Okay, tao if you rewrite a tao for limp work case we saw there is minus j omega m plus 2 z0 and a 2 z0. Okay. This expresses the motion of a roar, and the roar has only mass. And the reason why we have a minus j and an m, is because, simply because, we are seeing the motion in terms of impedance, and impedance is. Force divided by velocity, therefore this is minus j omega. The contribution of mass is simply minus j omega m. Then let me ask you, what is the contribution of stiffness? Then that is k. Because face difference between mass like behavior, sorry mass like behavior and spring like behavior should be J omega or what about the math like behavior and damping like behavior? Those should be J omega M,- J difference. Again, another- J difference with mass. So this is pretty much associated with the partition. So let me me call this is partition impeders, Jp. Okay? And this term over there, 2Z0 is the impedance that is seen that is, see, saw, seen, right? That is seen by the wall If there is only one medium, then it look like Z zero, but it look at the medium in both side, and therefore we have two. So, this is a freed loading impedance, and the free loading. And in fact, this form holds for everything. So we are now succeeded to look at any kind of partition problem in terms of fluid loading impedance and partition impedance. This is quite. In terms of my this afternoon expression, this is quite sexy expression. So rest of job, you will do for any climb a partition, for example, if we have partition that has a spring, and a damping, or if it has some finite, you know, play to do. Only job we have to do is to find out the partition impedance of each case. Then, we can get transmission. Coefficient or transmission ratio if you like. And we do know, this is fluid loading impedance. Okay? So let's move on to study what's going on if you have partition that has a spring mass and a case. Keeping that in our mind. All right, so let's consider the case when we have spring and a damping in the mass. And suppose there is an instant wave then they should have reflected and transmitted. Okay, we assume that Pi(x,t) is equal to pi capital complex explanation minus j omega t minus k. X assuming for simplicity we have the same medium over there, and Pr xt is Pr explanation minus j omega t + kx because we have left going way and we can write pt exponential minus j omega t plus kx. Same k because k equal omega over c and we have it here. So we assume that there is the same medium for if you have another medium then you have to use K1 and K2, same frequency because we are handling linear acoustics, small and pressure and apply a boundary condition and everybody can, Understand what boundary condition you can use, so pressure on left hand side would pi plus pr. And the right-hand side is pr. Therefore we are considering that force acting positive x direction. And that has to be equal to what? Not just math contribution which is -omega squared my, but we do also have something related with the motion of I mean due to the damping force that is this, and now we have contribution of k plus. I don't know whether this is minus or plus. Is it minus? Okay. Pt. Sorry. Thank you. Maybe, I owe you some money? [LAUGH] This is first valance, and another one is velocity continuity on this side, that would be, PR, over z zero, minus PR, over z zero. That has to be equal to the velocity of what? What is this? -j omega Y, all right. And also the same thing on the right-hand side, the velocity of the fluid particle over there is Pt over z0, and that is also minus j omega Y. And we solve this equation, then we come out with this. And in this case artesian impedance look like Y. Partition impedance look like sum c related mass, and the sum c related with, damping. Something related with spring. Right, so the only change is the partition impedance, and there may be some correction of this partition impedance. Okay, let's look at what we have by using the, our PowerPoint. It is okay? Okay. So as I said before, this problem can be regarded equivalent as we have blocked pressure and radiated the pressure, okay. And then I explained this problem and then we found this, transmission equation looked like that which is not same as I showed before. The reason why we made a mistake is because transmission laws in terms of mp does that is pressure divided by velocity. And that the formula shows, is in terms of displacement. So we have to multiply and this will give this. And this is s is the stiffness. And the reason I use S instead of K is because we use K as a wave number. And the RD term over here. So in this expression, figure color just means there's two terms. Expresses damping. What does damping do to linear damping, over there? And then, the other one. This one is damping due to radiation. There are two kinds of damping in this case, one is damping due to the damper, a bi-rotary system, and the other is damping, damping, what is damping physically? Damping is energy loss. So you can also lose your energy by radiating the radiated, Pressure is something you are losing. So that's why we have 2 Z0, and this is again, fluid loading. Okay, that is interesting. So again, if omega M increases compare, If omega m is getting large large compare to sob omega. That's the case when we increase omega, this will increase linearly and this will increase 1 over omega. So, term getting smaller and smaller compare with omega M Then this formula actually follows the mass law. What about if omega is decreased? Very small frequency, then this won't decrease, but this one will increase. So when omega is getting smaller and small, this term will follow, dominated by this, s over omega. That recalls the finish control region. And this is just another look of transmission coefficient and if you look at this term like that then this is partition impedance again. So, again, we can see that the transmission quotient is composed by partition quotient plus fluid loading impedance. All right. This just shows some of the case of no absorption coefficient. So this is the highlight. So for high frequency, it follows the mass law, 6 dB per octave. At low frequency, 6 dB for octave, as we decrease the frequency. But when over here. What kind of physical element will dominate. This is mass controlled region, and this is stiffness controlled region. This range, damping controlled region. So if you increase damping, the transmission loss will increase. The transmission loss increased. So, over here, if we increase the damping it will increase as you increase the damping. But as you may easily see, this is most expensive way to increase the transmission loss. Right? For example, you have partition, you increase damping, of course you can Increase the transmission loss, because you increase the damping, the movement of your partition will be decreased, therefore radiation damping, radiated power will be decreased. So you can use this kind of behavior if you have a lot of money. But if you are clever, of course you will use this control region. Very obvious, that's why I say the mass law plays a central role in partition design, okay. So this summarize everything related with a partition design. This is the highlight.
