I'm Walter Lewin I will be your this term make sure you have a handout and make sure you read it it tells you everything you want to know about the course this course is about waves and vibrations about oscillations periodic and not so periodic events when you look around in the world you see them everywhere for one thing your heartbeat that's a periodic oscillation at least I hope that for most of you in this periodic your breathing is some kind of a periodic motion the blinking of your eyes your daily routines and your habits your eating your sleeping taking a shower your classes and occasionally doing some work all those are periodic actions when you drink I drink some orange juice now notice as I try to move the liquid down into my stomach that is not a steady stream but it's a periodic motion look at my throat in fact even if I don't want to swallow the liquid but simply have a bottle with liquid and I turn it over then we all know that the water doesn't come out like a steady stream but it goes clock clock clock clock that's some kind of a periodic motion I have here a toy which I used to entertain my dinner guests particularly physicists is interesting there's liquid here and the idea is to get the liquid there and then the problem is how can you do that in the fastest possible way well if you turn it over you see that phenomenon that I just mentioned which is that clock clock clock it's not a steady stream it's almost pathetic the way that it runs from one side to the other I will take minutes before it's there but it can be done in 17 seconds and during the five minute intermission that we have you may give that a try and I hope you won't break it and see whether any of you can think of a way that you can transfer the liquid in 17 seconds you have breakfast in the morning and you casually put your breakfast plate on the table what are you here some kind of a periodic motion and two things can happen to this plate it can move as amusingly I call this as immutably because I'm an astronomer but it can also wobble in fact something can wobble without moving at the Meuse of it and something can move Azzam usually without wobbling in this case it does both and a fabulous example of that is what's called the Euler's disk which is a metal disk you will see it shortly there and this metal disk we're going to wobble in the similar way that I warble plates and then we'll follow it motion as amusingly and the wobbling frequency and what is interesting as you will see that the azimuthal motion which has a certain period that period gets longer in time but the wobble motion the frequency goes up so I'll start it here and then I'll show it to you in a way that is more appealing and you can follow that it's an amazing toy to work out to physics it's very very difficult I was told that professor will check at MIT ones gave a one-hour lecture exclusively on the explanation of this Euler's disk so try to see the azimuthal motion it will become clearer as it slows down further you may be able to hear the wobble motion I will hold my microphone close up can you hear it very high frequency already did you hear it it's quite amazing isn't it when you look at this so the wobbling frequency increases quite rapidly look how the azimuthal motion slows down and how the frequency of the wobble goes up ah now it comes to a stop that's a very difficult piece of physics right there if you take a tennis ball this is a Super Bowl and you bounce it Oh this called student involvement thank you then you also get some kind of a periodic motion whereby again the frequency increases just like in the case of the Euler disk and the breakfast plate here is an object that is floating in a liquid in water and if I push that a little further in and I let it go it wobbles and there is a very unique frequency that you will be able to calculate in 803 a very unique period of one complete oscillation as this object goes up and down even wind steady winds can generate periodic or almost periodic motions which all of you have experienced you walk outside it's windy and your hair goes like this your hair doesn't go flat like this always has this tendency just like a flag does the same thing if I generate wind here and I have here some aluminum now you will see that this wind doesn't make the aluminum just go straight out but it wobbles there is a certain period to that after work if you want to have some fun what is more fun than writing your own walking north that's a periodic motion falling in love can be a periodic event now don't do it too often because as most of you know quite exhausting the motion of electrons atoms molecules periodic and oscillatory the motion of the moon the planets and the stars periodic oscillatory sound is a beautiful example I produce sound I produce sound by oscillating my vocal cords I produce thereby pressure waves my vocal cords push on the air suck on the air push on the air which produces a pressure wave and the pressure wave propagates out in the lecture hall reaches your eardrum your eardrums start to move back and forth and you your brains tell you that you hear a sound I have here a tuning fork which is designed so that if I give it a hit that the prongs move 256 times per second we call that 256 Hertz a Hertz is 1 oscillation per second and all of you can hear it fresher waves I generated we will deal with them in 803 they travel through the air reach your eardrum and your eardrums starts to shake this is a higher frequency 440 Hertz most human beings can here in the range from 20 Hertz to 20 kilohertz there are animals who can go way beyond 20 kilohertz and to be nice to you for the first time this first lecture I would like to test your hearing and that will be free of charge I'm not so much interested in knowing what your lowest frequency is but what your highest frequency is so I'm going to generate here sound I will start with hundred Hertz and then we'll go up higher and higher and then we'll see where you hearing stops so let's start with hundred Hertz I'm not going to ask you who can hear it because clearly all of you can let's now go to kill over a thousand Hertz piece of cake right mm no problem I have to change now mind my scale 4,000 I look saying that this was going to be a pleasant test okay 5,000 is when violins come in 6,000 anyone by audience who cannot be a 6,000 7,000 anyone in my audience who cannot hear 7,000 I cannot hear seven times I hear nothing with age you lose ability to hear high frequencies you will experience that in your lifetime you won't escape that that for some people lose more than not I cannot hear about 6,000 is nothing okay 10,000 12,000 14,000 now I want to hear I want to see hands if you cannot hear it any longer who cannot hear 14,000 don't be ashamed of it is not your fault 14,000 all right we're slowly going up 15,000 who cannot hear it raise your hands ah professor Marvel voila you're also getting old marriages 16,000 who cannot hear 62 of course the ones who have already raised their hands you don't have to raise your hands again who cannot 16,000 who cannot 17,000 18,000 ok so now we're going to change it now I want you to raise your hands if you can hear it and so I first now go to 20,000 19,000 I'm going to 19,000 oh sorry I was only off by a factor of 10 19,000 who can hear it fantastic 20,000 21 ah you see how to cut off very sharp 22 very good 22 23 25 27 some of you have amazing news because I already turned it off at 21,000 all right key absolutely key in this course will be simple harmonic oscillations because they are extremely common in nature a simple harmonic oscillation and you've seen this of course in 801 can be written as follows x equals x0 cosine Omega T plus Phi you can write a sign here if you want to x0 is the amplitude that's the largest displacement from equilibrium Omega is the angular frequency angular frequency Omega which we express in terms of radians per second the period t 2 pi divided by omega is then expressed in terms of seconds and the frequency F which is 1 over T is what we call Hertz number of cycles per second do not confuse Omega with F there is a factor of 2 pi difference if I have a uniform circular motion and I project that uniform circular motion onto any line in the blackboard then I get a simple harmonic motion so I take for simplicity just this horizontal line that I could take any other line and let's call this the X direction and let this point be x0 and I take an object which is rotating around here is the object it's going around uniform circular motion if I project this onto the x-axis and this angle is Theta then this position here is x0 cosine theta and if I make theta a function of time theta equals Omega T this Omega is what we call not angular frequency but we call it angular velocity it's very awkward in physics that we have the same symbol for angular velocity and for angular frequency in this case they happen to be the same numerically because it's a uniform circular motion that's an accident so now you see that x0 then becomes cosine Omega T because the two are the same I do not have to call the position T equals zero here I can choose T equals zero anywhere along the circumference and that introduces then phase angle Phi we call that the initial condition so x0 is amplitude Omega is the angular frequency and Phi has to be adjusted so that at time T equals zero you get the right angle at the right position an easy example of a simple harmonic motion is a spring system if I have here a spring and this is in a relaxed position the spring constant is K the mass is M and x equals zero here and I bring it further out I bring it to a position X then there is a spring force that wants to drive it back to equilibrium it's a restoring force that's the spring force let's arbitrarily call this direction plus the spring force we call minus KX minus because if X is positive then this figure force is in the opposite direction if the mass of the spring can be ignored if it is negligibly small compared to the mass of the object I can write down Newton's second law F equals MA you may remember that from your good old days and so MA is MX double dot is now minus KX it's really a vector notation but since it's a one-dimensional problem the minus sign takes care of the directions and so I can massage this a little further and I can write this as X double dot plus K over m times x equals zero and what is the solution to this differential equation this is a differential equation X double dot and X this is the solution the simple harmonic motion provided that Omega is the square root of K over m so I advise you to take this function substitute it in here and you will see that out pops yes you can satisfy this equation provided that Omega is the square root of K over m notice which is not so intuitive that this angular frequency Omega and therefore also the period of oscillation 2pi divided by omega is independent of x0 so it's independent of how far I move it away from equilibrium if I move it far out it will take the same amount of time for one oscillation then if I move without teeny-weeny little bit not so intuitive so Omega is independent of my initial conditions its independent on how I start the experiment off it's independent of Phi its independent on what I call T equals zero nature doesn't give a damn word I call T equals zero nature has one answer for the frequency and that's only determined by K and by M not by my initial conditions not so intuitive if I take the same spring and if I hang the spring vertically there's the spring due to gravity the object will come to a halt equilibrium a little lower obviously if now I displace it from this equilibrium position and I let it oscillate I get exactly the same frequency maybe that's not so intuitive either and you can work that out for yourself it's an 801 problem what it means is that you can define this as x equals zero ignore gravity completely and set up your differential equation as if there was no gravity and this is x equals zero so you offset it over a distance X from that equilibrium position you only allow four and spring force minus KX and everything works and of course you should be able to prove that that is correct if it spring oscillates in the simple harmonic fashion and we have such a spring here Marko's if you can do me a favor and get it up here then I should be able to demonstrate that a uniform circular motion project it on the wall we call shadow projection should be able Thank You Marcos should be able to have the same motion as my spring provide it of course that we very carefully make the period of oscillation of the spring exactly the same as the time for this object to go around we then shadow projected on there and then I will even try to lease this one this is very difficult at the same time that this one is here and what you will see then you will see the uniform circular motion projected becomes a simple harmonic motion and you'll see the spring simple harmonic and so we'll try to do that in shadow projection we'll make it a little darker and for that I need some light here that's here okay and we - Oh someone already turned it off so here you see the spring and there you see this object which is rotating in a circle but you think it's a simple harmonic motion and that's of course my objection my objective and so now that is difficult I will have to block you for a few seconds I will try now to release this at the same time and also at the same amplitude oh boy that wasn't my best day was it no no oh this is perhaps the best I can do today so they don't go exactly next to each other but you see they have the same period and they both represent simple harmonic oscillation the spring because we just calculated that and the projection of the uniform circular motion so if we return to the spring here maybe we should remove this if we return to the spring then we have a period for the spring system which is 2pi times the square root of M over K and I want to bring this to a test to a quantitative test how accurate is is I'm going to double the mass that I'm going to hang on that spring we're going to measure the period and then I want the mass which is twice as high I want that period to be the square root of two times higher because that's what this equation predicts now whenever you want to do a measurement in physics whereby you want to compare numbers so you have a certain goal in mind a measurement without the uncertainty in the measurement is completely meaningless you must know the accuracy of your measurement so n1 is 500 plus or minus 0.2 grams and m2 is thousands plus or minus 0.2 grams that's the best we can do that's an extremely small error this is an error of only 0.04 percent and this is only half as large now comes the question if I measure the period of oscillation which the 500 grams hanging on the spring how accurately can I do that on a good day I can do it 2.1 second accuracy I have to start it they have to stop it and if I do that ten times obviously you get different answers and they vary by about a tenth of a second on a good day now on a bad day 2/10 of a second I don't know whether today is a good day or whether it's a bad day but let's say it's in between so let's say I can do it to 0.15 seconds which I cannot guarantee you but I'll try so I can measure the period to plus or minus 0.15 seconds this is with m1 however I can get a very accurate measurement for the time for the period if I oscillate if I make ten oscillations because if I make ten oscillations the error in T goes down by a factor of 10 because the point 1 5 is 0.15 that's not going to change so I'm going to oscillate it ten times and then we're going to make a prediction about what we should measure for the higher mass so let's first measure the period of this is the spring and here is the 500 gram plus or minus point - I'm going to oscillate it we already notice that it's independent of amplitude and then I'm going to start it when it is down that's the easiest for me and I will count to ten you will count to ten and then we'll stop it so let's give it a oscillation yeah one two three four five six seven eight nine ten fourteen point nine six so let's write this down we have fourteen point nine six if you want to see whether this is a good day or whether this is a bad day I can measure it again one two three four five six seven eight nine ten fourteen point nine eight so this is not a bad day but it's luck that it comes out so close of course so now we can make a prediction that 210 times tm2 must be 1 point 4 1 4 which is the square root of 2 times 14 point nine seven okay so I take the one point four one point four one four I multiply that by fourteen point nine seven and I find twenty-one point one seven twenty one point one seven and of course this has to be multiplied also by one point four four that becomes plus or minus 0.2 seconds that is a prediction this is predict and now comes the observation this is a thrilling moment for you because what is at stake is the integrity of physics and this is going to be measured plus or minus 0.15 right every time I make a measurement plus or minus 0.15 I'm nervous so I'm going to add 500 grams period will indeed increase I'm going to oscillate it yeah one two three four five six seven eight nine yeah oh boy oh boy what have I done what have I done what is it 20.5 - we have a problem physics is not working any one of you have an idea whether there is something wrong with the equation or whether there's something wrong with Walter Lewin any idea come on give it a try in the worst case your suggestion is not correct yeah ah you accused me right what's your name exceeding my point 1 5 you say man you couldn't even do better than point 4 seconds maybe and then of course the tool would be consistent with each other thank you very nice of you pretty okay now that's a very good suggestion another what you begin to think like a physicist now you also thought like a physicist because indeed if my uncertainty is higher than 0.15 you could be right friction in this case we will deal with friction later in the course has such a negligibly small effect that it couldn't be measured in either one of these two in any case it's almost the same for both because the shape has not changed much it's a very good suggestion friction doesn't come near the proper explanation but you tried and that's good one more try math we have said earlier you may not have heard it but I did say it we can replay the tape I can prove it to you I said if the mass of the spring is negligible then that is the equation now what do we do when the mass cannot be ignored that's not so easy but I request it so mandatory reading and I'm sure all of you have done that before this lecture and the mandatory reading was French paid 6261 among others and French says that if the mass of the spring itself is capital m and if Capital m divided by three is substantially less than the mass at the end of the spring then a very very good approximation is that the period of oscillation is than this and he actually derives it you period is higher so we can bring this to a test now in other words the mass of the spring we have weighed that in our case is 175 point 6 plus or minus 0.2 grams and so M divided by 3 is 58 point five plus or minus point O seven grams that's a very small error by the way so 0.1 percent error 0.1 percent error and so we can now do the following test we can now take the ratio of these two and eliminate there by K so we can write now 10 T m2 divided by 10 T m1 is now the square root of M 2 plus M divided by 3 divided by M 1 plus M divided by 3 and that number is easy to calculate because you know M - you know M 1 you know these numbers and I have calculated it for you and it is one point three seven seven and the uncertainty is so small compared to my timing uncertainty that I don't even have to allow for any uncertainty in that number because remember the uncertainties in these masses was of the order of 0.1 percent compare that with the uncertainty in the observations of the time which were closer to 1 percent so we can bring this now to a test and all I must do now is multiply if I want to find now 10 t + 2 then I take one point three seven seven and I multiply it by T and one times ten T M one so I take this number and now I'm really getting nervous not joking fourteen point nine seven multiplied by one point three seven seven that is twenty point six one and the uncertainty would be the same uncertainty as in there which is a one percent uncertainty so that is 0.2 seconds this number you can now compare with this number on the button within the error of measurements they now agree this is what we observed and this is what we predict if we apply the proper relation and take the mass of the spring into account so you see that physics works except that this equation was too simple to be used for our observation notice by the way that this one point four one four in our case is lower all right in 803 we will often do not always use complex notation and the reason why we do that is that it can at times simplify your life and you are completely free to choose when you want to use it and when you don't want to use it you can be the judge so let's talk a little bit about complex numbers I start with a circle and this is the complex plane the blackboard is a complex plane that's quite a promotion for the black sport and here I call this axis the real axis so all the real numbers are on this axis and let this be plus one let this be minus one and I call this axis the imaginary axis so this one is plus J and this one is minus J and J is the square root minus one we don't call it I in general because I is sense for current so we pick J I now pick a position here which now represents a complex number call this angle theta and I project this this is position Z complex number this is the real part of that complex number and this is the imaginary part of that complex number so you can see that indeed Z can be written since this length is one is the cosine of theta plus J times the sine of theta so that is this part which is real and this is the sine of theta because this is one I have to multiply that by J and this now according to Hoyle ur great mathematician Hoyle Euler Euler after whom this disk was also mentioned already in 1748 he proved that this is the same as e to the power J Phi sorry theta J theta this equality is mind-boggling and when I saw this equality for the first time I didn't believe it number one and I could hardly sleep at night because I couldn't prove it see I hadn't had any Taylor expansion yet so I couldn't prove it was my teacher in high school said this is this is the case and I said why he said this is the way this but we now can prove this you can do the Taylor expansion of the cosine theta Taylor expansion of the sine theta and the Taylor expansion of e to the power J theta and it's exactly correct not an approximation so why would we ever want to use this well if you make this thing go around going back to my uniform circular motion here if I make that point go around and I only look at the real part I have a simple harmonic motion and so if I change theta into Omega T then I get that Z equals the cosine of Omega T plus J times the sine of Omega T the real part of which is a simple harmonic motion and of course I'm not stuck to an amplitude of one I can easily make the amplitude a times larger and of course there is nothing wrong depending upon my initial conditions to have here a phase angle Phi and this then is a times e to the power J Omega T plus Phi according to Euler so what that means is that if you use this as your trial function to solve a differential equation and you can manipulate this very easily you can take first derivative second derivative of Exponential's extremely easy and then when you're done you take the real part of Z and out pops X as a function of time and you're done as I said it's up to you when you want to use it next lecture I will give you an example whereby it's clearly the way to go I wouldn't even know how to do it in any other way but often you do have a choice so we are interested in in the real part of that which is then our acceptable solution so if we have a complex number Z equals a plus JB then we should always be able to write it as an amplitude times e to the power J theta and then the amplitude a is the square root of a squared plus B squared and tangent of theta is B over a that follows immediately from that figure and so in problem set one you will get some chance to practice we'll give you a few interesting cases and a classic case that all of you in your lifetime have to be able to do once is the very non-intuitive problem J to the power J when I saw for the first time J to the power J I said to myself well what on earth can be more complex than J to the power J but is real it is not complex and you will wrestle with this there's an infinite number of solutions not one all of them are correct and I will help you a little because the first time I want to be nice to you it's only the first time I can also write J as e to the power J times pi over two do you agree because it simply means that the angle is PI over 2 here so I end up here that's J I'm not saying it is a very nice way of expressing J but it is J but not only is this J I can also rotate an integer number times 360 degrees whereby n 0 1 2 3 rotate either clockwise or counterclockwise and it's again J because if I rotate 90 degrees it's J but if I rotate another 360 degrees it's again J or if I rotate back 360 degrees and so you see that this is also a way to write J and that will help you believe I will always have a 5-minute break during this 85 minute lecture so that you can stretch your legs if you can manage to make it back and forth to the bathroom that's fine that that's your problem I will start exactly after 5 minutes however every Tuesday during part of these five minutes we will have a mini quiz it's really mini this small and we will collect it after the lecture and you will even get some credit for that before but only on Tuesdays but not today before we go into this 5-minute break today I want you to see something so that you have something to think about believe me it's healthy the MIT student to sleep but it's also healthy sometimes to not sleep sleepless nights and worried just the way that I had sleepless nights in high school about Euler's equation it's healthy the reason why that's healthy is because once you see the solution you say ah of course and you never forget it whereas if someone tells you from the start you say you are of course and you forget it and the next day you don't remember so what I want you to see is a remarkable example of an oscillation that can be produced not by wind as we have seen but by heat and by cooling I have here a nice pipe and there is a grid here I can touch it I'm touching it now that's all there is it's an open pipe when there is a grid here and when I heat that grid and cool it somehow it generates a hundred ten Hertz oscillation a proper way pressure wave which you will be able to hear and I'll give you until the end of December maybe mid December to come up with a solution why it's doing that I'm heating the grit now 110 Hertz roughly if you want to play with this don't break it try to transfer the liquid in 17 seconds I will start I will resume this lecture exactly 5 minutes from now if you turn this it in tornado you rotate it then you open up a funnel of air and so there is never the problem that the liquid cannot go through there's always pressure equilibrium and I don't remember how long it takes but I thought it was 17 seconds but if you want to weaken time then may even be less I now want to address the issue of simple harmonic oscillation of a pendulum as you will remember from 801 if you have a pendulum length L mass m and if the mass of the string is negligibly small compared to the mass that is hanging here then the period of oscillations is 2pi times the square root of L over G G in the Boston area being to a high degree of accuracy 9.8 0 meters per second squared if you simply take L approximately 1 meter then you can see that you get a period of about two seconds and if you make the length about 25 centimeters that is four times shorter then you would expect this period which is two times shorter which is about one second and without any pretense of accuracy just eyeballing not really testing if I just eyeball this to be about a meter and if I also laid this back and forth it's about two seconds for one oscillation one - one two one two if I ever make it 25 centimeters four times shorter then it is very close to one second no interference here one one one one one one remarkable when you look at this equation is that just like in the case of the spring it is independent of the amplitude in other words whether I have a large amplitude or a small amplitude it would take the same amount of time to go back and forth well not quite for a pendulum when we make when we derive this period you remember that you have to assume what we call small angle approximation you will see that again and again with a dopey called small-angle approximations with small-angle approximations the sine of theta is always the same as theta in radians now if you ask me how small is small that's a matter of taste in 26100 we have the model of all pendulums 5.18 meters long quite impressive so we have a pendulum with L is 5.1 plus or minus 0.05 meters we cannot measure it any better than five centimeters because it has to be under stretch when we measure it and now you have to go all the way to the ceiling and all the way down Marcos does that risking his life and he claims that the best he can do is five said we had 31 pounds hanging under there we try during this summer believe me we tried with technicians of MIT to have that pendulum and at one day it looked good but finally they said no we can do it they can't install it here as a safety issue so unfortunately we don't have the mother of all pendulums here in 26100 when I lectured Newtonian mechanics I demonstrated that the period that this pendulum produces is extremely close within the error of measurement which you predict in other words the mass of the string is indeed negligibly small compared to the mass of the object we even MIT we raid the string ones I remember I don't remember what it was but it was such a small fraction of em that indeed could be ignored and so the prediction then is if you simply put this L in there T predicted purely on the basis of that simple equation equals four point five seven plus or minus 0.02 seconds and this 0.02 is the result of this 0.05 is a one percent error in here right five out of five hundred and eighteen is one percent and so the error in T is half a percent because it's d'squared and so you get a half a percent error and I rounded that off so that is the prediction and then I made two measurements one at five degree angle and one a 10 degree angle and I did that ten times so 10 T at five degrees and ten T at ten degrees now this was in 1999 those were my good days there were my good times right past is always the good and so I then claimed that I could do this to an accuracy of 0.1 second I had a lot of courage in those days and I measured the 5 degree and what did I find unbelievable truly unbelievable purely lucky I found exactly that number which of course is an accident because my accuracy was no better than 0.1 seconds and then I did it at a 10 degree angle and then I found this and so I demonstrated that indeed 5 and 10 degrees I still considered small angles for that approximation and it is within the uncertainty of my measurement what you expect then I wanted to demonstrate which is not so intuitive that the period is independent of mass which is not the case for the spring so now if you change the mass and you don't change L you expect no change in period and that's what I really wanted to show you here but I can't and therefore I've decided to show you what I did in 1999 if you can show is that 2 minute version of my video lectures then you can judge for yourself to what extent the mass does not influence the one of the most remarkable things I just mentioned to you is that the period of the oscillations is independent of the mass of the object that would mean if I joined the Bob and I swing down with the Bob that you should get that same period or should you not I'm asking you a question before we do this awful experiment would the period come out to be the same or not some of you think it's the same have you thought about it that I'm a little bit taller than this object and that therefore may be effectively the length of the string has become a little less if I set up like this and if the length of the spring is a little less the period would be a little shorter yeah be prepared for that on the other hand I'm also pre well I'm not quite prepared for it I will try to hold my body as horizontal as I possibly can in order to be at the same level as the Bob I will start when I come to a halt here there we go now you count this hurts I want to hear your heart thank you ten tea with Walter Lewin 45.6 plus or minus 0.1 seconds physics works I'm telling you all right so I think this was convincing at least for the freshmen that indeed a period of a pendulum is independent of the mass provided that you can ignore the mass of the string itself which is the case for that pendulum many pendulums and some we will see in 803 are more complex more complicated than simply a massless string with an object at the end and those pendulums we call a physical pendulum for instance I could have this pair compasses and just let it oscillate like this that is not just a simple pendulum or I could have a ruler like this put a pole through here and have a pin and have it oscillate like this but I can also have it oscillate here it's a different period if I also let it ride in the middle then doesn't oscillate at all so now comes the question how do we deal with that and most of you must have seen that in 801 that I do want to address that in quite some detail so a physical pendulum then looks like this this is an object and I drill a hole in here at Point P and I put a pin in the wall and it can without friction it can oscillate back and forth and at the center of mass be here position o and this separation between B P and O is B and O is the center of mass and you can choose P anywhere you want to there's no restriction on P so you see that this pendulum is offset over an angle theta and it will start to oscillate back and forth and the question is what is the period so clearly we may put the entire gravitational force at point O in the center of mass so this is force acting at that point and now comes the question either any other force is acting on this object or is this the only because when you study it you've got to take all forces into account who is happy that we have taken all forces into account raise your hand most of you are getting scared right who says no there has to be at least one other force and which force is that yeah I'm not even wear it where is that where does it act at one location yeah so there must be somehow a force at P to hold it up otherwise it would just start to accelerate down now I'm not even sure that it is trade-up I doubt that it may simply be at a direction I don't want to think about that but surely there has to be a force up now remember F equals MA when you deal with rotation of objects and this is going to be a rotational then this equation changes into tau the torque is moment of inertia times alpha whereby alpha is Theta double dot it's the angular acceleration and so if I pick P as my point of origin then the torque due to this force does not contribute to my torque equation because the torque is R cross F it's a cross-product between the position vector and the force and this is the position vector to the center of mass and the position vector from P to P is zero so if we deal with the torque relative to point P that force is of no consequence so I'm going to take P as my origin and so now is the question what is the torque relative to point P well it's R cross F R is this distance which is B F is mg but I have a cross product so I have to take the sine of this angle into account so that is the magnitude of the torque and the magnitude of that torque then according to my rotational equivalent of F equals MA equals the moment of inertia for rotation about that point P times theta double dot however is a restoring torque the torque and you can do that with your right hand wherever where you have learned how to do that the torque is in the blackboard perpendicular to the blackboard in the blackboard our cross F is in the blackboard I have rotated it counterclockwise which is a vector out of the blackboard so one is like this and the other is like this that is like saying the torque is restoring the same reason why we wrote down Fe calls minus KX which the spring is why we now write this equals minus this taken to account the direction of the vectors and so this is the differential equation that you would have to solve and if now we go to small angle approximation then the sine of theta goes to theta if theta is in radians and so I can rewrite down this theta double dot plus B M G divided by the moment of inertia about point P times theta equals zero and now small angle approximation we have a differential equation which is again a piece of cake simple harmonic oscillation and so the simple harmonic oscillation solution must be that theta is some maximum angle theta zero times the cosine of Omega T plus Phi by this Omega is the square root of this number just like we earlier had the square root of K over m Omega now must be this Omega is the square root of BMG divided by I of P that means T the period of oscillation is the moment of inertia about point P divided by B and G I want to repeat what I said earlier this Omega is called angular frequency the angular frequency is a given that's the angular frequency do not confuse that with theta dot which we also call Omega which is called angular velocity and the angular velocity in this case is a strong function of time when the object comes to a halt the angular velocity is zero because theta dot is zero it is unfortunate that we give them the same symbol so this is independent of time but theta dot does depend on time and theta dot is the angular velocity and in the case of the uniform circular motion the two omegas are the same so now we have all the ingredients in hand to calculate for absurd looking objects what the period of oscillation is provided that we are able to calculate the moment of inertia about the point of rotation and of course we have to know B and the mass of the object you have a wonderful example in your problem set I will solve that equation will calculate this T for a hoop this is the hoop all the mass is at the circumference so it should be very easy to calculate the moment of inertia and we have a hole in here and so we're going to oscillate it right at the rim and so our geometry is easy but we should be able to bring this equation to a rigid test provided that we take into account the uncertainty of our measurements and so let me put here this circle this is this hope so all the mass - very good approximation is at the circumference and the oscillation is about an axis perpendicular to the blackboard Point P this is the center of mass oh and I'm going to offset this hoop so this is when it is an equilibrium and this is offset over an angle theta so point always now here call it o prime and so in analogy with what we did there we have here the force mg and the derivation is identical we don't have to go over that again and the radius is R and M is given if you need it and R is given I'll show you what these numbers are later so all I have to do now is go to this equation and calculate the moment of inertia for rotation of an axis like this through point P who remembers how to do that 801 come on in the worst case it's wrong see one hand there who remembers let me ask you this suppose it were rotating through an axis right through the center of mass that's difficult because nothing to hold on to would you know then what the moment of inertia is what is it then yeah you say yes but now you're quiet yeah okay moment of inertia is never mr it's never the time it's dimensionally wrong but you try it which is better than not trying yeah mr square is what the moment of inertia would be if the axis were straight through oh I'm slowly working you now now we move the axis from o to P what happens now well how do we call it what we call their theorem parallel axis theorem now we have to add the mass times the distance between the center of mass and that point square that's the parallel axis theorem and so the moment of inertia about point P is mr squared for rotation about this point we take the same axis we move it to P and we have to add M distance squared so we have to add plus M R squared so we get to M R squared and then we have B what is B what is the distance from P to the center of mass that's R so we come now with a prediction that T is 2 pi times the square root of 2m R squared divided by r mg m goes and always goes with pendulums you never have to worry about M if you do it right M always goes not with Springs but with pendulums one R goes and so you get 2 pi times the square root of 2r over G before we bring this to a test since this is a remarkable answer what does that make you think of excuse me it makes you think of a single pendulum whereby the length is 2r which is by no means obvious is it in other words if I had a pendulum here and I would hang here an object M that would have the same period because it has a length 2r so T is 2 pi times the length divided by G by no means obvious absolutely not clear why that is but that's the way it is so now comes the acid test and so we don't have to measure the mass but we did measure as accurately as we can the radius that's really all we have to do and the measurement of the radius is a little uncertain because it's not a perfect circle so we measured it at various places and we find that R equals 40 point zero plus or minus 0.5 centimeters so that's a 1 percent uncertainty and so we make a prediction out t we get the square root of R so the 1 percent uncertainty becomes half a percent because of the square root you take 2r u divided by G and you'll find that the prediction this is a prediction is that T is one point seven nine five plus or minus that is your half a percent 0.01 seconds that's because of the square root so this becomes half a percent error and now we do the observation and you guessed it of course we're going to do 10 T and if this is a good day Oh point one that will give myself a little bit extra leeway today 0.15 I'm fairly sure I should be able to do that and so we bring this now to a test if you're ready for this Oh still long I always like to start the timer when the object comes to a halt that is a better criterion and when it go through equilibrium and I will not look at the even if I did look at it there's no way I can stop that when I want to so we'll give it a an offset I first wanted to swing in a way that it's not wobbling because I make a very strong prediction I want to get that number seventeen point nine five so I better make sure that it's or that it's oscillating happily no this is not happy I don't want any wobbling like this maybe a little more okay I think this looks good if you're ready I'm ready now one two three four five six seven eight nine now yeah 17 point eight zero ah man point one five so what wasn't such a bad day after all for me okay see you Tuesday and work on your problem sets you
