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courses, visit MIT OpenCourseWare at ocw.mit.edu, YEN-JIE LEE: So
welcome, everybody. My name is Yen-Jie Lee. I am a assistant professor
of physics in the physics department, and I will
be your instructor of this semester on 8.03. So of course, one first
question you have is, why do we want to learn
about vibrations and waves? Why do we learn about this? Why do we even care? The answer is really, simple. If you look at
this slide, you can see that the reason you
can follow this class is because I'm producing sound
wave by oscillating the air, and you can receive
those sound waves. And you can see me-- that's really pretty
amazing by itself-- because there are
a lot of photons or electromagnetic waves. They are bouncing
around in this room, and your eye actually receive
those electromagnetic waves. And that translates
into your brain waves. You obviously, start to think
about what this instructor is trying to tell you. And of course, all those
things we learned from 8.03 is closely connected
to probability density waves, which you will learn
from 8.04, quantum physics. And finally, it's
also, of course, related to a recent discovery
of the gravitational waves. When we are sitting
here, maybe there are already some space-time
distortion already passing through our body and
you don't feel it. When I'm moving
around like this, I am creating also the
gravitational waves, but it's so small
to be detected. So that's actually really cool. So the take-home message is
that we cannot even recognize the universe without using
waves and the vibrations. So that's actually why we
care about this subject. And the last is actually
why this subject is so cool even without quantum,
without any fancy names. So what is actually the
relation of 8.03 to other class or other field of studies? It's closely related to
classical mechanics, which I will use it
immediately, and I hope you will still
remember what you have learned from 8.01 and 8.02. Electromagnetic force is
actually closely related also, and we are going to use
a technique we learned from this class to understand
optics, quantum mechanics, and also there are many
practical applications, which you will learn from this class. This is the concrete goal. We care about the future
of our space time. We would like to predict
what is going to happen when we set up an experiment. We would like to design
experiments which can improve our understanding of nature. But without using the
most powerful tool is very, very difficult
to make progress. So the most powerful tool
we have is mathematics. You will see that it
really works in this class. But the first thing
we have to learn is how to translate physical
situations into mathematics so that we can actually include
this really wonderful tool to help us to solve problems. Once we have done
that, we will start to look at single
harmonic oscillator, then we try to couple all
those oscillators together to see how they interact
with each other. Finally, we go to an infinite
number of oscillators. Sounds scary, but it's
actually not scary after all. And we will see waves
because waves are actually coming from an infinite number
of oscillating particles, if you think about it. Then we would do Fourier
decomposition of waves to see what we can
learn about it. We learn how to put
together physical systems. That brings us to the issue
of boundary conditions, and we will also enjoy
what we have learned by looking at the
phenomenon related to electromagnetic waves
and practical application and optics. Any questions? If you have any questions,
please stop me any time. So if you don't stop me, I'm
going to continue talking. So that gets started. So the first example,
the concrete example I'm going to talk about is a
spring block, a massive block system. So this is actually what
I have on that table. So basically, I have a
highly-idealized spring. This is ideal spring
with spring constant, k, and the natural length L0. So that is actually what I have. And at t equal to 0, what I am
going to do is I am going to-- I should remove this
mass a little bit, and I hold this mass still and
release that really carefully. So that is actually
the experiment, which I am going to do. And we were wondering what
is going to happen afterward. Well, the mass as you
move, will it stay there or it just disappear,
I don't know before I solved this question. Now I have put together a
concrete question to you, but I don't know how
to proceed because you say everything works. What I am going to do? I mean, I don't know. So as I mentioned before, there
is a pretty powerful tool, mathematics. So I'm going to use
that, even though I don't know why mathematics can work. Have you thought about it? So let's try it and see
how we can make progress. So the first thing which you
can do in order to make progress is to define a
coordinate system. So here I define a
coordinate system, which is in the horizontal direction. It's the x direction. And the x equal
to 0, the origin, is the place which the
spring is not stressed, is at its natural length. That is actually what I
define as x equal to 0. And once I define
this, I can now express what is actually the
initial position of the mass by these coordinates is x0. It can be expressed
as x initial. And also, initially, I said
that this mass is not moving. Therefore, the
velocity at 0 is 0. So now I can also formulate
my question really concretely with some mathematics. Basically, you can see
that at time equal to t, I was wondering
where is this mass. So actually, the question
is, what is actually x as a function of t? So you can see that once I have
the mathematics to help me, everything becomes
pretty simple. So once I have those
defined, I would like to predict what is going
to happen at time equal to t. Therefore, I would like to make
use physical laws to actually help me to solve this problem. So apparently what we are
going to use is Newton's law. And I am going to go
through this example really slowly so that
everybody is on the same page. So the first thing
which I usually do is now I would like to do
a force diagram analysis. So I have this mass. This setup is on Earth,
and the question is, how many forces are
acting on this mass? Can anybody answer my question. AUDIENCE: Two. We got the-- So acceleration and
the spring force. YEN-JIE LEE: OK, so
your answer is two. Any different? Three. Very good. So we have two and three. And the answer
actually is three. So look at this scene. I am drawing in and
I have product here. So this is actually the most
difficult part of the question, actually. So once you pass this step,
everything is straightforward. It's just mathematics. It's not my problem any more,
but the math department, they have problem, OK? All right. So now let's look at this mass. There are three forces. The first one as you mentioned
correctly is F spring. It's pulling the mass. And since we are
working on Earth, we have not yet
moved the whole class to the moon or somewhere
else, but there would be gravitational
force pointing downward. But this whole setup is on a
table of friction, this table. Therefore, there will
be no more force. So don't forget this one. There will be no more force. So the answer is that
we have three forces. The normal force is, actually,
a complicated subject, which you will
need to understand that will quantum physics. So now I have three force,
and now I can actually calculate the total
force, the total force, F. F is equal to
Fs plus Fn plus Fg. So since we know
that the mass is moving in the
horizontal direction, the mass didn't suddenly
jump and disappear. So it is there. Therefore, we know that the
normal force is actually equal to minus Fg, which is
actually Ng in the y direction. And here I define y is
actually pointing up, and the x is pointing
to the right-hand side. Therefore, what
is going to happen is that the total force
is actually just Fs. And this is equal
to minus k, which is the spring
constant and x, which is the position of the little
mass at time equal to t. So once we have those
forces and the total force, actually, we can
use Newton's law. So F is equal to m times a. And this is actually equal
to m d squared xt dt squared in the x direction,
and that is actually equal to mx double dot t x. So here is my notation. I'm going to use each of the dot
is actually the differentiation with respect to t. So this is actually equal to
minus kxt in the x direction. So you can see that here is
actually what you already know about Newton's law. And that is actually coming
from the force analysis. So in this example,
it's simple enough such that you can write
it down immediately, but in the later
examples, things will become very complicated
and things will be slightly more difficult. Therefore,
you will really need help from the force diagram. So now we have everything
in the x direction, therefore, I can drop the x hat. Therefore, finally, my equation
of motion is x double dot t. And this is equal to
minus k over n x of t. To make my life
easier, I am going to define omega equal to
square root of k over n. You will see why afterward. It looks really weird why
professor Lee wants to do this, but afterward, you will see that
omega really have a meaning, and that is equal to
minus omega squared x. So we have solved this problem,
actually, as a physicist. Now the problem is
what is actually the solution to
this differential second-order
differential equation. And as I mentioned,
this is actually not the content
of 8.03, actually, it's a content of 18.03, maybe. How many of you actually
have taken 18.03? Everybody knows the
solution, so very good. I am safe. So what is the solution? The solution is x of t equal
to a cosine of omega t plus b sine omega t. So my friends from
the math department tell me secretly that this
is actually the solution. And I trust him or her. So that's very nice. Now I have the
solution, and how do I know this is the only solution? How do I know? Actually, there
are two unknowns, just to remind you
what you have learned. There are two unknowns. And if you plug this
thing into this equation, you satisfy that equation. If you don't trust me,
you can do it offline. It's always good
to check to make sure I didn't make a mistake. But that's very good news. So that means we will
have two unknowns, and those will
satisfy the equation. So by uniqueness
theorem, this is actually the one and the only one
solution in my universe, also yours, which satisfy
the equation because of the uniqueness theorem. So I hope I have
convinced you that we have solved this equation. So now I take my
physicist hat back and now it is actually my job again. So now we have the
solution, and we need to determine
what is actually these two unknown coefficients. So what I'm going to use is to
use the two initial conditions. The first initial condition
is x of 0 equal to x initial. The second one is that since
I released this mass really carefully and the initial
velocity is 0, therefore, I have x dot 0 equal to 0. From this, you can solve. Plug these two conditions
into this equation. You can actually figure out
that a is equal to x initial. And b is equal to 0. Any questions so far? Very good. So now we have the solution. So finally, what is
actually the solution? The solution we get is x of t
equal to x initial cosine omega t. So this is actually the
amplitude of the oscillation, and this is actually
the angular velocity. So you may be
asking why angular? Where is the
angular coming from? Because this is actually
a one-dimensional motion. Where is the angular
velocity coming from? And I will explain that
in the later lecture. And also this is actually
a harmonic oscillation. So what we are
actually predicting is that this mass is going
to do this, have a fixed amplitude and it's
actually going to go back and forth with the
angular frequency of omega. So we can now do an experiment
to verify if this is actually really the case. So there's a small difference. There's another spring here,
but essentially, the solution will be very similar. You may get this
in a p-set or exam. So now I can turn on the air
so that I make this surface frictionless. And you can see that now
I actually move this thing slightly away from the
equilibrium position, and I release that carefully. So you can see that really it's
actually going back and forth harmonically. I can change the amplitude
and see what will happen. The amplitude is
becoming bigger, and you can see that the
oscillation amplitude really depends on where you put
that initially with respect to the equilibrium position. I can actually make a small
amplitude oscillation also. Now you can see that now the
amplitude is small but still oscillating back and forth. So that's very encouraging. Let's take another example,
which I actually rotate the whole thing by 90 degrees. You are going to get a
question about this system in your p-set. The amazing thing is that
the solution is the same. What is that? And you don't believe me,
let me do the experiment. I actually shifted the position. I changed the position, and I
release that really carefully. You see that this mass is
oscillating up and down. The amplitude did not change. The frequency did not change
as a function of time. It really matched with the
solution we found here. It's truly amazing. No? The problem is that we are
so used to this already. You have seen this maybe
100 times before my lecture, so therefore, you
got so used to this. Therefore, when I say,
OK, I make a prediction. This is what happened, you
are just so used to this or you don't feel
the excitement. But for me, after I teach
this class so many times, I still find this
thing really amazing. Why is that? This means that
actually, mathematics really works, first of all. That means we can use the same
tool for the understanding of gravitational waves, for the
prediction of the Higgs boson, for the calculation
of the property of the quark-gluon plasma
in the early universe, and also at the
same time the motion of this spring-mass system. We actually use always the
same tool, the mathematics, to understand this system. And nobody will understands why. If you understand
why, please tell me. I would like to know. I will be very proud of you. Rene Descartes said
once, "But in my opinion, all things in nature
occur mathematically." Apparently, he's right. Albert Einstein also once said,
"The most incomprehensible thing about the universe is
that it is comprehensible." So I would say this
is really something we need to appreciate
the need to think about why this is the case. Any questions? So you may say, oh, come on. We just solved the problem
of an ideal spring. Who cares? It's so simple, so easy,
and you are making really a big thing out of this. But actually, what
we have been solving is really much more than that. This equation is much more
than just a spring-mass system. Actually, if you think about
this question carefully, there's really no
Hooke's law forever. Hooke's law will give you
a potential proportional to x squared. And if you are so far away, you
pull the spring so really hard, you can store the energy
of the whole universe. Does that make sense? No. At some point, it
should break down. So there's really no Hook's law. But there's also
Hook's law everywhere. If you look at this system,
it follows the harmonic oscillation. If you look at this
system I perturb this, it goes back and forth. It's almost like everywhere. Why is this the case? I'm going to answer this
question immediately. So let's take a
look at an example. So if I consider
a potential, this is an artificial
potential, which you can find in
Georgi's book, so v is equal to E times L
over x plus x over L. And if you practice as
a function of x, then basically you get
this funny shape. It's not proportional
to x squared. Therefore, you
will see that, OK, the resulting motion for the
particle in this potential, it's not going to
be harmonic motion. But if I zoom in,
zoom in, and zoom in and basically, you will see
that if I am patient enough, I zoom in enough, you'll
see that this is a parabola. Again, you follow Hooke's law. So that is actually really cool. So if I consider an
arbitrary v of x, we can do a Taylor
expansion to this potential. So basically v of x
will be equal to v of 0 plus v prime 0 divided
by 1 factorial times x plus v double prime
0 over 2 factorial x squared plus v
triple prime 0 divided by 3 factorial x to the third
plus infinite number of terms. v 0 is the position of where
you have minimum potential. So that's actually where
the equilibrium position is in my coordinate system. It's the standard,
the coordinate system I used for the solving
the spring-mass question. So if I calculate the
force, the force, f of x, will be equal to
minus d dx v of x. And that will be
equal to minus v prime 0 minus v double
prime 0 x minus 1 over 2 v triple prime 0 x
squared plus many other terms. Since I have mentioned
that v of 0-- this will be x. v of 0 is actually the
position of the minima. Therefore, v prime of
0 will be equal to 0. Therefore. This term is gone. So what essentially is left over
is the remaining terms here. Now, if I assume that x is very
small, what is going to happen? Anybody know when x is very
small, what is going to happen? Anybody have the answer? AUDIENCE: [INAUDIBLE]. YEN-JIE LEE: Exactly. So when x is very small, he said
that the higher order terms all become negligible. OK? So that is essentially correct. So when x is very
small, then I only need to consider the
leading order term. But how small is the question. How small is small? Actually, what you can
do is to take the ratio between these two terms. So if you take the
ratio, then basically you would get a condition
xv triple dot 0, which will be much smaller
than v double prime 0. So that is essentially
the condition which is required to satisfy it
so that we actually can ignore all the higher-order terms. Then the whole
question becomes f of x equal to minus
v double prime 0 x. And that essentially,
Hooke's law. So you can see that
first of all, there's no Hooke's law in general. Secondly, Hook's law
essentially applicable almost everywhere when you
have a well-behaved potential and if you only perturb
the system really slightly with very small amplitude,
then it always works. So what I would like
to say is that after we have done this exercise, you
will see that, actually, we have solved all the
possible systems, which have a well-behaved potential. It has a minima, and if I have
the amplitude small enough, then the system is going to do
simple harmonic oscillation. Any questions? No question, then
we'll continue. So let's come back to
this equation of motion. x double dot plus omega
squared x, this is equal to 0. There are two
important properties of this linear
equation of motion. The first one is that if x1 of t
and x2 of t are both solutions, then x12, which is
the superposition of the first and second
solution, is also a solution. The second thing, which is very
interesting about this equation of motion, is that there's a
time translation invariance. So this means that if
x of t is a solution, then xt prime equal to xt
plus a is also a solution. So that is really
cool, because that means if I change t equal to
0, so I shift the 0-th time, the whole physics
did not change. So this is actually
because of the chain law. So if you have chain
law dx t plus a dt, that is equal to d t plus a
dt, dx t prime dt prime evaluated at t prime
equal to t plus a. And that is equal to dx t prime
dt, t prime equal to t plus a. So that means if I have
changed the t equal to 0 to other place, the
whole equation of motion is still the same. On the other hand, if the
k, or say the potential, is time dependent, then that
may break this symmetry. Any questions? So before we take a
five minute break, I would like to discuss further
about this point, this linear and nonlinear event. So you can see that
the force is actually linearly dependent on x. But what will happen
if I increase x more? Something will happen. That means the higher-ordered
term should also be taken into account carefully. So that means the
solution of this kind, x initial cosine omega t,
will not work perfectly. In 8.03, we only consider the
linear term most of the time. But actually, I would
like to make sure that everybody
can at this point, the higher-order
contribution is actually visible in our daily life. So let me actually give
you a concrete example. So here I have two pendulums. So I can now perturb
this pendulum slightly. And you you'll see that it goes
back and forth and following simple harmonic emotion. So if I have both
things slightly oscillating with
small amplitude, what is going to happen is that
both pendulums reach maxima amplitude at the same time. You can see that very clearly. I don't need to
do this carefully. You see that they always
reach maxima at the same time when the amplitude is small. Why? That is because the higher-order
terms are not important. So now let's do a experiment. And now I go crazy. I make the amplitude
very large so that I break that approximation. So let's see what will happen. So now I do this then. I release at the same time
and see what will happen. You see that originally
they are in phase. They are reaching
maxima at the same time. But if we are patient
enough, you see that now? They are is
oscillating, actually, at different frequencies. Originally, the
solution, the omega, is really independent
of the amplitude. So they should,
actually, be isolating at the same frequency. But clearly you
can see here, when you increase the
amplitude, then you need to consider also
the nonlinear effects. So any questions before we
take a five-minute break. So if not, then we would
take a five-minute break, and we come back at 25. So welcome back, everybody. So we will continue
the discussion of this equation of motion, x
double dot plus omega square x equal to 0. So there are three
possible way to like the solution to this equation. So the first one as
I mentioned before, x of t equal to a cosine
omega t plus b sine omega t. So this is actually
the functional form we have been using before. And we can actually also
rewrite it in a different way. So x or t equal to capital
A cosine omega t plus phi. You may say, wait a second. You just promised me that
this is the first one, the one is the one
and only one solution in the universe, which actually
satisfy the equation of motion. Now you write another one. What is going on? Why? But actually, they are the same. This is actually A
cosine phi cosine omega t minus A sine phi sine omega t. So the good thing
is that A and phi are arbitrary constant
so that it should be you can use two initial
conditions to determine the arbitrary constant. So you can see that one and
two are completely equivalent. So I hope that solves
some of the questions because you really
find it confusing why we have different
presentations of the solution. So there's a third one, which
is actually much more fancier. The third one is
that I have x of t. This is actually
a real part of A-- again, the amplitude--
exponential i omega t plus phi, where i is equal to
the square root of minus 1. Wait a second. We will say, well, professor,
why are you writing such a horrible solution? Right? Really strange. But that will explain you why. So three is actually
a mathematical trick. I'm not going to prove anything
here because I'm a physicist, but I would like to share with
you what I think is going on. I think three is really
just a mathematical trick from the math department. In principle, I can drive it
an even more horrible way. x of t equal to a real part of
A cosine omega t plus phi plus i f of t. And f of t is a real function. In principle, I can do that. It's even more horrible. Why is that? Because I now have
this function. I take the real
part, and I actually take the two out
of this operation. So f of t is actually
the real function. It can be something arbitrary. And i can now plot the locus
of this function, the solution on the complex print. Now I'm plotting this solution
on this complex print. What is going to happen is
that you're going to have-- That is what you
are going to get. If I am lucky, if this
f of t is confined in some specific
region, if I not lucky, then it goes out
of the print there. I couldn't see it. Maybe it go to the
moon or something. But if you are smart
enough, and I'm sure you are, if I choose f
of t equal to A sine omega t plus phi, can anybody tell
me what is going to happen? AUDIENCE: [INAUDIBLE]. YEN-JIE LEE: Would
you count a circle? Very good. If I plot the locus
again of this function, the real axis, imaginary axis,
then you should get a circle. Some miracle happened. If you choose the
f of t correctly, wisely, then you can actually
turn all this mess into order. Any questions? So I can now follow
up about this. So now I have x of t is equal to
the real part of A cosine omega t plus phi plus iA
sine omega t plus phi. And just a reminder,
exponential i theta is equal to cosine
theta plus i sine theta. Therefore, I arrive this. This is a real part of A
exponential i omega t plus phi. So if I do this
really carefully, I look at this the position of
the point at a specific time. So now time is equal to t. And this is the real axis, and
this is the imaginary axis. So I have this circle here. So at time equal to t,
what you are getting is that x is actually-- before taking the real part, A,
exponential i omega t plus phi, it's actually here. And this vector actually
shows the amplitude. Amplitude is A. And the angle
between this vector pointing to the position of this
function is omega t plus phi. So this is actually the
angle between this vector and the real axis. So that's pretty cool. Why? Because now I understand why
I call this omega angular velocity or angular frequency. Because the solution to
the equation of motion, which we have actually
derived before, is actually the real part of
rotation in a complex print. If you think about
it, that means now I see this particle
going up and down. I see this particle
going up and down. You can think about
that, this is Earth. If there is an extra dimension
which you couldn't see. Actually, this particle
in the dimension where we can see into the extra
dimension, which is hidden is actually rotating. And while we see
that reality, it's a projection to the real axis. You see? So in reality, this particle
is actually rotating, if you add the image
and the extra dimension. So that is actually pretty
cool, but the purity artificial. So you can see that
I can choose f of t to be a different function,
and then this whole picture is different. But I also would
create a lot of trouble because then the mathematics
become complicated. I didn't gain anything. But by choosing this
functional form, you actually write a
very beautiful picture. Another thing, which
is very cool about this is that if I write this thing
in the exponential functional form, since we are dealing
with differential equations, there is a very good property
about exponential function. That is it is essentially
a phoenix function. Do you know what is a phoenix? Phoenix is actually some kind
of animal, a long-beaked bird, which is cyclically called
the regenerated or reborn. So basically, when
this phoenix die, you will lay the eggs in the
fire and you were reborn. This is actually the
same as this function. I can do differentiation,
still an exponential function, and differentiate,
differentiate, differentiate. Still exponential function. So that is very
nice because when we deal with
differential equation, then you can actually
remove all those dots and make them become just
exponential function. So essentially, a
very nice property. So the first property, which is
very nice is that it cannot be killed by differentiation. You will see how useful this
is in the following lectures. The second thing,
which is really nice is that it has a
very nice property. So basically the exponential
i theta 1 times exponential i theta 2, and that will give
you exponential i theta 1 plus theta 2. So what does that mean? That means if I have a solution
in this form, A exponential i omega t plus phi. And I do a times translation,
t become t plus A. Then this become A exponential
i omega t plus A plus phi. So this means that times
translation in this rotation is just a rotation
in complex print. You see? So now t becomes t plus
A. Then you are actually just changing the angle between
this vector and the x-axis. So as time goes
on, what is going to happen is that this thing
will go around and around and around and the physics is
always the set, no matter when you start counting, and
the translation is just the rotation in this print. Any questions? So I think this is actually a
basic slide just to remind you about Euler's formula. So basically, the
explanation i phi is equal to cosine
phi plus i sine phi. And I think it will be useful if
you are not familiar with this. It is useful to actually
review a little bit about exponential
function, which will be very useful for this class. So I'm running a
bit faster today. So let's take a look at
what we have learned today. We have analyzed the physics
of a harmonic oscillator. So basically, we start by asking
really just a verbal question, what is going to
happen to this mass on the table
attached to a spring. And what we have learned is that
we actually use mathematics. Basically, we translate all what
we have learned about this mass into mathematics by first
define a coordinate system. Then I'd write everything
using that coordinate system. Then I use Newton's law to
help us to solve this question. And we have analyzed the physics
of this harmonic oscillator. And Hooke's law, we found
that he actually, not only works for this
spring-mass system, it also works for all kinds of
different small oscillations about a point of equilibrium. So basically, it's actually
a universal solution what we have been doing. And we have found out a
complex exponential function is actually a beautiful
way to present the solution to the equation of
motion we have been studying. So everything is nice and good. However, life is
hard because there are many things which actually,
we ignored in this example. One apparent thing,
which we actually ignore, is the direct force. So you can see that before I was
actually making this pendulum oscillate back and forth. What is happening now? There are not
oscillating anymore. Why? Well, they stopped being. Apparently,
something is missing. When I actually
moved this system, if I turn off the air so
that there's friction, then it doesn't really move. If I increase a bit, the
air so that the slide have some slight freedom,
then actually, you can see that you move
a bit then you stop. If I increase this
some more, you can see that the amplitude
becomes smaller and smaller. So in the following lecture,
what we are going to do is to study how to actually
include a direct force into it again and of course, using the
same machinery which we have learned from here and
see if we can actually solve this problem. Thank you very much. We actually end
up earlier today. Sorry for that. And maybe I will make the
lecture longer next time. And if you have any
questions about what we have covered today, I'm
here available to help you.
