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PROFESSOR: So Professor Gossard
gave the lecture last week.
00:00:24.680 --> 00:00:26.990
I'm going to pick up
where he left off.
00:00:26.990 --> 00:00:31.260
But let's talk about the concept
questions from the homework
00:00:31.260 --> 00:00:33.200
you've been working on.
00:00:33.200 --> 00:00:35.850
So the first one is our cart.
00:00:35.850 --> 00:00:38.520
You'd expect to be able
eliminate the terms involving
00:00:38.520 --> 00:00:41.070
gravity in the
equations of motion
00:00:41.070 --> 00:00:43.090
by choosing coordinates
with respect
00:00:43.090 --> 00:00:45.180
to the static
equilibrium position.
00:00:45.180 --> 00:00:47.060
So we've talked about that.
00:00:47.060 --> 00:00:53.000
And with this one does
the restoring force
00:00:53.000 --> 00:00:56.200
on the pendulum, what
makes it come back
00:00:56.200 --> 00:01:01.510
to zero after it's a damped
out and hanging straight down.
00:01:01.510 --> 00:01:02.260
AUDIENCE: Gravity.
00:01:02.260 --> 00:01:02.926
PROFESSOR: What?
00:01:02.926 --> 00:01:04.349
Gravity does.
00:01:04.349 --> 00:01:09.760
And that gravity term
varies with the torque
00:01:09.760 --> 00:01:17.780
that the gravity puts around the
pivot is MgL over 2 sine theta.
00:01:17.780 --> 00:01:21.930
So the gravity term is involved
with the motion variable theta.
00:01:21.930 --> 00:01:24.080
So in this case,
gravity is going
00:01:24.080 --> 00:01:27.270
to be involved in
the natural frequency
00:01:27.270 --> 00:01:30.420
and in the equations of
motion, no matter what,
00:01:30.420 --> 00:01:36.090
So you will not be able to
eliminate the gravity terms.
00:01:36.090 --> 00:01:41.220
Next, this one, this is a
vibration isolation question.
00:01:41.220 --> 00:01:44.430
Will the addition of
damping increase or reduce
00:01:44.430 --> 00:01:46.570
the vibration of the
table in response
00:01:46.570 --> 00:01:49.520
to the floor motion at 30 Hertz?
00:01:56.490 --> 00:02:00.480
I guess this depends on what the
natural frequency of the system
00:02:00.480 --> 00:02:00.980
is.
00:02:00.980 --> 00:02:03.300
But we're trying to do
vibration isolation.
00:02:03.300 --> 00:02:05.310
And presuming, if
you read the problem,
00:02:05.310 --> 00:02:09.880
you're supposed to
find a a stiffness such
00:02:09.880 --> 00:02:15.060
that you can reduce the
response of the table by 12 dB,
00:02:15.060 --> 00:02:19.460
I think it said, from
the motion of the floor.
00:02:19.460 --> 00:02:23.010
So that's something
substantially less than 1.
00:02:23.010 --> 00:02:31.240
And you will be-- this one
takes a, best described
00:02:31.240 --> 00:02:32.040
with a picture.
00:02:35.650 --> 00:02:42.760
The transfer function
for response of the table
00:02:42.760 --> 00:02:44.985
over the motion of the
floor, the magnitude
00:02:44.985 --> 00:02:46.360
of that transfer
function, that's
00:02:46.360 --> 00:02:49.350
just the ratio of x to y.
00:02:49.350 --> 00:02:55.560
That looks like this,
if the damping is zero.
00:02:55.560 --> 00:03:01.150
And as you add damping, all
points cross right here.
00:03:01.150 --> 00:03:03.450
Some damping does that.
00:03:03.450 --> 00:03:05.800
More damping because of this.
00:03:05.800 --> 00:03:13.260
And in order to accomplish
what's been described--
00:03:13.260 --> 00:03:16.380
this is 1.0 here.
00:03:16.380 --> 00:03:19.580
If you're trying to make
this table respond less
00:03:19.580 --> 00:03:22.710
than the floor, you must
be somewhere out here
00:03:22.710 --> 00:03:24.490
where you're below 1.
00:03:24.490 --> 00:03:27.970
So this is omega over omega n.
00:03:27.970 --> 00:03:30.720
And right here at
resonance, you're at 1.0.
00:03:30.720 --> 00:03:34.200
So this is, you know, two or
three or four for this value
00:03:34.200 --> 00:03:35.210
out here.
00:03:35.210 --> 00:03:38.470
Let's say here's where
you find the answer to B.
00:03:38.470 --> 00:03:40.730
And without damping,
you're there.
00:03:40.730 --> 00:03:44.570
And that's 12 dB down.
00:03:44.570 --> 00:03:47.620
If you add damping, it
pushes you up these curves.
00:03:47.620 --> 00:03:52.830
Does make the response larger,
the undesirable response,
00:03:52.830 --> 00:03:55.080
the motion of the
table larger or smaller
00:03:55.080 --> 00:03:59.885
as you add damping at
that operating point?
00:03:59.885 --> 00:04:02.070
It increases it, right?
00:04:02.070 --> 00:04:05.150
OK, so in this case, will
the addition of damping
00:04:05.150 --> 00:04:06.540
increase or reduce
the vibration?
00:04:06.540 --> 00:04:07.630
It'll increase it.
00:04:07.630 --> 00:04:10.080
But damping's a necessary evil.
00:04:10.080 --> 00:04:12.009
You need some damping
in the system.
00:04:12.009 --> 00:04:14.550
So if you bump it, it doesn't
sit there and oscillate all day
00:04:14.550 --> 00:04:15.049
long.
00:04:15.049 --> 00:04:16.760
Next.
00:04:16.760 --> 00:04:17.960
OK, this is a platform.
00:04:17.960 --> 00:04:19.459
Do you think I could
actually do it?
00:04:19.459 --> 00:04:21.250
Did you read this?
00:04:21.250 --> 00:04:24.440
So this is a Coast Guard
light station off of Cuttyhunk
00:04:24.440 --> 00:04:27.190
down off of Woods
Hole-- basically,
00:04:27.190 --> 00:04:30.840
I was doing this-- in time with
the motion of the platform.
00:04:30.840 --> 00:04:32.750
Resonance is a wonderful thing.
00:04:32.750 --> 00:04:35.230
If you can make
the force be right
00:04:35.230 --> 00:04:37.070
at the natural frequency
of the structure,
00:04:37.070 --> 00:04:39.530
it actually doesn't
take a lot of force
00:04:39.530 --> 00:04:42.950
to drive the amplitude to
pretty large amplitudes,
00:04:42.950 --> 00:04:44.690
if the damping is small.
00:04:44.690 --> 00:04:48.080
So I think the damping in
this cases is about 1%.
00:04:48.080 --> 00:04:50.250
And that means
the amplification,
00:04:50.250 --> 00:04:53.020
the dynamic amplification
1 over 2 zeta is about 50.
00:04:53.020 --> 00:04:54.970
So I actually could do this.
00:04:54.970 --> 00:04:56.190
This is a true story.
00:04:56.190 --> 00:04:58.566
OK, next.
00:04:58.566 --> 00:05:00.190
For small motions
about the horizontal,
00:05:00.190 --> 00:05:04.670
you expect the natural frequency
to be a function of gravity.
00:05:04.670 --> 00:05:09.330
So this is, oh, some of
you, about equal, yes no.
00:05:09.330 --> 00:05:13.680
But it's just horizontal, the
torque that gravity provides
00:05:13.680 --> 00:05:18.170
is some Mg pulling down
on the center of mass
00:05:18.170 --> 00:05:20.330
somewhere in that body.
00:05:20.330 --> 00:05:24.950
Not at the pivot, but let's
say some distance A away.
00:05:24.950 --> 00:05:27.630
So the torque, the gravity,
the restoring torque
00:05:27.630 --> 00:05:34.940
is some Mg cross r cross Mg,
Mgr if r is the distance.
00:05:34.940 --> 00:05:36.330
And that's the torque.
00:05:36.330 --> 00:05:39.400
And that length of that
moment arm might vary.
00:05:39.400 --> 00:05:42.230
It's going to vary
like cosine theta.
00:05:42.230 --> 00:05:46.790
Around horizontal line, if theta
is what's, for small angles,
00:05:46.790 --> 00:05:48.612
what's cosine?
00:05:48.612 --> 00:05:49.990
It goes to 1.
00:05:49.990 --> 00:05:52.550
So you find out that
this just looks like Mgr.
00:05:52.550 --> 00:05:54.190
It's for small
angle of vibration.
00:05:54.190 --> 00:05:56.070
And you can in fact
get it out of-- it
00:05:56.070 --> 00:06:01.672
doesn't enter into the equation
for the natural frequency.
00:06:01.672 --> 00:06:03.130
So the natural
frequency of a thing
00:06:03.130 --> 00:06:08.960
won't be a function of gravity
because of this small angle
00:06:08.960 --> 00:06:12.540
vibration around a
horizontal point.
00:06:12.540 --> 00:06:13.754
OK.
00:06:13.754 --> 00:06:15.420
One, when the
acceleration of the system
00:06:15.420 --> 00:06:19.220
is one half that required to
make the mass slide, what's
00:06:19.220 --> 00:06:21.555
the magnitude of
the friction force?
00:06:24.180 --> 00:06:28.020
So friction is one of
those things that is only
00:06:28.020 --> 00:06:30.670
as big as you need it to be.
00:06:30.670 --> 00:06:34.470
So even the largest friction
that this thing can sustain
00:06:34.470 --> 00:06:41.640
is in fact mu mg--
answer A. But f equals ma
00:06:41.640 --> 00:06:43.910
if the acceleration
is half of what
00:06:43.910 --> 00:06:47.150
is required to have
that thing be just slip.
00:06:47.150 --> 00:06:56.980
It will just slip when you
are at a force which is mu mg.
00:06:56.980 --> 00:07:02.430
And so that force is equal
to mass times acceleration.
00:07:02.430 --> 00:07:05.190
The acceleration then
you can figure out
00:07:05.190 --> 00:07:07.100
what that will be
just when it slips.
00:07:07.100 --> 00:07:09.211
But now if you reduce
acceleration to half that,
00:07:09.211 --> 00:07:11.710
the friction force required to
keep it in place is only half
00:07:11.710 --> 00:07:13.854
as big.
00:07:13.854 --> 00:07:15.270
And it will be
that friction force
00:07:15.270 --> 00:07:17.830
and it will be half of mu mg.
00:07:17.830 --> 00:07:19.370
So it's actually B.
00:07:19.370 --> 00:07:21.110
And next-- is that it?
00:07:21.110 --> 00:07:21.910
No.
00:07:21.910 --> 00:07:24.350
OK.
00:07:24.350 --> 00:07:30.890
This is a simple but
actually sometimes
00:07:30.890 --> 00:07:33.310
hard to see through question.
00:07:33.310 --> 00:07:35.710
What initial conditions
will be required?
00:07:35.710 --> 00:07:38.060
This problem can be solved
by initial conditions.
00:07:38.060 --> 00:07:42.100
This mortar launches its shell.
00:07:42.100 --> 00:07:45.980
And the trick to this question,
the key to this question
00:07:45.980 --> 00:07:48.750
is for your mathematical
model of the system--
00:07:48.750 --> 00:07:52.070
your equation in motion-- is
write the equation of motion
00:07:52.070 --> 00:07:52.975
without the shell.
00:07:55.500 --> 00:07:58.760
Because once it shoots this
thing, the shell's gone.
00:07:58.760 --> 00:07:59.620
And it's vibrating.
00:07:59.620 --> 00:08:02.840
It's now a system without
that 25 kilogram mortar
00:08:02.840 --> 00:08:04.210
shell part of the system.
00:08:04.210 --> 00:08:05.810
It's gone.
00:08:05.810 --> 00:08:07.580
Now it's just the
mass of the system
00:08:07.580 --> 00:08:09.790
without the mortar shell.
00:08:09.790 --> 00:08:11.470
And there are two
initial conditions
00:08:11.470 --> 00:08:15.672
that then you can say when you
shot the mortar shell, that was
00:08:15.672 --> 00:08:16.880
a certain amount of momentum.
00:08:16.880 --> 00:08:19.150
And from conservation of
momentum you can figure out
00:08:19.150 --> 00:08:24.400
what the momentum of the
main mass has to be-- equal
00:08:24.400 --> 00:08:26.850
and opposite to
the shell you shot.
00:08:26.850 --> 00:08:29.240
So that gives you
an initial velocity.
00:08:29.240 --> 00:08:32.770
But there's also an initial
displacement in this problem.
00:08:32.770 --> 00:08:34.510
So that's the key to
figuring this out.
00:08:34.510 --> 00:08:37.860
So what initial conditions
would be required?
00:08:37.860 --> 00:08:40.260
And it's C-- both
an initial velocity
00:08:40.260 --> 00:08:42.010
and an initial displacement.
00:08:42.010 --> 00:08:44.940
But the key is to make
your mathematical model
00:08:44.940 --> 00:08:48.630
about the system
without the shell.
00:08:48.630 --> 00:08:49.440
OK?
00:08:49.440 --> 00:08:50.408
Good.
00:08:50.408 --> 00:08:51.344
Is that it?
00:08:51.344 --> 00:08:53.690
All right.
00:08:53.690 --> 00:08:56.580
So today we're going
to pick up where
00:08:56.580 --> 00:08:59.200
Professor Gossard left off.
00:08:59.200 --> 00:09:03.290
But I'm also going to
do a bit of a summary
00:09:03.290 --> 00:09:07.390
right now about
vibration and modeling
00:09:07.390 --> 00:09:10.130
the different kinds of systems
that we talk about when
00:09:10.130 --> 00:09:11.330
we talk about vibration.
00:09:11.330 --> 00:09:17.510
They vary from simple, single
degree of freedom oscillators,
00:09:17.510 --> 00:09:19.560
like a simple
pendulum-- one degree
00:09:19.560 --> 00:09:24.910
of freedom-- to continuous
systems-- beams and vibrate.
00:09:24.910 --> 00:09:26.410
So I'm going to try
to give you just
00:09:26.410 --> 00:09:29.340
sort of an overview
of vibration just
00:09:29.340 --> 00:09:32.465
to sort of give you a
little map of information.
00:09:38.528 --> 00:09:44.200
Kind of to let you know what the
body of vibration analysis is
00:09:44.200 --> 00:09:48.860
and what part of it we're
covering in this course.
00:09:48.860 --> 00:09:55.723
So I think I will use
a little more board.
00:10:18.700 --> 00:10:22.330
So we classify
dynamics problems into,
00:10:22.330 --> 00:10:33.450
for convenience, rigid
bodies-- rigid body dynamics
00:10:33.450 --> 00:10:35.282
and flexible bodies.
00:10:35.282 --> 00:10:36.240
One way to think of it.
00:10:41.360 --> 00:10:44.310
And this course is basically
about rigid body dynamics.
00:10:44.310 --> 00:10:48.640
And under this we then
have two categories
00:10:48.640 --> 00:10:53.720
that are convenient-- single
degree of freedom systems
00:10:53.720 --> 00:10:58.045
and multiple degree
of freedom systems.
00:11:01.420 --> 00:11:03.670
For the purposes of vendors--
talking about vibration.
00:11:06.880 --> 00:11:12.030
Single degree of freedom systems
have one equation of motion.
00:11:12.030 --> 00:11:18.300
And if they vibrate, they have--
and if I'll put over here--
00:11:18.300 --> 00:11:28.195
if vibration occurs then you
have one natural frequency.
00:11:33.800 --> 00:11:37.877
And it's sort of silly
to talk about a mode
00:11:37.877 --> 00:11:39.710
shape for a single
degree of freedom system,
00:11:39.710 --> 00:11:42.090
because it's only
relative to itself.
00:11:42.090 --> 00:11:45.920
So one natural frequency and one
sort of degenerate mode shape.
00:11:45.920 --> 00:11:47.840
Multiple degree
of freedom systems
00:11:47.840 --> 00:11:52.140
have n equations of motion
for the number of degrees
00:11:52.140 --> 00:11:53.370
of freedom.
00:11:53.370 --> 00:12:02.510
And if they vibrate they
have n omega i's, or m values
00:12:02.510 --> 00:12:08.650
of omega i for i equals 1 to n.
00:12:08.650 --> 00:12:12.640
You get n natural
frequencies of the system.
00:12:12.640 --> 00:12:18.155
And you will get with
it n mode shapes.
00:12:20.920 --> 00:12:23.240
So a n degree of
freedom, this is
00:12:23.240 --> 00:12:27.320
equal to the number
of degrees of freedom.
00:12:27.320 --> 00:12:28.830
An n degree of
freedom system will
00:12:28.830 --> 00:12:30.570
have n natural
frequencies and n mode
00:12:30.570 --> 00:12:33.220
shapes that go along with it.
00:12:33.220 --> 00:12:35.820
Now, what about what
about flexible bodies?
00:12:35.820 --> 00:12:42.500
So a taut string
like a guitar string.
00:12:47.490 --> 00:12:50.080
And actually I should say
over here these rigid body
00:12:50.080 --> 00:12:52.920
things-- we have found what
kind of equations of motion?
00:12:52.920 --> 00:13:06.815
These are ordinary
differential equations.
00:13:11.230 --> 00:13:13.960
And there's a finite number
of them and so forth.
00:13:13.960 --> 00:13:17.840
The flexible bodies
like taut strings
00:13:17.840 --> 00:13:24.440
are described by partial
differential equations.
00:13:34.050 --> 00:13:36.460
The number of
degrees of freedom n
00:13:36.460 --> 00:13:40.960
here is the number
of degrees of freedom
00:13:40.960 --> 00:13:42.635
actually goes to infinity.
00:13:45.300 --> 00:13:53.070
And you get an infinite
number of omega i's,
00:13:53.070 --> 00:14:00.210
the natural frequencies,
and an infinite number
00:14:00.210 --> 00:14:01.675
of corresponding mode shapes.
00:14:10.020 --> 00:14:13.740
So just about everything in the
world can be made to vibrate.
00:14:19.860 --> 00:14:25.520
So how do you tell if a--
you've got a mechanical system,
00:14:25.520 --> 00:14:28.007
rigid bodies, you've got
three degrees of freedom.
00:14:28.007 --> 00:14:30.215
How do you know whether or
not it's going to vibrate?
00:14:32.577 --> 00:14:33.660
It will exhibit vibration?
00:14:40.500 --> 00:14:43.540
Well, one thing you
could do is figure out
00:14:43.540 --> 00:14:45.420
all the equations of
motion and solve them
00:14:45.420 --> 00:14:48.840
and see if cosine
omega t is a solution.
00:14:48.840 --> 00:14:50.460
Right?
00:14:50.460 --> 00:14:51.960
That's the hard way.
00:14:51.960 --> 00:14:55.010
The other way is to go up
to it and give it a smack
00:14:55.010 --> 00:14:57.450
and see if it vibrates.
00:14:57.450 --> 00:14:58.499
That's the simple way.
00:14:58.499 --> 00:15:00.790
If you have the mechanical
system just give it a whack.
00:15:00.790 --> 00:15:05.850
And if it oscillates around some
stable equilibrium position,
00:15:05.850 --> 00:15:07.070
it exhibits vibration.
00:15:07.070 --> 00:15:08.430
So this is a flexible system.
00:15:08.430 --> 00:15:10.620
You can actually probably
see this from there.
00:15:10.620 --> 00:15:14.150
Just by giving this frame a
smack, it will sit and vibrate.
00:15:14.150 --> 00:15:16.860
And it does it at some
natural frequency.
00:15:16.860 --> 00:15:18.360
But that's a continuous system.
00:15:22.060 --> 00:15:25.340
This continuously
improving little demo--
00:15:25.340 --> 00:15:28.060
so Professor Gossard for
his lecture last weekend
00:15:28.060 --> 00:15:31.280
had done this really
neat embellishment,
00:15:31.280 --> 00:15:33.770
which allows you to figure
out and excite the two
00:15:33.770 --> 00:15:35.344
different natural modes.
00:15:35.344 --> 00:15:37.510
But this system you have
equations of motion for it.
00:15:37.510 --> 00:15:38.301
You could write it.
00:15:38.301 --> 00:15:41.190
And if you come up and give
it a whack, it oscillates.
00:15:41.190 --> 00:15:44.280
And you could also find out that
sure enough cosine omega t sine
00:15:44.280 --> 00:15:48.090
omega t are solutions to
the equations of motion.
00:15:48.090 --> 00:15:52.320
So systems that
vibrate are systems
00:15:52.320 --> 00:15:57.070
that oscillate about static
equilibrium positions.
00:15:57.070 --> 00:15:59.020
And another way
you can say that is
00:15:59.020 --> 00:16:01.050
when mechanical
vibration occurs,
00:16:01.050 --> 00:16:03.960
there's always an
exchange of energy
00:16:03.960 --> 00:16:07.860
between kinetic and potential,
kinetic and potential.
00:16:07.860 --> 00:16:12.650
So our pendulum
it goes-- when it
00:16:12.650 --> 00:16:16.260
reaches zero velocity up here,
it's all potential energy.
00:16:16.260 --> 00:16:19.170
It reaches maximum velocity
down here, it's all kinetic.
00:16:19.170 --> 00:16:20.905
And it goes back and forth.
00:16:20.905 --> 00:16:23.540
As it sloshes back and
forth the energy system
00:16:23.540 --> 00:16:25.550
from kinetic to
potential and back again.
00:16:25.550 --> 00:16:29.850
All vibration has that property.
00:16:29.850 --> 00:16:34.310
So that sets some basic
properties of vibration.
00:16:34.310 --> 00:16:51.660
And now there's a whole body
of knowledge about vibration.
00:16:51.660 --> 00:16:56.160
And we choose, or for the
purposes of this class,
00:16:56.160 --> 00:16:58.510
we choose to break it down
into two kinds of vibration.
00:17:19.003 --> 00:17:21.694
And one is what we
call free vibration.
00:17:30.620 --> 00:17:34.490
And that we've learned already
is response, only a response.
00:17:34.490 --> 00:17:46.970
It's a response to
initial conditions
00:17:46.970 --> 00:17:49.135
and what we call
forced vibrations.
00:17:56.700 --> 00:17:58.820
Now, forces can
come of all kinds.
00:17:58.820 --> 00:18:01.890
And for the purposes
of this course
00:18:01.890 --> 00:18:05.730
we look at a particular
kind of force.
00:18:05.730 --> 00:18:11.310
So we focus on
harmonic excitation.
00:18:18.860 --> 00:18:32.740
So excitation that is of
the form cosine omega t,
00:18:32.740 --> 00:18:35.260
or e to the i omega t.
00:18:35.260 --> 00:18:41.483
These are external excitations.
00:18:46.270 --> 00:18:48.700
So we choose to break
down the analysis
00:18:48.700 --> 00:18:53.550
of the vibration of
systems into response
00:18:53.550 --> 00:18:55.430
to initial conditions
called free vibration,
00:18:55.430 --> 00:18:58.400
no external forces,
and force vibration.
00:18:58.400 --> 00:19:00.940
But we focus on a
particular kind-- harmonic.
00:19:00.940 --> 00:19:04.120
And we go even one step
further and say we're only
00:19:04.120 --> 00:19:05.450
going to study steady state.
00:19:11.009 --> 00:19:13.050
And steady state means
you've waited a long time.
00:19:13.050 --> 00:19:14.960
Turned it on, let it
shake for quite awhile.
00:19:14.960 --> 00:19:18.190
All the initial
startup transients
00:19:18.190 --> 00:19:19.460
have been damped out.
00:19:19.460 --> 00:19:21.930
And you're left with a
steady state vibration.
00:19:21.930 --> 00:19:24.900
And that leads to things
like the transfer functions
00:19:24.900 --> 00:19:26.400
for single degree
of freedom systems
00:19:26.400 --> 00:19:27.500
that we've talked about.
00:19:30.670 --> 00:19:36.500
Now, there's one other
breakdown or subdivision
00:19:36.500 --> 00:19:38.450
that we need to talk about.
00:19:38.450 --> 00:19:41.680
And that is whether systems
are linear or non-linear.
00:19:45.680 --> 00:19:49.350
And this is all set
up so you can see it.
00:19:49.350 --> 00:19:51.670
This is a double pendulum.
00:19:51.670 --> 00:19:55.130
How many degrees of freedom?
00:19:55.130 --> 00:19:56.414
Two.
00:19:56.414 --> 00:19:58.580
And in general, do you think
the equations of motion
00:19:58.580 --> 00:20:01.770
of this thing are
going to be non-linear?
00:20:01.770 --> 00:20:03.610
Right.
00:20:03.610 --> 00:20:06.540
Just a simple pendulum
is the restoring torque
00:20:06.540 --> 00:20:09.320
is Mgl sine theta.
00:20:09.320 --> 00:20:11.470
So you know it's got
sine theta and that.
00:20:11.470 --> 00:20:13.362
And this one gets quite messy.
00:20:13.362 --> 00:20:15.320
And especially if you
give it large amplitudes.
00:20:23.020 --> 00:20:26.060
And that really isn't vibration.
00:20:26.060 --> 00:20:26.600
It's not.
00:20:26.600 --> 00:20:29.410
It's looping all over itself
and then doing other things.
00:20:29.410 --> 00:20:32.640
So cosine omega t
is not a solution.
00:20:32.640 --> 00:20:34.090
It's not a solution to this.
00:20:34.090 --> 00:20:36.600
It's got to be more
complicated than that.
00:20:36.600 --> 00:20:39.550
So when this thing is
exhibiting large motions,
00:20:39.550 --> 00:20:41.792
the equations of motion
are completely non-linear.
00:20:41.792 --> 00:20:43.250
And you're going
to need a computer
00:20:43.250 --> 00:20:46.620
to crank out the full
solution to integrate
00:20:46.620 --> 00:20:48.770
these non-linear
equations of motion.
00:20:48.770 --> 00:20:58.400
But as the amplitude settles
down to something pretty small,
00:20:58.400 --> 00:21:02.155
now it's vibrating about
an equilibrium position.
00:21:02.155 --> 00:21:05.060
The equilibrium position
is straight down.
00:21:05.060 --> 00:21:06.990
And the damping of
it has made it such
00:21:06.990 --> 00:21:09.270
that the only motion
left is what's called
00:21:09.270 --> 00:21:12.650
its first mode of vibration.
00:21:12.650 --> 00:21:16.970
And so if we linearize
the equations of motion,
00:21:16.970 --> 00:21:21.830
assuming small amplitudes around
static equilibrium positions,
00:21:21.830 --> 00:21:25.870
then we can find a
vibration solution
00:21:25.870 --> 00:21:29.220
and work it out
by hand probably.
00:21:29.220 --> 00:21:30.940
That's first mode
for this system.
00:21:30.940 --> 00:21:36.110
And if I'm careful--
there's second mode.
00:21:39.390 --> 00:21:42.100
And for small
oscillations it has
00:21:42.100 --> 00:21:45.170
a very clear single frequency
that it vibrates at.
00:21:45.170 --> 00:21:49.425
The amplitude decays over
time because of damping.
00:21:49.425 --> 00:21:51.600
And for every natural
frequency there
00:21:51.600 --> 00:21:54.100
is a particular mode
shape that goes along
00:21:54.100 --> 00:21:55.800
with that natural frequency.
00:21:55.800 --> 00:21:59.040
The first one for
this system-- I
00:21:59.040 --> 00:22:02.800
have to wait for this
thing to damp out.
00:22:02.800 --> 00:22:04.410
It's got a little
mix of the two,
00:22:04.410 --> 00:22:09.250
but as it-- the second
natural frequency motion
00:22:09.250 --> 00:22:11.010
dies out faster than
the first because it
00:22:11.010 --> 00:22:13.769
has more cycles per unit time.
00:22:13.769 --> 00:22:14.560
So it settles down.
00:22:14.560 --> 00:22:16.310
This is now mostly
first mode vibration.
00:22:16.310 --> 00:22:20.480
And you can see that both
move in the same direction,
00:22:20.480 --> 00:22:23.330
the bottom one a little
more than the top.
00:22:23.330 --> 00:22:24.500
And that's the first mode.
00:22:24.500 --> 00:22:26.200
It has a unique
natural frequency.
00:22:26.200 --> 00:22:30.424
And a mode shape that is
specific that goes along
00:22:30.424 --> 00:22:31.590
with that natural frequency.
00:22:48.280 --> 00:23:04.560
So this further break
down here, I'll call it,
00:23:04.560 --> 00:23:14.215
is basically into
non-linear and linearized.
00:23:20.900 --> 00:23:24.820
So in our discussions of
vibration in this course
00:23:24.820 --> 00:23:28.400
we basically only
talk about this.
00:23:28.400 --> 00:23:49.810
So we're only doing-- so
that's quite a breakdown.
00:23:49.810 --> 00:23:53.220
You start at all possible
vibration systems,
00:23:53.220 --> 00:23:54.900
rigid bodies, single
degree of freedom,
00:23:54.900 --> 00:23:57.920
multidegree of freedom, finite
number of degrees of freedom
00:23:57.920 --> 00:24:00.490
or continuous.
00:24:00.490 --> 00:24:03.330
They can have linear or
non-linear equations of motion.
00:24:03.330 --> 00:24:06.590
But if we require
them to be linear,
00:24:06.590 --> 00:24:10.550
and that's what we're
going to look at then
00:24:10.550 --> 00:24:13.490
you sort of narrow this down
what we're looking at to this.
00:24:13.490 --> 00:24:15.720
So there's lots of other
things possible to look
00:24:15.720 --> 00:24:18.650
at it, like that really
non-linear motion of that two
00:24:18.650 --> 00:24:19.800
dimensional thing.
00:24:19.800 --> 00:24:22.796
But our study of
vibration is here.
00:24:22.796 --> 00:24:25.320
So this is what
we're doing in 2003.
00:24:25.320 --> 00:24:27.930
But there's a lot of
important problems
00:24:27.930 --> 00:24:29.100
that are covered by that.
00:24:31.760 --> 00:24:33.710
Lots of real things
in nature that
00:24:33.710 --> 00:24:36.675
are problematic for engineers
and problematic for design
00:24:36.675 --> 00:24:41.410
can be analyzed with
linear equations of motion.
00:24:41.410 --> 00:24:44.560
And even if they're not linear,
if you do the linear solution
00:24:44.560 --> 00:24:46.280
first, it gives you
a starting point
00:24:46.280 --> 00:24:48.080
to think about
what's the behavior
00:24:48.080 --> 00:24:50.360
of the non-linear systems.
00:24:50.360 --> 00:24:52.680
But this is our
study of vibration.
00:24:52.680 --> 00:24:56.890
And we're going to do
that for-- and what we had
00:24:56.890 --> 00:25:03.750
started doing-- in two ways.
00:25:03.750 --> 00:25:06.220
We look at the response to
initial conditions called
00:25:06.220 --> 00:25:07.450
free vibration.
00:25:07.450 --> 00:25:09.840
And we look at
response, steady state
00:25:09.840 --> 00:25:15.720
response of now these linear
systems to force vibration.
00:25:15.720 --> 00:25:20.630
And last week you were
looking for the first time
00:25:20.630 --> 00:25:24.850
at Professor Gossard's lecture
about the free vibration
00:25:24.850 --> 00:25:29.840
response of basically a
two degree freedom system.
00:25:29.840 --> 00:25:31.550
So why do two degree
of freedom systems?
00:25:31.550 --> 00:25:35.830
Well, it's the simplest next
step up from a single degree.
00:25:35.830 --> 00:25:38.050
And they're sort of
mathematically tractable.
00:25:38.050 --> 00:25:39.210
You can do them on paper.
00:25:39.210 --> 00:25:42.084
We emphasize looking at
two degree freedom systems
00:25:42.084 --> 00:25:43.750
because we can do the
math on the board,
00:25:43.750 --> 00:25:44.991
do the math on the paper.
00:25:44.991 --> 00:25:46.740
But as you get to more
degrees of freedom,
00:25:46.740 --> 00:25:49.470
you basically are going to
have to do-- it's easier
00:25:49.470 --> 00:25:51.460
to do it using the computer.
00:25:51.460 --> 00:25:54.960
And in order to do that you need
to know some linear algebra.
00:25:54.960 --> 00:25:57.860
So I'm kind curious.
00:25:57.860 --> 00:26:01.130
In terms of linear algebra,
like multiplying two matrices
00:26:01.130 --> 00:26:04.560
together, or finding the
determinant of a matrix,
00:26:04.560 --> 00:26:07.780
or inverting a matrix,
how many of you
00:26:07.780 --> 00:26:11.980
actually have been taught that?
00:26:11.980 --> 00:26:14.690
How may perhaps have it?
00:26:14.690 --> 00:26:16.770
Do you do that in 1803 now?
00:26:16.770 --> 00:26:17.970
Is that where you do it?
00:26:17.970 --> 00:26:18.470
OK.
00:26:18.470 --> 00:26:18.970
Good.
00:26:18.970 --> 00:26:19.830
So that's helpful.
00:26:19.830 --> 00:26:21.038
I wasn't sure whether or not.
00:26:21.038 --> 00:26:23.670
I can assume that you
at least know what
00:26:23.670 --> 00:26:27.040
the determinant of a matrix is.
00:26:27.040 --> 00:26:27.590
That's great.
00:26:27.590 --> 00:26:29.291
That's really helpful.
00:26:29.291 --> 00:26:29.790
OK.
00:26:32.840 --> 00:26:35.630
Let's talk.
00:26:35.630 --> 00:26:38.250
So we want linear
equations of motion.
00:26:38.250 --> 00:26:44.200
And I've done a little bit about
linearization but not much.
00:26:44.200 --> 00:26:49.600
So let's talk a little bit
about that for a second.
00:26:52.530 --> 00:26:59.001
For a pendulum we know the
equation of motion for it.
00:26:59.001 --> 00:27:01.500
And actually we could make this
a more complicated pendulum.
00:27:01.500 --> 00:27:05.920
It could be a stick or any rigid
body swinging about this point
00:27:05.920 --> 00:27:09.140
A.
00:27:09.140 --> 00:27:12.530
We know that we can write
the equation of motion Izz
00:27:12.530 --> 00:27:19.270
with respect to A, theta
double dot, plus Mgl-- omega L.
00:27:19.270 --> 00:27:32.020
The distance here to wherever
g is, Mgl sine theta.
00:27:32.020 --> 00:27:34.780
And for free vibration
that's all there is to it.
00:27:34.780 --> 00:27:38.790
And to linearize this
equation we just say,
00:27:38.790 --> 00:27:44.660
well, we know that sine theta
is equal to theta minus theta
00:27:44.660 --> 00:27:48.990
cubed over 3 factorial plus
theta to the fifth over 5
00:27:48.990 --> 00:27:51.080
factorial.
00:27:51.080 --> 00:27:53.890
And the cosine theta-- just
to have it available here--
00:27:53.890 --> 00:28:00.930
is 1 minus theta squared over
2 factorial and so forth.
00:28:00.930 --> 00:28:02.470
And when we say
linearize, we really
00:28:02.470 --> 00:28:06.890
mean we want our equations to
involve the motion variables
00:28:06.890 --> 00:28:09.580
at most to first order.
00:28:09.580 --> 00:28:12.280
So the first order
term for sine is theta.
00:28:12.280 --> 00:28:14.370
There's no first
order term for cosine.
00:28:17.610 --> 00:28:18.940
Theta squared is non-linear.
00:28:18.940 --> 00:28:23.190
So the small angle
approximation to cosine
00:28:23.190 --> 00:28:24.930
is it's approximately 1.
00:28:24.930 --> 00:28:27.480
And to sine is
it's approximately
00:28:27.480 --> 00:28:30.340
theta for small motions.
00:28:30.340 --> 00:28:31.090
Theta-- small.
00:28:33.650 --> 00:28:35.540
So when we linearize
this equation,
00:28:35.540 --> 00:28:39.925
we just substitute in for sine
theta its linear approximation.
00:28:39.925 --> 00:28:42.200
And we get Mgl theta.
00:28:42.200 --> 00:28:43.920
So we've seen that
one many times.
00:28:50.970 --> 00:28:53.330
And that's your linearized
equation of motion.
00:28:53.330 --> 00:28:57.122
But on this week's homework
you've got a harder problem.
00:28:57.122 --> 00:28:57.955
And that's our cart.
00:29:08.480 --> 00:29:13.620
And here you have a theta and
an x are your two equations.
00:29:13.620 --> 00:29:15.710
And you've worked
this problem before.
00:29:15.710 --> 00:29:20.520
And you know with the
previous homeworks you've
00:29:20.520 --> 00:29:21.840
gotten the equation of motion.
00:29:21.840 --> 00:29:24.670
I'll write one of
them down here.
00:29:24.670 --> 00:29:26.490
So one of the
equations of motion
00:29:26.490 --> 00:29:34.480
is-- this is m1, m1, k, b.
00:29:39.960 --> 00:29:46.130
And this is a stick, it's
l long, g in the middle.
00:29:46.130 --> 00:29:54.050
So the equation of motion for
this looks like m1 plus m2, x
00:29:54.050 --> 00:30:08.130
double dot plus m2 l
over 2, theta double dot,
00:30:08.130 --> 00:30:18.510
minus m2 l over 2 theta
dot squared theta,
00:30:18.510 --> 00:30:27.370
and then plus bx dot,
plus kx, and equals,
00:30:27.370 --> 00:30:32.580
and in fact, this one
has a force on it.
00:30:32.580 --> 00:30:33.790
It's equal to F of t.
00:30:36.680 --> 00:30:42.480
Now, is that a
non-linear equation?
00:30:54.760 --> 00:30:56.690
So this is the
force equation mass
00:30:56.690 --> 00:30:58.050
times acceleration or forces.
00:30:58.050 --> 00:30:59.450
You know you've got
another equation of motion
00:30:59.450 --> 00:31:01.150
in here, which is
the torque one.
00:31:01.150 --> 00:31:02.210
This is just one of them.
00:31:02.210 --> 00:31:03.525
So is it linear or non-linear?
00:31:07.890 --> 00:31:09.620
How many think it's non-linear?
00:31:09.620 --> 00:31:11.060
OK.
00:31:11.060 --> 00:31:19.820
If I have number the terms
here-- one, two, three, four,
00:31:19.820 --> 00:31:20.340
five.
00:31:20.340 --> 00:31:22.000
And that's not a motion.
00:31:22.000 --> 00:31:23.260
This doesn't involve motion.
00:31:23.260 --> 00:31:26.424
So if one through five, which
ones have a non-linear term?
00:31:26.424 --> 00:31:27.090
AUDIENCE: Three.
00:31:27.090 --> 00:31:27.670
J. KIM VANDIVER: Three.
00:31:27.670 --> 00:31:28.510
OK.
00:31:28.510 --> 00:31:31.080
So how do you
linearize that thing?
00:31:36.738 --> 00:31:40.240
AUDIENCE: You make it zero,
because it's second order.
00:31:40.240 --> 00:31:42.480
J. KIM VANDIVER: In
fact, it's third order.
00:31:42.480 --> 00:31:46.190
So the reason I wanted
to mention this today--
00:31:46.190 --> 00:31:49.210
if you haven't thought
about being confronted
00:31:49.210 --> 00:31:52.100
with linearization
problems before-- we're
00:31:52.100 --> 00:31:55.740
trying to linearize
the system so that we
00:31:55.740 --> 00:32:00.900
can by making it linear we can
make cosine omega t a solution.
00:32:00.900 --> 00:32:01.800
Right?
00:32:01.800 --> 00:32:06.120
We want cosine omega t to
be a solution to this thing.
00:32:06.120 --> 00:32:15.240
So if you let-- so
you've got a term that
00:32:15.240 --> 00:32:24.860
looks like ml over m2l over
2, theta dot squared theta.
00:32:24.860 --> 00:32:28.180
Well, theta in this problem
of some function of time
00:32:28.180 --> 00:32:30.460
we're hoping-- we want
to find a solution that
00:32:30.460 --> 00:32:37.260
has some amplitude times,
say, a cosine omega t.
00:32:37.260 --> 00:32:46.310
And theta dot is minus omega
theta naught sine omega t.
00:32:46.310 --> 00:32:48.790
And so that expression
up there, the magnitude
00:32:48.790 --> 00:32:52.540
of that expression or the
magnitude of theta dot
00:32:52.540 --> 00:32:58.100
squared theta is proportional
to-- you get a theta
00:32:58.100 --> 00:33:02.740
naught here and you
get omega squared
00:33:02.740 --> 00:33:08.590
theta naught squared here.
00:33:08.590 --> 00:33:12.700
So this term is proportional
to theta naught cubed.
00:33:12.700 --> 00:33:18.140
And if the angle theta is
small, then a small angle cubed
00:33:18.140 --> 00:33:20.090
is really small.
00:33:20.090 --> 00:33:23.870
And so the way you
linearize this equation
00:33:23.870 --> 00:33:27.200
is to throw this out.
00:33:27.200 --> 00:33:30.635
So when you've done all
your tricks you can,
00:33:30.635 --> 00:33:35.090
like replacing sine theta
with theta and cosine theta
00:33:35.090 --> 00:33:39.630
with one, and you still end
up with terms have a higher
00:33:39.630 --> 00:33:44.170
order than one in the
motion variable, theta or x,
00:33:44.170 --> 00:33:44.970
you throw it out.
00:33:50.510 --> 00:33:52.700
So if you throw that
term out then you
00:33:52.700 --> 00:33:55.580
end up with a nice linear
equation in motion.
00:34:09.060 --> 00:34:09.730
OK.
00:34:09.730 --> 00:34:12.100
So now for the
rest of today we're
00:34:12.100 --> 00:34:26.380
going to talk about
free vibration solution.
00:34:26.380 --> 00:34:29.170
So we're not going to worry
for the moment about the force
00:34:29.170 --> 00:34:32.230
vibration steady state
transfer function stuff.
00:34:32.230 --> 00:34:34.790
We're talking just
about free vibration.
00:34:34.790 --> 00:34:42.130
And this is of linear
equations of motion.
00:35:00.490 --> 00:35:03.220
So vibration is a pretty
big body of knowledge.
00:35:03.220 --> 00:35:06.990
And we're doing an introduction
to vibration in about half
00:35:06.990 --> 00:35:08.930
a dozen lectures here.
00:35:08.930 --> 00:35:13.180
So there's lots of
things that I'm not
00:35:13.180 --> 00:35:15.730
going to have time to
teach you, but there
00:35:15.730 --> 00:35:19.320
are a few things I really
want you to go away with
00:35:19.320 --> 00:35:20.800
and understanding.
00:35:20.800 --> 00:35:29.160
And one of these key concepts
is that the vibration
00:35:29.160 --> 00:35:32.950
of a multiple degree
of freedom system--
00:35:32.950 --> 00:35:35.950
say this is a two degree
of freedom system.
00:35:35.950 --> 00:35:38.510
That the vibration of this
system, the free vibration,
00:35:38.510 --> 00:35:45.670
can be made up of
the sum of two parts
00:35:45.670 --> 00:35:49.040
at any vibration of
this system at all.
00:35:49.040 --> 00:35:51.240
So at any arbitrary set
of initial conditions
00:35:51.240 --> 00:35:53.480
I give it-- I let it go.
00:35:53.480 --> 00:35:56.640
The key concept is
the response will
00:35:56.640 --> 00:36:01.410
be made up of two
pieces-- vibration
00:36:01.410 --> 00:36:03.940
in each of the two modes.
00:36:03.940 --> 00:36:05.845
And if you can solve
the vibration that's
00:36:05.845 --> 00:36:07.970
in first mode-- first mode
is the one where they're
00:36:07.970 --> 00:36:09.850
going kind of
together, second mode
00:36:09.850 --> 00:36:11.640
they're opposite one another.
00:36:11.640 --> 00:36:13.750
That the total
solution can be made up
00:36:13.750 --> 00:36:16.880
of a contribution for mode
one and a contribution
00:36:16.880 --> 00:36:18.960
from mode two.
00:36:18.960 --> 00:36:29.200
So this is this concept
called mode superposition.
00:36:29.200 --> 00:36:30.650
It's really quite powerful.
00:36:30.650 --> 00:36:34.630
So you can figure out the
response of the first mode
00:36:34.630 --> 00:36:37.670
in the system, figure out the
response of the second mode's
00:36:37.670 --> 00:36:42.230
contribution, add them together,
and that's the total solution.
00:36:42.230 --> 00:36:51.480
And this concept works--
there's all sorts of caveats
00:36:51.480 --> 00:36:57.040
that one gets into-- but
basically this is true
00:36:57.040 --> 00:37:11.860
for all lightly damped systems.
00:37:11.860 --> 00:37:15.570
You get into heavy damping
and strange damping,
00:37:15.570 --> 00:37:17.230
you have to make
some adjustments.
00:37:17.230 --> 00:37:19.770
But for lightly
damped systems you'll
00:37:19.770 --> 00:37:21.890
find that this concept
of mode superposition
00:37:21.890 --> 00:37:25.970
works out just fine.
00:37:25.970 --> 00:37:31.070
So an illustration of this, a
really simple illustration--
00:37:31.070 --> 00:37:33.500
in some ways easier
than this one.
00:37:33.500 --> 00:37:35.840
I don't know if I can get
this where you can see it
00:37:35.840 --> 00:37:37.090
in the picture or not.
00:37:37.090 --> 00:37:38.210
Maybe not really.
00:37:38.210 --> 00:37:41.190
This is just two
little lead weights.
00:37:41.190 --> 00:37:42.470
This is a double pendulum.
00:37:42.470 --> 00:37:46.080
It has two natural frequencies.
00:37:46.080 --> 00:37:48.190
One is that one.
00:37:48.190 --> 00:37:51.730
You can see the two weights
going the same direction.
00:37:51.730 --> 00:37:54.740
The bottom weight at a
little bit more larger angle
00:37:54.740 --> 00:37:57.540
than the top weight.
00:37:57.540 --> 00:37:59.960
And it's at a
particular frequency.
00:37:59.960 --> 00:38:03.940
And that's the mode shape that
goes through this frequency.
00:38:03.940 --> 00:38:07.790
So another key concept is
that for free vibration
00:38:07.790 --> 00:38:12.490
the total solution is made
up of the free vibration
00:38:12.490 --> 00:38:13.420
of each mode.
00:38:13.420 --> 00:38:15.775
And each mode has a
particular frequency
00:38:15.775 --> 00:38:18.740
and a particular shape to it.
00:38:18.740 --> 00:38:23.020
So that's the first mode
frequency and the first mode
00:38:23.020 --> 00:38:23.520
shape.
00:38:23.520 --> 00:38:26.440
The second mode-- I have a
little harder time getting it
00:38:26.440 --> 00:38:29.560
started-- it looks like that.
00:38:29.560 --> 00:38:31.930
Masses move in
opposite directions.
00:38:31.930 --> 00:38:35.225
It's kind of rotating around
where you can't see it.
00:38:35.225 --> 00:38:38.920
I have to do it in the plane.
00:38:38.920 --> 00:38:39.870
It's hard to do here.
00:38:39.870 --> 00:38:42.455
It doesn't want to behave
like it's confined to a plane.
00:38:47.420 --> 00:38:49.570
They're going
opposite directions.
00:38:49.570 --> 00:38:51.520
And the frequency is higher.
00:38:51.520 --> 00:38:53.600
But this motion,
that mode shape,
00:38:53.600 --> 00:38:58.530
is a fixed feature of
this mode of vibration
00:38:58.530 --> 00:39:00.470
along with this
natural frequency.
00:39:00.470 --> 00:39:04.540
So this idea of
mode superposition--
00:39:04.540 --> 00:39:27.310
and a second concept here
is that for free vibration
00:39:27.310 --> 00:40:00.740
of each mode it oscillates at
a unique frequency for this two
00:40:00.740 --> 00:40:01.970
degree of freedom system.
00:40:01.970 --> 00:40:04.060
You have two natural
frequencies--
00:40:04.060 --> 00:40:07.450
omega 1 and omega 2.
00:40:07.450 --> 00:40:23.220
And at each omega n there is
a corresponding mode shape.
00:40:25.790 --> 00:40:31.960
So any vibration of a linear
system, free vibration of it,
00:40:31.960 --> 00:40:36.870
any vibration at all is composed
of a superposition of the two
00:40:36.870 --> 00:40:37.870
modes.
00:40:37.870 --> 00:40:39.930
Part of this motion
is in the first mode
00:40:39.930 --> 00:40:42.370
at its natural frequency
and in its shape.
00:40:42.370 --> 00:40:47.440
And part of the motion
has a second contribution,
00:40:47.440 --> 00:40:50.030
which is at the natural
frequency of the second mode
00:40:50.030 --> 00:40:51.105
and in its shape.
00:40:53.670 --> 00:40:57.360
So I'm going to give
you a quick demo
00:40:57.360 --> 00:41:02.450
and ask you-- let's if
you can use what I just
00:41:02.450 --> 00:41:08.676
said to analyze a motion.
00:41:16.970 --> 00:41:21.330
So this is just a
block on some strings.
00:41:21.330 --> 00:41:23.910
And I'm going to
show you a motion.
00:41:23.910 --> 00:41:27.430
And I want you to
tell me whether or not
00:41:27.430 --> 00:41:36.580
it could possibly be a natural
frequency motion in one mode,
00:41:36.580 --> 00:41:40.552
or the other answer is it's
a sum of multiple modes.
00:41:40.552 --> 00:41:42.010
But I'm going to
show you a motion,
00:41:42.010 --> 00:41:44.070
and I want you to
tell me and argue
00:41:44.070 --> 00:41:47.160
on the basis of what I've
just told you whether or not
00:41:47.160 --> 00:41:49.260
you are seeing a single
mode of vibration.
00:41:52.810 --> 00:41:54.779
And maybe I'll
use the clamp here
00:41:54.779 --> 00:41:56.570
so I don't have to
stand there and hold it.
00:42:25.802 --> 00:42:27.010
I'm going to just place this.
00:42:27.010 --> 00:42:28.720
So the way you do
free vibration is
00:42:28.720 --> 00:42:30.360
you give it an
initial displacement,
00:42:30.360 --> 00:42:32.680
some initial
conditions, and let go.
00:42:32.680 --> 00:42:39.975
So I'm going to pull this
over and back and let go.
00:42:43.935 --> 00:42:46.480
And just watch closely
what you see it do.
00:43:14.794 --> 00:43:16.710
All right, now it's doing
more of what I want.
00:43:16.710 --> 00:43:19.224
It's like it's going
in a circle right now.
00:43:19.224 --> 00:43:21.390
And now it looks like it's
just going back and forth
00:43:21.390 --> 00:43:23.140
on a diagonal.
00:43:23.140 --> 00:43:25.940
And then it's going to start
circling the other way.
00:43:25.940 --> 00:43:27.650
It's going in a circle.
00:43:27.650 --> 00:43:31.320
And now it goes to on the
diagonal-- left and right.
00:43:31.320 --> 00:43:36.210
And then it starts back
into a circle again.
00:43:36.210 --> 00:43:40.950
Are you observing a
natural mode of vibration?
00:43:40.950 --> 00:43:42.780
It looks like it's
single frequency, right?
00:43:42.780 --> 00:43:45.340
This looks like it's all
happening at one frequency.
00:43:45.340 --> 00:43:50.340
But is it a natural mode,
a unique natural mode?
00:43:57.150 --> 00:44:00.070
Who wants to make a case
for whether it is or isn't?
00:44:00.070 --> 00:44:01.600
How many believe
that you're seeing
00:44:01.600 --> 00:44:03.240
a natural mode of vibration?
00:44:03.240 --> 00:44:03.740
None.
00:44:03.740 --> 00:44:05.810
How many think you're not seeing
a natural mode of vibration?
00:44:05.810 --> 00:44:06.770
Let's see if you're awake.
00:44:06.770 --> 00:44:07.270
OK.
00:44:07.270 --> 00:44:09.150
So you don't believe
that it's natural mode.
00:44:09.150 --> 00:44:10.830
Make the case.
00:44:10.830 --> 00:44:12.460
Why?
00:44:12.460 --> 00:44:16.090
How do you use sort of this
definition of a natural mode
00:44:16.090 --> 00:44:19.676
to tell me why this can't be?
00:44:19.676 --> 00:44:21.652
AUDIENCE: It looks
like a superposition
00:44:21.652 --> 00:44:25.110
of at least two different
kinds of vibration.
00:44:25.110 --> 00:44:26.592
J. KIM VANDIVER: OK.
00:44:26.592 --> 00:44:29.145
The evidence that you see
is because what does it do?
00:44:29.145 --> 00:44:30.436
AUDIENCE: It circled sometimes.
00:44:30.436 --> 00:44:33.407
And sometimes it goes straight
back and forth on a diagonal.
00:44:33.407 --> 00:44:34.240
J. KIM VANDIVER: OK.
00:44:34.240 --> 00:44:35.490
So it circled around
part of the time
00:44:35.490 --> 00:44:38.060
and then goes straight back
and forth part of the time.
00:44:38.060 --> 00:44:44.580
Is it a constant mode
shape of vibration?
00:44:44.580 --> 00:44:45.220
No.
00:44:45.220 --> 00:44:47.680
And that's all you
need to observe.
00:44:47.680 --> 00:44:52.500
If the thing doesn't keep
a constant single shape
00:44:52.500 --> 00:44:55.400
at a single frequency,
it's not a natural mode.
00:44:55.400 --> 00:44:57.230
So let's do a different case.
00:44:57.230 --> 00:45:00.690
I'll deflect it just this way.
00:45:00.690 --> 00:45:02.510
And ignore that
little bit of torsion.
00:45:02.510 --> 00:45:05.670
So it's just going
back and forth in line.
00:45:05.670 --> 00:45:09.900
Other than slowly damping out,
that has just one motion to it,
00:45:09.900 --> 00:45:12.190
and it's at one
natural frequency.
00:45:12.190 --> 00:45:14.740
So do you think that's a mode?
00:45:14.740 --> 00:45:16.100
That probably is two.
00:45:16.100 --> 00:45:17.160
And so is this one.
00:45:20.279 --> 00:45:21.570
And ignore that high frequency.
00:45:21.570 --> 00:45:22.778
Now it's just back and forth.
00:45:22.778 --> 00:45:24.160
It's just a pendulum.
00:45:24.160 --> 00:45:27.480
And it just stays in just
pendular motion, no circling
00:45:27.480 --> 00:45:28.790
around or any of that.
00:45:28.790 --> 00:45:31.210
So that's also
natural, and it occurs
00:45:31.210 --> 00:45:33.530
at a particular frequency.
00:45:33.530 --> 00:45:37.610
So these are two individual
pendular emotions--
00:45:37.610 --> 00:45:39.330
one this way and one that way.
00:45:39.330 --> 00:45:41.310
And what I was doing
at the beginning
00:45:41.310 --> 00:45:44.580
is I pulled it to
the side, which
00:45:44.580 --> 00:45:46.470
would start one of those modes.
00:45:46.470 --> 00:45:48.340
And I pulled it
back, which will put
00:45:48.340 --> 00:45:51.670
some energy into the
other mode, and let it go.
00:45:51.670 --> 00:45:55.540
And now what you have is the sum
of these two different motions
00:45:55.540 --> 00:45:57.230
adding up.
00:45:57.230 --> 00:46:00.020
It goes in circles and
then in straight lines.
00:46:00.020 --> 00:46:05.700
And the fact that they-- this
is a phenomenon called beading.
00:46:05.700 --> 00:46:09.240
And it is because
these two pendulums,
00:46:09.240 --> 00:46:13.410
even though they have
strings of the same length,
00:46:13.410 --> 00:46:16.654
they actually have slightly
different natural frequencies.
00:46:16.654 --> 00:46:18.570
They're each single
degree of freedom systems.
00:46:18.570 --> 00:46:21.040
They're two independent single
degree of freedom systems,
00:46:21.040 --> 00:46:22.990
each with their own
natural frequency.
00:46:22.990 --> 00:46:27.789
But if you mix them then they're
going to exhibit this motion.
00:46:27.789 --> 00:46:29.830
So that's something really
important to remember.
00:46:29.830 --> 00:46:34.900
A quiz question that I like
to ask is-- it's easy to grade
00:46:34.900 --> 00:46:38.850
and it's no math required--
is to literally-- I've often
00:46:38.850 --> 00:46:40.870
done this in exams--
walk in with something
00:46:40.870 --> 00:46:45.330
like that block of wood and
say, is this a natural mode?
00:46:53.880 --> 00:47:00.910
Time to do one--
let me see here.
00:47:29.460 --> 00:47:31.930
So now let's pick up where
Professor Gossard left off.
00:47:31.930 --> 00:47:34.380
Let's go talking about
natural frequencies and mode
00:47:34.380 --> 00:47:39.100
shapes of linearized two
degree of freedom systems.
00:47:39.100 --> 00:47:43.880
But I want to generalize a
little bit on what he did.
00:47:43.880 --> 00:47:49.400
So he, in his lecture,
analyzed this system like this.
00:47:52.710 --> 00:47:54.470
I'll just kind of put
the highlights here.
00:48:09.860 --> 00:48:22.100
And this is now solving for
natural frequencies and mode
00:48:22.100 --> 00:48:22.600
shapes.
00:48:26.460 --> 00:48:28.845
He came up with a set of
equations of motion for this.
00:48:28.845 --> 00:48:33.605
This was, I guess, M1, M2.
00:48:33.605 --> 00:48:40.580
And the equations of motion
for this are m1 in matrix form.
00:48:49.436 --> 00:48:53.000
Now I'm going to do this
to emphasize something.
00:48:57.480 --> 00:48:59.040
In general there
could be damping
00:48:59.040 --> 00:49:01.590
in our linearized system.
00:49:01.590 --> 00:49:08.520
And we have a stiffness
matrix-- K1 plus K2 minus K2.
00:49:19.380 --> 00:49:23.570
And in general there
could be forces,
00:49:23.570 --> 00:49:28.130
which are functions of
time on that system.
00:49:28.130 --> 00:49:30.850
Now, if we want to find natural
frequencies in mode shapes,
00:49:30.850 --> 00:49:33.980
we go looking for what we
call with the undamped natural
00:49:33.980 --> 00:49:35.700
frequencies in mode shapes.
00:49:35.700 --> 00:49:37.710
So this problem doesn't
even have dampers in it.
00:49:37.710 --> 00:49:40.290
But if it did for the
purpose of finding
00:49:40.290 --> 00:49:45.280
natural frequencies in mode
shapes, you just set to 0.
00:49:45.280 --> 00:49:47.130
And with the forces
you do the same thing.
00:49:52.030 --> 00:49:57.850
And now you have undamped,
unforced equations of motion.
00:50:01.610 --> 00:50:05.630
And this is then of the
form of mass matrix times
00:50:05.630 --> 00:50:11.720
an acceleration vector, X1,
X2, plus a stiffness matrix,
00:50:11.720 --> 00:50:15.815
times a displacement
vector equals 0.
00:50:15.815 --> 00:50:18.250
So in matrix notation
it looks like that.
00:50:20.950 --> 00:50:26.670
This is the way you would do any
rigid body vibration problem.
00:50:26.670 --> 00:50:28.490
This is two degrees of freedom.
00:50:28.490 --> 00:50:29.940
But this is the
general expression
00:50:29.940 --> 00:50:33.681
for an n degree
of freedom system.
00:50:33.681 --> 00:50:36.770
If we had three masses here,
then these would be 3 by 3
00:50:36.770 --> 00:50:40.390
matrices instead of 2 by 2's.
00:50:40.390 --> 00:50:44.160
So that's the basic formulation.
00:50:44.160 --> 00:50:47.660
And you went through
last time with 2 by 2.
00:50:47.660 --> 00:50:52.890
You can actually go through and
find the fourth order equation
00:50:52.890 --> 00:50:59.180
in omega and solve for two
roots of omega squared.
00:50:59.180 --> 00:51:02.370
And you've got the two natural
frequencies, plug them back in.
00:51:02.370 --> 00:51:05.050
You've got the two mode shapes
that go along with them.
00:51:05.050 --> 00:51:08.020
So you did it that
way by hand so you
00:51:08.020 --> 00:51:10.650
can see how you can work
out the natural frequencies.
00:51:10.650 --> 00:51:14.850
How can you do-- I'm going to
show an approach that you'd
00:51:14.850 --> 00:51:16.790
more likely use on a computer.
00:51:16.790 --> 00:51:20.850
And if you get the larger order
n degree of freedom systems,
00:51:20.850 --> 00:51:23.630
you're going to want to do
this-- instead of by hand--
00:51:23.630 --> 00:51:25.750
have a computer do
the work for you.
00:51:25.750 --> 00:51:30.495
So this is the generic form.
00:51:30.495 --> 00:51:34.580
And let's just
assume for a minute
00:51:34.580 --> 00:51:37.050
it's an n degree
of freedom system.
00:51:37.050 --> 00:51:41.470
So these are n by n matrices.
00:51:45.750 --> 00:51:48.100
How would we find the
natural frequencies and mode
00:51:48.100 --> 00:51:52.490
shapes of this general system?
00:51:52.490 --> 00:52:05.920
So you assume
solutions of the form
00:52:05.920 --> 00:52:09.760
of what I've been
describing-- a natural mode.
00:52:09.760 --> 00:52:15.110
Any natural mode of the
system has a particular shape
00:52:15.110 --> 00:52:18.519
to it and a
particular frequency.
00:52:18.519 --> 00:52:19.310
And that's the key.
00:52:19.310 --> 00:52:21.150
That's the key assumption here.
00:52:21.150 --> 00:52:26.647
You assume solutions of the
form that this vector x--
00:52:26.647 --> 00:52:28.480
instead of writing it
in brackets like this,
00:52:28.480 --> 00:52:30.532
I'm going to make this.
00:52:30.532 --> 00:52:37.460
So X here is just with
a line underneath it.
00:52:37.460 --> 00:52:48.080
So x is of the form
X1 of t down to Xn
00:52:48.080 --> 00:52:50.950
if you have n
degrees of freedom.
00:52:50.950 --> 00:52:53.890
You're looking for a
solution for that thing.
00:52:53.890 --> 00:52:57.900
And it's going to have
an amplitude to it--
00:52:57.900 --> 00:53:02.770
A1 down to An.
00:53:02.770 --> 00:53:06.590
And this is any one mode.
00:53:06.590 --> 00:53:10.710
So any one mode will look
like a set of amplitudes
00:53:10.710 --> 00:53:14.610
that govern its mode shape.
00:53:14.610 --> 00:53:17.110
And it will oscillate.
00:53:17.110 --> 00:53:21.110
We can write the
oscillation as cosine omega.
00:53:21.110 --> 00:53:24.070
And I'll put an i here, it's
the i-th natural frequency
00:53:24.070 --> 00:53:26.970
minus some phase angle.
00:53:26.970 --> 00:53:29.380
So in general, each
mode-- assume solutions--
00:53:29.380 --> 00:53:34.560
I'll say here for each mode.
00:53:37.520 --> 00:53:41.210
So each mode, any mode,
mode I will look like this.
00:53:41.210 --> 00:53:43.759
It will have a shape
to it governed by this.
00:53:43.759 --> 00:53:45.300
And these are
basically on constants.
00:53:45.300 --> 00:53:47.630
Once determined, this
is just a constant.
00:53:47.630 --> 00:53:49.510
And here's your time dependence.
00:53:49.510 --> 00:53:53.660
And it's going to-- I
left out my t-- oscillate
00:53:53.660 --> 00:53:54.950
at some natural frequency.
00:53:54.950 --> 00:53:57.770
So we know this is what the
solution has to look like.
00:53:57.770 --> 00:54:01.485
And we can take this and
plug it in to this equation.
00:54:11.710 --> 00:54:18.230
This vector of responses
is some vector of amplitude
00:54:18.230 --> 00:54:24.690
times the cosine
omega t minus phi.
00:54:24.690 --> 00:54:28.180
And just plug that into this
set of matrix equations.
00:54:28.180 --> 00:54:37.590
Note that x double dot is just
a-- you get minus omega squared
00:54:37.590 --> 00:54:41.910
a cosine omega t.
00:54:45.720 --> 00:54:49.455
And we now substitute
these into here.
00:54:56.040 --> 00:55:01.570
You get minus omega
squared for the first term.
00:55:01.570 --> 00:55:04.190
Minus omega squared
times the mass
00:55:04.190 --> 00:55:12.280
matrix a cosine
omega t minus phi,
00:55:12.280 --> 00:55:19.410
plus this stiffness matrix
a-- better consistent notation
00:55:19.410 --> 00:55:26.157
here, excuse me-- a
cosine omega t minus phi.
00:55:26.157 --> 00:55:27.365
And all that's equal to zero.
00:55:30.060 --> 00:55:32.120
So these go away.
00:55:32.120 --> 00:55:34.770
You can cancel them out.
00:55:34.770 --> 00:55:38.950
And we can factor
out this a quantity.
00:55:38.950 --> 00:55:56.740
And we have minus omega squared
m, plus k times a equals 0.
00:55:56.740 --> 00:55:59.067
So you can do this
with any linearized n
00:55:59.067 --> 00:56:00.650
degree of freedom
system that you know
00:56:00.650 --> 00:56:02.740
has a vibration solution to it.
00:56:09.110 --> 00:56:12.100
These are the
unknown mode shapes.
00:56:12.100 --> 00:56:15.010
And so in order to
satisfy this equation,
00:56:15.010 --> 00:56:19.290
either this a has to be 0,
which is a trivial solution.
00:56:19.290 --> 00:56:21.770
There's no motion,
no mode shape--
00:56:21.770 --> 00:56:35.700
or which this is
trivial, not too useful.
00:56:35.700 --> 00:56:48.520
Either a has to be 0, or the
determinant of this quantity
00:56:48.520 --> 00:56:49.450
has to be 0.
00:56:58.970 --> 00:57:00.660
But the way you do
that on a computer--
00:57:00.660 --> 00:57:02.220
so that would be
beginning the way
00:57:02.220 --> 00:57:05.120
you would analyze this by hand.
00:57:05.120 --> 00:57:07.160
You find the determinant
of that matrix.
00:57:07.160 --> 00:57:12.550
And if it is a two
degree of freedom system,
00:57:12.550 --> 00:57:16.810
you'll get an equation n
omega to the fourth, which has
00:57:16.810 --> 00:57:18.180
two roots for omega squared.
00:57:18.180 --> 00:57:19.888
If it's a three degree
of freedom system,
00:57:19.888 --> 00:57:22.930
you'll get an equation of omega
to the sixth when you write out
00:57:22.930 --> 00:57:24.180
that determinant.
00:57:24.180 --> 00:57:27.020
And it has three roots
for omega squared.
00:57:27.020 --> 00:57:28.740
An n degree of
freedom system has
00:57:28.740 --> 00:57:32.850
an equation that's of order
2n omega to the 2nth power.
00:57:32.850 --> 00:57:36.110
And it'll have n
solutions or roots
00:57:36.110 --> 00:57:39.684
for the natural frequency
for omega squared.
00:57:39.684 --> 00:57:41.100
But that would be
if you're trying
00:57:41.100 --> 00:57:42.290
to grind this out by hand.
00:57:46.450 --> 00:57:52.552
The way you do this on
a computer-- maybe I
00:57:52.552 --> 00:57:54.111
can get a little
bit more on here.
00:57:56.700 --> 00:57:58.080
Come back here.
00:58:15.370 --> 00:58:21.730
So I'll go back to the earlier
form I had here plus ka
00:58:21.730 --> 00:58:23.850
equals 0.
00:58:23.850 --> 00:58:35.110
And I'm going to
multiply by m inverse.
00:58:35.110 --> 00:58:39.200
So if I invert the
mass matrix, if I
00:58:39.200 --> 00:58:43.230
multiply a matrix by its
inverse, what do you get?
00:58:43.230 --> 00:58:45.720
So if I multiply
m times m inverse?
00:58:45.720 --> 00:58:46.860
AUDIENCE: A unit matrix.
00:58:46.860 --> 00:58:48.776
J. KIM VANDIVER: You get
a unit matrix, right?
00:58:48.776 --> 00:58:50.390
It has ones on the diagonal.
00:58:50.390 --> 00:58:53.360
So I'm going to multiply
through here by this.
00:58:53.360 --> 00:58:56.240
And so this gives me
a minus omega squared.
00:58:56.240 --> 00:59:00.180
And m times m inverse
gives me the unit matrix--
00:59:00.180 --> 00:59:02.460
ones on the diagonal.
00:59:02.460 --> 00:59:15.910
Times a plus M inverse
times ka equals 0.
00:59:15.910 --> 00:59:19.210
And this product is just a
matrix product-- m inverse
00:59:19.210 --> 00:59:20.110
times k.
00:59:20.110 --> 00:59:21.693
And I'm going to
call it the a matrix.
00:59:25.060 --> 00:59:27.790
And I'm going to move
this to the other side.
00:59:27.790 --> 00:59:32.440
So I have a linear
algebraic expression
00:59:32.440 --> 00:59:42.500
of the form a times the
vector equals omega squared
00:59:42.500 --> 00:59:46.500
times the unit matrix times a.
00:59:49.040 --> 00:59:55.300
And I could go ahead
and multiply this out.
00:59:55.300 --> 00:59:58.770
For example, this times
that and I'll get a vector.
00:59:58.770 --> 01:00:02.100
So this just looks
like omega squared
01:00:02.100 --> 01:00:04.250
a if you multiply it out.
01:00:06.940 --> 01:00:11.580
The vector times the matrix
is a vector on the left side.
01:00:11.580 --> 01:00:15.550
A vector times a matrix
gives me back a vector.
01:00:15.550 --> 01:00:16.720
It's just the unit matrix.
01:00:16.720 --> 01:00:19.630
So it gives me back the
vector times omega squared.
01:00:19.630 --> 01:00:24.140
This is in what is known
as standard eigenvalue
01:00:24.140 --> 01:00:26.150
formulation.
01:00:26.150 --> 01:00:28.760
It's a standard
eigenvalue problem now.
01:00:28.760 --> 01:00:36.020
It's a problem of the form
a times a vector equals
01:00:36.020 --> 01:00:40.310
something lambda times a.
01:00:40.310 --> 01:00:42.060
A parameter which
we know happens
01:00:42.060 --> 01:00:44.130
to be the frequency squared.
01:00:44.130 --> 01:00:49.580
But this is standard
eigenvalue formulation.
01:00:49.580 --> 01:00:50.771
Yeah?
01:00:50.771 --> 01:00:52.645
AUDIENCE: I was just
asking if you wrote down
01:00:52.645 --> 01:00:55.340
omega squared a because it's
equal to the length up there.
01:00:55.340 --> 01:00:58.660
J. KIM VANDIVER: Ah, good.
01:00:58.660 --> 01:01:00.320
Omega squared a.
01:01:00.320 --> 01:01:04.190
So a times the unit vector
you get a back as a vector.
01:01:04.190 --> 01:01:06.060
And I got the omega
squared in front of it.
01:01:06.060 --> 01:01:09.640
And oftentimes in a manual
for Matlab or something
01:01:09.640 --> 01:01:12.410
they'll describe this
as some parameter.
01:01:12.410 --> 01:01:15.460
It's a constant times a.
01:01:15.460 --> 01:01:20.160
And this is standard what they
call eigenvalue formulation.
01:01:20.160 --> 01:01:38.720
And in Matlab if you say, for
example, E equals EIG of A.
01:01:38.720 --> 01:01:49.455
This returns a vector, which is
the natural frequency squared.
01:01:52.392 --> 01:01:54.080
It will return these lambdas.
01:01:54.080 --> 01:01:56.835
And the first one is
omega 1 squared down
01:01:56.835 --> 01:01:58.060
to omega n squared.
01:02:00.780 --> 01:02:02.820
And if you go with
this function,
01:02:02.820 --> 01:02:12.170
if you go a little further,
if you say V comma D is EIG A,
01:02:12.170 --> 01:02:16.570
then this gives you
two matrices back.
01:02:16.570 --> 01:02:21.910
It gives you V. And
V is a matrix, which
01:02:21.910 --> 01:02:25.310
its columns are the mode shape.
01:02:25.310 --> 01:02:36.420
So A1 to An, this is mode 1
over to A1 to An for mode n.
01:02:36.420 --> 01:02:38.530
It gives you two matrices.
01:02:38.530 --> 01:02:40.070
One that's that.
01:02:40.070 --> 01:02:49.330
And another one a D matrix,
which has the lambdas-- lambda
01:02:49.330 --> 01:02:55.650
1 lambda n on the diagonals.
01:02:55.650 --> 01:02:58.210
And it's a diagonal matrix.
01:02:58.210 --> 01:03:00.120
So it gives you
two matrices back.
01:03:00.120 --> 01:03:04.270
One that has the
eigenvectors, the mode shapes.
01:03:04.270 --> 01:03:06.970
And another matrix whose
diagonal elements are
01:03:06.970 --> 01:03:10.190
the natural frequency squared.
01:03:10.190 --> 01:03:14.620
And that's all there is to it
if you do this numerically.
01:03:14.620 --> 01:03:17.450
And there's lots of
different programs.
01:03:17.450 --> 01:03:20.340
There's multiple ways
of doing this in Matlab.
01:03:20.340 --> 01:03:23.720
When you do it this way it
doesn't come out sometimes
01:03:23.720 --> 01:03:26.670
nicely ordered and
what I call normalized.
01:03:26.670 --> 01:03:31.067
But it does produce
the eigenvalues.
01:03:31.067 --> 01:03:32.900
They're called eigenvalues
and eigenvectors.
01:03:32.900 --> 01:03:36.210
The eigenvalues are the
lambdas, the natural frequencies
01:03:36.210 --> 01:03:37.180
squared.
01:03:37.180 --> 01:03:40.580
And the eigenvectors are
these mode shapes that
01:03:40.580 --> 01:03:45.510
go with each natural frequency.
01:03:45.510 --> 01:03:47.760
Once you know the natural
frequencies and mode shapes,
01:03:47.760 --> 01:03:50.310
now we want to get back to
talking about solutions.
01:03:50.310 --> 01:03:51.950
This idea of mode superposition.
01:03:51.950 --> 01:03:55.070
And if you give it a set
of initial conditions, what
01:03:55.070 --> 01:03:55.950
is the response?
01:03:55.950 --> 01:03:59.380
How do you add these
two modes together?
01:03:59.380 --> 01:04:01.220
So let's go back now.
01:04:01.220 --> 01:04:08.670
We'll return to two
degree of freedom systems
01:04:08.670 --> 01:04:11.415
like this one to do an example.
01:04:14.140 --> 01:04:27.840
And we assume that the solution
was sum A1, A2, cosine omega
01:04:27.840 --> 01:04:31.020
t minus a phase angle.
01:04:31.020 --> 01:04:34.750
That each mode would have
this character to it.
01:04:34.750 --> 01:04:39.645
And I'm going to
normalize my mode shapes.
01:04:51.790 --> 01:04:54.940
So for each mode shape
of the system-- so
01:04:54.940 --> 01:04:58.290
this could be for mode one.
01:04:58.290 --> 01:04:59.990
This is the mode
shape for mode one.
01:04:59.990 --> 01:05:03.720
This is natural frequency
one and phase angle one.
01:05:03.720 --> 01:05:06.610
Each mode shape-- I
could write this then
01:05:06.610 --> 01:05:10.460
as-- I could factor out the A1.
01:05:10.460 --> 01:05:13.090
Just pull out A1, divide
each member by A1.
01:05:13.090 --> 01:05:22.430
So this can be written as
A1 times 1 and A2 over A1.
01:05:22.430 --> 01:05:23.720
So I've just factored out.
01:05:29.670 --> 01:05:33.920
So for mode one it's
normalized mode shape--
01:05:33.920 --> 01:05:41.370
by normalize you just pick some
way that you repeatedly use,
01:05:41.370 --> 01:05:42.550
consistent in its use.
01:05:42.550 --> 01:05:47.100
I often say let's make the
top element of the vector 1.
01:05:47.100 --> 01:05:49.990
And to make the top
one 1 you factor out
01:05:49.990 --> 01:05:52.365
whatever its value is that
you get back from the computer
01:05:52.365 --> 01:05:54.080
or from your calculation.
01:05:54.080 --> 01:05:56.930
You factor that out
of every member.
01:05:56.930 --> 01:05:59.430
Now you have a normalized mode
shape whose top element is 1.
01:05:59.430 --> 01:06:01.263
There's lots of other
normalization schemes.
01:06:01.263 --> 01:06:03.380
That's just one way to do it.
01:06:03.380 --> 01:06:12.340
And that's one of
the mode shapes.
01:06:12.340 --> 01:06:26.790
The total solution
is X1, X2-- and this
01:06:26.790 --> 01:06:29.690
is where the mode
superposition part comes in--
01:06:29.690 --> 01:06:38.080
is some undetermined constant
A1 times the mode shape
01:06:38.080 --> 01:06:47.836
A2 over A1 for mode 1 cosine
omega 1 t minus phi 1.
01:06:47.836 --> 01:06:49.210
And I'm going to
run out of room.
01:07:14.455 --> 01:07:17.780
And now the responses to
initial conditions-- this
01:07:17.780 --> 01:07:20.040
has got another term.
01:07:20.040 --> 01:07:21.460
I'm just going to
rewrite it here.
01:07:21.460 --> 01:07:24.090
So we're looking at--
our total motion response
01:07:24.090 --> 01:07:39.515
now by mode superposition will
be A1, 1 A2 over A1, mode 1.
01:08:01.160 --> 01:08:05.840
So the free vibration
response of any two degree
01:08:05.840 --> 01:08:09.920
of freedom system,
linearized equation, any two
01:08:09.920 --> 01:08:12.940
degree of freedom linear
system can be made up
01:08:12.940 --> 01:08:16.050
of the sum of two terms.
01:08:16.050 --> 01:08:20.810
The motion at its first natural
frequency in its first mode
01:08:20.810 --> 01:08:22.990
shape.
01:08:22.990 --> 01:08:26.250
And another term is the motion
at a second natural frequency
01:08:26.250 --> 01:08:28.220
and its second mode shape.
01:08:28.220 --> 01:08:31.740
But now you have two
undetermined constants
01:08:31.740 --> 01:08:33.350
out here-- A1 and A2.
01:08:33.350 --> 01:08:34.775
Where do they come from?
01:08:38.340 --> 01:08:43.140
You have to use your initial
conditions to get those.
01:08:43.140 --> 01:08:44.460
I'll write down-- let's see.
01:08:44.460 --> 01:08:48.875
I have just maybe enough
time to write this down.
01:08:51.560 --> 01:08:52.810
These are functions of time.
01:08:55.340 --> 01:09:07.740
So A1 and A2 come from the
ICs, the initial conditions.
01:09:07.740 --> 01:09:16.310
So at t equals 0, for example,
plug in t equals 0 into here.
01:09:16.310 --> 01:09:20.380
You get cosine phi.
01:09:20.380 --> 01:09:23.890
And over here another phi 1
and another way over here.
01:09:23.890 --> 01:09:26.680
Cosine of minus phi
1 is cosine phi 1.
01:09:26.680 --> 01:09:31.740
So if you put in t equals
0, you find out the X1
01:09:31.740 --> 01:09:39.420
at 0, which I'll write
X1 0, and X2 of 0
01:09:39.420 --> 01:09:45.069
is equal to-- I'll
actually write them out.
01:09:45.069 --> 01:09:57.750
This is going to be A1 times
1 cosine phi 1, plus A2 times
01:09:57.750 --> 01:10:03.050
1 cosine phi 2.
01:10:03.050 --> 01:10:08.920
And the second equation
that this gives you is A1--
01:10:08.920 --> 01:10:12.030
and now to keep from writing
these many, many times,
01:10:12.030 --> 01:10:21.275
I'm going to let this
first A2 over A1 for mode 1
01:10:21.275 --> 01:10:28.130
be R1, and the second one
A2 over A1 for mode 2.
01:10:28.130 --> 01:10:30.360
I'll just call it R2.
01:10:30.360 --> 01:10:31.710
And then I can write this out.
01:10:31.710 --> 01:10:40.430
So the second equation,
this is R1 cosine phi 1,
01:10:40.430 --> 01:10:48.300
plus A2 R2 cosine phi 2.
01:10:48.300 --> 01:10:51.350
And now I have
initial conditions
01:10:51.350 --> 01:10:53.790
that are normally given.
01:10:53.790 --> 01:10:56.220
This is an initial
displacement on 1
01:10:56.220 --> 01:11:00.180
and a initial displacement on 2.
01:11:00.180 --> 01:11:01.379
The phis I don't know.
01:11:01.379 --> 01:11:02.420
And the A's I don't know.
01:11:02.420 --> 01:11:05.370
I have four unknowns
and two equations.
01:11:05.370 --> 01:11:07.215
How do I get two more equations?
01:11:07.215 --> 01:11:11.520
I take the derivative of this
expression to get velocity.
01:11:11.520 --> 01:11:16.620
And I get an A omega
here and an A omega here.
01:11:16.620 --> 01:11:21.440
And I plug in t equals 0.
01:11:21.440 --> 01:11:24.310
And I get two more equations.
01:11:24.310 --> 01:11:41.050
So X1 dot and X2 dot
equal at t equals 0.
01:11:41.050 --> 01:11:44.700
This gives me two
more equations.
01:11:44.700 --> 01:11:53.560
And if I have a place to
write them-- for example,
01:11:53.560 --> 01:12:00.750
X1 dot-- or X1 0 dot, the
initial condition on velocity.
01:12:00.750 --> 01:12:03.875
X2 dot-- this is two equations.
01:12:06.420 --> 01:12:08.790
I've only got time to
write down one of them.
01:12:08.790 --> 01:12:11.680
And you could do the
other for exercise.
01:12:11.680 --> 01:12:23.970
You find this is A1 omega
1 sine phi 1, plus A2 omega
01:12:23.970 --> 01:12:27.940
2 sine phi 2.
01:12:27.940 --> 01:12:35.430
And the second equation is you
get A1 R1 omega 1 sine phi 1,
01:12:35.430 --> 01:12:43.390
plus A2 R2 omega 2 sine phi 2.
01:12:43.390 --> 01:12:47.390
So now you have one, two.
01:12:47.390 --> 01:12:49.865
And this is the initial
conditions on velocity.
01:12:49.865 --> 01:12:52.850
These are initial
values of velocity.
01:12:52.850 --> 01:12:55.760
Now you have one, two, three,
four equations and one,
01:12:55.760 --> 01:13:02.750
two, three, four unknowns--
the A1, A2, phi 1, phi 2.
01:13:02.750 --> 01:13:13.070
And I guess I will give you the
answer, so you have it once.
01:13:30.810 --> 01:13:34.980
So a little tedious but this
is sort of in the spirit of we
01:13:34.980 --> 01:13:36.940
do two degree of
freedom systems so
01:13:36.940 --> 01:13:39.680
that we can see how it works.
01:13:39.680 --> 01:13:43.360
And then for larger
degrees of freedom systems
01:13:43.360 --> 01:13:45.580
you would do this
with a computer.
01:13:45.580 --> 01:13:52.555
But the solution for A1
is 1 over R2 minus R1.
01:13:55.485 --> 01:13:57.990
It's all in terms now
of things you know.
01:13:57.990 --> 01:14:00.910
These R2's and R1's are
part of the mode shape,
01:14:00.910 --> 01:14:06.270
and the rest is
initial conditions.
01:14:06.270 --> 01:14:23.020
R2 x1 0, minus X2 0
squared, plus R2 V1 0,
01:14:23.020 --> 01:14:31.550
minus V2 0 quantity
squared, over omega 1
01:14:31.550 --> 01:14:36.080
squared, the whole
thing square root.
01:14:36.080 --> 01:14:37.630
But now this is
all stuff you know.
01:14:37.630 --> 01:14:39.790
The given initial
conditions on velocity,
01:14:39.790 --> 01:14:43.530
given initial conditions
on displacement.
01:14:43.530 --> 01:14:44.930
You know the natural frequency.
01:14:44.930 --> 01:14:47.020
You know the pieces
of the mode shapes.
01:14:47.020 --> 01:14:50.940
Just plug it in, you're
going to get a number.
01:14:50.940 --> 01:14:54.350
The magnitude of A1 for the
given initial conditions.
01:14:54.350 --> 01:15:10.600
A2-- a very similar expression--
minus R1 X1 0 plus x2 0
01:15:10.600 --> 01:15:35.770
squared And you'd
solve for A2 and phi
01:15:35.770 --> 01:15:41.240
1-- it's a little simpler--
minus tangent inverse,
01:15:41.240 --> 01:16:00.540
phi 2 0 minus R2 V1 0, all over
omega 1 R2 x1 0, minus x2 0.
01:16:00.540 --> 01:16:02.210
That's one of the phase angles.
01:16:02.210 --> 01:16:06.570
And the other phase angle,
a very similar expression.
01:16:06.570 --> 01:16:18.650
Minus tangent inverse, minus
R1 V1 0, plus V2 0 omega
01:16:18.650 --> 01:16:25.650
2 R1 x1 0, minus x2 0.
01:16:25.650 --> 01:16:30.100
So just expressions in terms of
the initial conditions and you
01:16:30.100 --> 01:16:31.869
can get all four quantities.
01:16:31.869 --> 01:16:33.410
You can also do this
on the computer.
01:16:33.410 --> 01:16:37.040
But in the few short
lectures that we have
01:16:37.040 --> 01:16:38.967
we're not going
to get into that.
01:16:38.967 --> 01:16:40.550
But this just shows
you where it goes.
01:16:40.550 --> 01:16:41.590
You could do this now.
01:16:41.590 --> 01:16:44.040
There are straightforward
ways of doing it
01:16:44.040 --> 01:16:47.420
with matrix algebra
on the computer.
01:16:47.420 --> 01:16:52.470
Next time I'll do maybe just a
quick example-- I didn't quite
01:16:52.470 --> 01:16:57.830
get to it today-- of a response
to initial conditions problem.
01:16:57.830 --> 01:16:59.410
Plug it in there.
01:16:59.410 --> 01:17:00.715
See what happens.
01:17:00.715 --> 01:17:03.260
But we're out of time.
01:17:03.260 --> 01:17:06.680
See you on Thursday.