Okay. Today, we will talk about, I believe today is lecture number 20. We would like talk about Helmholtz resonator. Okay, for example, this is quite famous, popular bottle that many Koreans enjoy, during the dinnertime. We would like to estimate the resonant frequency of this Helmholtz resonator, because at this resonant frequency. As we discussed in the last lecture, the fluid particle over here is moving very fast with the frequency of Helmholtz resonator's frequency. Therefore there will be some big impedance mismatch over here. Because impedance Z is a P divided by velocity. And the velocity is very fast over here therefore there is an impedance mismatch. Let me first estimate by measuring the frequency of this Helmholtz resonator by using my, smart phone. Okay, all right. [SOUND] So as you can see here, the, you, you did, you did zoom that already? Okay, that's good. As you can see already, the, there are very high resonance peaks at about 180 Hz, okay? [BLANK_AUDIO] So, let us see or try to understand this problem as following. Okay, here is the resonator. And it is surrounded by rigid wall, in other words, the wall impedance is about infinity. And say the volume over here is very small compared with the wavelengths that I have interest. And of course, I excite over here by P out, I am exciting there by P out. [NOISE] This is my P out and, and then this- Say, this is the area of the neck, and say this is the length of neck. So in this case, length is about this, okay. Now. And let's recall that the representative acoustic variable is pressure, density change and velocity. Okay. We remember that the relation between pressure and velocity, which is the velocity of acoustic medium, is controlled by for, this governing equation, that is Euler's governing equation, simply says that the force of balance per unit area will induce the fluid accelerated by DU DT. And the relation between this and this, which we call conservation of mass. That says the change of density, has to be balanced by the mass flux, over here. Therefore, say this is three dimensional case. And the relation between pressure and density, which means that when I press the certain volume of fluid, the fluid will be squeezed. Therefore there is a relation between DP Drho for isentropic process, and that is the speed of sound. So we would like to see this problem by considering pressure, density and velocity, of course. So let's see how much the, when we apply the pressure. And if the pressure is changing, and say we are, there is some pressure in the neck that we call P in, in this case P is P in, then this fluid will be. Okay, this fluid will be squeezed if I apply the pressure, so Dp Dt will change the volume change, that is quite obvious. Okay. And of course we would like to find a certain, constant. They can make this proportionality to equality, right? This, this is, this is just come from our physical physical physical observation. I applied the pressure then this volume will be squeezed. Yeah? Because the wave lengths of interest is much, much larger than, than, the, the, the characteristic lengths of this, this resonator. Okay, and look at the dimension of this, okay. Look at the dimension, this is the pressure, divided by time, and this is volume divided by time. We invite certain proportionality, CA, which. We will explain later, dp dt can be equal to dv dt and the CA. If you look at the dimension then CA has to do with something volume over pressure. [BLANK_AUDIO] So, that means the volume change, volume change [BLANK_AUDIO] When I apply unit pressure, that we call compliance, okay. So this is, the combined, introducing compliance, we can make equality over here. So what is the next step? Next step, as you can see over here, this is volume. So, we have to relate somehow this volume with velocity. Yeah? Okay, that's what we attempt to, attemp to have in the next time. In the nest step. Okay. So what is dv dt So, if look at the dv dt, that is the volume change per unit time. So, volume change per unit time should be related with the something has to do here. Right? If there are some velocity acting on this area A, okay. Then, the, the du dt, which is the velocity per unit time. On this area we'll squeeze this volume right? So I can write the dv dt it has to be A times the velocity on the neck. And if the velocity is a positive then dv dt will decrease or if velocity has negative value then dv dt will increase. Okay, that's quite acceptable. Right, we just consider the sort of conservation of mass upon this volume, okay. Okay, this I could say. So the cons-, We apply the conservation of mass. Now this is related with U, therefore we can write. CA, dp dt, is equal to minus A, ut. Is it correct? I think it should be a minus over here. Right? If I apply the, the pressure, the pressure will squeeze this volume therefore, that is plus. So I have this equation. Okay, and here P is, of course, pressure inside, P in. All right? Now. So, I want to change this ut somehow to pressure, then I will have, governing equation, governing equation with respect to P, so want to change P, with respect to U. So, let's consider again what's going to happen over here. This is pressure, excitation pressure outside, and this is the pressure right over here, and the difference between these two pressure will move the mass inside of neck. Right? We only consider what's going to happen over here, but now we have to consider what's going to happen between P out and P in. So, I can say P out, minus P in, would move the mass in the neck, okay? I can write rho0, and the L is the length over here, and the A, that is area. Then that is the mass of neck. This is simply net pressure difference. And this will move with certain acceleration, and acceleration I can write du dt. Okay? This is acceleration. And I have to multiply A of course over here. This is force acting on, net force acting on this mass. I just applied Newton's second law. Okay, that's pretty straight forward. Then of course therefore, P out minus P in is equal to rho0 L, du dt. And we have this right? So okay, and if I take derivative with respect to time over here, then knowing that CA d square p dt is equal to A du dt there for combining this equation gives me P out equals to P in Plus rho0 L, and du dt is CA over A, d square P, dt square. Yeah. And this is Pin. So, what it says? In fact, the, this equation says P, pressure. Since, I mean, this pressure is P in plus rho0 L, divided by A, CA. d square P, dt square is equal to P out, that is the excitation pressure, excitation pressure. OK, now this is very exciting. I have an excitation and the inside of pressure follows this, this a kind of the differential equation. Okay? So what is natural frequency, then? Helmholtz resonator. [SOUND] Uh-huh. What is natural frequency, then? Okay, recall that, if you like When we have a mass spring dashpot system. What is natural frequency? How do you obtain natural frequency? When Ft is 0, that means we are exciting the system with the for example, unit impulse. And this will be oscillated? And we assume that xt can be some magnitude exponential minus j omega t, then, we have m minus m, omega square plus k is equal to 0. And that is natural frequency and omega square is K over m. Right? That is the sometimes we call Eigenvalue. So in this case, what is natural frequency? This is like m, and we have a one. So, the natural frequency in this case, we can say. Therefore, natural frequency of a Helmholtz resonator is one divided by CA. rho0 L over A, if I rearrange this this is A CA rho0 L. That is the natural frequency of Helmholtz resonator, okay. When we need to express CA in terms of what is related with resonator. Right? If you can express CA in terms of what it's re, related with resonator, then it's, it's more useful. And omega m square will be more useful. So starting with the definition of the, CA. Okay, CA is related how much volume change when I apply the pressure change, okay? So pressure, velocity, and the density. And the density change is related with the volume change. So let's apply conservation of mass, okay, in this volume. So, in the beginning the volume has this mass and then when the volume is changing would the, I mean, there's a density change. Then, we can write this has a volume change, V, plus delta V, and then I can write this is rho0, say delta rho, rho0. V plus rho0 delta V, plus I have delta rho V, then I have delta rho, delta V. And I have high order term, and these two terms go away, therefore what I get, is. What I get is rho0 delta V, plus V delta rho is zero, therefore I can write the following equality. That is, rho0 delta V is equal to minus V delta rho. Therefore, within short period of time. Which is dt, or delta t, this equation I can write rho0, dv dt, has to be equal to V, d rho dt. That is good. That is, I just wrote in terms of the V and d rho, and then, we want to have some relation that shows the compliance, the compliance in the beginning, it says recall dP dt CA is equal to dV dt minus, right? That's what we had before. Pressure change will induce the volume change and that is related to CA. So, to get this relation we have to convert rho to,to P right? So, rho and P relation. What is this relation? Okay, that is, the, state equation, right. And the state equation says d rho, dP d rho for isentropic process is C square. So, dP has to be C square d rho and d rho, is equal to dP over C square, therefore this will be changed to V multiply. I want to write that this is a dP dt. Then the rho is equal to dP divided by C squar.e so dP divided by C square, right. So, now, I have this and what is CA? Then CA has to be equal to this one has to, this one has to go over there. So rho0 C square divided by V, that is CA is it correct? Oh no, no dP dt, this term. So, CA has to be V rho0 c square. That is the compliance. Therefore, therefore. Therefore, natural frequency of Helmholtz resonator, huh? Therefore, omega n square is equal to A divided by CA. That is V, rho0 C square. Okay, rho0 L, right? We- rho0 L, so therefore what I have is A divide by volume and the rho0 go away and then I have L and C square. That is very interesting. [BLANK_AUDIO] So, therefore the omega n is equal to C square root A V L. Oh. So, if neck length is large, I mean, very long, the natural frequency will go down. If volume is large then natural frequency is going down. If A is getting large then natural frequency is going up, like this. [SOUND] In this case. [SOUND] So, let me measure again the natural frequency of this beer bottle by using our famous spectrum analyzer. Okay. Which is less than what we had. [SOUND] That is about 180, and this is about [SOUND] probably somewhere 150 or 130. Okay. Let's estimate whether or not this formula is correct, to estimate with this. I dont know whether it can reach within the acceptable error bound, but let's try. Okay, this has to be a omega n. And this is the natural frequency in terms of Hz. therefore, as you can see over here, we have to calculate A, that is the area of bottle, and then L, and this is the volume. And the volume, we estimated it is about 360, millilitres. So, and, so, so, in terms of the, mks unit that has to be, about 3.6 multiply by 10 to the minus 4 meter cube. And the area in this case is, has to be pie multiply by r square, and r here is about the diameter which we estimated was 2, 2 centimeters, so therefore the radius would be 10 to the minus 2. So in this case that has to be pi multiplied by 10 to the minus 4 meter square. So the, the frequency has to be C. That is the speed of propagation and two pi. And we have an area, which is about this, and the volume is about this, and L has to be the effective length, in this case, L should be the 7 centimeter that is the actually measured length of a bottle and one, this part has to be the un-flanged case, this is the un-flanged. And this is not really unflanged as well as not really flanged case, but if we assume that this is flanged case, therefore we are using two different length scale, so it come up with 8.46 10 to the minus 2 meter. And the plot these, all these value we got was 175.3 Hz which is quite close to the value we used. But actually, we can use the 180 Hz, okay? Which is quite easy to remember. In many, in many respect, 180 Hz, 180 degree, you can imagine some of the many interesting things, okay, as you can see here, the effective length for the flange case, we will use this value, and without flange we use this value. Okay. [BLANK_AUDIO]