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J. KIM VANDIVER: Today's lecture
is not mathematically hard,
00:00:26.150 --> 00:00:30.932
but it's really important to
establish vocabulary today.
00:00:30.932 --> 00:00:32.390
We're going to talk
about vibration
00:00:32.390 --> 00:00:34.690
for the rest of the term.
00:00:34.690 --> 00:00:40.110
And vibration is essentially
applied dynamics.
00:00:40.110 --> 00:00:44.740
So up until now, we've been
finding equations of motion,
00:00:44.740 --> 00:00:46.330
but not solving them.
00:00:46.330 --> 00:00:48.190
Did you notice that?
00:00:48.190 --> 00:00:52.660
I've almost never asked you to
solve the equation of motion
00:00:52.660 --> 00:00:55.890
that you've just discovered
using Lagrange or whatever.
00:00:55.890 --> 00:00:57.390
The rest of the
term, we're actually
00:00:57.390 --> 00:01:00.980
going to be talking mostly
about the resulting motion.
00:01:00.980 --> 00:01:03.170
The equations of motion
are pretty easy to find.
00:01:03.170 --> 00:01:07.020
You have all the techniques that
you need to know for finding.
00:01:07.020 --> 00:01:11.140
And now, we're going to talk
about how things vibrate.
00:01:11.140 --> 00:01:14.130
So why do we choose vibration?
00:01:14.130 --> 00:01:20.690
Vibration, one, is an
incredibly common phenomenon.
00:01:20.690 --> 00:01:23.010
We wouldn't have speech
without vibration.
00:01:23.010 --> 00:01:26.980
You wouldn't have musical
instruments without vibration.
00:01:26.980 --> 00:01:30.640
It's a positive thing when
it's making good music.
00:01:30.640 --> 00:01:33.190
It's a negative thing when
it's keeping you awake at night
00:01:33.190 --> 00:01:35.620
because the air conditioner
in the next room
00:01:35.620 --> 00:01:38.130
is causing something
to rattle in the room
00:01:38.130 --> 00:01:40.190
and it's driving you nuts.
00:01:40.190 --> 00:01:44.000
So you can want it,
it can be desirable,
00:01:44.000 --> 00:01:45.880
and you cannot want it.
00:01:45.880 --> 00:01:49.060
And you need to know ways
of getting rid of it.
00:01:49.060 --> 00:01:54.340
And so we're going to talk
about vibration, about making
00:01:54.340 --> 00:01:56.870
vibration, about
suppressing vibration,
00:01:56.870 --> 00:02:01.160
about isolating sensitive
instruments from the vibration
00:02:01.160 --> 00:02:03.240
of the floor, things like that.
00:02:03.240 --> 00:02:05.570
So that's the topic of
the rest of the term.
00:02:05.570 --> 00:02:09.190
And today, we're going to talk
about single degree of freedom
00:02:09.190 --> 00:02:11.490
systems.
00:02:11.490 --> 00:02:17.460
And you might think that we're
spending an awful lot of time
00:02:17.460 --> 00:02:19.720
on single degree
of freedom systems.
00:02:19.720 --> 00:02:24.500
But actually, there's
a reason for that.
00:02:24.500 --> 00:02:27.980
Lots of things in
real life, like-- this
00:02:27.980 --> 00:02:30.000
is just an aluminum rod.
00:02:30.000 --> 00:02:32.400
This will vibrate.
00:02:32.400 --> 00:02:35.470
And continuous
systems, which this is,
00:02:35.470 --> 00:02:40.390
have a theoretically infinite
number of degrees of freedom.
00:02:40.390 --> 00:02:44.950
Yet when it comes to
talking about its vibration,
00:02:44.950 --> 00:02:50.320
it is conceptually easy to
think about the vibration
00:02:50.320 --> 00:02:54.740
of an object like this,
one natural frequency,
00:02:54.740 --> 00:02:57.280
one natural mode at a time.
00:02:57.280 --> 00:02:59.950
And in fact, you can
model that natural mode
00:02:59.950 --> 00:03:03.600
with its single degree
of freedom equivalent.
00:03:03.600 --> 00:03:06.300
And that's the way I
approach vibration.
00:03:06.300 --> 00:03:08.600
So if you can isolate
one particular mode,
00:03:08.600 --> 00:03:11.560
you can literally model it
as a Mass-Spring-Dashpot.
00:03:11.560 --> 00:03:15.950
So you need to understand the
Mass-Spring-Dashpot behavior
00:03:15.950 --> 00:03:21.590
inside and out, because it's
the vocabulary we use to do
00:03:21.590 --> 00:03:23.470
much more complicated things.
00:03:23.470 --> 00:03:26.680
So a single degree
of freedom system,
00:03:26.680 --> 00:03:32.962
like the simple pendulum,
has a natural frequency.
00:03:32.962 --> 00:03:34.295
In this case, it has mode shape.
00:03:38.710 --> 00:03:44.170
Here's another one, kind of
fun, single degree of freedom.
00:03:44.170 --> 00:03:46.196
This obviously
involves rotation.
00:03:58.356 --> 00:03:59.730
And you can figure
that out using
00:03:59.730 --> 00:04:02.500
Lagrange or whatever, single
degree of freedom systems.
00:04:02.500 --> 00:04:05.365
But now, I'm going to excite
one mode of vibration of this.
00:04:08.638 --> 00:04:09.634
[CLANG]
00:04:09.634 --> 00:04:11.630
[HIGH-PITCHED TONE]
00:04:11.630 --> 00:04:13.225
Hear the real high pitch?
00:04:16.730 --> 00:04:19.079
I'll get it down here by the
mic so that people at home
00:04:19.079 --> 00:04:24.430
can hear it-- about a
kilohertz, way up there.
00:04:24.430 --> 00:04:27.300
And that's one natural
mode of this thing
00:04:27.300 --> 00:04:28.620
in longitudinal vibration.
00:04:28.620 --> 00:04:32.511
When I thump it sideways--
00:04:32.511 --> 00:04:33.010
[CLANG]
00:04:33.010 --> 00:04:33.900
[LOWER TONE]
00:04:33.900 --> 00:04:34.870
you hear a lower tone.
00:04:34.870 --> 00:04:35.750
Hear that?
00:04:35.750 --> 00:04:36.760
[HUMS LOW]
00:04:36.760 --> 00:04:37.647
rather than--
00:04:37.647 --> 00:04:38.581
[HUMS HIGH]
00:04:38.581 --> 00:04:40.920
[CLANG]
00:04:40.920 --> 00:04:43.250
That's bending
vibration of this thing.
00:04:43.250 --> 00:04:45.040
But each mode of
vibration I can think
00:04:45.040 --> 00:04:47.930
of in terms of its equivalent
single degree of freedom
00:04:47.930 --> 00:04:48.900
oscillate.
00:04:48.900 --> 00:04:51.430
So we'll get to talking
about these things
00:04:51.430 --> 00:04:54.140
a little bit-- continuous
systems-- in the last couple
00:04:54.140 --> 00:04:55.760
lectures of the term.
00:04:55.760 --> 00:04:57.940
But for today then,
we're really going
00:04:57.940 --> 00:05:01.090
to develop this vocabulary
around the vibration
00:05:01.090 --> 00:05:04.620
of single degree
of freedom systems.
00:05:04.620 --> 00:05:05.486
So let's start.
00:05:24.643 --> 00:05:27.640
All right.
00:05:27.640 --> 00:05:31.310
So to keep it from
being totally boring,
00:05:31.310 --> 00:05:35.280
I'm going to start with a
little Mass-Spring-Dashpot that
00:05:35.280 --> 00:05:36.305
has two springs.
00:05:45.280 --> 00:05:49.210
And they're of such a
length that unstretched,
00:05:49.210 --> 00:05:52.780
they just meet in the middle.
00:05:52.780 --> 00:06:00.050
And then, I'm going
to take a mass
00:06:00.050 --> 00:06:04.170
and I'm going to squeeze it
in between these two springs--
00:06:04.170 --> 00:06:08.170
I can't draw a spring very
well today-- and this is k1
00:06:08.170 --> 00:06:11.580
and this is k2 and here's m.
00:06:11.580 --> 00:06:14.150
And we'll put it on a roller
so it's obviously constrained
00:06:14.150 --> 00:06:17.910
to motion in one direction.
00:06:17.910 --> 00:06:24.830
And I'll pick this
point here as the place
00:06:24.830 --> 00:06:28.244
I'm going to put my
inertial coordinate.
00:06:28.244 --> 00:06:30.160
So my inertial coordinate's
just measured from
00:06:30.160 --> 00:06:32.740
or happens to be where the
endpoints of these two springs
00:06:32.740 --> 00:06:34.080
were.
00:06:34.080 --> 00:06:36.520
Now, to squeeze
the spring in here,
00:06:36.520 --> 00:06:40.720
I have this clearly
pre-compression
00:06:40.720 --> 00:06:41.990
in these springs.
00:06:41.990 --> 00:06:49.050
So we are no longer in a
zero-force state, right?
00:06:49.050 --> 00:06:51.660
So and I want to get the
equations of motion in this.
00:06:51.660 --> 00:06:53.772
And moreover, I
want to predict--
00:06:53.772 --> 00:06:55.730
I want to find out what's
the natural frequency
00:06:55.730 --> 00:06:58.320
of this spring.
00:06:58.320 --> 00:07:00.610
So let's check your intuition.
00:07:00.610 --> 00:07:02.690
So write down on
your piece of paper
00:07:02.690 --> 00:07:08.120
whether or not the natural
frequency will be different
00:07:08.120 --> 00:07:11.090
because there's pre-compression,
or whether or not
00:07:11.090 --> 00:07:12.810
that pre-compression
in the springs
00:07:12.810 --> 00:07:16.880
has nothing to do with
the natural frequency.
00:07:16.880 --> 00:07:19.640
So write down on your
paper "natural frequency
00:07:19.640 --> 00:07:25.442
is different" or "natural
frequency is the same."
00:07:25.442 --> 00:07:26.650
Let's have a prediction here.
00:07:30.379 --> 00:07:32.170
And then, we'll set
about figuring this out
00:07:32.170 --> 00:07:33.669
and in the course
of doing it, we'll
00:07:33.669 --> 00:07:35.680
develop a little vocabulary.
00:07:35.680 --> 00:07:37.610
All through the
course so far, when
00:07:37.610 --> 00:07:39.160
we've done equations
of motion, we've
00:07:39.160 --> 00:07:42.590
usually picked the
zero-spring-force position.
00:07:42.590 --> 00:07:46.310
And we sort of led you down
this rosy path that suggests
00:07:46.310 --> 00:07:47.500
that's the way we do it.
00:07:47.500 --> 00:07:50.340
But there are other
ways that you're
00:07:50.340 --> 00:07:52.800
going to find that are
preferable to that, sometimes.
00:07:52.800 --> 00:07:56.150
So that's one of the reasons
I'm doing this example.
00:07:56.150 --> 00:07:57.665
So let's do a free body diagram.
00:08:06.030 --> 00:08:08.520
And if I held this
mass, for example,
00:08:08.520 --> 00:08:10.970
right at the center when
I put the springs in,
00:08:10.970 --> 00:08:13.590
it's obvious that this
spring gets compressed
00:08:13.590 --> 00:08:16.200
by half of the
length of the mass
00:08:16.200 --> 00:08:18.010
and this spring gets
compressed by half
00:08:18.010 --> 00:08:20.160
of the length of
the mass, right?
00:08:20.160 --> 00:08:23.980
So this is going to be L long.
00:08:23.980 --> 00:08:26.340
So if I held it
right in the middle,
00:08:26.340 --> 00:08:28.715
it would compress L/2 and L/2.
00:08:28.715 --> 00:08:30.590
But then, when I release
it, if these springs
00:08:30.590 --> 00:08:32.580
are a different
spring constant, it's
00:08:32.580 --> 00:08:33.840
going to move a little bit.
00:08:33.840 --> 00:08:37.260
So the force on this
side pushing back
00:08:37.260 --> 00:08:43.330
is sum k1 times L/2
minus the distance
00:08:43.330 --> 00:08:46.660
that I move in that direction,
which would relieve it.
00:08:46.660 --> 00:08:49.690
And the force on this
side also pushes back.
00:08:49.690 --> 00:08:56.840
It's k2 times L/2 over
2 plus x, because when
00:08:56.840 --> 00:09:00.550
I go in that direction, I'm
compressing it even further.
00:09:00.550 --> 00:09:04.360
And those are the total
forces in the x-direction
00:09:04.360 --> 00:09:05.220
on this body.
00:09:05.220 --> 00:09:08.040
There's an N and an
mg, which we know
00:09:08.040 --> 00:09:09.930
we don't have to deal
with because we're only
00:09:09.930 --> 00:09:12.750
interested in motion
left and right.
00:09:12.750 --> 00:09:13.250
All right?
00:09:13.250 --> 00:09:18.980
So we can say sum of the
forces in the x-direction,
00:09:18.980 --> 00:09:22.810
mass times the acceleration.
00:09:22.810 --> 00:09:39.130
And those forces are k1 L/2
minus x minus k2 L/2 plus x.
00:09:39.130 --> 00:09:44.840
And that's the complete equation
of motion for this problem.
00:09:44.840 --> 00:09:47.060
And rearrange it so
that I get the functions
00:09:47.060 --> 00:09:50.006
of x together here.
00:09:50.006 --> 00:10:00.160
So mx double dot plus
k1 plus k2 times x
00:10:00.160 --> 00:10:10.690
equals L/2 times k1 minus k2.
00:10:10.690 --> 00:10:13.410
And that's your
equation of motion.
00:10:13.410 --> 00:10:15.050
It's non-homogeneous.
00:10:15.050 --> 00:10:19.620
This is all constants
on the right-hand side.
00:10:19.620 --> 00:10:25.672
And on the left-hand side are
the functions of x, right?
00:10:25.672 --> 00:10:27.930
So what's the natural
frequency of the system?
00:10:34.888 --> 00:10:37.380
AUDIENCE: Square root
of k1 plus k2 over m.
00:10:37.380 --> 00:10:40.530
J. KIM VANDIVER: I hear a
square root of the quantity k1
00:10:40.530 --> 00:10:44.300
plus k2, the stiffness,
divided by m, k over m,
00:10:44.300 --> 00:10:46.899
a usual
Mass-Spring-Dashpot system.
00:10:46.899 --> 00:10:48.440
Did the pre-compression
have anything
00:10:48.440 --> 00:10:50.490
to do with the
natural frequency?
00:10:50.490 --> 00:10:52.960
I won't ask you to embarrass
yourselves, but a few of you
00:10:52.960 --> 00:10:56.030
probably got that
wrong, all right?
00:10:56.030 --> 00:11:00.440
So there's a lesson in this
that I want you to go away with.
00:11:08.090 --> 00:11:09.740
and I'll say it once.
00:11:09.740 --> 00:11:17.090
And that is when
an external force
00:11:17.090 --> 00:11:23.000
has nothing to do with
the motion coordinates
00:11:23.000 --> 00:11:24.739
in the problem.
00:11:24.739 --> 00:11:26.405
It doesn't affect the
natural frequency.
00:11:29.900 --> 00:11:31.320
These come from external forces.
00:11:31.320 --> 00:11:34.340
These are these
pre-compressions, right?
00:11:34.340 --> 00:11:39.000
And I can separate them out and
they are not functions of x.
00:11:39.000 --> 00:11:41.215
The stuff on the right-hand
side of the equation,
00:11:41.215 --> 00:11:44.010
that's not a function
of the motion variable--
00:11:44.010 --> 00:11:47.820
cannot affect the
natural frequency.
00:11:47.820 --> 00:11:49.550
So I'll give you another one.
00:11:52.070 --> 00:11:59.910
This is our common thing
hanging from a stick.
00:12:02.890 --> 00:12:05.940
I've taken my system
I built the other day
00:12:05.940 --> 00:12:11.620
for a different purpose,
but now, it's just a mass
00:12:11.620 --> 00:12:13.910
hanging from a spring.
00:12:13.910 --> 00:12:16.420
And it's right now at
its equilibrium position
00:12:16.420 --> 00:12:21.862
or there's non-zero
force in the spring.
00:12:21.862 --> 00:12:23.320
It clearly has a
natural frequency.
00:12:26.070 --> 00:12:29.205
And is that natural frequency
a function of gravity?
00:12:32.420 --> 00:12:34.870
And so if you go to write
the equation to motion
00:12:34.870 --> 00:12:48.940
of this system, you would find
mx double dot plus kx equals mg
00:12:48.940 --> 00:12:49.630
g.
00:12:49.630 --> 00:12:51.350
But the mg is not
a function of x.
00:12:51.350 --> 00:12:53.600
The natural frequency's
again, the square root of k/m.
00:13:01.470 --> 00:13:04.360
Now, we want to talk about
solving this differential
00:13:04.360 --> 00:13:05.540
equation.
00:13:05.540 --> 00:13:07.380
And because it's got
this constant term
00:13:07.380 --> 00:13:09.350
in the right-hand side,
it's non-homogeneous,
00:13:09.350 --> 00:13:11.760
which is kind of a nuisance
term in terms of dealing
00:13:11.760 --> 00:13:13.390
with a differential equation.
00:13:13.390 --> 00:13:19.610
It'd be a lot nicer if the
right-hand side were 0.
00:13:19.610 --> 00:13:26.010
So I want to make the
right-hand side of this one 0.
00:13:26.010 --> 00:13:32.180
And draw a use of a
conclusion from that.
00:13:50.540 --> 00:13:51.920
First thing I need
to know is I'd
00:13:51.920 --> 00:13:56.060
like to know what is the static
equilibrium position of this.
00:14:07.740 --> 00:14:09.674
And when you go to compute
static equilibrium,
00:14:09.674 --> 00:14:11.090
you look at the
equation of motion
00:14:11.090 --> 00:14:13.630
and you say, make all
motion variables things
00:14:13.630 --> 00:14:15.560
that are functions of time 0.
00:14:15.560 --> 00:14:19.160
So no acceleration--
you're left with this.
00:14:19.160 --> 00:14:21.730
So you just solve this for
whatever the value of x is
00:14:21.730 --> 00:14:24.670
and I'll call it x
of s for x-static.
00:14:24.670 --> 00:14:26.290
And you'll find
that, oh, well, it's
00:14:26.290 --> 00:14:34.770
that term divided
by k1 plus k2, k1
00:14:34.770 --> 00:14:40.100
minus k2 all over k1 plus k2.
00:14:40.100 --> 00:14:41.465
And that's the static position.
00:14:47.970 --> 00:14:50.910
So now, let's say, ah,
well, we started off
00:14:50.910 --> 00:14:54.340
with this motion variable that
wasn't arbitrarily defined
00:14:54.340 --> 00:14:56.540
at the middle.
00:14:56.540 --> 00:15:00.380
And let's say that,
well, it's made up
00:15:00.380 --> 00:15:05.060
of a static component,
which is a constant, just
00:15:05.060 --> 00:15:09.720
a value, plus a
dynamic component I'll
00:15:09.720 --> 00:15:11.450
call x of d, which moves.
00:15:11.450 --> 00:15:14.310
This is the function of time.
00:15:14.310 --> 00:15:15.240
This is a constant.
00:15:15.240 --> 00:15:17.660
It's not a function of time.
00:15:17.660 --> 00:15:19.480
And that means if I
take its derivative,
00:15:19.480 --> 00:15:21.835
I might need a value for x dot.
00:15:21.835 --> 00:15:23.280
That goes away.
00:15:23.280 --> 00:15:25.220
It's just xd dot.
00:15:25.220 --> 00:15:31.440
And x double dot
is xd double dot.
00:15:31.440 --> 00:15:36.230
And let's substitute this
into my equation of motion.
00:15:36.230 --> 00:15:51.300
So it becomes m xd double
dot plus k1 plus k2 times--
00:15:51.300 --> 00:16:00.200
and now, this term has got two
pieces now-- times xd plus k1
00:16:00.200 --> 00:16:11.420
plus k2 times xs
equals L/2 k1 minus k2.
00:16:15.418 --> 00:16:17.760
All right?
00:16:17.760 --> 00:16:22.340
Now if I say, well, let's
examine the static case,
00:16:22.340 --> 00:16:24.900
then this goes away.
00:16:24.900 --> 00:16:27.740
For the static equilibrium
case, this term is 0.
00:16:27.740 --> 00:16:31.790
This term is 0 because
the dynamic motion
00:16:31.790 --> 00:16:35.710
is 0 in the static case.
00:16:35.710 --> 00:16:39.460
That xd is motion about the
static equilibrium position.
00:16:39.460 --> 00:16:41.860
So for static case,
these two terms go away
00:16:41.860 --> 00:16:44.920
and we know that
this equals that.
00:16:44.920 --> 00:16:47.110
But if that's true, we
can get rid of these.
00:16:47.110 --> 00:16:49.530
They cancel one another.
00:16:49.530 --> 00:16:55.360
These terms cancel and I'm
left with m xd double dot
00:16:55.360 --> 00:17:03.460
plus k equivalent, I'll
call it, xd equals 0.
00:17:03.460 --> 00:17:05.589
So the k equivalent's
just the total stiffnesses
00:17:05.589 --> 00:17:07.297
in the system, whatever
works out, right?
00:17:07.297 --> 00:17:10.960
In this case, it's k1 plus
k2 and the natural frequency,
00:17:10.960 --> 00:17:16.440
omega n, is the square root
of k equivalent divided by m.
00:17:19.450 --> 00:17:23.709
So most often, if you're
interested in vibration,
00:17:23.709 --> 00:17:25.500
you're interested in
natural frequencies,
00:17:25.500 --> 00:17:29.360
you're interested in solving
the differential equation,
00:17:29.360 --> 00:17:33.570
you will find it advantageous to
write your equations of motion
00:17:33.570 --> 00:17:37.065
around the static
equilibrium position.
00:17:40.030 --> 00:17:42.840
So I could have started
this problem by saying,
00:17:42.840 --> 00:17:44.990
whatever the static
equilibrium position is
00:17:44.990 --> 00:17:48.190
of this thing, that's
what I'm measuring x from.
00:17:50.850 --> 00:17:54.874
And then, I would have come
to this equation eventually.
00:17:54.874 --> 00:17:57.540
You'd have to figure out what is
the static equilibrium position
00:17:57.540 --> 00:18:00.560
and know what you're doing,
but once you know it,
00:18:00.560 --> 00:18:02.050
then you have the answer.
00:18:02.050 --> 00:18:06.830
Now, the same thing is
true of that problem.
00:18:06.830 --> 00:18:09.600
That's a non-homogeneous
differential equation
00:18:09.600 --> 00:18:14.610
for the hanging mass.
00:18:14.610 --> 00:18:19.610
And we derive the
equations of motion things
00:18:19.610 --> 00:18:22.780
for this many different
ways this term, all right?
00:18:22.780 --> 00:18:25.250
But we usually said,
zero-spring force.
00:18:25.250 --> 00:18:28.010
But now, if you started
from here and said,
00:18:28.010 --> 00:18:29.940
this is the static
equilibrium position,
00:18:29.940 --> 00:18:33.110
what's the motion
about this position,
00:18:33.110 --> 00:18:40.636
then you'd get the equation
with 0 on the right-hand side--
00:18:40.636 --> 00:18:42.260
lots of advantages
there to using that.
00:19:07.450 --> 00:19:10.620
All single degree of
freedom oscillators
00:19:10.620 --> 00:19:14.120
will boil down to this equation.
00:19:14.120 --> 00:19:16.540
This is one involving
translation,
00:19:16.540 --> 00:19:19.390
but for a simple pendulum.
00:19:19.390 --> 00:19:23.975
This object, for example, is a
pendulum, but it's rotational.
00:19:23.975 --> 00:19:26.100
So it's a pendulum, but
it's one degree of freedom.
00:19:28.650 --> 00:19:31.210
All pendulum problems,
if you do them
00:19:31.210 --> 00:19:33.700
about equilibrium
positions, boil
00:19:33.700 --> 00:19:39.130
down to some I with
respect to the point
00:19:39.130 --> 00:19:44.580
that they're rocking
about, theta double
00:19:44.580 --> 00:19:52.030
dot plus some Kt, torsional
spring constant theta,
00:19:52.030 --> 00:19:52.990
equals 0.
00:19:52.990 --> 00:19:55.300
They take the same form.
00:19:55.300 --> 00:19:57.545
So all translational
single degree
00:19:57.545 --> 00:20:00.020
of freedom systems, all
rotational single degree
00:20:00.020 --> 00:20:03.070
of freedom systems, it's the
same differential equation--
00:20:03.070 --> 00:20:06.000
just this involves mass
and linear acceleration.
00:20:06.000 --> 00:20:08.370
This involves mass moment
of inertia and rotational
00:20:08.370 --> 00:20:10.270
acceleration.
00:20:10.270 --> 00:20:13.260
So everything that I
say about the solution
00:20:13.260 --> 00:20:15.710
to single degree
of freedom systems
00:20:15.710 --> 00:20:19.390
applies to both
types of problems.
00:20:19.390 --> 00:20:27.900
So let's look into the solution
of this equation briefly.
00:20:33.040 --> 00:20:35.545
Mostly, I'm doing this to
establish some terminology.
00:20:38.960 --> 00:20:45.320
So a solution I know or
I can show that xd of t,
00:20:45.320 --> 00:20:47.590
the solution to this
problem-- notice,
00:20:47.590 --> 00:20:50.090
are there any external forces,
by the way, excitations,
00:20:50.090 --> 00:20:51.320
f of t's or anything?
00:20:51.320 --> 00:20:52.350
No.
00:20:52.350 --> 00:20:55.170
So this thing has no
external excitation
00:20:55.170 --> 00:20:56.620
that's going to make it move.
00:20:56.620 --> 00:21:02.640
So it's only source of
vibration or motion is what?
00:21:02.640 --> 00:21:08.040
Comes from-- I hear
initial conditions, right?
00:21:08.040 --> 00:21:10.670
You have to do
something to perturb it
00:21:10.670 --> 00:21:13.980
and then it will vibrate.
00:21:13.980 --> 00:21:15.050
So here it is.
00:21:15.050 --> 00:21:18.960
It's about its
equilibrium position.
00:21:18.960 --> 00:21:23.630
I give it an initial
deflection and let go.
00:21:23.630 --> 00:21:26.560
Or it's around its
initial condition
00:21:26.560 --> 00:21:29.540
and I give it an
initial velocity.
00:21:29.540 --> 00:21:32.130
It also responds to some
combination of the two.
00:21:32.130 --> 00:21:34.190
So initial conditions
are the only things
00:21:34.190 --> 00:21:37.300
that account for
motion of something
00:21:37.300 --> 00:21:39.500
without external excitation.
00:21:39.500 --> 00:21:42.430
And that motion, I can
write that solution
00:21:42.430 --> 00:21:49.360
as A cosine omega t.
00:21:49.360 --> 00:21:53.070
You'll find this is
a possible solution.
00:21:53.070 --> 00:21:59.000
B sine omega t is another
possible solution.
00:21:59.000 --> 00:22:06.690
Sum A cosine omega t minus
phase angle's also a solution.
00:22:06.690 --> 00:22:15.210
And sum A e to the i omega t
you'll find is also a solution.
00:22:15.210 --> 00:22:17.150
Any of those things
you could throw in
00:22:17.150 --> 00:22:20.350
and the precise values of these
things, the A's, the B's, the
00:22:20.350 --> 00:22:23.965
phi's, and so forth depend on--
00:22:23.965 --> 00:22:25.340
AUDIENCE: the
initial conditions.
00:22:25.340 --> 00:22:26.740
J. KIM VANDIVER: The
initial conditions.
00:22:26.740 --> 00:22:28.040
So let's do this one quickly.
00:22:37.950 --> 00:22:40.370
All right.
00:22:40.370 --> 00:22:54.600
And I'll choose And I'm going to
stop writing the x sub d here.
00:22:54.600 --> 00:23:02.356
This is now my position
from the equilibrium point.
00:23:02.356 --> 00:23:05.660
So x of t-- I'm
going to say, let
00:23:05.660 --> 00:23:16.690
it be an A1 cosine omega
t plus a B1 sine omega t
00:23:16.690 --> 00:23:18.460
and plug it in.
00:23:18.460 --> 00:23:28.670
When I plug it into
the equation of motion,
00:23:28.670 --> 00:23:31.120
x double dot requires you to
take two derivatives of each
00:23:31.120 --> 00:23:31.830
of these terms.
00:23:31.830 --> 00:23:34.940
Two derivatives of cosine
gives you minus omega cosine.
00:23:34.940 --> 00:23:39.690
Two derivative sine minus omega
squared cosine minus omega
00:23:39.690 --> 00:23:40.850
squared sine.
00:23:40.850 --> 00:23:46.390
So the answer comes
out minus m omega
00:23:46.390 --> 00:23:56.300
squared plus k equivalent
here times A1 cosine
00:23:56.300 --> 00:24:07.836
plus B1 sine-- omega t's
obviously in them-- equals 0.
00:24:07.836 --> 00:24:09.710
So I just plugged in
that equation of motion.
00:24:09.710 --> 00:24:11.790
I get this back.
00:24:11.790 --> 00:24:13.250
This is what I started with.
00:24:13.250 --> 00:24:15.670
That's x.
00:24:15.670 --> 00:24:22.990
In general, it is not
equal to 0, can take on all
00:24:22.990 --> 00:24:23.730
sorts of values.
00:24:23.730 --> 00:24:27.350
So that's not generally 0
and that means this must be.
00:24:30.470 --> 00:24:32.600
And from this, then,
when we solve this,
00:24:32.600 --> 00:24:37.907
we find that omega what we
call n squared is k over m.
00:24:37.907 --> 00:24:40.490
And that's, of course, where our
natural frequency comes from.
00:24:40.490 --> 00:24:54.349
This is called the
undamped natural frequency,
00:24:54.349 --> 00:24:56.390
because there's no dampening
in this problem yet.
00:24:56.390 --> 00:24:57.806
We get the square
root of k over m
00:24:57.806 --> 00:25:01.650
is the natural
frequency of the system.
00:25:01.650 --> 00:25:04.420
Let's find out
what are A1 and B1.
00:25:04.420 --> 00:25:13.864
Well, let's let x0 be
x at t equals 0 here.
00:25:16.850 --> 00:25:19.950
And if we just plug that in
here, put t equals 0 here,
00:25:19.950 --> 00:25:21.270
cosine goes to 1.
00:25:21.270 --> 00:25:23.140
This term goes away.
00:25:23.140 --> 00:25:30.080
So this implies
that A1 equals x0.
00:25:30.080 --> 00:25:33.970
So we find out right away
that the A1 cosine omega
00:25:33.970 --> 00:25:38.100
t takes care of the response
to an initial deflection.
00:25:38.100 --> 00:25:54.700
And we need a x dot here minus
A1 omega sine omega t plus B1
00:25:54.700 --> 00:25:58.440
omega cosine omega t.
00:25:58.440 --> 00:26:02.230
That's the derivative of x.
00:26:02.230 --> 00:26:04.550
You know the solution's that,
so its first derivative,
00:26:04.550 --> 00:26:06.640
the velocity, must
look like this.
00:26:06.640 --> 00:26:13.450
And let's let v0 equals
x dot at t equals 0.
00:26:13.450 --> 00:26:17.180
When we plug that in,
this term goes away
00:26:17.180 --> 00:26:23.570
and we get B1 omega
and cosine is 1.
00:26:23.570 --> 00:26:29.750
So therefore, B1
is v0 over omega.
00:26:29.750 --> 00:26:33.120
But in fact, the
omega's omega n,
00:26:33.120 --> 00:26:38.740
because we already found that,
that the only frequency that
00:26:38.740 --> 00:26:42.920
satisfies the equation
of motion when
00:26:42.920 --> 00:26:46.530
you have only initial
conditions in the system,
00:26:46.530 --> 00:26:49.360
the only frequency that
is allowed in the answer
00:26:49.360 --> 00:26:52.460
is the natural frequency.
00:26:52.460 --> 00:27:02.630
So we now know B1 is v0
over omega n and A1 is x0.
00:27:02.630 --> 00:27:08.300
So if I give you any combination
of initial displacement
00:27:08.300 --> 00:27:12.120
and initial velocity, you can
write out for me the exact time
00:27:12.120 --> 00:27:14.510
history of the motion.
00:27:14.510 --> 00:27:19.370
X0 to cosine omega t plus
v0 over omega n sine omega t
00:27:19.370 --> 00:27:23.740
is the complete solution for a
response to initial conditions.
00:27:55.780 --> 00:28:05.920
So any translational
oscillator one degree
00:28:05.920 --> 00:28:13.950
of freedom where you have
a translational coordinate
00:28:13.950 --> 00:28:18.730
measured from its
equilibrium position
00:28:18.730 --> 00:28:23.290
has the equation of
motion-- actually,
00:28:23.290 --> 00:28:24.290
you've done this enough.
00:28:24.290 --> 00:28:27.620
But if we added a force here
and we added some damping
00:28:27.620 --> 00:28:29.680
and I wanted the equation
of motion of this,
00:28:29.680 --> 00:28:34.830
you know that it's m
x double dot plus b x
00:28:34.830 --> 00:28:42.430
dot plus kx equals F of t.
00:28:42.430 --> 00:28:45.034
And so you're going to be
confronted with problems--
00:28:45.034 --> 00:28:46.700
find the equation of
motion in a system.
00:28:46.700 --> 00:28:47.950
It comes up looking
like that and they say,
00:28:47.950 --> 00:28:49.585
what's the natural frequency?
00:28:49.585 --> 00:28:50.850
And I've been a little sloppy.
00:28:50.850 --> 00:28:55.310
I really mean, what's the
undamped natural frequency?
00:28:55.310 --> 00:28:56.890
And so to find the
undamped-- when
00:28:56.890 --> 00:28:59.360
one says that, what's the
undamped natural frequency,
00:28:59.360 --> 00:29:11.500
you just temporarily let b
and F be 0, just temporarily,
00:29:11.500 --> 00:29:19.932
and solve then for omega n
equals square root of k/n.
00:29:19.932 --> 00:29:21.180
It's what you do.
00:29:21.180 --> 00:29:26.890
And then, so we know this
is a parameter that tells us
00:29:26.890 --> 00:29:29.700
about the behavior of the
system, which we always
00:29:29.700 --> 00:29:32.320
want to know for the single
degree of freedom systems.
00:29:32.320 --> 00:29:34.390
What is the natural
frequency of the system?
00:29:42.712 --> 00:29:49.050
And we know for b
equals 0 and F of 0,
00:29:49.050 --> 00:29:52.590
then the response can be only
due to initial conditions.
00:29:52.590 --> 00:29:54.636
So we have x of t.
00:29:54.636 --> 00:29:59.770
We know it's going to be
some x0 cosine omega n
00:29:59.770 --> 00:30:08.110
t plus v0 over omega
n sine omega n t.
00:30:08.110 --> 00:30:11.510
And every simple vibration
system in the world
00:30:11.510 --> 00:30:14.310
behaves basically like this
from initial conditions.
00:30:14.310 --> 00:30:17.570
It'll be some part responding
to the initial displacement,
00:30:17.570 --> 00:30:19.830
some part to the
initial velocity.
00:30:19.830 --> 00:30:23.780
And damping is going to make
it a little bit more complex,
00:30:23.780 --> 00:30:25.470
but not actually by much.
00:30:25.470 --> 00:30:29.180
The same basic terms appear even
when you have damping in it.
00:30:33.940 --> 00:30:45.640
This can be expressed as sum
A cosine omega, in this case,
00:30:45.640 --> 00:30:48.110
n t minus the phase angle.
00:30:48.110 --> 00:30:53.820
And it's useful to know
this trigonometric identity
00:30:53.820 --> 00:30:55.980
to be able to put
things together
00:30:55.980 --> 00:30:57.660
into an expression like that.
00:30:57.660 --> 00:31:07.100
And you'll find out that A is
just the square root of the two
00:31:07.100 --> 00:31:07.630
pieces.
00:31:07.630 --> 00:31:09.270
It's a sine and cosine term.
00:31:09.270 --> 00:31:16.910
So you have an x0 squared
plus a v0 over omega n squared
00:31:16.910 --> 00:31:18.970
square root.
00:31:18.970 --> 00:31:21.620
Remember, this is any A and
B. It's just a square root
00:31:21.620 --> 00:31:23.620
of A squared plus B squared.
00:31:23.620 --> 00:31:25.100
That's what we're doing here.
00:31:25.100 --> 00:31:34.360
And the phase angle,
the tangent inverse
00:31:34.360 --> 00:31:38.270
of this-- we've been calling
this like an A and this
00:31:38.270 --> 00:31:39.680
is the B quantity.
00:31:39.680 --> 00:31:46.230
So tangent inverse of--
get my signs right--
00:31:46.230 --> 00:31:49.480
B over A, which
in this case then
00:31:49.480 --> 00:32:12.710
is tangent inverse of
v0 over x0 omega n.
00:32:12.710 --> 00:32:15.310
That's all there is to it.
00:32:15.310 --> 00:32:20.030
And finally, another
trig thing that you
00:32:20.030 --> 00:32:22.910
need to know-- we're going
to use it quite a bit--
00:32:22.910 --> 00:32:29.090
is that if you have an
expression A cosine omega
00:32:29.090 --> 00:32:46.278
t minus phi, that's equal to the
real part of A e to the i omega
00:32:46.278 --> 00:32:46.778
t.
00:32:50.200 --> 00:32:57.710
And if A is real and-- I don't
want to write it that way--
00:32:57.710 --> 00:33:09.550
when A is real, it's A times
e to the i omega t minus phi,
00:33:09.550 --> 00:33:13.530
because Euler's formula
says e to the i theta
00:33:13.530 --> 00:33:19.850
equals cosine theta
plus i sine of theta.
00:33:19.850 --> 00:33:22.590
So if you have an
i omega t minus phi
00:33:22.590 --> 00:33:26.070
here, you get back
a cosine omega
00:33:26.070 --> 00:33:30.240
t minus phi and another term,
an i sine omega t minus phi.
00:33:30.240 --> 00:33:36.130
So you can always express
that as the real part of that.
00:33:36.130 --> 00:33:38.130
So we're going to need
that little trig identity
00:33:38.130 --> 00:33:39.230
as we go through the term.
00:33:44.320 --> 00:33:54.925
Now, I've found in many
years of teaching vibration
00:33:54.925 --> 00:33:58.800
that something
that many students
00:33:58.800 --> 00:34:02.810
find a little confusing is
this notion of phase angle.
00:34:02.810 --> 00:34:05.400
What does "phase
angle" really mean?
00:34:05.400 --> 00:34:09.080
So I'll try to explain it to
you in a couple different ways.
00:34:09.080 --> 00:34:15.159
So let's look at
what this vibration
00:34:15.159 --> 00:34:18.070
that we're talking about
here, x0 cosine omega
00:34:18.070 --> 00:34:24.179
t plus v0 over omega n
sine-- what's it look like?
00:34:24.179 --> 00:34:32.530
So that's-- we've just got our--
and we see what it looks like.
00:34:32.530 --> 00:34:37.520
But if you plot the motion of
this thing just versus time,
00:34:37.520 --> 00:34:42.899
what's it look like and where
does phase angle come into it?
00:34:42.899 --> 00:34:49.560
So this is now x
of t and this is
00:34:49.560 --> 00:34:58.894
t equals 0 and this
undamped system
00:34:58.894 --> 00:35:00.860
is essentially going
to look like that.
00:35:04.540 --> 00:35:07.730
And this is the value
x0, the amplitude,
00:35:07.730 --> 00:35:11.900
the initial condition on
x that you began with.
00:35:11.900 --> 00:35:21.420
And right here, the slope-- v0
is the slope, the initial slope
00:35:21.420 --> 00:35:27.140
of this curve, right, because
the time derivative is F x dot.
00:35:27.140 --> 00:35:32.320
If we were plotting x dot, the
initial velocity is omega x0.
00:35:32.320 --> 00:35:36.200
And so it's just the
slope is v0 here.
00:35:36.200 --> 00:35:38.450
So this is your
initial velocity.
00:35:38.450 --> 00:35:42.540
This is the-- and I
didn't-- yeah, that's right.
00:35:50.850 --> 00:35:54.570
This is the initial
displacement.
00:35:54.570 --> 00:35:56.290
The total written
out mathematically,
00:35:56.290 --> 00:35:58.100
it looks like that.
00:35:58.100 --> 00:36:04.040
And I'm plotting this function,
A cosine omega t minus phi.
00:36:04.040 --> 00:36:05.660
Yeah?
00:36:05.660 --> 00:36:06.510
Did I see a hand up?
00:36:06.510 --> 00:36:12.310
AUDIENCE: Does x0 at t equals
0 or is it a little bit after?
00:36:12.310 --> 00:36:14.050
J. KIM VANDIVER:
Well, I was just
00:36:14.050 --> 00:36:16.174
looking at it myself and
said, this can't be right.
00:36:18.160 --> 00:36:22.290
This has got to be the initial
condition on x and this
00:36:22.290 --> 00:36:24.250
has to be the initial
condition on v.
00:36:24.250 --> 00:36:27.380
Now, whatever this
turns out to be is
00:36:27.380 --> 00:36:28.589
whatever it turns out to be.
00:36:28.589 --> 00:36:29.880
You have some initial velocity.
00:36:29.880 --> 00:36:31.550
You have some
initial displacement.
00:36:31.550 --> 00:36:34.640
The system can actually
peak out sometime later
00:36:34.640 --> 00:36:36.550
at a maximum value, right?
00:36:36.550 --> 00:36:41.550
And that maximum value is that.
00:36:41.550 --> 00:36:46.800
So this over here
is the square root
00:36:46.800 --> 00:36:57.210
of x0 squared plus v0 over
omega n squared square root.
00:36:57.210 --> 00:36:58.570
That's what the peak value is.
00:36:58.570 --> 00:37:01.680
And this system's undamped,
so it just goes on forever.
00:37:01.680 --> 00:37:07.120
So the question is, though,
what is this gap here
00:37:07.120 --> 00:37:11.400
between when it starts and
when it meets its maximum?
00:37:11.400 --> 00:37:19.680
Well, when we use an
expression like-- we
00:37:19.680 --> 00:37:24.400
said we can express this as
some A cosine omega t minus phi.
00:37:24.400 --> 00:37:27.250
It's just the point at
which the cosine then
00:37:27.250 --> 00:37:28.185
reaches its maximum.
00:37:32.290 --> 00:37:40.870
So if this axis here is omega
t, if we plot this actually
00:37:40.870 --> 00:37:45.350
versus omega t,
then one full cycle
00:37:45.350 --> 00:37:51.150
here is 2 pi or 360 degrees.
00:37:51.150 --> 00:37:55.470
So if you plot it versus
omega t, then this gap in here
00:37:55.470 --> 00:37:58.510
is just phi.
00:37:58.510 --> 00:38:02.370
That's the delay in
angle, if you will,
00:38:02.370 --> 00:38:04.680
that the system goes
through between getting
00:38:04.680 --> 00:38:06.740
from the initial
conditions to getting
00:38:06.740 --> 00:38:08.160
to the peak of the cosine.
00:38:14.460 --> 00:38:20.250
And phi must also then
be equal to some omega
00:38:20.250 --> 00:38:28.080
n times a delta tau, I'll
call it, some time delay.
00:38:28.080 --> 00:38:35.500
So if this is plotted--
if this axis is time--
00:38:35.500 --> 00:38:50.580
not omega t, but time-- then x
the same plot, this delay here,
00:38:50.580 --> 00:38:52.840
this is a time delay.
00:38:52.840 --> 00:38:55.060
And when you plot
it against time,
00:38:55.060 --> 00:38:58.640
it's a delay in time
to get to the peak.
00:38:58.640 --> 00:39:04.750
And omega n delta
tau, this delay,
00:39:04.750 --> 00:39:07.390
must be equal to
the phase angle.
00:39:07.390 --> 00:39:13.730
So the delta tau, this time
delay, is phi over omega n.
00:39:16.560 --> 00:39:20.220
So you can think about
this as a delay in time
00:39:20.220 --> 00:39:23.890
or as a shift in phase angle,
depending on whether or not
00:39:23.890 --> 00:39:27.130
you want to plot this thing
as a function of omega t
00:39:27.130 --> 00:39:29.340
or as a function of time.
00:39:29.340 --> 00:39:33.820
But you're going to need
this concept of phase angle
00:39:33.820 --> 00:39:36.690
the rest of the term.
00:39:36.690 --> 00:39:39.035
Want to ask any
questions about phase?
00:39:45.190 --> 00:39:51.170
Because we're doing vibration
for the remainder of the term,
00:39:51.170 --> 00:39:55.510
this is an introduction to a
topic called linear systems.
00:39:55.510 --> 00:39:58.960
And so this is basically the
fundamental stuff in which you
00:39:58.960 --> 00:40:01.230
then, when you go
on to 2004, which
00:40:01.230 --> 00:40:03.040
is controls and
that sort of thing,
00:40:03.040 --> 00:40:06.900
this is the basic intro to it.
00:40:06.900 --> 00:40:11.350
And we'll talk more about linear
system behavior as we go along.
00:40:20.600 --> 00:40:23.210
Now, we're going to do something
that you've-- much of this
00:40:23.210 --> 00:40:24.920
stuff I know you've seen before.
00:40:24.920 --> 00:40:29.060
Some of the new parts is
just vocabulary and ways
00:40:29.060 --> 00:40:31.020
of thinking about
vibration that engineers
00:40:31.020 --> 00:40:33.880
do that mathematicians
tend not to.
00:40:33.880 --> 00:40:36.896
So you have seen most of
this stuff before where?
00:40:36.896 --> 00:40:37.651
AUDIENCE: 1.803.
00:40:37.651 --> 00:40:38.900
J. KIM VANDIVER: 1.803, right?
00:40:38.900 --> 00:40:40.340
You've done all this.
00:40:40.340 --> 00:40:49.210
And a year ago last May, in
May, I taught the 1803 lecture
00:40:49.210 --> 00:40:51.120
with Professor Haynes Miller.
00:40:51.120 --> 00:40:52.600
Now, if you had
1.803 last spring,
00:40:52.600 --> 00:40:54.799
I think you had
somebody different.
00:40:54.799 --> 00:40:56.090
But he invited me to come here.
00:40:56.090 --> 00:40:57.620
It was in the same
classroom and we
00:40:57.620 --> 00:41:02.210
taught the second-order ordinary
differential equation together.
00:41:02.210 --> 00:41:03.720
It was really a lot of fun.
00:41:03.720 --> 00:41:05.529
He said, well,
here's what we do.
00:41:05.529 --> 00:41:07.070
And then, I said,
oh, well, engineers
00:41:07.070 --> 00:41:08.890
look at it the following way.
00:41:08.890 --> 00:41:10.940
So what I'm going
to show you is what
00:41:10.940 --> 00:41:12.270
he and I did in class that day.
00:41:12.270 --> 00:41:13.940
You can go back and
watch that on video.
00:41:13.940 --> 00:41:15.150
It's kind of fun.
00:41:15.150 --> 00:41:17.080
But I'll give you
my take on it today.
00:41:17.080 --> 00:41:18.980
So this is the
engineer's view of what
00:41:18.980 --> 00:41:21.310
you've already seen in 1.803.
00:41:21.310 --> 00:41:25.100
So we have that system and we
have that equation of motion.
00:41:25.100 --> 00:41:27.270
And the engineers
and mathematicians
00:41:27.270 --> 00:41:31.840
would more or less agree to
that m x double dot plus bx.
00:41:31.840 --> 00:41:34.120
But I went and looked at
the web page last night.
00:41:34.120 --> 00:41:37.240
Last spring, the person
used c instead of b.
00:41:37.240 --> 00:41:40.010
Haynes Miller the
year before used b.
00:41:40.010 --> 00:41:45.730
So you can't depend on
any absolute consistency.
00:41:45.730 --> 00:41:51.020
So let's start off with our
homogeneous equation here.
00:41:51.020 --> 00:41:55.380
And I'm looking now for the
response to initial conditions
00:41:55.380 --> 00:41:56.480
with damping.
00:41:56.480 --> 00:41:58.470
You've done this in 1.803.
00:41:58.470 --> 00:42:02.180
You know that you can solve
this by assuming a solution
00:42:02.180 --> 00:42:05.310
of a form Ae to the st.
00:42:05.310 --> 00:42:10.060
Plugging it in gives
you a quadratic equation
00:42:10.060 --> 00:42:18.170
that looks like s squared
plus sb plus k equals 0.
00:42:18.170 --> 00:42:19.050
This has roots.
00:42:23.770 --> 00:42:27.700
I left out my m here, so it
starts off looking like that.
00:42:27.700 --> 00:42:30.210
You divide through
by the m. s squared
00:42:30.210 --> 00:42:36.875
plus b/m s plus k/m equals 0.
00:42:36.875 --> 00:42:38.500
And that's where
Haynes would leave it.
00:42:38.500 --> 00:42:41.910
And he'd give you the entire
answer in terms of b/m and k/m
00:42:41.910 --> 00:42:43.650
and that kind of thing.
00:42:43.650 --> 00:42:47.870
Engineers, we like to call that
the natural frequency squared.
00:42:47.870 --> 00:42:50.920
And this term, we'd
modify to put it
00:42:50.920 --> 00:42:56.060
in a terminology that is more
convenient to engineering.
00:42:56.060 --> 00:42:59.140
So I'll show you
how that works out.
00:42:59.140 --> 00:43:01.100
When you solve this
quadratic just using
00:43:01.100 --> 00:43:04.385
the quadratic equation,
you get the following.
00:43:15.167 --> 00:43:17.000
You get that the roots,
there's two of them.
00:43:17.000 --> 00:43:20.060
I'll call them S1 and 2.
00:43:20.060 --> 00:43:21.900
The roots to this
equation look like
00:43:21.900 --> 00:43:34.290
minus b over 2m plus or minus
square root of b squared
00:43:34.290 --> 00:43:39.490
over 4m squared minus k/m.
00:43:42.480 --> 00:43:51.811
And that's what you'd
get to do in 1.803.
00:43:51.811 --> 00:43:56.680
And an engineer would say, well,
let's change that a little bit.
00:43:56.680 --> 00:44:03.210
So my roots that I
would use for S1 and 2,
00:44:03.210 --> 00:44:07.120
I just factor out--
that's omega n squared.
00:44:07.120 --> 00:44:10.290
I can factor that out and it
becomes omega n on the outside.
00:44:10.290 --> 00:44:13.000
And I put an omega n in the
numerator and denominator here,
00:44:13.000 --> 00:44:14.310
as well.
00:44:14.310 --> 00:44:49.030
So I get roots that
look like-- so I've just
00:44:49.030 --> 00:44:50.610
manipulated that a little bit.
00:44:53.834 --> 00:44:57.175
I have a name for this term.
00:44:57.175 --> 00:45:04.240
I use the Greek letter
zeta is b over 2 omega n
00:45:04.240 --> 00:45:07.626
m is the way I remember
it in my brain.
00:45:07.626 --> 00:45:08.875
It's called the damping ratio.
00:45:19.900 --> 00:45:24.320
And if I say that, then the
roots, S1 and 2 for this,
00:45:24.320 --> 00:45:32.060
look like minus zeta
omega n plus or minus
00:45:32.060 --> 00:45:38.788
omega n times the square
root of zeta squared minus 1.
00:45:45.340 --> 00:45:48.810
And those are the roots
that a vibration engineer
00:45:48.810 --> 00:45:55.160
would use to describe
this second-order linear
00:45:55.160 --> 00:45:58.135
differential equation
solution homogeneous solution.
00:46:00.720 --> 00:46:03.870
Those are the roots
of the equation.
00:46:03.870 --> 00:46:07.970
And when you have no damping,
then this term goes away
00:46:07.970 --> 00:46:12.010
and you're left with-- and I
left an i out of here, I think.
00:46:15.300 --> 00:46:16.800
No, I'm fine.
00:46:16.800 --> 00:46:17.900
The i comes out of here.
00:46:33.730 --> 00:46:37.240
So for one thing to absolutely
take away from today
00:46:37.240 --> 00:46:38.810
is to remember this.
00:46:41.800 --> 00:46:44.850
That's our definition of damping
called the damping ratio.
00:46:44.850 --> 00:46:47.400
When that's 1, it's a number
we call critical damping.
00:46:47.400 --> 00:46:50.060
I'll show you what
that means in a second.
00:46:50.060 --> 00:46:55.100
And when it's greater than
1, the system won't vibrate.
00:46:55.100 --> 00:46:57.660
It just has exponential decay.
00:46:57.660 --> 00:46:59.630
If it's less than 1,
you get vibration.
00:46:59.630 --> 00:47:02.320
And that's why we like to use
it this way as it's meaningful.
00:47:02.320 --> 00:47:04.040
Its value, you
instantly know if it's
00:47:04.040 --> 00:47:05.080
greater than or
less than 1, it's
00:47:05.080 --> 00:47:06.830
going to change the
behavior of the system
00:47:06.830 --> 00:47:10.190
from vibrating to not vibrating.
00:47:10.190 --> 00:47:14.580
So now, there's four
possible solutions to this.
00:47:14.580 --> 00:47:20.690
I'm not going to elaborate on
all of them, but zeta equals 0,
00:47:20.690 --> 00:47:21.795
we've already done.
00:47:21.795 --> 00:47:23.870
We know the answer to that.
00:47:23.870 --> 00:47:26.210
Response to initial
conditions-- simple.
00:47:26.210 --> 00:47:27.840
We know that one.
00:47:27.840 --> 00:47:31.960
We have another solution
when zeta's greater than 1.
00:47:31.960 --> 00:47:34.200
When zeta's greater than
1, this quantity here
00:47:34.200 --> 00:47:38.780
is the inside is greater than 1,
so it's a real positive number.
00:47:38.780 --> 00:47:42.580
And both the roots of this
thing are completely real.
00:47:46.840 --> 00:47:52.200
And you know that the--
remember the response,
00:47:52.200 --> 00:47:54.920
we hypothesize in the
beginning that response
00:47:54.920 --> 00:47:58.210
looks like some Ae to the st.
So now, we just plug back in.
00:47:58.210 --> 00:47:59.890
This is our st value.
00:47:59.890 --> 00:48:01.690
We can plug them back
in and we will get
00:48:01.690 --> 00:48:04.880
the motion of the system back.
00:48:04.880 --> 00:48:11.420
So for zeta greater
than 1, st comes out
00:48:11.420 --> 00:48:21.490
looking like minus zeta
omega n t plus or minus
00:48:21.490 --> 00:48:29.040
square root of zeta
squared minus 1 times t.
00:48:29.040 --> 00:48:32.590
And you just plug this in,
and x is just e to the st.
00:48:32.590 --> 00:48:35.750
But these are just
pure real values.
00:48:35.750 --> 00:48:40.930
And you'll find out that the
system from initial conditions
00:48:40.930 --> 00:48:43.280
on velocity and
displacement just--
00:48:43.280 --> 00:48:44.039
[WINDS DOWN]
00:48:44.039 --> 00:48:44.580
and dies out.
00:48:48.010 --> 00:48:51.180
Zeta equals to 1.
00:48:51.180 --> 00:49:00.940
Then, st is just minus-- you get
a double root-- minus omega nt,
00:49:00.940 --> 00:49:01.440
twice.
00:49:04.340 --> 00:49:07.710
And the solution for this, I
can write out the whole thing.
00:49:07.710 --> 00:49:16.700
x of t here is just
some A1 plus t A2 e
00:49:16.700 --> 00:49:21.200
to the minus zeta omega n t.
00:49:21.200 --> 00:49:24.320
And again, it looks--
it's just some kind
00:49:24.320 --> 00:49:28.740
of damp, not very interesting
response, no oscillations.
00:49:28.740 --> 00:49:31.590
And then finally,
zeta less than 1.
00:49:31.590 --> 00:49:34.216
And this is the only one--
this one produces oscillation.
00:49:37.670 --> 00:49:52.000
And the solution for st is plus
or minus-- minus zeta omega n
00:49:52.000 --> 00:50:04.950
t, a real part, plus
or minus i omega n t
00:50:04.950 --> 00:50:07.710
times the square root
of 1 minus zeta squared.
00:50:07.710 --> 00:50:13.780
Now, I've turned around
this zeta squared minus 1.
00:50:13.780 --> 00:50:15.220
This is now a negative number.
00:50:15.220 --> 00:50:18.270
A square root of a
negative number gives me i.
00:50:18.270 --> 00:50:20.220
And now, I turn this
around, so this is just
00:50:20.220 --> 00:50:22.240
a real positive number.
00:50:22.240 --> 00:50:24.510
So when you get i
into this answer,
00:50:24.510 --> 00:50:29.274
what does it tell you that
the solution looks like?
00:50:29.274 --> 00:50:30.440
AUDIENCE: Sines and cosines.
00:50:30.440 --> 00:50:32.460
J. KIM VANDIVER: Sines
and cosines, right?
00:50:32.460 --> 00:50:36.490
So now, this gives you sines
and cosines with a decay.
00:50:36.490 --> 00:50:39.215
This is an exponential to
e to the minus zeta omega n
00:50:39.215 --> 00:50:42.420
t multiplied by a
sine and a cosine.
00:50:42.420 --> 00:50:44.410
And so this is the
interesting part.
00:50:44.410 --> 00:50:47.740
So most of the work of
the rest of this term,
00:50:47.740 --> 00:50:50.585
we're only interested
in this final solution.
00:50:53.340 --> 00:51:13.336
And what it looks like for this
one-- so for zeta less than 1,
00:51:13.336 --> 00:51:20.820
x of t is some Ae to
the minus zeta omega n
00:51:20.820 --> 00:51:32.310
t times a cosine omega
d t-- make it d times
00:51:32.310 --> 00:51:36.770
t minus a phase angle--
come out looking like that.
00:51:36.770 --> 00:51:41.720
And if you draw it, depends on
initial conditions, so again,
00:51:41.720 --> 00:51:44.290
a positive velocity and
a positive displacement.
00:51:44.290 --> 00:51:46.240
It does this, but
then it dies out.
00:51:49.120 --> 00:51:50.900
It's very similar to
the undamped case,
00:51:50.900 --> 00:51:53.900
except that it has
this damping that
00:51:53.900 --> 00:51:55.690
causes it to die out with time.
00:51:55.690 --> 00:51:59.160
But this right here, this is
still the initial slope is v0
00:51:59.160 --> 00:52:01.040
and the initial
displacement here is x0.
00:52:15.050 --> 00:52:19.750
And I'm now going to give you
the exact expressions for this
00:52:19.750 --> 00:52:20.780
and we'll talk about it.
00:52:27.810 --> 00:52:30.410
Another way of writing
this then in terms
00:52:30.410 --> 00:52:41.170
of the initial conditions is
this looks like x0 cosine omega
00:52:41.170 --> 00:52:49.680
d t plus v0 over omega d.
00:53:46.110 --> 00:53:49.560
So expanding this out,
this result clearly
00:53:49.560 --> 00:53:52.360
has to depend on the
initial displacement
00:53:52.360 --> 00:53:54.470
and on the initial velocity.
00:53:54.470 --> 00:53:55.230
Now, what's this?
00:53:55.230 --> 00:53:57.470
I keep writing this omega d.
00:53:57.470 --> 00:53:59.530
So notice in here
in the solution,
00:53:59.530 --> 00:54:03.370
it's omega n times the square
root of 1 minus zeta squared.
00:54:03.370 --> 00:54:06.570
So the frequency that's in
here isn't exactly omega n.
00:54:06.570 --> 00:54:10.470
It's omega n altered by a bit.
00:54:10.470 --> 00:54:23.250
Omega sub d is called the
damped natural frequency.
00:54:28.050 --> 00:54:32.250
And it's equal to omega
n times the square root
00:54:32.250 --> 00:54:33.995
of 1 minus theta squared.
00:54:36.880 --> 00:54:40.790
The system actually oscillates
at a slightly different
00:54:40.790 --> 00:54:41.400
frequency.
00:54:41.400 --> 00:54:47.140
And for most systems
that vibrate at all,
00:54:47.140 --> 00:54:49.260
this damping term
is quite small.
00:54:49.260 --> 00:54:52.090
And when you square it,
it gets even smaller.
00:54:52.090 --> 00:54:58.310
So this is usually a number
that's 0.99, oftentimes,
00:54:58.310 --> 00:54:59.930
or even bigger than that.
00:54:59.930 --> 00:55:04.960
This is very close to 1 for
all small amounts of damping.
00:55:04.960 --> 00:55:08.350
But being really
careful about this
00:55:08.350 --> 00:55:11.190
in including it
everywhere, that's
00:55:11.190 --> 00:55:13.610
what this result looks like.
00:55:13.610 --> 00:55:18.080
And this little thing, psi,
this little phase angle here,
00:55:18.080 --> 00:55:30.860
is tangent inverse of theta
over the square root of 1
00:55:30.860 --> 00:55:33.220
minus theta squared.
00:55:33.220 --> 00:55:36.550
And this number--
when damping is small,
00:55:36.550 --> 00:55:39.530
this is a very small number.
00:55:39.530 --> 00:55:43.160
And most of the time of
problems that we deal with,
00:55:43.160 --> 00:55:44.510
the damping will be small.
00:55:44.510 --> 00:55:54.230
So let's say, for
small damping--
00:55:54.230 --> 00:55:58.640
and by that, I mean
zeta, say, less than 10%,
00:55:58.640 --> 00:56:01.990
what we call 10%, 0.1.
00:56:01.990 --> 00:56:06.560
And if you have a little
more-- you don't care too much
00:56:06.560 --> 00:56:09.480
about the precision,
it might even be 20%.
00:56:09.480 --> 00:56:16.470
Actually, if it were 0.2,
squared is 0.04, right?
00:56:16.470 --> 00:56:25.090
1 minus 0.04-- 0.96
square root, 0.98.
00:56:25.090 --> 00:56:27.890
So even with 20%
damping, the difference
00:56:27.890 --> 00:56:30.370
between the undamped
natural frequency
00:56:30.370 --> 00:56:31.972
and the damped natural
frequency's 2%.
00:56:34.480 --> 00:56:39.740
So for most cases with any
kind of small damping at all,
00:56:39.740 --> 00:56:42.870
we can write an approximation
which is easier to remember.
00:56:42.870 --> 00:56:45.410
And it's all I carry
around in my head.
00:56:45.410 --> 00:56:47.800
I can't remember
this, quite frankly.
00:56:47.800 --> 00:56:51.290
Don't try to and I
would instead express
00:56:51.290 --> 00:57:01.160
the answer to this as
just x0 cosine omega
00:57:01.160 --> 00:57:15.530
d t plus v0 over omega d
sine omega damped times time
00:57:15.530 --> 00:57:21.260
times e to the minus
zeta omega n t.
00:57:21.260 --> 00:57:24.630
So why do I bother to
carry the omega d's along
00:57:24.630 --> 00:57:29.590
if I just said that they're
almost exactly the same.
00:57:29.590 --> 00:57:33.582
For light damping, then omega
n's approximately omega d.
00:57:33.582 --> 00:57:36.760
Well, you need to
keep this one in here
00:57:36.760 --> 00:57:43.840
because even though it's only
2% difference at 20% damping,
00:57:43.840 --> 00:57:50.530
if you say the solution is omega
n when it's really omega d,
00:57:50.530 --> 00:57:56.040
this thing will accumulate
a phase error over time.
00:57:56.040 --> 00:58:00.380
So it's gets bigger and
bigger, this error here,
00:58:00.380 --> 00:58:02.720
because you haven't taken
care of that little 2%.
00:58:02.720 --> 00:58:06.550
That 2% can bite you after
you go through enough cycles.
00:58:06.550 --> 00:58:10.092
So I keep omega d in
the expression here.
00:58:10.092 --> 00:58:14.300
But other than that, it's almost
exactly the same expression
00:58:14.300 --> 00:58:17.280
that we just came up to
for the simple response
00:58:17.280 --> 00:58:20.100
of an undamped system
to initial conditions,
00:58:20.100 --> 00:58:25.740
x0 cosine plus v0
over omega n sine.
00:58:25.740 --> 00:58:30.000
And now, all we've added to
it is put the transient decay
00:58:30.000 --> 00:58:33.770
and the fact that it
decays into the expression
00:58:33.770 --> 00:58:38.030
and changed the frequency
it oscillates at to omega
00:58:38.030 --> 00:58:39.490
d instead of omega n.
00:59:13.770 --> 00:59:17.280
So I'm going to try to
impress something on you.
00:59:17.280 --> 00:59:33.750
If I took this pendulum
and my stopwatch,
00:59:33.750 --> 00:59:39.690
measured the natural
frequency of this thing,
00:59:39.690 --> 00:59:43.080
I could get a very accurate
value if I do it carefully.
00:59:43.080 --> 00:59:47.010
Then, I take the same object
and I dunk it in water
00:59:47.010 --> 00:59:49.404
and it goes back and forth.
00:59:49.404 --> 00:59:51.070
And it conspicuously
goes back and forth
00:59:51.070 --> 00:59:52.944
but dies down now after
a while, because it's
00:59:52.944 --> 00:59:54.890
got that water damping it.
00:59:54.890 --> 01:00:00.180
But I measure that
frequency and it's
01:00:00.180 --> 01:00:05.120
10% different, 20% different.
01:00:05.120 --> 01:00:09.670
And I have seen people make
this mistake dozens of times.
01:00:09.670 --> 01:00:12.150
You say, that's the experiment.
01:00:12.150 --> 01:00:14.396
Explain why.
01:00:14.396 --> 01:00:21.250
What's the reason that that
measured frequency has changed?
01:00:21.250 --> 01:00:25.469
Got any ocean engineers
in the audience?
01:00:25.469 --> 01:00:27.380
All right.
01:00:27.380 --> 01:00:30.799
So why does-- if you put
the pendulum in water--
01:00:30.799 --> 01:00:32.090
and it's still oscillating now.
01:00:32.090 --> 01:00:34.330
So it isn't so damp that it's--
01:00:34.330 --> 01:00:35.029
[BLOWS]
01:00:35.029 --> 01:00:36.070
So it's got some damping.
01:00:36.070 --> 01:00:38.050
It's dying out and the
natural frequency's
01:00:38.050 --> 01:00:39.820
changed by 15% or 20%.
01:00:39.820 --> 01:00:42.040
What's the explanation?
01:00:42.040 --> 01:00:49.020
And the answer you always
get from people is, damping.
01:00:49.020 --> 01:00:49.520
Why?
01:00:49.520 --> 01:00:55.665
Because everybody's been
taught this thing, right?
01:00:55.665 --> 01:00:58.040
And they all then assume that
the change in the frequency
01:00:58.040 --> 01:00:58.914
is caused by damping.
01:00:58.914 --> 01:01:01.500
But damping couldn't
possibly be the reason,
01:01:01.500 --> 01:01:03.720
because with 20% damping,
this thing'll die out
01:01:03.720 --> 01:01:06.750
in about two swings
and it's done.
01:01:06.750 --> 01:01:08.890
That's a lot of
damping, actually,
01:01:08.890 --> 01:01:11.730
but it only accounts for 2%
change in natural frequency,
01:01:11.730 --> 01:01:14.160
not 15%.
01:01:14.160 --> 01:01:14.710
Hmmm.
01:01:14.710 --> 01:01:16.571
So what causes the
change in the frequency?
01:01:20.892 --> 01:01:22.350
AUDIENCE: Buoyancy
of the pendulum?
01:01:22.350 --> 01:01:23.766
J. KIM VANDIVER:
No, not buoyancy.
01:01:27.220 --> 01:01:30.000
That could actually
have an effect.
01:01:30.000 --> 01:01:33.957
That's actually-- I should
say, yes, you're partly right.
01:01:33.957 --> 01:01:34.915
There's another reason.
01:01:39.190 --> 01:01:43.230
When the thing is swinging back
and forth there in the water,
01:01:43.230 --> 01:01:46.110
it actually carries
some water with it.
01:01:46.110 --> 01:01:48.740
Effectively, the
kinetic energy-- you
01:01:48.740 --> 01:01:51.546
now know how to do
vibration problems.
01:01:51.546 --> 01:01:53.170
Find the equations
of motion accounting
01:01:53.170 --> 01:01:56.420
for the potential energy
and the kinetic energy.
01:01:56.420 --> 01:01:59.250
The kinetic energy
changes, because some water
01:01:59.250 --> 01:02:03.010
moves with the object and
it's called added mass.
01:02:03.010 --> 01:02:05.450
It literally-- there is water
moving with the object that
01:02:05.450 --> 01:02:07.670
has kinetic energy
associated with the motion
01:02:07.670 --> 01:02:10.270
and it acts like
it's more massive.
01:02:10.270 --> 01:02:11.940
It is dynamically more massive.
01:02:11.940 --> 01:02:15.400
There's water moving with it.
01:02:15.400 --> 01:02:17.820
So trying to impress
on you that damping
01:02:17.820 --> 01:02:21.920
doesn't cause much of
a change in systems
01:02:21.920 --> 01:02:22.940
that actually vibrate.
01:02:22.940 --> 01:02:24.430
Really observe the vibration.
01:02:24.430 --> 01:02:26.200
If you can observe
the vibration,
01:02:26.200 --> 01:02:29.290
damping cannot possibly
account for a very large shift
01:02:29.290 --> 01:02:30.830
in frequency.
01:02:30.830 --> 01:02:34.252
What's the motion look like?
01:02:34.252 --> 01:02:36.130
Let's move on a little bit here.
01:02:44.110 --> 01:02:46.160
So that's what this
solution looks like.
01:02:46.160 --> 01:02:48.870
We know it depends on
initial conditions.
01:02:48.870 --> 01:02:57.020
The distance from here to here
will make this a time axis.
01:02:57.020 --> 01:02:59.110
This is one period.
01:02:59.110 --> 01:03:00.640
So this is tau d.
01:03:00.640 --> 01:03:03.810
That's the damped
period of vibration.
01:03:07.060 --> 01:03:20.820
And we know that x of t is some
Ae to the minus theta omega n t
01:03:20.820 --> 01:03:26.740
cosine omega d t
minus a phase angle.
01:03:26.740 --> 01:03:29.370
We can write that
expression like this.
01:03:31.880 --> 01:03:39.230
And this term, this
is just a cosine.
01:03:39.230 --> 01:03:44.780
This term repeats
every period, right?
01:03:47.560 --> 01:03:52.210
If it's at maximum value here,
exactly one period later,
01:03:52.210 --> 01:03:54.090
it's again at its maximum.
01:03:54.090 --> 01:04:00.020
So the cosine term
goes to 1 every 2 pi
01:04:00.020 --> 01:04:03.530
or every period
of motion, right?
01:04:03.530 --> 01:04:11.440
So I want to take-- I'm going
to define this as the value
01:04:11.440 --> 01:04:16.110
at x at some time t.
01:04:16.110 --> 01:04:18.970
I'll call it t0.
01:04:18.970 --> 01:04:32.220
And out here is x at t0 plus
n tau d, n periods later.
01:04:32.220 --> 01:04:34.280
So this is the period,
defined as period.
01:04:36.960 --> 01:04:44.900
Remember, omega d
is the same thing
01:04:44.900 --> 01:04:51.090
as 2 pi times the
frequency in hertz.
01:04:51.090 --> 01:05:01.980
And frequency is 1 over
period, 2 pi over the period.
01:05:01.980 --> 01:05:04.170
So remember, there's
a relationship
01:05:04.170 --> 01:05:08.180
that you need to remember
now that relates radian
01:05:08.180 --> 01:05:12.450
frequency to frequency in
cycles per second in hertz
01:05:12.450 --> 01:05:15.815
to frequency
expressed in period.
01:05:15.815 --> 01:05:17.030
All right?
01:05:17.030 --> 01:05:21.170
This would be tau d here
and this would be an f d.
01:05:21.170 --> 01:05:23.910
For any frequency,
you can say that.
01:05:23.910 --> 01:05:27.652
At omega is 2 pi f
is 2 pi over tau.
01:05:27.652 --> 01:05:29.110
So you've got to
be good with that.
01:05:29.110 --> 01:05:35.190
But now, so here we are, two
peaks separated by n periods.
01:05:35.190 --> 01:05:44.050
And I want to take the ratio of
x of t to x of t plus n tau d
01:05:44.050 --> 01:05:45.680
here.
01:05:45.680 --> 01:05:49.600
And that's just
going to be then my--
01:05:49.600 --> 01:05:54.260
when I take that ratio, x
of t has cosine omega d t
01:05:54.260 --> 01:05:56.150
minus phi in it.
01:05:56.150 --> 01:05:59.810
And n periods later, exactly
the same thing appears, right?
01:05:59.810 --> 01:06:02.300
So the cosine term
just cancels out.
01:06:02.300 --> 01:06:05.390
This just is e-- and
the A's cancel out.
01:06:05.390 --> 01:06:07.135
That's the initial conditions.
01:06:07.135 --> 01:06:11.660
It's e to the minus
zeta omega n t--
01:06:11.660 --> 01:06:20.480
and I guess I called
it t0-- over e
01:06:20.480 --> 01:06:31.225
to the minus zeta omega n
t0 plus n damped periods.
01:06:33.950 --> 01:06:36.500
And if I bring this
into the numerator,
01:06:36.500 --> 01:06:38.540
the exponent becomes positive.
01:06:38.540 --> 01:06:45.630
The t0 terms, minus zeta omega
and t0 plus, those cancel.
01:06:45.630 --> 01:06:53.530
And this expression is just
e to the plus zeta omega
01:06:53.530 --> 01:06:56.844
n times n td.
01:07:02.640 --> 01:07:06.390
And the last step that I want to
do to this, what I'm coming up
01:07:06.390 --> 01:07:10.570
with is a way of
estimating-- purposely
01:07:10.570 --> 01:07:20.520
doing this-- is this transient
curve we know is controlled
01:07:20.520 --> 01:07:22.850
by a damping, by zeta.
01:07:22.850 --> 01:07:28.340
I want to have an experimental
way to determine what is zeta.
01:07:28.340 --> 01:07:31.530
And I do it by computing
something called
01:07:31.530 --> 01:07:33.300
the logarithmic decrement.
01:07:33.300 --> 01:07:41.060
So if I take the natural
log of x of t over x of t
01:07:41.060 --> 01:07:50.270
plus n periods, it's the
natural log of this expression.
01:07:50.270 --> 01:07:53.580
So I just get the exponent back.
01:07:53.580 --> 01:08:07.526
This then is n zeta
omega-- I guess
01:08:07.526 --> 01:08:13.530
I better to do it
carefully-- omega n n tau d.
01:08:13.530 --> 01:08:18.439
The tau d is 2 pi over omega
and I get some nice things
01:08:18.439 --> 01:08:19.640
to cancel out here.
01:08:29.630 --> 01:08:38.689
So this natural log over the
ratio-- this is n zeta omega n
01:08:38.689 --> 01:08:44.040
and this is 2 pi
over omega d, which
01:08:44.040 --> 01:08:49.649
is omega n times the square
root of 1 minus zeta squared.
01:08:49.649 --> 01:08:52.260
Omega n's go away.
01:08:52.260 --> 01:09:09.149
And for zeta small, this term's
approximately 1, in which case
01:09:09.149 --> 01:09:19.120
this then becomes n 2 pi zeta.
01:09:19.120 --> 01:09:27.200
And zeta equals 1 over
2 pi n natural log
01:09:27.200 --> 01:09:36.729
of this ratio of x of
t over x of t plus nt.
01:09:44.060 --> 01:09:50.310
So experimentally, if you
just go in and measure your--
01:09:50.310 --> 01:09:55.960
if you plot out the response,
you measure a peak value,
01:09:55.960 --> 01:09:58.990
you measure the peak
value n periods later,
01:09:58.990 --> 01:10:03.950
compute the log of that
ratio, divide by 1 over 2
01:10:03.950 --> 01:10:06.570
pi n, the number of
periods, you have
01:10:06.570 --> 01:10:08.840
an estimate of the
natural frequency--
01:10:08.840 --> 01:10:10.752
estimate of the damping
ratio, excuse me.
01:10:15.380 --> 01:10:20.000
And to give you one quick
little rule of thumb here,
01:10:20.000 --> 01:10:23.790
so this is an experimental
way that very quickly, you
01:10:23.790 --> 01:10:29.360
can estimate the damping
of a pendulum or whatever
01:10:29.360 --> 01:10:31.380
by just doing a
quick measurement.
01:11:05.070 --> 01:11:08.550
So if it happens
that after n periods,
01:11:08.550 --> 01:11:12.610
this value is half
of the initial value,
01:11:12.610 --> 01:11:15.960
then this ratio is 2, right?
01:11:15.960 --> 01:11:18.570
So x of t-- some n periods
later, this is only half
01:11:18.570 --> 01:11:19.930
as big.
01:11:19.930 --> 01:11:20.950
This value's 2.
01:11:20.950 --> 01:11:24.910
The natural log of 2 is some
number you can calculate.
01:11:24.910 --> 01:11:27.030
So there's a little rule.
01:11:27.030 --> 01:11:29.700
If you just work
that out, you find
01:11:29.700 --> 01:11:44.300
that zeta equals 1 over 2 pi n
50% times the natural log of 2.
01:11:44.300 --> 01:11:49.150
And you end up here
was 0-- let me do this
01:11:49.150 --> 01:11:54.960
carefully-- 1 over 2 pi,
n 50%, natural log of 2.
01:11:54.960 --> 01:12:03.450
And that is 0.11 over n 50%.
01:12:03.450 --> 01:12:07.130
That's a really handy
little engineer tool
01:12:07.130 --> 01:12:09.140
to carry around in your head.
01:12:09.140 --> 01:12:15.190
So if I have an oscillator,
this little end here,
01:12:15.190 --> 01:12:17.920
I can do an experiment.
01:12:17.920 --> 01:12:22.020
Give it initial deflection
and it starts off
01:12:22.020 --> 01:12:24.940
at six inches or three
inches amplitude.
01:12:24.940 --> 01:12:27.040
And you let it oscillate
until you see it die down
01:12:27.040 --> 01:12:28.800
to half of that value.
01:12:28.800 --> 01:12:33.795
So let's say, one,
two, about four cycles
01:12:33.795 --> 01:12:38.710
this thing decays by about 50%.
01:12:38.710 --> 01:12:42.580
Four cycles-- plug
4 into that formula.
01:12:42.580 --> 01:12:47.240
You get about 0.025.
01:12:47.240 --> 01:12:48.762
Agree?
01:12:48.762 --> 01:12:52.490
2 and a 1/2% damping.
01:12:52.490 --> 01:12:55.720
Really very convenient
little thing
01:12:55.720 --> 01:12:58.260
to carry around with
you-- measure pendulum,
01:12:58.260 --> 01:12:59.510
how much damping does it have?
01:12:59.510 --> 01:13:00.960
And now, this is
what I'm saying.
01:13:00.960 --> 01:13:03.850
Most things that have
any substantial amount
01:13:03.850 --> 01:13:08.078
of vibration, the damping is
going to be way less than 10%.
01:13:11.330 --> 01:13:15.950
If it dies, if it takes
one cycle for the amplitude
01:13:15.950 --> 01:13:20.530
to decrease, one cycle for the
amplitude to decrease by 50%,
01:13:20.530 --> 01:13:22.222
how much damping does it have?
01:13:22.222 --> 01:13:23.550
AUDIENCE: 11%.
01:13:23.550 --> 01:13:25.930
J. KIM VANDIVER: 11%.
01:13:25.930 --> 01:13:28.320
So 11% damping is
a lot of damping.
01:13:28.320 --> 01:13:32.189
The thing starts out here and
the next cycle, it's half gone,
01:13:32.189 --> 01:13:34.230
and the next cycle after
that, it's half of that.
01:13:34.230 --> 01:13:37.520
And so in about three
cycles, it's gone.
01:13:37.520 --> 01:13:40.060
So if you see anything that's
vibrating any length of time
01:13:40.060 --> 01:13:43.990
at all, its damping
is way less than 10%
01:13:43.990 --> 01:13:50.520
and this notion of small
damping is a perfectly good one.
01:13:50.520 --> 01:13:52.730
And I'll close by just
saying one other thing.
01:13:52.730 --> 01:13:56.705
If something vibrates a
lot, the damping's small.
01:14:01.030 --> 01:14:03.160
You need small
damping for things
01:14:03.160 --> 01:14:04.890
to actually vibrate very much.
01:14:04.890 --> 01:14:06.770
This thing, this is vibrating--
01:14:06.770 --> 01:14:07.630
[HIGH TONE]
01:14:07.630 --> 01:14:12.320
that high-pitched one,
that's about a kilohertz.
01:14:12.320 --> 01:14:14.730
How many cycles do you
think it's gone through
01:14:14.730 --> 01:14:19.270
to get down to 50% of
that initial amplitude
01:14:19.270 --> 01:14:21.062
that you could hear?
01:14:21.062 --> 01:14:23.730
A few thousand?
01:14:23.730 --> 01:14:27.900
How much damping do
you think this rod has?
01:14:27.900 --> 01:14:32.080
Really tiny, really tiny.
01:14:32.080 --> 01:14:32.640
All right.
01:14:32.640 --> 01:14:34.290
So even though all we
talked about today was
01:14:34.290 --> 01:14:35.831
single degree of
freedom oscillators,
01:14:35.831 --> 01:14:39.840
I hope you learned a few
things that we'll carry now
01:14:39.840 --> 01:14:41.230
through the rest of the term.
01:14:41.230 --> 01:14:44.750
We'll use all these
concepts that we did today
01:14:44.750 --> 01:14:47.150
to talk about more
complicated vibration.
01:14:47.150 --> 01:14:51.820
Good luck on your 2.001 quiz.
01:14:51.820 --> 01:14:55.500
See you on Tuesday.