 So, what we will talk about today is about mufflers. So, we have talked a lot of different applications of sound, we have talked about speakers, microphones, you have also seen how shakers were accelerometers, which use principles, which are similar to what is used in X-ray seismometer and seismometer. Seismometers, we have also seen how sound moves in architectural spaces. So, mufflers is another thing I wanted to talk about and to understand mufflers, we have to understand how sound propagates as it moves through three different medium, three different media. So, what we will do is we will initially see how sound moves through three different media and then we will use those principles in context of mufflers and see how mufflers work. What you can have is you have sound and this is medium one, medium two, medium three. These could be physically separated boundaries or you could have three different units where properties of the material are significantly different. So, let us say z naught is rho 1 c 1 in first medium, rho 2 c 2 in the second and rho 3 c 3 in the third. Now, we know that as sound moves from one medium to the other medium, there is some reflection. So, let us say the incident wave is a 1 e j omega t minus a 1 x and then when it hits the first boundary, it gets reflected and the reflection is a there should be b 1 e j omega t minus a 1 x plus k 1 x. Similarly, here now when it moves to second medium and it hits the second boundary, let us say that the incident wave is a 2 e j omega t minus k 2 x and it gets reflected part of it. So, what you get is b 2 e j omega t plus k 2 x and finally, the final wave which emanates from the third into the third medium is a 3 e j omega t minus k 3 x, k 1 k 2 what are k 1 k 2 k 3. So, this is omega over c 1 omega over c 2 omega over c 3. Now, think about it as this wave hits the first boundary, what is the boundary condition at that point? The velocity at that point, velocity at that point whether you consider boundary 1 or boundary 2 the same. Velocity just inside the boundary and just outside the boundary is the same. So, velocity is continuous, velocity is continuous across the boundary, what else? Displacement, that will be a consequence of velocity itself, but pressure will also be continuous across the boundary and why will pressure be constant because if pressure is jumpy. So, this is my boundary and just inside you have p a and just here you have p b or p 1 and p 2 and with that film is extremely thin which of size thickness is it is vanishingly thin. So, the mass of it is 0, but you have across a film of 0 thickness a non-zero pressure difference and what that means is that it will have infinite acceleration, but in reality it does not happen. So, pressure is continuous. So, we use these two boundary conditions and get several equations and from those equations we compute a 1 a 2 a, a 1 we already know whatever is the incident wave. At the end of the day what we are trying to find is what is the value of a 3, we know a 1 we have what we do not know is b 1 b 2 a 2 and a 3 4 unknown. So, from these two boundary conditions if we can generate 4 equations then we should be able to compute all these unknowns b 1 a 2 b 2 and a 3. For the pressure boundary condition what you have is a 1 plus b 1 at x equals 0. So, here I am saying x equals l. So, at x equals 0 my boundary condition is a 1 plus b 1 equals a 2 plus b 2 and at x equals l I get a 2 I just put x equals l in these expressions and what I get is a 2 e minus j a l a 2 l plus b 2 e and here I will have a positive j a 2 l equals a 3 exponent minus j a 3 l. So, this is my pressure boundary and for velocity what I get is a 1 over rho 1 c 1 minus b 1 over rho 1 c 1 equals a 2 over rho 2 c 2 minus b 2 over rho 2 c 2 at x equals 0 and then the other boundary condition at x equals l is a 2 exponent minus j a 2 l over rho 2 c 2 minus b 2 e j a 2 l over rho 2 c 2 equals the velocity on the other side of the layer rho 3 c 3 e minus j a 3 l. So, I have 4 equations 1 2 3 and 4 and using these 4 equations I can compute all the unknowns in terms of a 1. So, what we are interested in understanding is that how much power is transmitted and how much power is lost as sound moves through this media. So, we will define a ratio of t and t is power transmission coefficient and what is power transmission coefficient? It is the ratio of a 3 square divided by 2 rho 3 c 3 over whatever is coming in 2 rho 1 c 1. So, if I use all these equations and I find a 3 in terms of a 1 at the end of the day what I get is transmission coefficient of power is 4 a number r 1 3 divided by r 1 3 plus 1 entire thing squared r 2 3 minus 1 square minus r 1 3 this entire thing divided by r 1 3 plus 1 whole square times sin square of a 2 l. So, how did I get this equation? By solving a 3 in terms of a 1 and plugging those values in this equation 3 and these r's are ratios essentially r 1 3 is rho 3 c 3 over rho 1 c 1 r 2 3 is rho 3 c 3 over rho 2 c 2 and r 1 2 equals rho 2 c 2 over rho 1 c 1. So, I know the properties of the medium. So, I can find these ratios I can plug them in this relation and I know how much power is getting transmitted through the medium and how much is getting lost. So, now what we will do is we will do some 2 special cases and then we will the third special case will be a form of special cases. First special case is that what happens when this l vanishes and becomes 0 l equals 0 once that happens then t equals what we call t 0 and that is essentially 4 r 1 3 over r 1 3 plus the entire thing is a square this value of t for now if r 1 3 equals 0 then t is 0 and also if r 1 3 equals infinite then t is again 0. This is the case when you have perfect relief of pressure. So, as sound is moving from first medium to the third medium because second medium is just not there then the pressure just gets perfectly relieved then r 1 3 0 this is the case when you have perfect rigid. So, you have a wall and it is infinitely rigid it just does not move then nothing gets. That is and if I plot this I plot it against r 1 3 this is my t and my curve looks something like this and this is t max and this is 1 for r 1 3 equals 1 I get maximum value of t max and I can find that also analytically by doing this d r 1 is equal to 0 which gives me if I differentiate t with respect to r 1 3 what I get is 4 over r 1 3 plus 1 whole square minus 8 r 1 3 over r 1 3 plus 1 cube yeah equals 0. So, if I solve it what I get is for extreme of condition r 1 3 equals 1 a good so bear in mind this is the definition of r 1 3 it is rho 3 over C 3 divided by rho over C 1. Suppose you have an acoustic transducer and let us say it is under water and it is generating sound and that sound has to be transmitted let us say in the in water, but also because of the fact that it is immersed in water you want to protect it from leakage of moisture into the system electrical circuit has to be protected salt is there so it has to be protected in a mechanical sense so that there is no corrosion and all that. If you apply some sort of a layer on top of it there will be loss of transmission if you want to maximize the loss of transfer you want to minimize that transmission loss the property of the material has to be such chosen that r 1 3 equals 1. So, they are special rubber for which they have actually tuned the material properties of that particular rubber to see water and they are called rho C rubber. So, for those specific materials if they use it to encapsulate the whole transducer the impedance is perfectly matched and there is no transmission loss so you get T L because your r 1 3 is 1. So, that was the first case is 2 is 2 could be where the middle medium medium number 2 is of a thickness which is very small it is not a very thick. So, in that case sin k 2 L is extremely less than 1 which means I can approximate sin k 2 L as approximately equal to k 2 L and this condition is valid only when L is very small compared to lambda over 2 and lambda is the wavelength of the sound where into the second medium. So, in medium 2 so if that is the case now we take a special case where rho 1 C 1 equals rho 3 C 3 equals. So, such that r 1 3 equals 1 so this is a special case of case 2 so then my T becomes 1 minus r 2 3 minus 1 r 1 2 minus 1 a 2 L square and this entire thing is inverted. Now, we know that r 2 1 is very small compared to 1 and for the same reason r 1 2 is very large compared to 1. So, my transmission factor is 1 over 1 minus r 1 2 square times minus 1 times k 2 L square over 4 and this becomes 1 over 1 plus r 1 2 square k 2 L square over 4. So, now if I put k 2 as omega over C 2 then what I get is T equals 1 over 1 plus rho 2 C 2 square over 4 rho 1 C 1 square times omega square L square over C 2 square and that upon simplification I get is 1 plus rho 2 L square omega square over 4 rho 1 C 1 everything is square. Now, what is physically rho 2 L see I have a medium which is very thin it has a density rho 2 it has a length L. So, physically what does it mean mass per unit area. So, rho 2 L is mass per unit area and I call that upper case m. So, this is basically 1 plus m square omega square over 4 rho 1 C 1 square. Now, on this one if I do a transmission loss this is the power transmitted. So, if I want to compute transmission loss which I call T L and if I want to do that in decibel scale then that will be 10 logarithm on base 10 I incident over I reference minus 10 log I transmitted over I reference and then this essentially becomes 10 log of 10 log in base 10 I incident over I. So, this is basically inverse of T. So, my transmission loss is basically 10 log 10 over 1 plus m omega over 2 rho 1 C 1 then that is I mean this relation a lot of people call it limp mass relation or lock for limp mass lock for limp mass relation and why it is called limp is because you are not assuming any stiffness. Suppose you have sound and there is a thin wall in between you have room 1 big room then you have a thin wall and then you have another big room. What you are not assuming is what you are not factoring in is the stiffness of the wall. So, that is why it is called limp mass lock and typically if I plot this law against logarithm of frequency or omega and here I am plotting in decibels transmission loss then I get some sort of a straight line. So, this is limp now in reality the structure does something like this which I am going to plot it in real. These peaks are related to the resonance modes of the wall itself. So, whenever you have a resonance you lose a little bit more energy which just happens that the shape of the wall is out of phase with the pressure wave. So, you lose a little bit more of energy and this is a place which is called at this point for this frequency this frequency is called coincidence frequency. And what is this frequency physically what it means is that when the frequency in the air matches the frequency of the bending waves in the wall then you have this thing called coincidence frequency. So, rather than going into the map at this you have a dip in transmission loss, but otherwise this limp mass lock is a fairly good estimate of how much energy gets transmitted or lost to the medium. So, with this background now we will move to another case of mufflers. So, what is the definition of the coincidence frequency? That coincidence frequency is when the bending flexural waves of the wall coincide with the incident frequency that is the frequency of this. So, now we will talk about mufflers. So, in a very broad sense you have two types of mufflers. First type are reactive and the other type are dissipative. In dissipative type of mufflers the reliance in terms of reducing the noise coming out of engine or whatever is the system is reliance is on the dissipative characteristics of the materials inside the muffler. So, that is why it is called dissipative. And here it is family how you are tuning and the bigger reliance is how to cancel pressure waves inside the muffler. So, that the total noise emitted out is minimized. In reality mufflers are a hybrid they rely on both reactive plus dissipation rates. So, they are a mix of these two. So, within reactive what we will today talk about is a particular type of muffler which is called expansion type of muffler. And in general the construction is something like this. So, you have air coming in from engine of course, goes to some catalytic converter where all the poisonous gases are catalyzed and emissions are made safe. So, it comes from a pipe then you have an expansion chamber which is this and then it gets emitted out to another tube a short tube. Let us say the cross sectional area of the inlet tube is A 1 and we also assume outlet tube is still A 1 and the middle tube is A 2 such that A 2 over A 1 equals m a ratio m. So, this is a three medium problem you have the first medium where the cross sectional area is A 1 then you have A 2 and then you have another one A 1. It is a we cannot idealize this directly as a lumped volume because in this in case of a lumped volume what is the property of sound? Velocities are 0 in that here velocities are not 0. We cannot also idealize it as lumped mass because in lumped mass it moves in a rigid way air moves in a rigid way it is not happening here. So, pressure gradient in the lumped mass is negligible it is not the case here. So, we use it use the three medium approach which we talked about just now and understand this problem. From the earlier discussion we know that transmission coefficient is 4 r 1 3 over r 1 3 plus 1 whole square 1 minus r 2 3 plus 1 minus 1 r 1 2 minus 1 square over r 1 3 plus 1 sin square A 2 l and I mentioned that this distance is l length of the expansion chamber. Now, what we will do is we will compute what are the different values of r 1 3 r 2 3 r 1 2. So, r 1 3 equals 1 also we know that rho 2 A 2 equals rho 1 A 1 because of conservation of mass that gives me rho 2 equals rho 1 over m. Similarly, rho 3 equals rho 2 times m I also want to find how are c is behaving c velocity of sound behaving in each of these chambers. So, we know that e 1 v 1 equals e 2 v 2 which gives me e 2 equals e 1 v 1 over v 2 equals e 1 over m and similarly e 3 equals e 2 times m. Now, velocity of sound is what gamma does not change e over rho. Sir, why do you consider isothermal process? We have to understand this the flow of the air may have a different nature compared to flow of the sound. Air may be going through a different process, but as sound is propagating it is going through adiabatic. Air is essentially going through this also being this. So, we are valid in making one set of using one set of rules for air and another set of rules for sound. Sir, what is the sound is travelling in air only? Sound travels in air, but just because air is moving like that sound does not travel with air. In the first lecture what we had seen is that the motion of particle is not same as like you drop a stone in a wave the wave travels, but in that case the lake is still sitting at one place. So, we are using a similar thought process. Sir, why are you considering the travel of air? Because we have to find pressure. These are not, these are not in these p 1, p 2 and p 3. They are basically average pressures, bulk pressures in each of these chambers. First chambers around these mean values of pressure there are fluctuations in the pressure because of sound, but even if sound is not travelling through this whole expansion chamber still p 1, p 2, p 3 will hold. See is gamma p over rho. So, if I use these this relation in conjunction with 3 and 4 what I get is r 1 3 equals 1, r 2 3 equals m, r 1 2 equals 1 over m. So, now I put these values of r's in first relation which is the relation for transmission factor and what I get is p equals 4 divided by 4 1 minus m square minus 1 1 over m square minus 1 over 4 sin square a 2 l and then what I get is 1 minus m m square minus 1 1 minus m square divided by 4 m square sin square a 2 l. I can sort this out as 1 over 1 plus square minus 1 whole square over 4 m square sin square a 2 l and then I further modify as 1 plus m minus 1 over m square divided by 4 sin square a 2 l and a 2 is again omega over c 2 and what we found is that c 2 is same as c 1 as the same as c 3 which is essentially omega over c. So, if I use this particular relation and then find transmission loss in decibels then my relation for transmission loss is 10 log to the base 10 1 plus m minus 1 over m whole square divided by 4 times sin square a 2 l. So, in cars what is the typical rpm at which the engine moves around 1500 start you know most of the times it set out 1500. So, the range is 1500 rpm to what is 4000 let us say that is 25 hertz 67 hertz. So, if I have to maximize T l I have to improve this factor m minus 1 over m we can I can make it as high as possible and I also have to make sin square k 2 l maximum the maximum value it can assume is 1 for 25 hertz. So, l is basically c over 2 pi times 1 over f I have to make k 2 l equal to 90 degrees basically. So, if I get this and this is there. So, that translates to l equals 1.37 meters for 40 hertz and it is it is 2 centimeter or 67 hertz. So, if I know what type of what frequency I am trying to minimize then based on that I have to choose the right value of l that is what I am trying to do couple of other things as m goes up so does T l. So, if you have a bigger expansion chamber which has a big much bigger area compared to a 2 a 1 a 2 over even a 5 I maximize it my transmission loss will be maximized. And the other thing we have already mentioned is that if k 2 l equals pi over 2 then again this is the condition for maximizing what I will do is I will show you a few slides. So, this is one picture I pulled from internet that is the source no at omuffler.net and what you are seeing here is from the engine you have a pipe then it comes goes through a yellow box which is what they call a catalytic converter where they reduce the poisonous content of the gases by through some catalytic process. And then you have a long pipe and then that is your muffler this is another one. So, in the analysis which we did that muffler is tuned to a particular frequency right. So, as I shift from that frequency slowly the transmission loss will not be as great as I would like it to be if I am very far away and once sin k 2 l equals 0 then that will proposes loss. So, in this figure which I pulled from this acoustics for engineers book you have the upper picture A 14.10 you have a series of chambers and each chamber is tuned for a particular frequency. So, you can tune it accordingly. Now, this is the picture of two types of mufflers where the reliance is not that much on expansion or reactive mufflers, but these are dissipative one. So, you have lining inside the tube which absorb sound and in the second one what you have is serigated. So, the absorption area in the expansion chamber gets enlarged and that enlarged area surface is coated with absorptive material. So, that absorbs sound. So, your B is like a hybrid you have an expansion chamber and also absorptive dividers are there. This is probably the last one. So, what this picture shows is that how my transmission loss changes as k l changes. So, k l starts from 0 goes up to pi and then for different values of m I keep on increasing my transmission losses as my m becomes higher and higher and again at 90 degrees for a given value of m my transmission loss is maximum. Last one. So, one limitation of these reactive mufflers is that they generate a back pressure and what that does is that it has an impact on the engine's efficiency. If you have a very high back pressure then the engine efficiency goes. So, in a lot of cars which are used on racetracks where they do not care that much about noise pollution, but are more worried about speed. They tend not to use so much of reactive type of mufflers, rather they tend to use have more reliance on absorption as the muffling agent. So, this is a particular type of a muffler and the name is called cherry bomb. I do not know why they call it cherry bomb muffler, but where you have is an inner pipe which has a lot of holes in it and that is covered with thick sound insulating layer which is basically glass, wool and other absorptive materials and then outside of that you have this red sheet which is basically made of steel. So, this is another one. So, what we will do is now we will go back and do a little bit more about mufflers. So, we have talked about reactive type of mufflers and if also if we want to understand how absorption plays a role in terms of absorbing sound then we already have the necessary background based on previous class lecture, how to figure out how much sound is absorbed through the walls in a muffler, because if we know the alpha absorption coefficient of the material and if we know S then we can figure out how much sound is getting absorbed. So, what Sabine had done was he developed this relation e is a is attenuation is 12.6 alpha 1.4 e over S and this is attenuation in decibels per feet alpha is again absorption sound absorption coefficient e is not pressure e is here perimeter of lined duct and this is an inches. So, you can convert it into meters and then S equals cross section area of duct. So, it is an inch square. So, it is trivial to figure out how to reframe this particular relation in SI units. So, what we will do is we will do an example. So, let us say I have a long pipe after it comes out of the catalytic converter then I have an expansion box or expansion chamber and then song gets out. I was looking at some of our cars and one car I found had a length of about 2 meter this length 6 feet and the diameter of this was about 3 inches and this diameter was about 12 inches and again this diameter is again 3 inches this length was about a meter. So, I can convert it into. So, m is what for this 16 right m is 16 T L equals 10 log 10 1 plus 16 minus 1 over 16 square over 4 sin square k 2 L and that is in this box I have is 1 plus 63.5 sin square f over 55 I did the math and in terms of frequency what this comes out. So, now I will construct a matrix. So, frequency could be 40, 50, 60, 70, 80, 90 and 100 hertz. So, in terms of engine rpm this is something we see on the screen and on the dashboard is 2400, 3600, 4200, 86, 4800, 7 you know 54 and this is 6000. Transmission loss in decibels I calculated is 14.6, 16.1, 17.1, 10 to increment 0.7, 18, 18.1, 17.8. So, it starts going after 80, 90 hertz. Now, if I want to see how much less noisy it becomes in terms of what is the reduction in pressure. So, I say pressure reduction factor. So, in this case it goes down by a factor of 5.4, 6.4 what will be the value for 18 dB 8, 6 dB is twice 7.1, 7.7, 8, 7.8. So, you get significant reduction in pressure based on this. So, now if you below the car if you have more space then you can make the value of m larger or you do not necessarily have to make it round it could be a rectangular shape wherever you have more space. So, that your cross sectional area is larger m is basically cross sectional area ratio of cross sectional area. So, if I want further reduction in sound I can increase my m and I can also check see where my maximum noise is coming from what frequency and based on that I can change the value of L. So, I am getting 14 to 18 dB's of transmission loss. So, then I said for comparison purposes let us say that if I line this entire tube with some sort of a highly acoustic damping material what kind of an impact it has. So, in that case a equals 12.6 times and I assumed alpha of 0.08 which is fairly high times pi d is the perimeter, pi d square over 4 is my area. So, if I do the math what I get is 0.49 dB per feet. So, this is roughly 6 feet. So, total absorption attenuation will be of how much 3 dB attenuation. So, over 6 feet they will be 3 dB attenuation by absorption. So, the point being that absorptive mefflers are not that efficient their response is much wider in terms of frequency band, but they are not broadband, but they absorb their absorptive capacity is not as great as compared to reactive meffler. So, most of the reliance when you are trying to reduce the noise coming from engine is on reactive nature of the system. So, that is what I wanted to cover in the context of meffler that is another application. Now, today in another 15-20 minutes and in the next lecture we will talk a little bit about how noise gets generated and some very broad approaches to which noise is managed of course, one way is to meffler, but I wanted to cover it in a more philosophical way, how noise is managed and noise can come through several sources. There could be sources there is some noise happening outside the room I would I do not want it to get inside the room, there is noise from the fan, noise from the ducts, noise from electronic equipment, but even units which are supposed to make good sound they also generate noise if you have a loudspeaker and I send it a pulse of 40 hertz, it will not be always the case that it will only generate 40 hertz content, it will also generate its noise and most of the that type of noise comes through nonlinearity in the system. So, by a very simple example we will today see how nonlinearity generates noise content and for that we will have just a simple spring mass system. So, mass is m, this is k and this is f not e f and if k is linear then the relationship between f and x will be linear, but if k is nonlinear then there will be noise generated. So, what we will so let us say k equals k naught plus k 1 x cube no x square, this very simple model has direct application in transducer. So, you have force which is being generated by which part of the loudspeaker the voice value that is generating force on the membrane. So, that is your f naught e s t, then there is moving mass of the loudspeaker which is m and then there is a stiffness I have eliminated damping here for simplicity purpose and there is stiffness. Now, if that stiffness is nonlinear as in this case then what I am trying to understand is that if I excited by a frequency s what happens to x, if x has only s frequency content then there is no noise, but if it has something in addition then there is noise. So, the equation of motion is what m x double dot plus k where k is a function of x times x. So, we say that let we assume a solution for x, x 1 e s t plus 2 e 2 s t plus x 3 e 3 s t. So, you can take an infinite series and we plug this equation in the equation of motion. So, my equation of motion becomes f naught e s t equals and then I am going to collect different terms. So, mass times x 1 s square e s t plus x 2 s square e 2 s t plus x 3 e I am sorry 4 x 2 s square 2 s square 3 s square 2 s 3 4 x 2 s square e 2 s t plus 9. So, mass times acceleration, so when I differ you are saying here it should be 4 s t 4 s 2 4 into x 2 coefficient of x 2 and 9 e 3 s t that is mass plus a naught x 1 e s t plus x 2 e 2 s t plus x 3 e 3 s t plus this infinite number of terms here plus k 1 x 1 cube e 3 s t plus x 2 cube e what 6 s t plus x 3 cube e 9 s t plus and then they will be cross coupling terms. So, 3 x 1 square x 2 e 4 s t plus 3 x 1 square x 3 e 5 s t plus 3 x 1 square x 4 e 6 s t. So, I get an infinite number of terms here. So, if this equation has to hold true then all the terms associated with e s t when I take their coefficients they have to vanish they have to become 0 all the terms associated with e 2 s t their coefficients collectively they have to vanish and so on and so forth. So, that is what we will do. So, once we rearrange we get we collect terms in terms of e s t m s square plus a naught minus f naught then e 2 s t 4 m s square plus a naught x 2 plus e 3 s t 9 m s square plus a naught x 3 plus a 1 x 1 cube plus e 4 s t 16 m s square plus a naught x 4 plus from this guy I write my first relation that now there has to be an x 1 here in the first one x 1. So, x 1 equals f naught over k naught plus m s square what is x 2 x 2 is 0 x 3 is minus a 1 x 1 cube over a naught plus 9 m s square x 4. So, what you are seeing is that just because you had one cubic term in the stiffness you are generating infinite number of extra frequencies for x and that is all noise. So, this all the terms associated with x 2 x 3 x 4 they are all noise. So, in reality the system is not only linear in the sense that it is dependent on x square, but it is also dependent on x cube x and all powers of x. Stiffness by itself can is dependent only on cubic you know. So, the dependence of stiffness on x is only at 0 th order square 4 term 6 term and so on and so forth, but there are other terms damping that has other dependencies then there is also dependence on frequency properties of the material change with frequency. So, things are not very nice and linear and once they are not nice and linear then you get second or the third or the fourth or the fifth or so on and so forth harmonics. As long as your displacements are small the magnitude of these harmonics is small compared to the base tone to the input tone, but if your displacements are large for instance if x is large then k 1 also starts becoming important and once that happens then noise gets generated by the system. So, nonlinearity is a very big source of the system. So, as we are engineering systems and if we want to manage noise then one important consideration has to be that our system has to be more or less linear in the operating system.