l is less than significantly smaller than 1 6th of lambda this relation becomes z equals z naught j times omega l over c. So, we have seen that z which is the impedance is z naught times j times omega l over c. Now, we know that z naught is rho naught c. So, once I introduce this relation in this I get z equals j times omega rho naught l rho naught is the density of here and this is basically p over u. If I equate p such that it is similar to voltage u is similar to current then this relationship looks like the relationship for a for an inductor because there is a j omega here. If I invert these analogies such that p looks like current u looks like voltage which I did in mobility analogy. So, in such conditions a closed tube looks like a capacitor. So, here my thing looks like I have an inductor here of value rho naught l voltage across the inductor is p and the current flowing through is u and in mobility equivalence I have a small closed tube behaving like a capacitor rho naught l voltage across it is q and the pressure is. So, now we introduce not new term term which we have talked about earlier, but we will revisit that and that is z a or acoustic impedance. So, again it could be a function of x and omega and that is basically it is a function of pressure which is again dependent on x and omega and volume velocity. So, this is same as p over u both are functions of x and omega times 1 over a where a is the cross section in this case of the tube is the area cross sectional area of the tube and this is basically z x omega divided by area. So, the acoustic impedance for an open tube minus l omega at x equals minus l omega is j times omega rho naught l over a and I can use this acoustic impedance in mobility analogy as a capacitor and in impedance analogy as an inductor. Now, we will talk something called acoustic mass and we will try to develop some sense of we will see how acoustic mass behaves like a mass in mechanical area. So, we are right now is still in doing all this discussions in context of a small tube open at both ends and it is small in the sense that its length is significantly less than lambda over 2 pi. Again you have a small tube whose length is significantly small in this sense because it is very small. The pressure then the air inside that tube because based on some of the discussions which we have talked about earlier that when you look at the velocity profile in such a tube. The velocity profile in such a small tube which is open at both ends is such that it does not change significantly as I move on x axis. So, the air inside such a tube moves like a rigid mass moves back and forth. If I have a pressure here and the tube is extending and it is fairly small in length compared to the overall lambda then because of that vibrating piston the air will move more or less like a rigid mass back and forth. So, if that is the case then I can use this relation force equals mass times d u over d t. This Newton's law is applicable to rigid bodies that is why I am able to use that. Now force is pressure times area. So, p times area where area is the cross section of this tube is mass and mass itself is density times the volume of the tube which is l times a times d u over d t. Now I know that area times u is volume velocity. So, I can say p times a equals rho naught l times d v v over d t pressure equals rho naught l over a times d v v over d t. So, what this shows is that if a tube is short in length compared to 1 6 of the wavelength then because the air is more or less behaving in a very rigid way. So, it moves back and forth and it is governed by this relation. Now what you have here is d v over d t 2 term. So, depending on how you look at it if you move to a mobility equivalence where v v corresponds to voltage and your pressure corresponds to current then a small tube behaves like an acoustic mass in mechanical mass behaves like a capacitor. Similarly, this acoustic mass behaves like a capacitor because then what this tells me is that my v v equals 1 over 1 over rho naught l over a times integral of p d t. So, this is the equation for a capacitor. So, in mobility analogy for a short tube p maps to current and then a short tube then v v volume velocity maps to voltage and the capacitance is called acoustic mass and that is the value for that is 1 over the term itself rho naught l over a. This behaves like capacitance and the value of this is rho naught l over a. So, if I have a short tube and if I am using mobility equivalence then the total amount of air all I have to do is it behaves like a mass, but it is acoustic mass. So, it is not the physical mass and the value of that acoustic mass is rho naught l over a. So, I can construct a circuit and where I short tubes and these tubes physically can be fairly long for instance at 40 hertz. The length of a short tube will be anything which is less than lambda over 6 and the lambda for 40 hertz sound wave will be 340 over 345 350 over 40 which is a significantly large number. So, these are not like micro devices these can be fairly long. We talked about a closed tube an open tube now we will move to a closed tube. So, again I have a tube closed at 1 and where x equals 0. It is l long then based on earlier classes we had seen that z at minus l is minus z naught j cotangent of omega l over c and the condition here is p plus equals p minus because of reflection. So, again for small tubes such that omega l over c is very small that is l is very small compared to lambda over 2 pi my z minus l omega over c minus is equal to z naught j times c over omega l and then I when I put z naught equals rho naught c then I get this number as and I bring j in the denominator what I get is 1 over j omega l over rho naught c square. My acoustic impedance z a at minus l omega is basically z over a as we saw earlier is 1 over j omega times l a over rho naught c square. Now l times a is volume is j omega it is volume over rho naught c square when we look at the physics of this tube pressure will be as we saw in earlier lectures pressure will be maximum here at the closed end whenever you have a closed wall pressure is maximum here and from here it will start falling as I move away from the tube, but because the tube is not very long in the context of l being significantly less than lambda over 2 that fall in pressure will not be significant. What that means is that I can approximate the pressure is more or less constant in that tube it will not go to 0 because lambda because the length of the tube is significantly small compared to the wavelength 1 6th of the wavelength. So, what that also means is that if I have to understand how gas or the fluid is behaving in this then pressure is not changing significantly then I should be able to use the gas law adiabatic gas law p v to the power of gamma equals constant and see what it tells us how the fluid is behaving in this closed tube where pressure is more or less constant. So, that is what we will do now p t is the p t is total pressure which is p naught atmospheric pressure plus change in pressure p and v t is again total pressure and then now I differentiate this in time. So, differential of p t in time will be same as d p over d t because p naught is not changing with time. So, I can call it d p over d t times v t gamma plus now I differentiate v t to the power of gamma in time times v t gamma minus 1 times d v t over v t equals 0. v t is original volume plus change in volume which was tau and the original volume does not change with time. So, I can erase this and I can replace it with now d tau over d t equals area of the tube cross sectional area times d x over d t equals a u equals volume velocity. So, I can replace this whole term by volume velocity in this relation. So, we get d p over d t times v t to the power of gamma plus gamma p t p t gamma minus 1 times v v equals 0. If I solve for d p over d t I get minus gamma p t over v t minus gamma p t over v t minus volume velocity. Now, we know that in the ream of acoustics the fluctuations in pressure are of much lower order of magnitudes than p naught. Similarly, v t is more or less very closely equal to p naught. So, I can simplify this further minus gamma u naught over d t p naught times v v again. We know that c square is gamma p naught over it is p naught over density rho naught. So, I get p naught gamma equals c square rho naught. So, I put this in this relation. So, I get time rate of pressure equals minus c square rho naught over v naught times v v. The significance of negative sign is as you have it in a spring that when you put a pressure then the volume will reduce and vice versa. So, that is significance. So, based on that I can drop my negative sign and I can say volume velocity equals origin of volume over rho naught c square times d p over d p. What this is showing me is that a closed tube behaves like a spring and physically it makes sense. If you have a closed volume and you put a pressure on it it is not behaving like a mass because mass has no way to be it behaves like a spring and we had done some earlier some problems also on that. The stiffness of that is this number this relates to this v naught over rho c square the acoustic stiffness. The other thing is that as we were developing this relation we did not use the one dimensional wave equation very explicitly. We did use in the case of an open tube theory of one dimensional equation. In this case we had made one single assumption that the length of the tube is less than lambda over 2 pi. So, what that means is that even if I have not necessarily long a tube but it could be any closed volume it could be an arbitrary sac like this as long as the longest dimension of this sac is significantly less than lambda over 2 pi. The volume velocity at the exit point of that here will be related to change in pressure in such a way and it will behave as a spring as a mechanical spring. So, in mobility analogy for a closed sac v v again is similar to voltage pressure corresponds to current and v naught over rho c square is called acoustic compliance and this corresponds to inductance. Because the relationship exactly volume velocity is something like l times d p over d p which is rate of change of current. So, we have developed equivalence for acoustic mass acoustic compliance and now we will go to acoustic resistance and that is really straight forward. So, you can have a screen a fine mesh it could be the screen could be a piece of cloth or it could be crisscross of wires and as air flows through it because of viscosity it dissipates energy. So, that case p which is function of x n omega over u x n omega is called r a which is function of x n omega over u x n omega is called r a and in mobility analogy it is 1 over r a is my acoustic resistance and that is such that my voltage across this thing is v v and the current flowing through this is p. So, this is for mobility we have done all the three elements resistance mass acoustic mass acoustic spring or acoustic compliance and the last one is a transformer an acoustic transformer what could be an acoustic transformer basically you have what happens in a transformer electrical transformer current goes up steps up or steps down right voltage steps up or steps down it has to have a step cross section because the stepping function is it goes up it is not gradually going up or gradually. So, if I have a tube which has two different cross sections then that could work as an acoustic transformer acoustic transformer. So, physical device could be like this. So, I have area here a 1 here my area is a 2 my input parameters are p 1 and v v 1. So, again when we are talking about acoustic elements it is not velocity and pressure it is volume velocity and pressure remember that and here my output parameters are p 2 and v v 2. So, the equivalence relations between two sections of this tube would be what for conservation of mass or continuity I should call it u 1 velocity 1 times a 1 equals u 2 times a 2 which gives volume velocity 1 equals volume velocity 2 that is the first equivalence right. The second equivalence is that p 1 equals p 2 because at this interface the pressure have to equal p 1 has to equal p 2. So, my velocity can go up, but the volume velocity will remain same my pressures will also remain same. So, this is a little different than regular tubes where water is flowing, but when we are talking about sound it is little different. So, what will be the turn ratio in such a transformer 1 is to 1 volume velocity 1 is same as volume velocity 2 and p 1 is same as p 2 the turn ratio is 1 is to 1. So, what is changing step going up or down is the velocity, but not volume velocity volume velocity is present. And finally, so we have developed three elements in the area of acoustics, acoustic mass acoustic compliance acoustic resistance we have a four point transformer and then there are laws of like Karkov's voltage law and Karkov's current law. Similarly, here you have similar conservation of that if you have a closed circuit v v 1 plus v v 2 plus v v 3 equals 0 and then also at a node pressures are flowing then p 1 plus p 2 plus p 3 and so on and so forth they also equate to 0. So, what we have done in last couple of lectures is we have developed equivalence between electrical, mechanical and acoustic rings. What is missing is that how do you go jump from acoustic to mechanical there is a junction interface point. So, we have not talked about that that how do you jump from electrical to mechanical and how do you jump from mechanical to electrical meaning that if you have current coming in what is the transformation happening that we are current transforms to force what is that relationship in the next class we will talk about that.