8.03 - Lect 10 - First Exam Review, Breaking Wine Glass with Sound

I want to start with the physical pendulum who is exactly the same one that I discussed during my first lecture this is a hoop as mass m and radius R and we were calculating the period of this hoop as it oscillates and we did that using the famous torque equation from 801 the torque relative to point P is the moment of inertia relative to point P times the angular acceleration opera today I will do this again that I will use the conservation of energy to show you that in case there is no damping where mechanical energy is conserved that you can find the correct differential equations through the conservation of energy if the thing is swinging in general there are two components you have kinetic energy and you have potential energy and the kinetic energy K is one-half times the moment of inertia about point P times Omega squared you remember that from 801 and this Omega is Theta dot we call that the angular velocity this angular velocity changes with time when the object goes through equilibrium the angular velocity is a maximum and when the object goes - comes to a halt the angular velocity is zero do not confuse this Omega with Omega zero I will give that a zero now to distinguish it from omega which is the angular frequency the angular frequency is a constant of the motion and that is 2pi divided by t 0 if T 0 is the period of oscillation so this is the kinetic energy and this is the square of the angular velocity and then we have potential energy that this be point a and when we are here the center of mass is at point B and so the potential energy you or actually I should say the potential energy at point B minus the potential energy at point a it's always the difference in potential energy that matters that equals mg H H being the difference in height between point B and point a so this here is H MGH Massachusetts General Hospital that's the way to remember it now H is very easy H is the same as R times one minus the cosine of theta we went through that many times so you can easily confirm that R is the radius of this circle this is the radius now for small angles the cosine of theta is Theta squared divided by two and so I can rewrite this differential equation now that 'total I see nothing wrong with that I'm sorry is there anything wrong with it the cosine theta alone you're right is one minus theta squared over two thank you very much thank you so I'm going to write it now as the total energy is one-half IP times theta dot squared and then I get plus mg R times theta squared over two and what you do now since this is a constant of the motion you always do that if you work with the conservation of energy you take the time derivative of this equation which must be zero because mechanical energy is conserved and out pops the differ rental equation that we also found during my lecture number one when I use the different method so I take the time derivative so this key each of this half and so we get I of P times theta dot times theta double dot I have to use the chain rule and then I get plus MGR this to each of this two and so I get theta times theta dot and now this equals zero because the EDT is zero and whenever you do that you will always see that the theta dot term disappears or if you have the equation in X then the X dot term disappears and you see that this term goes you can divide it out and so your differential equation takes all now very familiar form let me write this to here so I get theta double dot plus MGR divided by I of P times theta equals zero and this differential equation you should recognize the solution to that equation is immediately obvious Omega zero squared I use now the Omega zero equals mg times R divided by I of P and so the general solution for this oscillation then becomes that theta is theta zero that is the amplitude times the cosine or the sine if you prefer that Omega 0 T plus some phase angle Phi that is the general solution and if you knew the initial conditions what the situation was at T equals zero if you know the velocity the angular velocity equals zero and if you know where this T equals zero you can solve for this theta zero and you can solve for this Phi but Omega zero is independent of the initial conditions very well now I would like to cover a case whereby I am going to introduce damping whenever we deal with damping there are two terms that are important in physics that is the speed itself there is a damping term which opposes the velocity which is linearly proportional with the speed and there is a damping term which opposes the velocity which is proportional with the square of the speed we will always leave this square out with 803 because the differential equations become impossible to solve maybe numerically you can do it but not analytically however if you're interested in all the physics which is wonderful with the v square and the v my lecture number 12 on OCW OpenCourseWare from 1999 Newtonian mechanics i deal with the v square and with the v term and i do many demonstrations to show you that there are certain domains where the V term is important we call that the viscous term and there are certain domains in physics where the v square term is important so I will now simply restrict myself then to the two the damping force which is linearly proportional with the velocity and we will write it down as F equals minus B times V we will use a shorthand notation that gamma equals B over m that is only a shorthand notation I will erase this we don't need this anymore we need so many blackboards today that I will use this blackboard for my damping problem when you deal with damping we recognize three different domains one domain whereby gamma is smaller than Omega 0 we call that under damped then we have a domain whereby gamma is larger than Omega zero we call that over them and then you have a very special case where gamma equals Omega zero and the behavior of these three different kinds is very different I will only discuss with you today the under damped case so I have a pendulum and this pendulum has mass m and length L and I will assume that the mass in the string is negligibly small I have damping and this is the case that gamma is smaller than Omega zero this is the equilibrium position of the pendulum I will therefore call this position X and I am going to put all the forces on this object in this picture which is mg and that is T and there are no other forces on that mass now if we deal with small angles as we have done before more than once then the tension T is very close to mg so the only force that is driving it back to equilibrium the only restoring force then is the horizontal component of T and so that is this horizontal component that's the only one that we are concerned about so the differential equation then in terms of Newton's second law becomes MX double dot and then we get minus T times the sine of theta minus B times X dot that's the damping and this is the restoring force due to the tension in the string and the sine of theta is X divided by L so I get M X double dot plus mg divided by times X that is the sine of theta plus B times X dot equals zero make sure I have this M double dot I have the plus sign here I have mg x over L that's fine so now I divide M out and I also use shorthand notation that Omega zero squared is G divided by L these are shorthand notations would give you a little bit more insight when you see the solutions and so that gives me now the differential equation that X double dot plus gamma times X dot plus I divide em out right plus Omega zero squared times x equals zero and that is the differential equation that you will recognize and I would never want you to derive the solution to this differential equation that's of course way too time-consuming to do that during an exam and the solutions to this are given on your formula sheet which will be part of your exam and so let me write down here what that solution is so X as a function of time is a certain amplitude times e to the minus gamma over 2 times T and then we have a cosine or a sine cosine Omega T plus some phase angle alpha if you knew the initial conditions then you can find what the amplitude is and you can find what the angle alpha is if you don't know the into initial conditions then you do not know this there is another way that you can write this form which is sometimes better and I cannot tell you when it is better and when it is not better depends on the initial conditions but I want you to appreciate that you can also write this for instance as e to the minus gamma over two times T and then B times cosine Omega T plus C times the sine of Omega T fit from the physics point of view there is no difference but from the math point of view there is a difference you now have these as the two adjustable constants depending upon the initial conditions and sometimes if you assume this it works faster than if you assume that and as I said it really depends on the initial conditions which goes faster but they are very similar of course Omega which is now the frequency with which this object is going to angular frequency is going to move that Omega is always lower than Omega zero and that shouldn't surprise you because when there is damping there is something that is opposing the motion and so it shouldn't surprise you that when you solve for Omega that Omega squared is omega zero squared minus gamma squared divided by 4 that is also something that we would probably give you on the formula sheet because you will only find that if you substitute this back into the differential equation we often particularly French likes to write down Q equals Omega divided by gamma this is the quality Omega zero divided by gamma and if you do that then Omega squared can also be written as Omega zero squared times one minus one over four Q squared that's all you see that if Q is for instance ten which is by no means absurdly high that Omega is only one eighth of a percent lower than Omega zero so they are that close and even when Q is two the difference is only three percent between Omega and Omega zero and so what happens here is that the amplitude is decaying at the time constant one ovary time constant which is two divided by gamma seconds if you put in T 2 divided by gamma the amplitude goes down by a factor of e since energy is always proportional to amplitude squared the decay time of the energy is not 2 over gamma but is 1 divided by gamma now I'm going to make a change I'm going to drive this system this is no longer the equilibrium position but in liquid equilibrium position was really here and I am driving the top of this pendulum with a function Etta is at a zero times the cosine of Omega T so I am driving it now so this Etta is in terms of inches millimeters this is the motion of my hand and this Omega is no longer negotiable this Omega has nothing to do with this Omega this is the frequency with which the system likes to oscillate that now is the frequency with which I want the system to oscillate they are totally unrelated this is my will this is non-negotiable I can make this anything I want that I can make it zero I can make it large I can make it all so that value of course so now the equilibrium position is not here and I will call now this distance from the equilibrium position X you're always using your coordinate system the equilibrium position as you zero and so what is the only thing that is going to change now is that the sine of theta is no longer X divided by L but the sine of theta is X minus Etta divided by L because notice if this is Etta which is the position of my hand then the sine of this angle is now X - Etta divided by L and this now you have to carry through in your differential equations and what you will see then if you do that that this now is not zero but this now becomes at a zero times Omega zero squared times the cosine of Omega T my Omega so you instead of sine theta X divided by L all you have to do is this you know that a is after zero cosine Omega T you carry that through and you will see that you differential equation now changes notice that X double dot is an acceleration notice that gamma times X dot is an acceleration gamma is one over second and X dot is meters per second this is an acceleration notice that this is an acceleration it has the dimension Omega squared which is one second square times meters that is an acceleration notice that this is an acceleration after zero is meters an Omega squared is one divided by second squared so these are apples these are apples these are apples and these are apples so the equation looks very kosher to me now the solutions that's a different story that is something that I wouldn't want you to derive either but the solutions now become as follows I now get X as a function of T is an amplitude which is very strong function of Omega but I will simply write it down as an amplitude a times the cosine of my Omega t minus some phase angle Delta and this is what we call the steady state solution steady-state and this a becomes then a rather complicated function upstairs I get the at the zero Omega zero squared I get this part and downstairs I get the square root of Omega zero squared minus Omega squared squared plus Omega gamma squared and that is the amplitude as a function of Omega and the tangent of Delta becomes Omega gamma divided by Omega zero squared minus Omega squared if we plot that function a as a function of Omega so here is Omega and here I'm going to plot the a then I can recognize that if Omega goes to 0 that means if I move this very slowly that for sure the amplitude of this object must be the same as my hand must be at a zero if it takes me one week to go from here to here then of course the pendulum is always hanging below my hand so when Omega goes to zero I can always check the result here that a must go to at a zero and indeed if you put in omega equals zero you will see that it's the case if Omega goes to infinity then a goes to zero and then there is a very special case that is when Omega happens to be Omega zero which is the frequency of the system in the absence of damping this is my Omega zero which is not this one but it is the Omega zero then you will get an amplitude which is Q times at a zero and again I just happen to remember that if you substitute this back in here this Omega equals Omega zero you will see that and that's always very nice to remember at that Omega zero you always get Q times more than you driver if Q is a hundred you'll get hundred times the amplitude of the driver and so here you get at a zero if Omega is zero and then when you reach that Omega zero frequency you come out high and then it goes back to zero and this point here is Q times at a zero and we discussed before I'll never make too much of a deal out of that that the actual maximum of this curve is not at Omega zero that is always a little lower that is more really more of an algebraic of algebraic interest than that it is of physical interest because if gamma is low if Q is high the to almost coincide so this steady-state solution has no adjustable constant in other words the system has lost its memory of what happened when we started driving it so at T equals zero if we know what X is and if we know what X dot is and if we know what the driving term is at T equals zero then the general solution for all times larger than zero is the sum of the transient solution and the steady-state solution and so here you see the trend solution you can write it in this form or you can write it in this form and here you see the steady-state solution this a has nothing to do with that a this a follows from the initial conditions just like that alpha this a does not follow from the initial conditions this a follows from after zero what my amplitude is of my driver and it follows from omega so the two A's don't confuse the two and so the general solution is then the the sum of the two the transient one will die out if you wait a few times 2 over gamma the transient is gone and so you and end up then only with the steady-state solution and you had some chance at your homework to to work with that if we are not driving the system and if we have one object on a spring or on a pendulum or a floating object in liquid then there is one frequency one normal mode frequency one resonance frequency we call it also the natural frequency the moment that you couple oscillators if you couple to you get to normal mode frequencies you get two resonant frequencies and if you have three objects you get three normal mode frequencies and so now I would like to discuss with you a case whereby I couple two oscillators if I gave you on an exam three coupled oscillators that would be very nasty because it's extremely time consuming if I gave you four coupled oscillators that would be criminal because you cannot finish that in eighty five minutes so two is certainly within reason three is marginally within reason four is out of the question when we have coupled oscillators we always leave damping out and yet we will learn a lot from it even without damping we will learn a lot and so what I have chosen to do with you is a spring system with two objects two masses and the springs have no mass negligible mass the spring constant are K and the masses of the objects are M so this is the equilibrium position of each two object and the ends are fixed of the string I will introduce a shorthand notation that Omega 0 squared equals K over m now what I do I offset both these objects from equilibrium just as I did that here I always offset them in the same direction and I call that positive is that a must no is that useful yes so I offset them and so this position now this is object number 1 and this is object number 2 this is now a distance X 1 away from equilibrium and this one is now at a position X 2 away from its equilibrium you always want to know what the displacement is relative to its own equilibrium and it's only Calabrian for one is here and the equilibrium for 2 is there I now make the following assumption that X 2 is larger than X 1 is that a must know is it useful yes but if you want to assume that X 2 is smaller than X 1 be my guest I will assume that X 2 is larger than X 1 now follow me closely now so we have an object here and we have an object here and this is now one spring this is the other spring and this is the third spring it is immediately obvious that this spring is longer than it wants to be so there is a force that drives it back to equilibrium if x2 is larger than x1 this spring is also longer than it wants to be so it wants to contract so there is a force in that direction if this spring is longer than it wants to be it wants to contract so there is a force on x2 in this direction this spring is clearly shorter than it wants to be so it's pushing so there is a force in this direction and so now I'm going to write down the differential equation first for object number 1 so I get M x1 double dot so this one is minus K times x1 and this one is plus K times x2 minus x1 is it a disaster if it turns out that x2 is not larger than x1 not at all this equation now is correct for all situations the fact that I have assumed that x2 is larger than x1 gave me a plus sign here and so my differential equation is safe no matter what x1 is relative to x2 so you can always make that assumption and you don't have to worry later anymore about science I can divide now M out and then I get that x1 double dot plus 2 Omega zero squared times x1 the in order to have one here and I have one here they have both a minus sign and then I have here plus so that becomes minus Omega 0 squared times x2 that is zero so that is my first differential equation so I put a 1 here so now I go to the second one I get em x2 double dot now I have two forces both in the negative direction first I have the one that drives this one away from equilibrium so I get minus K times x2 minus x1 and then I have the force due to the fact that this one is shorter than it wants to be by an amount K 2 so I get minus K times x2 it's shorter by an amount x2 then it wants to be and so I can divide em out and so I get em x2 double dot and then I get plus Omega zero squared times x2 and then I get minus Omega zero squared times X 1 equals zero now compare this one with this one notice the incredible symmetry I could have found this one by changing a 1 to 2 here and by changing the 2 to 1 of course the system is so symmetric that I yes thank you very much for the - right yeah thank you the system is so symmetric that clearly nature cannot make any distinction between 1 & 2 so it is in this particular case because of the beautiful symmetry it is obvious that these two differential equations look very very similar now I'm going to make an important step I'm going to substitute in here x1 is C times cosine Omega T and X 2 this is C 1 is C 2 times cosine Omega T I want to know what the normal mode frequencies are for this system I want to solve for Omega I want to find Omega I'm not driving the system since I have no damping in which case at normal mode solutions either the two objects are in phase with each other or they're out of phase with each other but the they're out of phase we can always take care of that with a minus sign so I'm now going to substitute that in here and I may want to continue working on the center board in which case I might as well erase this so that it stays a little compact so I'm going to substitute this trial function into my differential equations and I every term will have a cosine Omega T so I dumped all the cosine Omega T's so I go to the equation which has a big one there and so that one becomes c1 now x1 double dot gives me a minus Omega squared right if I take cosine Omega T and I do the second derivative I get minus Omega squared in front of it so I get c1 times two Omega zero squared that is that two Omega zero squared and then I get minus Omega squared and then I get plus not plus then I get minus Omega zero squared times C 2 and that equals zero and then I go to my second differential equation which is this one but I'm going to rearrange the C ones and the C twos so I'm going to put the C ones here so I'm going to get minus Omega zero squared times c1 and here I'm going to get plus two Omega zero squared minus Omega zero squared times C 2 and that equals zero so I can leave this plus out what do I have here now I have here two equations with three unknowns I want to know what Omega is in the normal modes but I also would like to know what c1 and c2 is well there is tough luck you can't have it both ways we do not know at all what c1 and c2 is separately but you can always find in the normal modes without knowing anything else what the ratios are of those amplitudes c1 and c2 and you can always solve for Omega and you will see how that works so with two equations and three unknowns you can only solve for Omega and for the ratio c1 over c2 so now you may do that any way you want to this is not so difficult to solve this I will however use Cramer's rule and I will introduce ad which is the determinant of the following two Omega zero squared minus Omega squared and then I have here minus Omega zero squared and then I have here minus Omega zero squared and here I have two Omega zero squared minus Omega zero squared and this now must be zero if I am looking for solutions which are not C 1 is 0 and C 2 is 0 it's clear that if C 1 and C 2 is 0 that you equations are satisfied 0 is 0 but that's not an interesting case and the only way that you can get a solution which is interesting with values for c1 and c2 which are not 0 is by making this determinant 0 and so that means that D which now becomes 2 Omega 0 squared minus Omega squared squared minus Omega 0 to the fourth you have to make that 0 and when you do that you find two values for Omega you'll find that Omega squared it's two Omega zero squared plus or minus Omega zero squared those are the two solutions and when we evaluate those two solutions you find the result which is so embarrassing ly simple that you could almost have said that without any work almost notice that Omega minus which is the lowest one of the two is the same as Omega zero because that's the minus sign what does that mean if Omega minus is omega zero well it means of course that the two objects are just oscillating like this the inner spring is never stretched it's never longer than it wants to be is never shorter than it wants to be so each one is only driven so to speak by the outer spring so that's immediately obvious that that's a normal mode and you can make the prediction that c1 over c2 must be plus one they must go in unison like this and you can confirm that by substituting this Omega one if you substitute that in this equation you will find the ratio c1 over c2 if you prefer this equation be my guest you can do that too and you will find no matter which one of the two you take that it is plus one now Omega plus is less obvious is the square root of three times Omega zero that's when you take the plus even though it's not obvious that it is the square root of three it is clear what is happening here are the two objects and now they are doing this they are exactly 180 degrees out of phase with each other so now you can make comfortably the prediction that c1 over c2 is minus one and indeed if you substitute the square root of three Omega zero in either this or in that equation this indeed comes out the general solution now for a given initial condition so if now I give you the initial condition then the general solution would be 4 X 1 would be X 0 minus the minus makes reference to that lowest frequency times the cosine or we got minus T plus some phase angle - all these minus signs make reference to this mode and then I have plus X 0 plus times the cosine of Omega plus T plus some phase angle plus of course if you prefer instead of cosine signs that's fine of course and you have 4 adjustable constants 1 2 3 4 and if you know the initial conditions if you know what X 1 is and what X 1 dot is and what X 2 is and what X $2 at time T equals 0 you can solve for o 4 in principle there is no longer any freedom for number 2 because all four adjustable constants have now been consumed and so therefore you can write down out a solution for X 2 by simply making this one a 2 and this whole term here is the same the only difference is that this plus sign now becomes a minus sign nothing else is different the frequencies are the same otherwise there wouldn't be normal mode frequencies and it is the ratio here of the c1 over c2 that is plus 1 and it is the ratio here that is the minus 1 that's the reason why this becomes a minus and why the plus here remains a plus so that now is the general solution if you also know the initial conditions so now we're going to derive the system so now we go back to where we were and I'm going to drive this end and I'm going to drive this end with my hand this is Etta and Etta equals at a zero times cosine Omega T there is only one term in all of this on the blackboard that is going to change and that is this term this spring here on the left is no longer shorter by the amount x1 but it sure the body amount x1 - Atta and so it's only that term it's only this - K x1 that now have to be changed into minus K times x1 - Atta nothing else changes but that has major consequences of course for one thing if you're going to substitute now these trial functions Omega is no longer negotiable Omega is my Omega now I set this Omega you're not going to solve for that Omega that would be an insult to me I dictate what Omega is so that means I now get two equations with two unknowns c1 and c2 Omega is no longer unknown so now I get solutions for c1 and I get solutions for c2 and of all the equations on the blackboards that you have the one that is going to change is the first one and when you carry through that etta which is at a zero cosine Omega T this zero here changes into at a zero times Omega zero squared the cosine Omega T is gone because I've divided cosine Omega T out in all those other terms so now I use Cramer's rule and now I can actually come up with a solution for c1 no longer ratio c1 over c2 no I can actually come up now with solutions so c1 now using Cramer's rule this now is my first column at a zero Omega zero squared zero and this now becomes my second column minus Omega zero squared and I get here two Omega zero squared minus Omega squared divided by D if I pick a random value for Omega D is not zero D is only zero at those two resonance frequencies and then c2 becomes the first column it's like this that is two Omega zero squared minus Omega squared minus Omega zero squared and now the second column becomes at a zero Omega zero squared and a zero divided by D I worked see one out a little further than I get c1 equals at a zero Omega zero squared times two Omega zero squared minus Omega zero squared divided by D and here I get that C 2 this is zero so I get plus Omega zero to the fourth times at a zero divided by D and D now is this determinant that is not zero it's only zero for those two special frequencies if you want to see what the amplitudes now are as a function of Omega is a very very interesting behavior then what helps is you go Omega first to zero and then you see what happens and it's not so intuitive what happens if you put in Omega equals zero you'll find that c1 is plus two-thirds at a zero you can confirm that by substituting that into c1 you will see that that's what happens and you'll find that C 2 is plus one third at a zero so that is at Omega equals zero and when you go to infinity with frequencies then it should not surprise you that c1 is 0 and that c2 is 0 and then there is one case which is pathetic and that is the case that I make Omega squared 2 Omega 0 squared at that frequency C 1 becomes 0 but C 2 is not 0 and I spent almost a whole lecture with you on that demonstrating that in three different ways that indeed there is this bizarre solution so when Omega squared equals 2 Omega 0 squared then C 1 becomes 0 but C 2 is not 0 and so now you can make a plot of seas as a function of Omega and during that lecture I showed you three of those plots and I use the convention then that when the object is moving in phase with the driver I put it positive and out of phase negative and I plot here C divided by F a0 and I will use a color code I will do see one in red and I will do c1 is read and c2 I will do in in white chalk so here is zero Omega here is omega zero which is my Omega minus and then my Omega plus was at three the square root of three so that's about one point seven so here is my Omega plus and then things go nuts that is when D goes to 0 so C 1 is 2/3 here and then it goes to infinity there and then here when it is 1.4 times Omega 0 the square root of 2 then it goes through 0 it comes up here and then I get a curve here without being too precise and for C 2 I get 1/3 and then it goes up and then C 2 something like this and then I get something like that yeah I can live with that now we ignore damping and because we ignore damping we get these unphysical infinities when you hit the frequency Omega minus and Omega plus these resonance frequencies now if you include damping then the solutions become extremely complicated but of course you avoid the infinity values but the resonance amplitude can still be very high and so these plots are still very useful provided that you don't interpret the infinities as being real but something that is very large if Q is high then of course the amplitudes are enormous lehigh and this can lead to destruction we've seen the movie of the Tacoma bridge and then we have seen the dramatic experiment of the breaking wineglass you remember that wineglass demonstration because of the catastrophic success to use Bush's words of that demonstration students have asked me for an encore they would like to see it again and that's a very reasonable thing to do it fits very nicely into this concept of coupled oscillators a wineglass would have a huge number of oscillators coupled and a wineglass I have one here can be made to oscillate easily in its normal in its lowest normal mode I have to wash my hands to make you listen to it so that because my hands are now a little greasy from the chalk and this is the frequency that is the lowest mode if it is a circle like this looking from above it becomes an ellipse and then it becomes in your lips like this and the ellipse like this and in the lips like that and if now you drive that with sound that's exactly that frequency and you put in enough power another word way of saying is if you make at the zero large enough then the system can break it's making a lot of noise I've warned you those who are sitting here I really think you should protect your ears the sound can be deafening so be careful and those who are a little bit further away make sure that you close your ears I have the luxury that I can turn my hearing aids off but without my hearing aids believe me I can still hear a lot so I also will have to protect myself so one hearing aid is off the other is off and I'm going to put this on any I'm going to stroke this glass with a frequency which is only slightly different from this frequency of the sound that is about 427 Hertz which you can very easily here and then you will see the glass in slow motion because of the fact that the stroboscopic light has a slightly different frequency then the that should be coming up now and you say it will thank you very much are going to make it very dark now because this is important that you can see this very well so we'll make it completely dark in the room I'm going to protect my ears I hope you can still hear me I can hardly hear myself and I'm going to drive the system now at very low amplitude and the frequency close to its resonance you can already see that the glass is moving I'm going to increase the amplitude you see it moving I can go a little bit over the resonance and under the resonance you will see it stopped moving then so I do that purposely now now I'm off resolute I'm over it and now I'm off resonance I'm on the Reverend I'm below I now it's 423 Hertz where the resonance is at 427 so I'm going to put it back at 427 and now I'm going to increase the sound so I warn you and that's it it broke and it broke very fast so this is an ideal moment for a break since you had such a good time with the breaking glass let's settle for four minutes we reconvene in four minutes okay I now want to discuss with you continuous media if you have n coupled oscillators then you get n normal modes when you make n infinitely high then you get continuous media and if we have one dimensional continuous media like a string or a pipe with air sound then we have studied the transverse motion of a string and then we derived that now you have to use the wave equation we derive the wave equation and then you get that d2y DX squared is 1 over V squared times D 2 y DT squared Y being now the displacement away from equilibrium in this position if X is in this direction and ya that is fine and so excuse me uh-uh it's amazing how you can look at it and think that it's fine it is not alright so a V in the case of a string is the square root of T over mu that was just a special case for the string we also examined a longitudinal motion again 1d sound is the longitudinal wave and if you deal with pressure you think of sound as being a pressure wave then you get D to P the x squared equals 1 over V squared times D to P DT squared P then being positive would be overpressure over and above the ambient one atmosphere and if it's negative then it is below so it's not the total pressure but it is the overpressure and V then is the speed of sound which in air room temperature is about 340 meters per second if you prefer not to work in pressure but in terms of the actual position of the air molecules in analogy with the position of the string then you can write down the same equation in terms of size D 2 psi DX squared 1 over V Square D 2 X I DT squared that then gives you the position the actual motion of the air molecules I often prefer the pressure and I will follow that also today so our number an infinite number of normal modes and the ratios of the amplitudes of two adjacent oscillators they are coupled now reflects itself of course in terms of the overall shape which you can best see when you deal with a string as I said it's easiest to see with transverse oscillations what the displacements look like it's harder to see that with sound I also discussed with you a special situation that I connected to media medium 1 and medium 2 and I gave these two media different masses per unit length when you set up a traveling wave on a string or you set it up you can set up a pulse it reflects at the end and how it reflects depends on the boundary conditions at the end and when you connect them to media then you get not only a reflection but you also get some of the pulse some of the wave goes into medium too and so let's assume that we have an incident wave coming in like this and we have here mu 1 and we have here V 1 given by this equation they both have the same tension T and here you have mu 2 and you have V 2 we discussed that we use the wave equation to solve what happens when we have an incident harmonic wave coming in we derived even this speed of propagation using the wave equation none of this came out of the blue we always derived that and then we found by using the boundary conditions at the junction namely that the string is not breaking and that dydx on the left side is the same as dy D X on the right side with no boundary conditions we found that the amplitude of the reflected wave and the same would hold for a pulse divided by the amplitude of the incident wave was V 2 minus V 1 divided by V 1 was v2 and I called that our reflectivity and we found that the amplitude of the transmitted wave or pulse divided by the amplitude of the incident one was 2 V 2 divided by V 1 plus V 2 and I call that shorthand notation transmittivity and when we had done that I put in some simple test cases where our intuition is very good the first thing I did was I said suppose mewtwo is infinitely high so mu 2 is infinitely high in other words that medium 2 is a wall that means number 1 the string number one cannot move is fixed at the end so that means V 2 is 0 and we go to this equation and we find that R equals minus 1 V 2 is 0 you get minus V 1 over V 1 and we like that because what it means is that when a mountain rolls in it comes back as a valley and then a very rolls in it comes back as a mountain and I demonstrated that with strings it was very pleasing that T of R is then 0 well better be 0 right if everything comes back at you upside down with never the less everything comes back at you you expect that nothing goes into the wall and you see when V 2 is 0 that TR is 0 and we were all very happy and we could all sleep but then and you guess it then I said let's suppose you to become 0 so I attached that string to nothing to NT space is that practical yes I can do it I can take a nickel wire and I have here a magnetic field very strong and I can pull on that wire and the end of the nickel wire ends up in nothing in empty space but I keep the tension on so it is completely practical it can be done it's not just nonsense so mute goes mute too close to 0 that means that v2 goes to infinity and then we looked at R and we say well if R goes if v2 goes to infinity then R equals plus 1 and we were all very happy a mountain comes back as a mountain and I even demonstrated that we were still able to sleep at that point but then then came the awful thing that if we substitute v2 equals infinity in this equation that TR becomes plus 2 and now we can no longer sleep because this is absurd all the energy that rolls in comes back but there is something in addition that goes into that second medium now admit it who could not sleep that night you should all fail this course by the way but in any case I don't have to feel guilty right who thought about this and said there is something weird I must find an explanation for that in my case I had to find an explanation because I couldn't sleep who found an explanation who could say oh yes don't worry about it what was your solution here's me very good that's a very nice way of looking at it so that's probably why you could sleep yeah well actually I'll tell you why I could sleep but your solution is even shorter but the reason why I want to show you what I'm going to show you is that I want to also expose you to the idea which you already alerted to namely that there is energy involved when we deal with pulses and when we deal with with waves do remember when we have a traveling wave that the total energy per wavelength ladder-like only did it per wavelength that that total energy perhaps you remember that equals to a squared times pi squared times the tension T divided by lambda a was the amplitude energy is always proportional to amplitude squared this was the tension and this was the wavelength and if V 2 goes to infinity then the wavelength goes to infinity that's obvious right if something moves with the speed of light even faster infinity is even faster than the speed of light then let that goes to infinity so this goes to zero and so we came to the same conclusion some decadent solution TR equals plus 2 has no meaning because there is no energy in there and so I was able to sleep let's now tour and turn to normal modes of continuous media and I suggested I suggest that we go wrong Kate O'Donnell because we did so much transfer stuff let's go along here to little I have here a pipe has length L and it is open here and it is open there so that means the overpressure or under pressure here can never build up it's connected to the universe so at these two boundary conditions this P that I have there must be zero if I write down the general equation for a standing wave because normal mode solutions are standing waves I can write down P equals some amplitude times the sine or the cosine let me take a sine 2pi divided by lambda times X I will be very general I will introduce some face angle alpha for which I will find the solution very shortly and then cosine Omega T or sine Omega T if you prefer that this is a standing wave in very general terms everything here is in terms of is the space X this is X and here every all the information here deals with time which is typical for a standing wave I prefer always to write down for two PI over lambda K and I must observe the situation that P must be 0 at x equals 0 and also at x equals L if P is 0 at x equals 0 immediately you see that alpha is 0 so I'm going to rewrite it now I'm going to write down p is p 0 times the sine of KX so I'm going to replace this by K I know that alpha is zero times the cosine of Omega T and now I must meet the boundary condition that when P when x equals L that P is again 0 and that now breaks open a whole spectrum of possibilities in which I introduce this normal mode number n as in Nancy whereby n can be 1 2 3 etc and then I get solutions when K of M equals n pi divided by L you see that immediately because if I make X now L then I get the sine of n pi no matter what n is I always get 0 and so my lambda of n which is 2 pi divided by K is then 2l divided by n so I can rewrite now this equation as p 0 times the sine of n PI X divided by L and now I have cosine Omega NT because Omega M is now and now the frequencies which are associated with the ends mode and being n as in Nancy what now is the connection between this Omega n and this K well that connection you will find through the wave equation you now have to substitute this result back into the wave equation which is this one to solve for Omega and I want to do that with you it is not that much work I go to D to P DX squared so here it is d to P DX squared so all I get is I get this out twice so I get n squared PI squared over l squared n square PI squared over l squared I get a minus sign because if I take twice the derivative I always end up with a minus sign but the sign comes back as a sign and not only does the sign come back but all the rest comes back and so I will just write P here so this P is exactly that P so this is the only thing that was added by taking the second derivative against X now what is d 2p DT squared now I have to go to this function where partial derivatives X is constant here but T is the one that is changing whereas here we had that X was changing the T was constant so now we're going to get minus Omega N squared again times P the whole function comes back and now we are in business because now the wave equation will tell us the connection between the two it tells us that N squared times pi squared divided by l squared is now 1 over V squared that is the 1 over v square that I have there and there's a minus sign here and there's a minus sign here times Omega N squared so this is not connected and so you see the veloute solution for omega n just is being presented to you on a silver plate Omega n is now n PI times V divided by L that follows from the wave equation and so if you prefer the frequency in Hertz then you have to divide this by 2 pi so you get n v divided by 2l so if I want to plot now so this is x equals 0 and this x equals L I can plot now here the pressure in terms of this overpressure under pressure B and you get a curve which look very similar to a transverse solution but of course it is not transverse it's really longer to denote that you would get then that for N equals 1 you would get this mode there must be a pressure node here and there because we can never build a pressure it's connected with the universe you can never built up overpressure and so the pressure build up here positive and then later in time it will be negative and then positive and then negative and when you go to N equals 2 then you get another node in pressure here and so now you get this always a pressure node here always a pressure node there but now you end up with another pressure node there and if you have any difficulties to see what the air molecules are doing I would recommend you go back to sigh space which is the actual position of the molecules and when you do that you will always find that where the pressure has an anti node which is here sy always has a node and where the pressure has a note sy always has an anti node of course the molecules can freely move in and out there is no problem so the molecules can freely move in and out here so where is its largest amplitude is antinodes that is where the pressure cannot build up but going to size space actually often helps me to see precisely what is going on with the motion of the molecules I have here a linear system which is a sound cavity it's made of magnesium it does not air but it is magnesium and it is open open on both sides you can't have it any better I'll make a drawing for you so here is my magnesium this is a rod it's one dimensional and the length is 122 centimeters and the speed of sound in magnesium is about five thousand meters per second when I hit that magnesium rod on the side it wants to go into standing waves it prefers the lowers mode it almost always does but it may also create second and third harmonics and so the lowest mode f1 is then V divided by 2l so the lowest frequency f1 is V divided by 2l which when I calculate it is about 250 Hertz and the actual value we measured is about two thousand forty four and there may also be when I hit it with a hammer there may also be some that is twice as high so that may be 40 100 Hertz that would be double the frequency that would be the second mode and this is quite remarkable so this is not filled with air but this is filled with magnesium but by exciting it here it's like blowing on the flute there it goes into these normal mode solutions and it is this one that you will hear loud and clear it's a beautiful code and this one you may hear in the beginning that the higher harmonics often die out faster than the lower harmonics so you ready for this open open open on both sides and I just bang the hell out of it 2044 they're beautiful it's oscillating like this exactly oscillating the way that I derived for you for air and here you see it that it had holds for only the fundamental is there okay I wish you luck on Tuesday on Thursday I'll see you then

Channel: Lectures by Walter Lewin. They will make you ♥ Physics.

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Id: J2qfKaK3N6Q

Length: 77min 20sec (4640 seconds)

Published: Wed Feb 11 2015