8.03 - Lect 11 - Fourier Analysis, Time Evolution of Pulses on Strings

so today we're going to discuss Fourier analysis you see a lot of math but I will try to physics imagine that I pluck a string which is fixed at both ends so here is X and here is y and this is the string at T equals zero and imagine I let it go then we know from what we have learned that it will start to oscillate in the superposition of its normal modes which are standing waves and so the sixty-four million dollar question today is now in which modes will it oscillate and what are the amplitudes of those modes if you write down Y as a function of X and T then we know from what we have learned to date that you should be able to write this N equals one to infinity and platoon's which I indicate with a B now rather than with an A but B is of course an amplitude has a dimension length times the sine of K and x times the cosine of Omega and T and because of the boundary conditions K of n because the spring the string is fixed at both ends K of n is n pi divided by L and Omega n is the velocity propagation times K of N and n then is one two three four five so we know this this is nothing new we've had this recently so I can also write down then that at T equals zero I can write down this shape in terms of a series of science I believe the time off now so T equals zero then we have Y X at time T equals zero would be b1 times the sine of PI x over L plus b2 times the sine of 2 pi X over L plus b3 and so on and our task now today is what values of n are necessary to make that pulse and what I then the amplitudes be that we have here and that's what Fourier analysis is all about Fourier was born in 1768 and he died in 1830 so this is work that was done a long time ago brilliant a brilliant mathematician now I will give you the general approach first and then I will come back to this special case of a string which is fixed at both ends but I really think I owe you the general format the general procedure first so the idea behind Fourier then is that any periodic single valued function with period of 2 pi radians can be represented by a Fourier series and I will write down this Fourier series here so this is now a function of X X is now in radians it's not that X I will come back to that so a function of X can now be written as a constant we call it a 0 divided by 2 plus the sum 1 to infinity of a.m. cosine M X plus the sum 1 to infinity of B M times the sine of M X and this X that you see here is also called X in back affair and Barrett is in radians I will call this first term 1 I will call this one 2 and I will call this one 3 so our task now is to find the values for a and B if you know the function FX and I will build up with you the recipe that allows you to calculate the values for a and the values for B what you do the first thing you do to find a0 you take the integral from minus PI to plus PI DX on both sides of the aisle this is one side of the aisle and this is the other side of the arrow and when you do that you will see that for all values of X excuse me for all values of M miri all terms here will be 0 and the same is true for 3 for any value of M that you take if you do an integral over a sign over a complete period from minus PI to PI that is a complete period you get 0 of course so all values for 3 will be 0 and so what you're left with is then that the integral from minus PI to plus PI of XX DX is then the integral from minus PI to plus PI of a 0 divided by 2 DX and that is pi times a 0 that's immediately obvious so we now already have our first way of calculating a 0 a 0 if I use this result is nothing but 1 over pi times the integral from minus PI to plus PI times the function X of DX so you tell me about the function of X's and I can calculate a 0 for you keep in mind that a 0 divided by 2 is nothing but the average value of the function over the period 2pi and i will come back to that later during this lecture how are we going to calculate now the other values for a well what you do now is you take the integral for minus PI to plus PI times the cosine of Nancy X DX and you do that on both sides of the aisle you do it on the left side and you do it on the right side and Nancy now is one two three and so on when you do that you will see that this term is going to be zero that's immediately obvious we do an integral over the cosine function times the constant you get zero so the first one you don't worry about the third one for any value of Mary and for any value of Nancy the third one is also zero and if you don't believe that check that on your own so then you would get a cosine of Nancy x times the sine of Mary X for any value of an any integer of an any integer of M if you integrate it over one whole period you get zero so this is also zero if we now go to term number two it is also always zero except when Mary is the same as Nancy so number two is also zero except when M equals n when N equals n you get here a cosine squared and when you integrate a cosine squared you do not get 0 in other words what you have to do if Mary equals Nancy you're going to get the cosine squared of M X DX because Mary is Nancy and I must integrate integrate that between minus PI and PI and that equals PI so now we have a recipe for a of M because a of n now is 1 divided by PI that is this PI that comes from the integral of the cosine squares integrate between minus PI and plus PI a function x times the cosine of Miri X DX so we know now how to calculate a zero and we know how to calculate the values for a am a of M and we will of course do an example together it's now clear what you're going to do to find two values for B you're now going to integrate both sides of the aisle between minus PI and plus PI times the sine of Nancy X DX and when you do that you will see exactly the same that you had here this will always be 0 this will now always be 0 and this will always be 0 except when Mary is Nancy then you get here the sine square and the integral of the sine square will be PI just like the integral here of the cosine square was PI and so you see now that we also have now the recipe to find all the values for B of M that's going to be 1 divided by PI times the integral of minus PI to plus PI that is the period of the function times f of X times the sine of Mary X DX and here you see in its most general form the formalism of Fourier analysis keep in mind that whenever you have an integral from minus PI to PI if you prefer 0 to 2 pi that's fine because the function is periodic so you can always replace this 0 to 2 pi if that suits your purpose if you look at these three recipes you can do away with this because if you make em equals zero here originally M goes one two three four five but if you also include M equals zero here you get exactly the same that you have here because when M equals zero the cosine is one so this is then identical that is the only reason why we called this code this constant term a zero divided by two we could have called that a C because after all a constant is a constant why would you want to call it a zero divided by two if you had called this C which is the average value of the function then you would need this C here and you would have here 1 over 2 pi and therefore you need 3 recipes to do Fourier analysis whereas now if we define this constant as a 0 over 2 you only need 2 so that's the only reason it's just for practical purposes but a 0 over 2 is a constant and it is the average value of the function over the period now all this may look a little bit opaque to you now and it will become clear I hope during this lecture when I put this Fourier analysis to work what you've seen here and that's the way I want you to look at it is that is simply a recipe and you and I when we use this recipe we execute it and we do not always ask ourselves why is the recipe the way it is if I apply Kramer's rule I do not every time say to myself why is it really the way it is I have seen once why it is the way it is and I apply it of course you have to know when you can use it and when you cannot use it so now I would like to return to the plucked string but I first want to take a close look at it at time T equals 0 and what I'm going to do now is to pluck it in a very unusual way there is a reason why I do that so unusual because that is doable in one lecture to do the Fourier analysis on it so I have a string now which is fixed between 0 and L and I'm going to plug it in and extremely obnoxious way namely like this so I'm forcing it like a square very painful for the string but that's not my problem and that this be an amplitude a if I want to use Fourier analysis and you will see that I do and you will also see how I'm going to do that I need a periodic function and this is not periodic so I'm going to make it periodic and the way I'm going to make it periodic is the following I'm going to pretend that the function is really this so I'm adding this I make it to L but I go much further I go on and go on and go on and go on so I'm going to define it between 0 and 2l and that's periodic notice that this pattern now up squared down square is periodic so that means my period is 2l and also notice that by doing that my a 0 divided by 2 is 0 because the average value of this function which is now defined between 0 and 2 L and much beyond 2l and below 0 that that function has an average value of 0 and so my function now my function of X you can write down Y of X if you want to it's plus a Forex being between l and zero and it is - a Forex between 2 L and L because this here is minus a now I want to express this shape into Fourier series but before I can do that I have to make some changes in the recipe because in the recipe we have radians and radians our apples here we have X but that's not radians that is meters and meters are coconuts you have a question between 2 L and L it is - a between zero is that what you want thank you very much I appreciate Corrections because it's awkward to do it later thank you so radians are apples and meters are coconuts and so what should I do now to make the coconuts into apples I now have to take the old X which was in radians which is the one that I have here and that is the one that I have here and I'm going to replace that in that formalism by PI x over L and this X now is in meters and if I have done that you can see that if my X now becomes 2l then this has moved over to PI radians so that's exactly what you have to do it is unfortunate that we call this X we could have called this formalism Z or some other symbol but in general we give X we could have called this Z well you still have written you will have to write down that the X there is then PI Z divided by L now we have apples here and we have apples there so that means that now my Fourier series are going to look very much like what I have here but I prefer to write them again so my function Y as a function of X which of course is my function of X I know that it is in the Y Direction that function should now be written somewhat differently a zero divided by two plus the sum from 1 to infinity of a of M times the cosine of n PI x over L this is now our X this is X in meters plus the sum 1 to infinity of B of M times the sine of M X times pi divided by L and so now we have made a modification to the general idea which is in in radians to be applicable to a case whereby we don't have radians but why we have X in meters before we can execute the errata phase I also have to change the recipe in terms of a of M and B of M first of all I don't have an integral over a period 2pi but i have to do an integral from 0 to 2 L or from minus L to plus L so where earlier I had 1 over pi integral minus PI to plus PI I have to change this boundary either from 0 to 2 L or if you prefer 4 minus L to L I don't care it's the same thing because the function is periodic but I also have to change this 1 over pi which we have in front here because that was the result of this integral and if you do the integral in our space you would have found L and not pi this is half the period and half the period of our function is L so this a 1 over PI from minus PI to plus now must be changed into 1 divided by L integral 0 to 2 L and so to make sure that you're not going to get confused I'm going to rewrite the recipe for you and I might as well do that again in in red so we got now that our a of M is 1 divided by L times an integral let's just go from 0 2 to L and then we get our function of X and then we're going to get the cosine of M PI X divided by L DX and B of M is then 1 divided by L times the integral from 0 to 2 L of my function X which is my plucks string times the sine of M PI x over L times DX and so now we have not changed anything in terms of the formalism of Fourier analysis but we have adapted the recipe so we now have a new series written in terms of our new X and we have the new recipe to calculate the a values in terms of our new X and in terms of the new period which is now 2l so now you may think that I am ready to plunge into the math and to start executing it but I'm not going to do that yet I would like to take a look first at the values of a we already agree that a 0 is 0 so that's that's not an issue anymore but look for instance at the function a 1 times the cosine of PI x over L that would be the first term of the cosine series and the second term would be a 2 times the cosine of 2 pi X over L well let us assume that we found the value for a 1 so I'm going to plot now into this function of hours the a 1 value so it's here here here here and here and so here is that function and this amplitude then is a 1 that's unacceptable why is this unacceptable look at my function and then look at the red curve which is my a 1 cosine function that red curve is supposed to help construct this function here it is positive so I demand that it must be negative here otherwise I can never make this trough whereas it's going to add or positive here so this a 1 out of the question can never contribute to my function but not only can a 1 not contribute but any cosine function that you draw will always look here like this it will always be positive here and positive there whereas I will demand that it has to be positive here and negative there otherwise I can never build up this function to put it in a more intellectual way a way that mathematicians would use it a cosine function is an even function that means that the function of X is the same as the function of minus X we call that an even function here X is zero a cosine function of X has exactly the same value as at minus X but our function is odd our function the way we defined it zero here is odd and an odd function the function of x equals minus the function of minus X and to put it in a nutshell you can never fit an odd function with even functions in Fourier neither can you fit an even function with alt functions so that's a perhaps a better way of looking at it so we can already conclude that all cosine functions are out for this specific case of course let's now take a look at the values for B so keep in mind bi the sine functions so a 0 is 0 all values for a are 0 now let's turn to the B values so I'm going to make draw the function again here we realize of course it's only defined from 0 to L the rest is only a mathematical way of doing the Fourier analysis so we are going to define it all the way up to 2l and then we make it periodic so let us put in there the function be 1 sine PI be 1 times the sine of PI x over L that is the very first one in the series of the B's that one is wonderful that is exactly what we want that would be this is sinusoid which would have an amplitude be 1 it couldn't be better here it is a mountain and here it is a valley we need a mountain here and we need a value there no surprise because assigned function is an odd function and our function is odd so clearly alt functions will do very well let's now take a look at B to be two times the sine of two pi x over L let's put it in there I'm full of expectations that b2 will do a great job so we have here zero crossings so here is my b2 and this amplitude is b2 out of the question there is no way that b2 can do me any good why yeah exactly there are various ways of looking at it one is easy here you see a mountain that helps building this up but look what you see here you want a valley and it's not adding to the valley that's one way of looking at it there's another way of looking at it to say look everything has to be symmetric about the line 1/2 L if this builds up it got to build up there and it's not doing that so we must conclude that b2 must become 0 and by analogy all even values of M B will be 0 so this is a prediction that I make we haven't done any fancy Fourier analysis we have already concluded that for our specific function for our specific strain which is fixed at both ends that all values for a will be 0 and that only odd values for B for which the M value is odd will we get non 0 answers so now we are ready to actually execute the Fourier recipe and if you insist that you want to do the A's be my guest you can do all the A's and you will see that all of them are 0 so I will not waste your time on that but I will certainly do the B's so we now execute our recipe and here is adjust it to our X and so we're going to get that B of M now the function is defined between 0 and L is plus a so I can I can bring the a outside and put it here and then I get the integral from 0 to 2 L and then I get the sine of M PI X divided by L DX that is not between 0 and 2 L you should have screamed that is only between 0 and L it's only between 0 and L that it is plus a but when we integrate from L to 2l then it becomes minus a so the next part of the integral is minus a so I bring the a outside so I get minus a divided by L times the integral from L to 2 L times the sine of M PI x over L DX so the function comes in two parts boy there are Fourier analysis where the function comes in more than two parts believe me this is an easy case that's why I chose it this integral is trivial of course if I do an integral I get minus the cosine so I get an A divided by L out and I get M PI downstairs and I get L upstairs and then I get the cosine of M PI x over L and I have to evaluate that between 0 and L and here I get an minus sign out so I get plus a divided by L I get M PI downstairs I get L upstairs and then I get the cosine of mary pi X divided by L and I have to evaluate that between l and 2l that's all I have to do that is the integral a very simple one now when n is odd this one alone don't look at this this one alone is -2 you have enough knowledge to confirm that you can see if you substitute x equals L you for instance you take M equals 1 which is odd you get the cosine of pi that's minus 1 and then the cosine of 0 is 1 that you have to subtract that you get another minus 1 so you get minus 2 and so this one if you do that you get plus 2 check it and you will see that it is plus 2 because the borders the boundaries L 2 to LR different it's not the same as 0 to L but if M is even you get a 0 here and you get a 0 there and so what comes out is exactly what we predicted that all even values of Mary will give 0 values for B and that only the odd values will give me values that are not 0 before I write this out in a complete fool a for him you should appreciate the fact that this entire integral here gives me exactly the same answer as this entire integral here because minus x minus 2 is plus 2 and this plus x plus 2 is also plus 2 and there is always the case when you have a string which is fixed at both ends that the 0 to L integral always give you the same answer as this imaginary L to 2l which you introduced to make the function periodic it is for that and only for that reason that Tony French gives you a much easier way to calculate Fourier components in strings and he says all you have to do is say that B of M is 2 divided by L he multiplied the recipe by 2 so instead of having where is my recipe instead of having 1 divided by L he says no you should really do 2 divided by L but then he says all you have to do is now integrate between 0 and L of the function of x times the sine times Mary pi X divided by L times DX and this is equation 6 32 in French if that's all we knew about Fourier analysis we would have a very narrow picture because it is an extremely special case but it's true that whenever you have a string which is fixed at both ends this will do I felt an obligation to you to show you the Fourier formalism which is a beautiful formalism in more general terms so we are now ready to write down the complete Fourier series notice that the sum of these two become four a divided by M PI so B of M is going to be 4 a divided by M PI but M is only odd and so with that in mind I can write down now here our function which was one of our big steps that why I put a 0 here because this is a time equals 0 when I have this string in this crazy shape can now be written as 4 a divided by M pi I could write it down as simply a pie and then I got here the sine Phi X divided by L plus 1/3 that is that M equals 3 which is 3 times lower because you have an M 3 here times the sine of 3 PI X divided by L plus 1 fifths and then you get 2 sine of 5 PI X divided by L and so on and so on so now I would like to put in my function I would like to put in the first two terms and I will do that here on the blackboard so I'm going to concentrate now only on my function 0 to L in reality the string is only here and now I would like to put in there and this is a and now I would like to know what is b1 well b1 is going to be for a divided by PI which is approximately one point two seven a well let's put it in there this is a so I have to put in 1/3 more so it comes up to this point roughly so there it is I love it it's not quite a square yet but we're getting there we're on our way to building up the square the next one is b3 b3 is for a divided by 3 pi so it is one third of 1.27 which is about 0.4 to a and b5 is 1/5 times b1 let's put in b3 so b3 has an amplitude which is about 0.4 which is about this much that's about the height point for a little lower and so here we have the maximum and then we have a zero here we have a zero here and so we have here and then we go through this point so the curve is something like this so this now is b3 and this is b1 now look at b3 look at the beauty of Fourier analysis this is higher than it should be I'll be three says I'll take care of that I will subtract something there we'll take that off there isn't that beautiful so are you going to flatten this out already already with two terms b1 and b3 you're really beginning to see the building up of this crazy square and look here B 1 says sorry I cannot fill this up for you this is the best I can do b3 says I'm going to help you I'm going to help stuff there and it's going to do the same there and so you see that if you now keep adding all values of M that you will gradually approach that square which is an amazing concept and what I will demonstrate to you is making a square and we will make it like this just like we did here we can do that on an oscilloscope we can have a signal that makes this trace admittedly this is time that the way you will see it it is space so you can think of it as being space this is no different from the function that we had and we will show you the b1 value in this case since it is in time it actually has a frequency it's 440 Hertz we can make you listen to it and then we will show you to b3 value which has three times higher frequency and then we will show you the b5 and the b7 and the b9 and then we will add them all up and you will see it begin to look like a square and so let us set it up supposed to see that on the central there what you see here is simply the b1 so you don't see very much do you you just see a sine curve well you're looking at this sine curve now I'm going to show you actually can we hear it to Marco's here's the 440 Hertz now I'm going to show you the b3 notice it is three times smaller in amplitude that's the way we set it as a three times higher frequency and now I'm going to show you b5 which is five times smaller then the hold it no no no no no okay it's five times smaller than v1 and and the frequency of course is five times higher you can hear it yeah this this there it is and now I'm going to show you seven seven times smaller than v1 and here is B nine this is nine times smaller than v1 and now I'm going to show them all five to you and it's really beginning to look like a square the square pulse it is clear that b1 is very very important so if I take b1 out which I will do now it doesn't even look like a square that shows you how important that first term is if I take b3 out that is less disastrous but it is still pretty disastrous but you're ready beginning to see something that looks like a square slowly and it should be clear to you now that in order to get the real sharp edges here you need very high harmonics you have to go to and values of 49 101 201 they're all odd you have to go to very high values and the higher you go the closer you will get to the square so this is a nice moment to have a break and I know you're dying to do you fifths mini-quiz I will hand it out now and then we will all start at the same time with the mini quiz so each one of you has the same amount of time I owed you the histogram of the exam you see that here I think it's a wonderful histogram as far as I'm concerned it is too early to talk passing and failing and ABCs indeed but if I knew nothing else no other grades then I would have to put you in the danger zone if you score less than 45 that's all I can say I can add nothing more than what I already wrote you by email so that is now returned to our string our string is still at t equals zero so our string is still in this position dying to be released that was the goal remember we would pluck it and we would let it go at time T equals zero this is the Fourier analysis this these are the Fourier components of that string but now I say okay go and what happens now is that this one is going to oscillate with Omega 1t this one going to oscillate Omega 2t and this one Omega 3 T and this one is going to correlate with cosine Omega 5 T each one of these are standing waves and so each one of these are going to do this with their own amplitude and with their own frequency and Omega M is going to be V times K of M and the speed of propagation is the square root of T divided by mu and that's what's going to happen now you may wonder what you're going to see if you let this spring go and chances are if I ask you what do you think will happen that you may think that this will happen with the string as a whole but that's not true and the best way actually to answer the question what you're going to see is to go back to a simple case of a triangular pulse on a string if I have a triangular I'll do that here if I have a triangular pulse on a string I exaggerate of course the amplitude highly fixed here at 0 and fixed here at L and I let it go yes I can decompose that in terms of the Fourier components and each of these Fourier components are going to extending waves up and down doesn't give me much insight but it is intriguing that that's what happens I can also think of it that nature is seeing a disturbance and it says well we're going to propagate this disturbance Nature doesn't know the difference we left and right and so clearly what will happen if this has an amplitude a there will be a pulse with an amplitude 1/2 which goes in this direction and one with the same 1/2 which goes into that direction and that is exactly what you're going to see when you let it go so a little later in time you will see this going in this direction going in this direction at the moment that the top of this mountain reaches L and the top of this mountain reaches 0 you're going to see nothing absolutely nothing because the reflected one and the incident won't exactly cancel each other and then a little later in time they will be on their way back and then a little later in time you will see this one but completely flipped over when these two triangles meet each other and so that's when you will see the thing upside down so there are two very different ways of looking at what happens when you release a plucked strength one way is to say there are oscillations of standing waves Allah fool a and the other way is to say well we're going to always cut it in two pulses each of half the amplitude and let them go back and forth and let them reflect at the ends now keep in mind that standing waves are the result of the superposition of travelling waves so the two different ways of looking are of course connected they have the same underlying physics and now I want to demonstrate this to you using a program that was developed by Professor we Klaus when he was lecturing a doe through and also Nargis I think she's in the audience Nargis are you hiding are you hiding Nargis has also worked on this program it's a wonderful program I'm going to put a triangle first on a string the width of the triangle is about one third of the length and then I will release it and I will let it go and we will see the Fourier components and at the end when the whole show is over after one half the time for the fundamental period we will wait one half period of the fundamental we will inspect very closely the Fourier components because of the symmetry of this problem all even values of M will have zero values of B and you will be able to see that now where are we going to change the light setting for this or were we not no I thought we were why don't you turn that off also oh that's too difficult perhaps yeah suppose the bar no the bar the bar the bar alright so what you're going to see first is the triangle the number of Fourier components that we have is 25 do I do something wrong excuse me we have no what you broke the line oh that's nice thank you the number of terms we have is 25 we have M equals 1 M equals 3 M equals 5 all the way up to 49 that's when we cut it off all even values for M have no B value and if you are ready I'm ready so let's first run this one there you see the triangle and the blue lines are the Fourier components they are each doing this thing their own thing standing waves up and down but the net result is something that you acquired familiar with you see two pulses with half the amplitude of the original one they move through each their own site they reflect and they come back and then we will stop when we have half the period of the fundamental now look look at the individual Fourier components this is B 1 this one is B 3 you see it helps building up that triangle this is B 5 it helps building up the triangle this is B 7 and this is benign and here they all conspire amazing there are 25 sinusoids that are conspiring there and they show you nothing and the same here and it's hard to believe that all these sinusoids can add up to zero here and how beautifully they connect here and the only reason why this tip is not very sharp is we only have 25 terms if I would run it with 400 terms then you would see this sharper but it would take a lot more time let's look at the Fourier components which you will see here so this is b1 which we have arbitrarily called 1 you see all the even values this is b2 b3 sorry b2 before b6 b8 all the even values are 0 the b3 is negative in this case that was not the case for our square remember in the case of the square b1 b3 b5 b7 were all positive if they all had the same sign that's not the case with the triangle they alternate the next one that I want to run for you is a square which has a width half the length of the string and it is centered at the middle of the string and again two pulses are each going their own direction with half the amplitude they're being reflected and strange things happen at the ends and now there comes a time 1/4 of a period of the fundamental that you see nothing and now they are coming back from the reflection very different from what you expect isn't it now look at b1 that's a real biggie look at b1 boy here is that a surprise no because clearly if b1 is not quite a square but it's almost a square right you just have to subtract something here well b3 says that's okay I will subtract there and b3 will subtract here and b1 is a little little shallow here it's a little bit too not generous enough and then b3 says okay I'll help you building it up and so if we look at the Fourier components you see now that again all the even values are 0 b1 but definition 1 and now you see this one is negative and b5 is also negative not so obvious and b7 is positive and benign is positive and now I'll show you one whereby I'm going to offset the square the square is only one quarter the length and I'm going to offset it and when I offset it from the center you need both alt and even values and you're going to see that so now you cannot just get away with only even values of Miri so here you see the wave splitting up into just like you expect it it has a width a quarter of the length of the string this one has already reflected this one is now on its way to being reflected now this one is coming back there it is and this one is already on its return and of course they're going to meet again and now we are one half period of the fundamental later and now look this now is the first harmonic but now there is also a second harmonic M equals two has a B value here it is this is the b2 value large amplitude you notice that and look how important b4 is I think this is the before value yes indeed here is the b3 this is the before has a huge amplitude and look how important it is because right here where the pulses you need this bulge to push this further out because the b1 alone cannot reach that point let's take a look at the Fourier spectrum here you see Bobe one which we by definition call one you see b2 is there it's large it's a little larger than 0.7 and then you have a beat which is negative and before which is negative and b5 which is negative here before is off scale and now you get b6 positive b7 positive b8 is 0 and B 16 is 0 well that's probably the result of the way we set up the the square which has a width of one quarter of a length and we offset it from the middle by one quarter of the length remarkable example of hue analysis now which is way more complicated than if you nicely Center it all values of B now all values of M are important you will now understand when I said if we pluck the string of a harp or the string of a violin and you pluck it this way fixed here and fixed here the tone that you hear is different than when you pluck it this way you now understand that because the Fourier components of this one and this one are very different than the Fourier components from this one and so when you let the string go the frequencies that will be produced are the frequencies of these normal modes of these B values but the various B values are very different here from there and so the sound that you hear is distinctly different when you pluck a string on the site and when you pluck it in the middle you may remember that during that lecture I told you that the hammer of a piano hits the string about one-seventh of the length of the string from one side and that is done to suppress the seventh harmonic for some reason that beats me people don't like the seventh harmonic so that's the way they kill it so it does depend on where you pluck and where you hit and now you can put that in context because now you understand that you analyze the thing here in terms of Fourier components and each one will then oscillates at their own frequency which their own amplitude so what we have discussed now at length is a Fourier analysis in space in X in meters and we decomposed a function in the sum of many harmonics which together make up than the shape of the string now if we look at sound sound of course is something that is a function in time suppose I look at the sound signal made by a tuning fork of 440 Hertz and I have here that signal in time suppose I have one second of data 440 Hertz so this is now time but if you want to think of this as being in space be my guest now there are special programs special algorithms that takes this one second of data perform a Fourier analysis they have a name we call them fast Fourier transforms and it's going to tell me what the amplitude of B one is an a one of B 2 and a 2 of B 3 and a 3 and so on if I now think in terms of Hertz rather than in terms of Omega there is nothing at 1 Hertz so though A's and B's are zero there is nothing at 2 Hertz so those eighth and B's are zero but by the time I reach 440 Hertz this system says yeah P there's a lot of power there at 440 Hertz and what we now do we make a plot of the square of a plus the square of B and the reason why we square them is that energy is always proportional to the amplitude squared and so you take the a squares and the beat squares and you add them we call as a power then the spectrum a Fourier spectrum and here you have your Omega or you have your F value in Hertz the difference is only two pi and if you now do a Fourier analysis of that signal you will see at 440 Hertz a huge value and you will see almost nothing anywhere else so you have now done a Fourier analysis of a time signal and you have decomposed it into its Fourier components which in this case is only 440 Hertz if you did the same with the middle a of the piano which is 440 Hertz you would see a big value here but you would see also something at 880 and you may even see something at the third harmonic and the same is true for a for a piano and it is also for a violin if you take a violin and you ask the violin is to produce the 440 you always get a little bit more of the 880 and you get some of three times the fundamental and this is something that we can demonstrate we have here a admittedly somewhat poor man's version of what's called the Fourier and air analyzer we take in sound for about one fifth of a second and then we ask the computer to fully analyze that for us and show is the Fourier spectrum in terms of where the frequencies were and we will see that here for which I think we also invented a special light condition didn't we the scale that you see from left to right is two kilohertz I'll first make you see the 440 Hertz see that dude spike that's at 440 Hertz so the end of the scale is 2 kilohertz 256 Hertz I have here a fruit and remember during the lecture on musical instruments we believed and I demonstrated that that you can really make this one resonate only in its fundamental I will now show you is this Fourier analyzer that there is also a little bit of the second harmonic maybe even the third harmonic look at it now you see that there was a fundamental second and third harmonic even look again ah so this is a beautiful way of analyzing the frequencies in in sound I will whistle for you amazing would you have guessed it very nicely separated what is the frequency it's about here that's about one kilohertz about 900 Hertz and the second harmonic not so obvious my whistle very high frequency it is so high that it's off scale and I did that purposely because I know you hate the whistle so I didn't want you to see what it really looks like we can listen to the radio and then we do can do a Fourier analyst analysis on what we hear hi Kenny have you got a question for mr. nadir what you're doing is not the right time for me people always hear things that are said and maybe this will give them the context to understand that and then also at the kill 2 complicated4me anyone if you want to sing come on be brave anyone wants to sing yeah come on give it a try it's nice to be here believe me you see all these ok your decision alright so you see how you can do a Fourier analysis of sound and then I decompose the signal in terms of its Fourier components neutron stars were discovered in 1967 by Jocelyn Bell Jocelyn was a graduate students in Cambridge England at the time and her supervisor Anthony Hewish had built a new type of radio telescope and she was in charge of analyzing the data and she was sure that she had received a periodic signal which came almost every 1.3 seconds and they realized that that could have been the discovery of the century because they thought that they were receiving signals from intelligent life elsewhere in the universe and so they called it the little green man and then a few months later Jocelyn discovered another one and so they called the first one little green man one and then little green man two lgm1 and LGM two and then when discovered a third one they abandoned the idea of LG M's we now call these pulsars and we know that the period is the spin period of the neutron stars in 1974 Anthony Hewish received the Nobel Prize for this discovery it is a shame and it is scandal that jocelyn did not share in the Nobel Prize because she actually made the discovery I've known her very well I've discussed it with her many times she was a graduate student maybe that was the reason why the Nobel Committee didn't think it was appropriate to give a Nobel Prize to a graduate student ridiculous but perhaps true she was a woman and there is perhaps a very sad case of sex discrimination again we will never know but she did not share in the Nobel Prize which he should have neutron stars have a mass about one and a half times that of the Sun they are a hundred thousand times smaller there are only 20 kilometers across so they have a density which is 10 to the 15 times higher than the Sun 10 to the 15 times higher than water which is higher than nuclear density and I had some email contact with Jocelyn just a few days ago she's now in Oxford and I said I really won't like to show the class your picture and so she showed me she sent me a picture which I'm going to show you we still connected where is Jocelyn it's also been hiding no there she is there she is she sent me a picture at the time that she made the discovery you see the radio telescope here and you see her standing there very modestly it's a very modest woman and she made one of the most important discoveries of the past century and did not share in the Nobel Prize we now know of hundreds of pulsars in the sky and so the question now is how do you find these pulsars you can observe the sky in radio way with radio telescopes you can do it also with x-ray observatories and what you do now is you take the data just like we took the data of our sound signal and you perform a Fourier transform you ask that data what are the Fourier components that are hidden that I cannot see that they're hidden and in 1998 rudi Winans in amsterdam and independently at morgan here at MIT were analyzing x-ray data from a known x-ray source in our galaxy and they were performing fillet analysis which is standard nowadays in astronomy and they discovered that the neutron star was rotating with a spin period of 2 and 1/2 milliseconds they noticed in the Fourier power spectrum a huge spike at 401 Hertz it means that at the equator of the neutron star the speed going around is about one tenth of the speed of light and I ask you really who worked with me for several years here at MIT to send me some of the data that he obtained from which he could then finally derive that we were dealing with a neutron star rotating around in 2 and 1/2 milliseconds and so he said well Walter why don't you show the class only 1/5 of a second of my data 1/5 of a second that you see here 2/10 per second so this is the time scale and when one x-ray arrives you see a vertical bar there and at this time they were so close together that you see their two x-rays if you count the total number of x-rays in that one fifth of a second he observed 33 x-rays during that time the neutron star was rotating 80 times around already because it rotates every 2 and 1/2 milliseconds so when you look at this data you have no idea that here is an underlying neutron star with a period of 2 and 1/2 milliseconds with a frequency of 401 Hertz but now you take 3,000 seconds of data and if you have taken 803 you know how to perform a fast Fourier transform those programs by the way are readily available on the market and here you see a power density spectrum vertically is the sum of a squared plus B squared and horizontally is the frequency in Hertz and look what you see at 401 Hertz you see a huge spike and that is the underlying two and a half millisecond neutron star this is the frontier of astrophysics nowadays you cannot even think of astronomy or astrophysics without Fourier analysis it has an enormous Lee important impact on our research my graduate students and my postdocs perform Fourier transforms every day and so what we have discussed today is not just intellectually interesting it is at the forefront of research see you Thursday you

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

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Keywords: Walter Lewin, Physics, Fourier Analysis, Fourier Synthesizer, High Resolution

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Length: 74min 22sec (4462 seconds)

Published: Wed Feb 11 2015