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visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Well let's see
if we can't get started. Everyone I trust can
hear me adequately. Welcome back. It's Tuesday. For those of you who are not
in my recitation section, I'm Dave Gossard, and I'll
be your lecturer for the day. Professor Vandiver
is out of town. It looks like some of
you may be as well. We probably could have held this
at the gate at Logan Airport and done a little better. But be that as it
may, glad you came. This should be fun. Today we have a new topic
and a demonstration, a real physical system. So unless there are any
outstanding questions? Anybody have any
questions or complaints to address to Vicente? No. All right, hearing none let's
go ahead and get started then. Today the topic is multiple
degree of freedom systems. Now to date, with a
couple of exceptions, all of the systems
that you've dealt with had a single degree of freedom,
either a linear displacement x or an angular
displacement theta. You know the concept
of equations of motion, or I should say the
equation of motion and the notion of undamped
natural frequency. Well, today we're going to
generalize, if you will, to systems that have not one
but multiple degrees of freedom and see how those
notions generalize. In particular, as
you might expect, the system that has
multiple degrees of freedom has multiple natural
frequencies, also known as eigenvalues as we
will explain here shortly. Multiple degrees
of freedom systems have a new property,
a new characteristic you haven't seen before. And that's what this
lecture is all about, is to illustrate that to
you and demonstrate it. It's the notion
of natural modes, also called eigenvectors. And then the general response
to initial conditions. So that is the plan for the day. And we'll start with this. This is kind of a
classic textbook case, two springs, two masses. A straightforward extrapolation
of what you've done before. You've got a spring K1,
mass M1, spring K2, mass M2. And the displacements
are indicated as shown there, X1 and X2. I want to hasten to point out
that the displacements we speak of here are defined with respect
to the static equilibrium position. This is a notion that Professor
Vandiver went over at least once. And for those of you who've
forgotten it or weren't there that day, I have for you a
reference, essentially reprised that notion over there. So in the meantime,
let me press on. If you have any questions, we
can go back and cover that. But assuming you agree, let
me simply say you've got two springs, two masses. The typical way we've
taught you to do it is if you're
going to generate the equations of motion
by the direct method, you generate two free
body diagrams, the sum forces in the x direction
for each of the masses, get f equals MA and
you'd get these. Conversely, you could
also do it by Legrange. You could generate
the expression for the kinetic energy,
for the potential energy, for the Legrangian, do this
Lagrange equation business, and you'd get the same thing. But either way you do it,
what comes out the other side looks like this. And it's not a bad exercise
for you to offline convince yourself that this is right. Not right now, but in
the comfort and leisure of another time. So there you have it. That's what the two equations
of motion would look like. Again, either by the direct
method or by Legrange, you end up in the same
place, so to speak. And now for today, we haven't
asked you to do this much, but let me simply say
the weapon of choice for multiple degrees
of freedom system, because there's a certain
repetitive quality to it, matrix notation is preferred. In these equations over
here, written in matrix form would look like this. And that looks like this. There's two matrices,
and let me hasten to point out that this is
exactly this and nothing more. There's no magic, no
additional derivation. This is simply a restructuring,
and reorganization of these equations. And this may seem
foreign to those of you who have not had any
or very much linear algebra. Do not be dismayed. It is not a difficult
thing to learn. As you probably know, a matrix
multiplies by a vector-- or multiplying a
vector by a matrix-- is done with two hands. The first item for example
is M1 X1 dot plus 0. That gives you this
term right here. Over here you get
X1 times K1 plus K2. That's this one. And here you have minus
K2 X2, that's that term. So matrix notation,
this becomes that. No problem. It's like a model train set. It's great. Everyone should
have one of these. All right, so what I would
like to do for our example here is because we're going
to be doing some algebra, for the express purpose of
simplifying the algebra let me consider a special case
where the masses are identical. And we can simply call
them M. And similarly, the springs are identical,
and we'll simply call them K. At that point, these
equations become simplified. That's just M, that's
just M. So this is the problem we're
going to-- we're going to tackle this problem
first, because as I say, it simplifies the algebra. Now here is-- this
is not an assumption. This is a-- I would call
this more a mechanism to get this job done here. Harmonic motion is
one where we assume that the masses oscillate
at the same frequency. So what this looks like is this. What we're basically saying
is that X1 is actually equal-- X1 has
amplitude A-- whoa. I'm being attacked here
by the second board. Basically, the situation
here is that we're assuming that both
of these masses move-- I wouldn't call
it exactly together. They're not in
complete synchrony, as you'll see here in a moment. But what they are
is they're going through a sinusoidal motion. And it is an oscillation
at the same frequency. Both of them are oscillating
at the same frequency. However, they differ
in their magnitudes. They are not the same magnitude. But that assumption right
there allows us to say this. If we differentiate those
twice, we get the following. That comes back out. Douglas, where's
that minus sign come? Can you-- first of all,
can everybody read that? Can you guys read that
in the back row here? For example, this says that X1
double dot, if x1 is A1 cosine, then X1 double dot
is A1 cosine preceded by a minus omega squared. Where does that come from? AUDIENCE: The minus sign came
from when you differentiated the cosine in the first one. PROFESSOR: Exactly. And the cosine returns. And that's what you get. Well, here is, shall we say,
the heart of the matter. When you substitute
this into this-- let me call this-- I'll try not
to get too obsessive over this. But these are our
equations of motion. So when we-- you
get this, equals 0. Excuse me. Many people, I think,
simply put a big 0 there, but I'll do it properly. It's two zeros, if you will. And forgive me for
writing this out, but I would like you
to be able to do this by yourself, to recreate
this after the fact. This becomes-- dividing
and collecting terms. OK, anybody unclear about
how this is obtained? Yes, ma'am. Emma. AUDIENCE: I have a question
about the previous one. PROFESSOR: Yeah? AUDIENCE: If the second
term on the left hand side, should it also be
multiplied by A1 A2? PROFESSOR: Yes it should. Thank you very much. Oh, hang on. Yes, thank you,
that's exactly right. AUDIENCE: [INAUDIBLE]. PROFESSOR: Hang on a second. We're fighting the boards here. Let's see, one thing at a time. We've got A1, A2 cosine
[INAUDIBLE] minus phi equals. Now Emma, does that
take care of you? Yeah? And you said, Vicente? AUDIENCE: Diagonal terms--
shouldn't there be a minus? PROFESSOR: I'm sorry,
that's absolutely correct. Wonderful. So there we have it. Any other questions? I hope I got it right. Yes sir. AUDIENCE: Are they plus K? PROFESSOR: I'm sorry? AUDIENCE: Are they plus K? PROFESSOR: Plus K? No. They're minus. Yeah, why is that? Everybody see that? This is a straight-- there's
less here than meets the eye. There's a straight segregation
collecting of terms. The minus got added to the
elements of the mass matrix, but not to the K metrics. The K metrics goes
shows through as is. Now, the question on the
floor is what we do with this? Can everybody appreciate
that-- get out of the spring mass
business and look at this from a math point of view? Does everyone
appreciate that this is a set of linear equations? There's the old AX
equal B kind of thing. And if you recall, to solve,
what we're basically going to do is solve for A1 and A2. That's the game
we're playing here. And if you recall from your math
course, the determinant of this has got to equal 0. So let me simply repeat it here. The determinant of
has got to equal 0. And you recall, the determinant
is for at least the two by two you can do it
by hand more or less. It's the cross products
with appropriate sign. And I'm sparing you
some algebra here, but trust me when you do
this, this is what you get. M squared omega 4 minus
3KM omega squared, plus K2. That's it right here. This little guy-- oh, all right. I can't do that anymore. I know, it's this one. That is called the
characteristic equation. So here's the first answer. AUDIENCE: [INAUDIBLE]
should that be K squared? PROFESSOR: I'm sorry. That's a typo. That's simply K. AUDIENCE: [INAUDIBLE] K squared? PROFESSOR: Or it's
K squared, rather. Sorry. Thank you. AUDIENCE: I think
in the above line, the determinant-- the upper
left-- should had a 2K. PROFESSOR: Oh, this
is 2K, absolutely. All right, 2K. Good enough? All right, thank you. So let's send this to the top. So the roots of
the characteristic are the natural frequencies. Let's do this this way. Did I do that right? Yeah, plus or minus. So the situation is that when
you apply the quadratic formula to that characteristic equation
to find the values of omega for which that
equation is satisfied, those omegas that come out
are the natural frequencies. They are the quantities we seek. And what that yields, as you
can see from the plus or minus here, there are two of them. I'll write the whole
thing out here. And these are numerically. OK, everybody see that? So here are our two
natural frequencies. Here's the first one-- excuse
me, that's not right either. That's the square. OK So these are our natural
frequencies, once again for this special case
where the masses are equal and the springs are equal. Anybody recognize
that number, 0.618, for all you fuss budgets? Ring any bells? Any number freaks here? No? I heard it. That's it. Exactly, nice job. The golden mean,
the golden ratio. Also, let me simply
say if there are-- as far as the number
of things-- if there are n degrees of freedom. There are n natural frequencies. What else? So that's that. So now it's time to get to this
notion of the natural modes. Let me say, we've got
to go all the way back to this set over here. If you take the first row
of this matrix equation-- that's the first of the
equations of motion-- and you make that assumption of
the harmonic motion in there. Does everybody see that? What we've done is we've
taken basically the first row of that expression
right up there and formed the amplitude
ratio A1 over A2. What we're doing is
we've found the omegas. You remember, just
review the bidding. Our original
assumption was harmonic motion, that is to say
all the displacements are moving in synchrony as it were. The same sinusoidal
frequency, we've just found what
frequencies those are. There are two of them,
and they're right there. Now we're after these guys. Now we're after the
relative magnitudes or the relative
amplitudes of A1 and A2. And we from one of the
equations isolated one of those. And let me just say, if
you plug these back in, plug in the first one, you'll
get oddly enough 1.618, These amplitude ratios. Are the so-called natural modes. And I think you can appreciate
that this is the first one, and this is the second one. Any questions so far? Wonderful. Hearing none. Yes ma'am, Sara? AUDIENCE: [INAUDIBLE] PROFESSOR: You see
this amplitude ratio. You saw how we got that. You see that the right hand side
has got system parameter, Ks and Ms, and stuff like that. But this is the ringer, omega. This amplitude ratio
is expressed in part in terms of omega. So what omega-- there's no
ambiguity as to Ks and Ms, but what omega? Well the answer is, when
we plug in this one, you get this answer. When you plug-in this
one, you get this answer. So while we're at it-- Sara,
want to hazard a guess? How many natural mode
do you think we've got? Yeah, exactly. So you're going to
have one of these for each degree of freedom. Let me just point out a
couple of elements here, and then I'll show you a
demonstration because we have to have some fun today. These are point of informations. They're ratios, not
absolute magnitudes. That's number one. The second is-- I already told
you, they got the same number. OK, each natural
mode is associated with a particular
natural frequency. This one goes with that one. This one goes with that one. And once again, they're
associated with-- yeah, let me say that. I need another board. Let's just go over here. So in a sense--
this is decouple. Decouple essentially into
independent subsystems. So in general, what the system's
response looks like is-- I'm talking about the
one in front of us here. This special case, where the
masses and springs are equal. I think there's a
minus sign in here. Does this come up? Wonderful. That's it. Does everybody see that? Yes, sir. AUDIENCE: [INAUDIBLE]. PROFESSOR: I'm sorry? AUDIENCE: What does it say
under that first bullet point? PROFESSOR: Here? AUDIENCE: These
describe the situation. PROFESSOR: These
describe the situation in which the entire
system is oscillating at. It's the second bullet here. Thank you. AUDIENCE: At what? PROFESSOR: At one frequency. Sorry, I'm just getting a
little tired of writing. So, any other questions,
problems, complaints? All right. AUDIENCE: I have a question. PROFESSOR: Yes, sir. AUDIENCE: [INAUDIBLE] PROFESSOR: That's correct. Then, let's see. Then there's a
mistake right here. Thank you. Yeah, because that's
the way it came out. When you plug omega 2
having this value into here, the amplitude ratio
comes out minus. Fair enough? It threw me there for a minute. I thought you were going to say,
why is the minus sign is there, rather than you could have
had minus 1.618 and plus 1. And the answer is no reason,
because these are ratios. Yeah, Kaitlin? AUDIENCE: But shouldn't-- when
we go back and look at what you wrote down, it's [INAUDIBLE]. I don't understand
how that [INAUDIBLE]. PROFESSOR: I'm sorry, say again? AUDIENCE: Never mind. PROFESSOR: Find it? Yeah, they're ratios, It's
just as simple as that. So you multiply them by any
number and it still works. I'll actually show
you here in a second. At least, I believe
that's the case. We'll just see here in a second. OK, questions? Comments? All right. Now is the time. Could I bring up
the side board here? Let me show you--
anybody here taken 2086? Wonderful. I've got one person? Great. Anyway, I believe in 2086,
don't they teach you MATLAB? Isn't that the weapon of choice? OK, that's the program
I'm using here, MATLAB. For those of you who
haven't seen it yet, it is definitely a mixed bag. I don't know how
you feel about it. It's very-- yeah--
it's very powerful. It stands for Matrix Laboratory. It was written, I don't
know, 20, 30 years ago here, I believe, at MIT by people who
were into matrices, into matrix algebra. And it's kind of
command line oriented. The good news,
it's very powerful. Whatever you want to do,
you can do in MATLAB. The bad news is, the
user interface stinks. The language is very
difficult to learn. It's even harder to remember. So with that
rousing endorsement, let me show you
what we've got here. This is a program I've-- is
that font readable by you guys? No? No? AUDIENCE: [INAUDIBLE]. PROFESSOR: I'm sorry? AUDIENCE: [INAUDIBLE]. PROFESSOR: Yes. Well, I believe I can--
here we go, fonts. Upping the fonts is
kind of a mixed bag, because you get bigger
letters but they're. OK, how's that? So here's the situation. This is a MATLAB program. And I'll explain to you
what it does as we go. Let me see if my little
cursor-- my cursor's here, but I can't see it. All right, here's the
system parameters. Once again, we're doing
a simple spring mass-- this simplified spring mass
system, exactly the one we've done here. When I wrote it, you'll see I
generalized it to do this guy. So we got M1 and M2, K1 and K2. But if you'll notice, you see
here their values are equal. We've got the mass
at one kilogram each. And we've got 10 newtons per
meter on each of the springs. Everybody appreciate that
this system's numbers that we're putting in here
match our case here, K over M? OK. And you can see
here, we've defined-- and again, let me
just say I'm not trying to sell you on MATLAB. I don't want to leave
you with the impression that we expect you to be
able to instantly become a user of MATLAB. This is simply to illustrate
the point of the lecture here. Here is the M,
the system matrix. There's the K matrix. And I'll show you the eigenvalue
and eigenvector thing later. But let me-- take
my word for it. See this here? Ode45 is a cryptic allusion
to the Runge-Kutta algorithm, fourth order Runge-Kutta that
is the workhorse for integrating differential equations. And so let me just run this. And what I've got here
is, here's the point I wanted you to get
here, because I'll bet you can't see
that cursor either. Yes, anyway, see
this right here? tspan is the time scale and
the time step, defined up here. But these are basically
the initial conditions. See it here? X1, X1 dot, X2, X2 dot. So here's the first one. This is a 0.618 is for the X1. And 1 is for X2. Everybody appreciate that? Got it? OK. If these are the initial
conditions, what I've done, I have artfully chosen
the initial conditions to have the same ratio. What do you expect
is going to happen? When I turn this thing--
I've got a simulation here. I'm going to run this,
and you're actually going to see it. What do you think
you're going to see? It's a two spring,
two mass system. What I've done is I've
displaced the two masses. AUDIENCE: [INAUDIBLE] PROFESSOR: They'll certainly
have an amplitude, because I'm putting it in there. That's the initial condition. The question is, what frequency
you think they'll oscillate at? AUDIENCE: [INAUDIBLE] PROFESSOR: Pardon? AUDIENCE: [INAUDIBLE] PROFESSOR: Each of
them will oscillate with the same frequency,
for sure, but what do you think it's going to be? AUDIENCE: That one. PROFESSOR: It's
going to be that one. So off we go. So let us hope that yours
truly's program worked. Here we go. Oh, look at that. [INAUDIBLE], please
interpret that for me. What do you see there? Hang on a second. Let me blow it up
so you can see it. Ooh, isn't that pretty? And I believe the blue is
X1 and the green is X2. See? Everybody agree? Everyone appreciate
what's going on? You pull them both at
slightly different-- you basically used
the first natural mode as the initial condition. And sure enough, they
oscillate together. They oscillate at
the same frequency. They oscillate at
that frequency. Let me just see--
I just want to make sure we get the full
value out of this thing. Well, of course you
can't see it anymore because our numbers are so big. Well, that's great. Anyway, take my word for
it at-- oh, here it is. The period for the first
natural frequency-- or I guess it's the second--
it should be like 1.2. Or is it 3? Yeah, I'm sorry,
the period is 3.2. And sure enough, there it is. It's about 3. 3.2. Fabulous. Everybody got it? OK, now watch closely. Let me see if I can do this. This requires a little
dexterity, which is always a short supply here. I have to hit this and this. Make sense? That's what it
actually looks like. They're both oscillating at
the same natural frequency, going up and down together. But they have
different amplitudes. So one's bigger than the other. So that's what it looks like. Questions? Christina, you good? Clear enough? Wonderful. So let's go to our program. And instead of that set
of initial conditions, we'll do the other. Read them to us here. What are the initial
conditions here? AUDIENCE: It's 1.618. PROFESSOR: That's
right, it's this guy. AUDIENCE: That guy. PROFESSOR: It's this guy. It's this ratio. So I basically
arbitrarily chose, is it the negative first? No. I chose that one
over there, 1.618. And then a minus 1
for the second one. Fair enough? OK, there it goes. We've got to save it
and make sure we got it. So again, you got a clue
what's going to happen here? Here we go. Boom. Look at that. What's going on there? Yikes. Explain me. Is that good, bad, indifferent? Is it right? Wrong? AUDIENCE: The way
the system acts, it has a higher frequency. PROFESSOR: Yeah, exactly. Two things. One is, they're out of phase. They're doing this. One's going this
way, and the other's going the other at
different amplitudes but the same frequency. But the frequency in question
is higher than the previous. AUDIENCE: Why are
they out of phase? PROFESSOR: I'm sorry? AUDIENCE: Why are they
opposite of each other? PROFESSOR: [INAUDIBLE], why
are they opposite each other? Because we made them that way. We said, that's the
initial condition. Does that make sense? That minus sign does. One starts out,
and one starts in. And they do that. Clear enough? Now what's going to happen
if we plain just choose any old initial condition? These were special. We worked like a dog
to compute these, so that the system would
decouple in that way. So what if we-- now
let me put that back. Now look at this one. All right, look at that. Read that to me. AUDIENCE: [INAUDIBLE] PROFESSOR: Yeah. So that says the initial
condition for the first mass is 1 and whatever that is. One whatever that is. The second masses'
initial condition is half that in the same direction. Both positive. So Christina, they're
going to go together. But [INAUDIBLE],
at what frequency? Any idea what it's
going to look like? If you do, you're a
better man than I, because what you're going
to see here is that. It's this thing right here. It's that expression
right there. And here's what it looks like. Did I stop the--
oh, wait a minute. Did I ever show you that before? I think I forgot to
show you the other. Anyway, not to worry. Hang on a second. I've got to stop this guy. First I have to find my finger. There it goes. That's the previous case. When they're out of phase,
different magnitudes, going in opposite directions. And you can see, they're
going at a higher frequency than before. Make sense? So now we are-- just to
refresh your memory-- now we're going for the
third case, in which there's nothing special. We just picked a couple
of initial conditions out of a hat. And here we go. Oops, I think not. I think that's
the previous case. So let's go here. This is another wonderful
thing about MATLAB is nothing happens
until you save it. So we were just running
the previous case. Nasty. Look at this. All right, can
everybody see that? If you can interpret this,
you're smarter than I am. But what this is, this is simply
this expression over here. It's this expression
for just some arbitrary initial condition. Do you see that that
behavior though? Each of them, they're
going together kind of, but they-- anyway, watch this. Here's what the simulation
of that looks like. What the heck is that? Well anyway, the
point of the story is that multiple
degrees of freedom system in general's response
can be arbitrarily complicated. It's not arbitrarily
complicated, but pretty complicated. You'll get, in general if it's
an nth order system, if you don't know anything
about the worst case, you'll see four
frequencies in there. And they're all mixed together
in some mystical way that's unknown to you. Fair enough? And it's only when you reach the
natural modes that you actually find out what is going on here. Well now I have to turn
your attention to this guy. This is made by Professor
Vandiver's machinist, a perfect example of
a second order system. And I bring it to
your attention here for two-- at the end of the
day what we're going to do is I'm going to demonstrate
exactly what I just did for the textbook
case, the textbook system. I want to demonstrate exactly
the same thing for this guy, only this is a real system. Very nice. We have a steel rod. It must be a half
inch in diameter. The whole thing
weighs several pounds. These sliding masses are right
circular cylinders with a hole drilled through them. It's ever so slightly
larger than these here. They're of different lengths. They're made of brass. They're serious masses. And the springs,
which extend from here to here, and from here to
here are wound on a lathe, and attached, and so forth. Pretty, no? Now look right off the bat. Did you see how
that thing operates? Would you agree you have some
complicated behavior here? Now also would you agree
that this is it like that? Everybody see that? Before we go too far, this
is a mixed message here. [INAUDIBLE], is this
exactly like that? In what way is it
similar to that? AUDIENCE: [INAUDIBLE] PROFESSOR: Well, what is clear
is that you've got two springs and you've got two masses. About that there is
very little argument. AUDIENCE: It's damped. PROFESSOR: It's damped. Can everybody see that? How does [INAUDIBLE]
know that it's damped? How's he know it's damped? I mean, that's just a wild
guess on his part, but. AUDIENCE: You can hear
it, and it slows down. And it slows down. This is the most important
part is that it stops. Eventually if you come back
in a minute or two, it's done. PROFESSOR: All
right, [INAUDIBLE]. You're on a roll. There's definitely
damping there. What kind of damping? AUDIENCE: Friction. PROFESSOR: Friction, yeah. Does that have another
name that you can think of? It's definitely friction. What it's not is
viscous friction. What it is not is a damper or
a dashpot which we've shown you before with the ideal expression
that generate a force that opposes the-- generation of an
opposing force that's linearly proportional to the velocity. What's going on
here, do you think? What kind of damping
do you think? It's called Coulomb. This is called Coulomb damping. And this is a digression. Now we're on the part where this
is really-- everything that's on the board is what I
wanted you to really come away from today with. So now we're out
kind of in the, I would call it the
winging it area right here, because this
is the part where I simply had fun with the demo. This is viscous. And this has got the
symbol-- well anyway, this is what it looks like. And this is the
force of the damper. We'll call it B. And this is the velocity. And this is for constant
of proportionality B. And it has this little
symbol, like that. And when equations of motion
are solved that contain that, the response looks like this. What we're talking about
here, the force put out is a constant. That just comes
from the sliding. And what it generates are
distinctly non-linear equations of motion. And what you get here is you
get this kind of behavior. If you really looked
at it, what you'll see is there's definitely
damping for large motions when the inertial forces and
so forth are large compared to the friction forces. It'll look a lot like
conventional viscous damping. It's just that when
motions get really small, and the forces get down there
to on the order of this, all a sudden you'll see on
one cycle it'll just stop. And were you up here
where you could see, or if we had a closeup of this--
you can't see it, but just watch this thing stop. Right there. Do you see that? That's a little hard for you
to see from there, but watch. Anyway, were you up
here, you'd see this. That's what we're looking at. Well here we go. I need some help here. Who's in a volunteering
frame of mind? Amy, all right. I appreciate the help here. Here's what I want to do. We just blew out some
wonderful theory. All this is just
solid as a rock. Yes, sir. AUDIENCE: For the Coulomb
friction, is that a linear [INAUDIBLE]? Or is it still exponential? PROFESSOR: I'm sorry? Oh, no. If I'm not mistaken, I
didn't really look this up, but I believe it's linear. I'd have to-- take that
with a grain of salt, but I believe it's linear. Yes, Amy, here's the situation. We have all this
marvelous theory. My goal is to-- and we have this
fabulous demo apparatus, though inherited. And what I'd like to do-- oh,
and we have computational means to. And in fact, we just went
through the exercise. We already know those
same equations that work for this work for this. Those are general. However, it's not my
piece of apparatus. And well, here's the deal,
what are the Ms and Ks. What are the values of-- I need
M1, K1, M2, K2 to put into the. AUDIENCE: [INAUDIBLE]. PROFESSOR: Yeah, Yeah. That's what I'd like
to do is I'd like to. AUDIENCE: Do I have
to just determine the it by looking at it? PROFESSOR: Oh no, no, no. No, no. AUDIENCE: I'm not that good. PROFESSOR: You're my assistant. I guess the question
is, how would you-- and I've got
to tell you, that's the math part of a program. Now we're in the engineering
part of the program, because somebody gave you
a real live demo apparatus. Works like crazy, or appears to. And I'd love to take
advantage of it, but I don't know
any of the numbers. AUDIENCE: [INAUDIBLE] PROFESSOR: No. That's the constraint
I'm operating on. It doesn't belong to me. I mean, I could take it apart. That's an absolutely
appropriate thing to do. I would have liked to. It would be easier if you could. You just go, take a
screwdriver to it. Here, put this on there
and pull this out. I didn't have the luxury
of any of that, so what's your next best suggestion? Nice suggestion, but no cigar. I'm sorry? AUDIENCE: Take it apart
anyway, put it back together before the person notices. PROFESSOR: Well yeah. Yeah, no. That's fudging. Yeah, they notice. Have you ever taken
apart anything made in modern manufacturing method? Oh, it's good because you
can't put them back together. They're assembled by machine. And once upon a time you
could disassemble one and reassemble things
without detection. But anymore, once
you take them apart, it's wicked hard to
get them back together. OK, the floor is open. I need another suggestion. What are you going to do? AUDIENCE: Do you need to
know the exact K and M, or do you just
[INAUDIBLE] another ratio? PROFESSOR: I thought
you were going to-- I need to know M or
K. I need to know them all. AUDIENCE: Do you know
the density of the-- PROFESSOR: I was going
to say, but I don't need to know anything exactly. All I need to know is as good a
guess as you can come up with. It's all an estimate. AUDIENCE: If you know the
density of the material, you can easily work
up [INAUDIBLE]. I'm assuming you're
about [INAUDIBLE]. PROFESSOR: Oh
absolutely, absolutely. AUDIENCE: [INAUDIBLE]. PROFESSOR: Wonderful. Absolutely. She hit the jackpot,
rang the magic buzzer. That's exactly what I did. Here's a little crummy sketch. Oh wait, you can't see that. Anyway, these are right circular
cylinders with holes in them. And they've got
measurements beside them. I can tell you, this
is 75 millimeters. This is 35 millimeters. This one is 37 millimeters long. So I did that. That's great. That's an excellent suggestion. And after I did exactly that,
I won't write out the formula. You know area, and
volume, and all of that. Let me get you the
right order here. M1 is 0.2929, and M2 is 0.5938. Everybody got that? This was obtained by taking
a ruler to these things, taking diameters, lengths,
and diameters of holes, multiply them times the density
of brass taken out of the book. Do you believe that? Do you believe that number? Well, you're a trusting soul. I don't. To me, I believe that number. This was done with a ruler. The little millimeter thingies. So I just say, don't fall in
the trap of false precision. OK Amy, you're on a roll. We've got the masses. What now? AUDIENCE: Free body diagram. PROFESSOR: Yeah,
we got all that. AUDIENCE: Yeah, you've got that. But then what you can
do for the spring, the forces of the
spring when static. Don't move it. Don't move it. So take the top mast. It's not moving,
which means that you know that the force going
upwards-- which is the spring-- is equal to the force going
downward, which is [INAUDIBLE]. So you can measure
the displacement from the start of the spring
to the bottom of the spring. Do you know the natural
length of the spring? PROFESSOR: No Anyway,
what I was going to say is, excellent idea. Can't do it. But what Amy was
basically saying is, you know the masses now. Why not simply take
from that expression right there, MH over K, right? What's the problem with that? How about it? Devin, how come I can't do that? AUDIENCE: [INAUDIBLE]. PROFESSOR: Yeah, I really don't
know the no load position. Is that it, Amy? Right there? Maybe. Devin, you're the one
that suggested it. Are those the no load
positions of the masses? And if not, why not? I did mean to give
you a clue there. Yeah, Nick? AUDIENCE: It can't be because
there's static friction. PROFESSOR: Exactly. So Nick, you brought it up. What's that number? You don't know that either. No. Like I said, Amy, nice idea. No cigar. What else? We're running out of time? Here we go, Douglas. AUDIENCE: Could you
displace each mast a certain [INAUDIBLE], and
then measure the time it takes for them to stop
and get the damping ratio? PROFESSOR: Hit
the damping ratio. Well, I'll tell you what--
you want to say that again? He said displace one or
both count oscillations and get the damping ratio. Nick? AUDIENCE: Do we have anything
like a force gauge or a spring scale? PROFESSOR: No. This is my living room
I'm talking-- or my study. Anyway, Douglas said-- I
forgot what you said now. He said-- oh, I know. You said, displace it and
count the oscillations. Get the damping ratio. First off, the damping ratio is
no help, even if we did get it. And the only formula for
which we've ever given you-- the only formula
we've ever given you to do that with pertains
to this kind of friction, which is not present. No cigar. Nick says, how
about force gauge? Now, don't have it. AUDIENCE: [INAUDIBLE]. PROFESSOR: Yeah, we are. AUDIENCE: [INAUDIBLE]
displace the other one and then find the frequency. PROFESSOR: Oh, what's your name? Sean? Or John? AUDIENCE: Sean. PROFESSOR: Say that out loud. Say it loud enough that
Devin can hear you. AUDIENCE: You hold
the first mass, and then you displace
the second one. PROFESSOR: Hang on. He says, hold the first mass
like this set screw right here. And? AUDIENCE: And then displace
the other one then. PROFESSOR: Like that? AUDIENCE: [INAUDIBLE]. PROFESSOR: Actually,
to answer your question Nick, the only instrument
I have is a clock. Hang on a second. Here it is. Of course, this is the
big task is finding it. AUDIENCE: [INAUDIBLE]. PROFESSOR: Count to 10,
remember like Vandiver told you. Skip 1, 1, 2, 3, count to 10. Stop. Excellent, excellent, excellent. When you do that,
that's the second one. I did that and right
here, right here it's TP. The period of 10 of them--
and then I divide to get 1-- is 0.83 seconds. What does that tell you, Sean? AUDIENCE: [INAUDIBLE]. PROFESSOR: Well
actually, these two-- you can get the frequency. But what this does, because
this gives you the frequency, you know in general
that-- in particular, you know that this second natural
frequency, which is just associated with this single
spring and a mass here. It's just this guy. It's not both of them. Anyway, this turns out
to-- I didn't graph that. That's square root of
K over M. Trust me, you can put those two
together, and you get K2 is equal to-- newton meters. This is exactly what you said. Freeze the first mass. Displace the second. Measure the period. You get the natural frequency
for basically K2 over M2. So you got K2 out of it. Yeah, Douglas? AUDIENCE: So how come it gives
you just the natural frequency and a damp natural frequency? PROFESSOR: Oh no. It absolutely is. It's all damped, no
question about it. But again, what
we're doing is we're going close enough, right? Because I have nothing. So even the damped
natural frequency is better than nothing. All right, so Amy back to you. You're back in business. What now? So now we've got M2 and K2. AUDIENCE: You just need K1. PROFESSOR: Yeah. Now we need K1. What do we do now? AUDIENCE: We want to do
the same thing that we just did for K2. [INAUDIBLE]. PROFESSOR: Exactly. Well, not quite. Let's see, now I'm going
to turn loose-- now we're back to our original system. It doesn't hurt anything. It's just ugly to look at. Now what? Sean, do that trick again. AUDIENCE: [INAUDIBLE]. PROFESSOR: Push this one up? AUDIENCE: [INAUDIBLE] PROFESSOR: I don't think so. Yeah, Nick. AUDIENCE: So just
fix the second mass. PROFESSOR: Fix the second mass. AUDIENCE: [INAUDIBLE] PROFESSOR: Yeah, you
see this frequency here? I'm going to overrule Chandler. I'm going to say, this is the
natural frequency of that's mass and these two springs. Do that same trick again. You get the equivalent
spring rate, subtract the second from it,
and you get the other one. Devin, is that what
you were going to say? Wonderful. That is exactly what I did. And you get out
of it, you get K1 is equal to 50.45
newton per meter. In the interest of time, I'm
going to short circuit this. I took exactly
these parameters, I put them into that same
computer program we had before, and what came out-- I have
to have a place to put it. Ah, wonderful. And now I have to find it. Here we go. Here are the two modes. Actually, let me
put it right here. For this system, because
here's the first one. And here's the second one. 0.9760 and minus 0.2177. Everybody appreciate that? This is by the same
computational procedure we spoke of earlier. And sparing no expense,
we have here a, made fresh from my basement, a custom
made initial condition setting device, which I can hopefully
avoid killing myself with. OK here's what we have. So I guess I didn't
show you this first. If you can see it,
what we have marked is the reference position. That's the rest position
that we couldn't find by laying it down. Or excuse me, this is the
static equilibrium position of mass number one,
static equilibrium position of mass number two. Mode one or there. What they said over there, 0.4. And .97 is down here. Mode two is over here. So Devin, while I'm
doing this, tell me how am I going to
know if this is right or if this is all just bogus? What observable's going to
tell me that I got it right? AUDIENCE: [INAUDIBLE] PROFESSOR: I'm sorry, speak up. AUDIENCE: [INAUDIBLE] PROFESSOR: How about it, Nick? AUDIENCE: [INAUDIBLE] PROFESSOR: Exactly. And you ought to be able to
see it from where you are. Can you appreciate that
they're not at the moment? All right, now hang on. Here comes mode number one. This takes two hands to do it. All right, you ready? This is mode number one. Again, it's a those numbers. How about it? Can you see it? Very good. How about the other? And so here, this is number two. And this is a little
more complicated, because the other one has
to be done from the bottom. Hang on a second. Now this one we're doing
is we're deflecting-- this one is positive downward. So X1 is down, but
X2 is negative. So it's displaced upward a bit. Are you ready? Nick, what do you
expect to see this time? AUDIENCE: The frequency
should be higher and they'll move in
opposite directions. PROFESSOR: That's the key. Once again, they're going to
move with the same frequency, albeit in different directions. But that new frequency is
going to be higher than before. And sure enough, stand back. So there you go. What did that tell us? That told us that
the first order we got the system parameters
identified correctly and the theory holds up. Questions? Comments? Complaints? Devin? AUDIENCE: [INAUDIBLE] PROFESSOR: I'm sorry? AUDIENCE: What was the
second set of conditions? PROFESSOR: The second
set of initial conditions were right here. This is the second mode,
X1, 0.97, X2, minus 0.2. OK, have a great
Thanksgiving holiday.
