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PROFESSOR: Are we ready
with the concept questions
00:00:26.430 --> 00:00:29.740
from the homework this week?
00:00:29.740 --> 00:00:32.284
How do we get
different-- there we go.
00:00:32.284 --> 00:00:33.700
I looked at one,
it was one thing.
00:00:33.700 --> 00:00:35.270
I looked at the
other, [INAUDIBLE].
00:00:35.270 --> 00:00:37.260
Does g enter into
the expression for
00:00:37.260 --> 00:00:39.950
the undamped natural frequency?
00:00:39.950 --> 00:00:45.705
And most people said no, but
about a third of you said yes.
00:00:45.705 --> 00:00:47.330
If you have worked
on that problem now,
00:00:47.330 --> 00:00:49.830
you have already
discovered the answer.
00:00:49.830 --> 00:00:56.520
So you'll find that g does
not come into the expression.
00:00:56.520 --> 00:01:02.200
When you do a pendulum,
g is in the expression.
00:01:02.200 --> 00:01:05.090
And there's a question
on the homework
00:01:05.090 --> 00:01:07.420
about what's the difference.
00:01:07.420 --> 00:01:09.830
How can you predict
when g is going
00:01:09.830 --> 00:01:12.880
to be involved in a natural
frequency expression
00:01:12.880 --> 00:01:14.180
and when it is not?
00:01:16.562 --> 00:01:18.270
I want you to think
about that one a bit,
00:01:18.270 --> 00:01:20.170
maybe talk about it
at-- if there's still
00:01:20.170 --> 00:01:21.670
questions about it,
we talk about it
00:01:21.670 --> 00:01:23.480
in recitation on
Thursday, Friday.
00:01:23.480 --> 00:01:25.020
OK.
00:01:25.020 --> 00:01:27.205
So does g enter into
the expression here?
00:01:27.205 --> 00:01:31.170
I'm sure you know this simple
pendulum, the natural frequency
00:01:31.170 --> 00:01:34.000
square root of g over l.
00:01:34.000 --> 00:01:35.910
For a simple
mass-spring dashpot,
00:01:35.910 --> 00:01:38.070
the natural
frequency is k over m
00:01:38.070 --> 00:01:41.070
whether or not it's
affected by gravity.
00:01:41.070 --> 00:01:44.840
So there's something
different about these two.
00:01:44.840 --> 00:01:46.510
OK, let's go onto the third one.
00:01:51.340 --> 00:01:54.180
"In an experiment, this system
is given initial velocity
00:01:54.180 --> 00:01:57.240
observed to decay in
amplitude of vibration
00:01:57.240 --> 00:01:59.620
by 50% in 10 cycles.
00:01:59.620 --> 00:02:01.230
You can estimate
the damping ratio
00:02:01.230 --> 00:02:02.742
to be approximately," what?
00:02:02.742 --> 00:02:05.200
Well, it gives a little rule
of thumb I gave you last week.
00:02:05.200 --> 00:02:06.930
I'll go over it again today.
00:02:06.930 --> 00:02:11.360
0.11 divided by the number
of cycles did decay 50%.
00:02:11.360 --> 00:02:14.290
So it took 10
cycles, 0.11 divided
00:02:14.290 --> 00:02:19.670
by 10, 0.01, 1.1% damping.
00:02:19.670 --> 00:02:22.460
OK, next.
00:02:22.460 --> 00:02:25.160
"At which of the three
excitation frequencies
00:02:25.160 --> 00:02:28.380
will the response
magnitude be greatest?"
00:02:28.380 --> 00:02:33.210
You've done oscillators
excited by cosine omega t
00:02:33.210 --> 00:02:34.820
kind of things before.
00:02:34.820 --> 00:02:38.142
So at the ratio 1,
most people said
00:02:38.142 --> 00:02:39.600
it's where it would
be the largest.
00:02:39.600 --> 00:02:40.880
Why at 1?
00:02:40.880 --> 00:02:43.870
Anybody want to
give a shout here?
00:02:43.870 --> 00:02:46.750
What happens when
you drive the system
00:02:46.750 --> 00:02:49.384
at its natural frequency?
00:02:49.384 --> 00:02:52.050
It's called resonance, and we're
going to talk about that today.
00:02:52.050 --> 00:02:54.540
So it's when the
frequency ratio is
00:02:54.540 --> 00:02:59.820
one that-- and for the system
being lightly damped that you
00:02:59.820 --> 00:03:01.540
get the largest response.
00:03:01.540 --> 00:03:05.749
Finally-- which
one are we on here?
00:03:05.749 --> 00:03:06.290
Oh, this one.
00:03:06.290 --> 00:03:10.680
Can all the kinetic energy be
accounted for by an expression
00:03:10.680 --> 00:03:12.765
of the form IZ omega squared.
00:03:17.040 --> 00:03:19.310
By the way, I brought this
system if you haven't.
00:03:22.190 --> 00:03:23.940
So there's a really
simple demo, but it
00:03:23.940 --> 00:03:26.180
has all sorts of--
so in this case,
00:03:26.180 --> 00:03:32.340
we're talking about that motion.
00:03:32.340 --> 00:03:36.740
It certainly has some I omega
squared kind of kinetic energy,
00:03:36.740 --> 00:03:40.255
but does the center of gravity
translate as it's oscillating?
00:03:43.841 --> 00:03:45.590
What's the potential
energy in the system?
00:03:48.970 --> 00:03:52.610
By the way, any time
you get an oscillation,
00:03:52.610 --> 00:03:57.060
energy flows from potential
to kinetic, potential kinetic.
00:03:57.060 --> 00:03:58.580
That's what oscillation is.
00:03:58.580 --> 00:04:01.790
So there has to be an exchange
going on between kinetic energy
00:04:01.790 --> 00:04:02.930
and potential energy.
00:04:02.930 --> 00:04:04.640
And if there's no
losses in the system,
00:04:04.640 --> 00:04:08.590
the total energy is constant.
00:04:08.590 --> 00:04:12.150
So the kinetic energy's
certainly in the motion,
00:04:12.150 --> 00:04:14.450
but when it reaches
maximum amplitude, what's
00:04:14.450 --> 00:04:17.459
its velocity when
it's right here?
00:04:17.459 --> 00:04:18.230
Zero.
00:04:18.230 --> 00:04:22.070
So all of its energy
must be where?
00:04:22.070 --> 00:04:22.880
In the potential.
00:04:22.880 --> 00:04:25.840
And where's the
potential in this system?
00:04:25.840 --> 00:04:26.960
Pardon?
00:04:26.960 --> 00:04:27.590
In the string.
00:04:27.590 --> 00:04:29.427
That's not stored in the string.
00:04:29.427 --> 00:04:31.260
There's only two sources
of potential energy
00:04:31.260 --> 00:04:33.360
we talk about in this class.
00:04:33.360 --> 00:04:33.920
Gravity and--
00:04:33.920 --> 00:04:34.670
AUDIENCE: Strings.
00:04:34.670 --> 00:04:35.461
PROFESSOR: Strings.
00:04:35.461 --> 00:04:38.700
Well, these strings
don't stretch,
00:04:38.700 --> 00:04:40.995
so there's no spring
kinetic energy.
00:04:40.995 --> 00:04:42.120
We've got potential energy.
00:04:42.120 --> 00:04:47.490
There must be gravitational
potential energy.
00:04:47.490 --> 00:04:49.215
How is it coming
into this system?
00:04:52.880 --> 00:04:58.180
So he says when it turns,
the center of gravity
00:04:58.180 --> 00:05:02.440
has to raise up a
little bit, and that's
00:05:02.440 --> 00:05:04.590
the potential energy
in this system.
00:05:04.590 --> 00:05:07.010
The center of gravity goes
up and down a tiny bit.
00:05:07.010 --> 00:05:10.560
So is there any velocity
in the vertical direction?
00:05:10.560 --> 00:05:12.270
Is there any kinetic
energy associated
00:05:12.270 --> 00:05:13.710
with up and down motion?
00:05:13.710 --> 00:05:16.480
Yeah, so that doesn't
entirely capture it,
00:05:16.480 --> 00:05:17.841
1/2 I omega squared.
00:05:17.841 --> 00:05:19.340
Is it an important
amount of energy?
00:05:19.340 --> 00:05:23.530
I don't know, but there is
some velocity up and down.
00:05:23.530 --> 00:05:26.170
My guess is that it
actually isn't important,
00:05:26.170 --> 00:05:28.270
that the answer is it
does move up and down.
00:05:28.270 --> 00:05:30.850
It has to, or you would not
have any potential energy
00:05:30.850 --> 00:05:32.810
exchange in the system.
00:05:32.810 --> 00:05:35.030
OK.
00:05:35.030 --> 00:05:37.130
Is that it?
00:05:37.130 --> 00:05:39.312
OK.
00:05:39.312 --> 00:05:40.270
Let's keep moving here.
00:05:42.960 --> 00:05:44.710
Got a lot of fun things
to show you today.
00:05:47.400 --> 00:05:50.862
So last time, we talked about
response to initial conditions.
00:05:50.862 --> 00:05:52.820
I'm going to finish up
with that and then go on
00:05:52.820 --> 00:05:57.360
to talking about excitation
of harmonic forces.
00:05:57.360 --> 00:05:59.110
So last time, we were
considering a system
00:05:59.110 --> 00:06:02.260
like this-- X is measured
from the zero spring
00:06:02.260 --> 00:06:04.510
force in this case.
00:06:04.510 --> 00:06:07.050
Give you an equation
of motion of that sort.
00:06:07.050 --> 00:06:12.320
And we've found that
you could express x of T
00:06:12.320 --> 00:06:19.150
as a-- I'll give you the exact
expression-- x0 square root
00:06:19.150 --> 00:06:20.893
of 1 minus zeta squared.
00:06:29.980 --> 00:06:38.050
Cosine omega damped times time
minus a little phase angle.
00:06:38.050 --> 00:06:50.830
And there's a second term here,
v0 over omega d sine omega dt.
00:06:50.830 --> 00:06:58.560
And the whole thing times e
to the minus zeta omega n t.
00:06:58.560 --> 00:07:02.790
So that's our response
to an initial deflection
00:07:02.790 --> 00:07:07.260
x0 or an initial velocity v0.
00:07:07.260 --> 00:07:11.000
That's the full kind
of messy expression.
00:07:11.000 --> 00:07:13.645
There's another way of writing
that, which I'll show you.
00:07:19.760 --> 00:07:34.060
Another way of saying is that
it's in x0 cosine omega dt
00:07:34.060 --> 00:07:53.140
plus v0 plus zeta omega n
x0, all over omega d sine
00:07:53.140 --> 00:07:58.110
omega d times t e to
the minus zeta omega
00:07:58.110 --> 00:08:01.330
nt, the same exponent.
00:08:01.330 --> 00:08:03.260
This is your
decaying exponential
00:08:03.260 --> 00:08:04.950
that makes it die out.
00:08:04.950 --> 00:08:07.020
And so I just rearranged
some of these things.
00:08:07.020 --> 00:08:08.978
There's another little
phase angle in here now.
00:08:08.978 --> 00:08:13.560
So you have just a cosine
term, this proportional x0,
00:08:13.560 --> 00:08:18.070
and a sine term, which
has both v0 and x0 in it.
00:08:18.070 --> 00:08:19.970
The x0 term, if
damping is small,
00:08:19.970 --> 00:08:25.540
this term is pretty small
because it's x0 times zeta.
00:08:25.540 --> 00:08:28.450
When you divide by omega
d, which is almost omega n,
00:08:28.450 --> 00:08:30.130
that goes away.
00:08:30.130 --> 00:08:34.039
So this term, the
scale of it is zeta x0.
00:08:34.039 --> 00:08:38.360
So if this is 1% or 2%,
that's a very small number.
00:08:38.360 --> 00:08:40.190
I gave you an
approximation, which
00:08:40.190 --> 00:08:48.870
for almost all practical
examples that you
00:08:48.870 --> 00:08:55.650
might want to do, make it
approximately sine here.
00:08:55.650 --> 00:09:14.080
So this is x0 cosine omega dt
plus v0 over omega d sine omega
00:09:14.080 --> 00:09:21.340
dte to the minus zeta omega nt.
00:09:21.340 --> 00:09:23.230
And this is the practical one.
00:09:23.230 --> 00:09:27.160
For any reasonable system that
has relatively low damping,
00:09:27.160 --> 00:09:31.940
even 10% or 15% damping, you
get part of the transient decay
00:09:31.940 --> 00:09:38.930
comes from x0 cosine, the other
part v0 over omega d sine.
00:09:38.930 --> 00:09:42.280
That's what I can remember in my
head when I'm trying to do it.
00:09:42.280 --> 00:09:44.930
Now the question, the thing
I want to address today
00:09:44.930 --> 00:09:48.140
is what's this useful for?
00:09:48.140 --> 00:09:50.410
My approach to
teaching you vibration
00:09:50.410 --> 00:09:54.160
is I want you to go
away with a few simple,
00:09:54.160 --> 00:09:57.010
practical understandings
so that you can actually
00:09:57.010 --> 00:10:01.820
solve some vibration
problems, and one of them
00:10:01.820 --> 00:10:05.412
is just knowing this allows
you to do a couple things,
00:10:05.412 --> 00:10:07.370
and we'll do a couple of
examples this morning.
00:10:11.120 --> 00:10:23.365
By the way, this form, this is
A cosine plus B sine expression.
00:10:26.570 --> 00:10:36.770
And I label them
A and B. A and B,
00:10:36.770 --> 00:10:44.830
they're both of the
form A1 cosine omega
00:10:44.830 --> 00:10:50.090
t plus B1 sine omega t.
00:10:50.090 --> 00:10:53.020
And you can always add
a sine and a cosine
00:10:53.020 --> 00:10:56.290
at the same frequency.
00:10:56.290 --> 00:10:57.670
If I put just any
frequency here,
00:10:57.670 --> 00:10:59.620
they just have to be the same.
00:10:59.620 --> 00:11:02.570
You can always take an
expression like that
00:11:02.570 --> 00:11:07.520
and rewrite it as some
magnitude cosine omega
00:11:07.520 --> 00:11:11.090
t minus a phase angle.
00:11:11.090 --> 00:11:14.100
And the magnitude is
just a square root
00:11:14.100 --> 00:11:21.060
of the sum of the squares--
A1 squared plus B1 squared.
00:11:21.060 --> 00:11:25.830
And the phase angle
is the tangent inverse
00:11:25.830 --> 00:11:28.600
of the sine term
over the cosine term.
00:11:28.600 --> 00:11:32.000
So you can always
rewrite sine plus cosine
00:11:32.000 --> 00:11:36.080
as a cosine omega t minus
v. We use that a lot,
00:11:36.080 --> 00:11:37.950
and that'll be used
a lot in this course.
00:11:43.629 --> 00:11:45.420
And then, of course,
if this whole thing is
00:11:45.420 --> 00:11:50.010
multiplied by an e to the minus
8 omega mt, then so is this.
00:11:55.010 --> 00:11:57.420
OK.
00:11:57.420 --> 00:11:59.340
OK, what are these
things useful for?
00:12:03.480 --> 00:12:08.470
And we've derived this all
for a mass spring system.
00:12:08.470 --> 00:12:12.810
Is that equation
applicable to a pendulum?
00:12:12.810 --> 00:12:20.280
So this expression is applicable
to any single degree of freedom
00:12:20.280 --> 00:12:23.430
system that oscillates.
00:12:23.430 --> 00:12:25.861
You just have to
exchange a couple things.
00:12:25.861 --> 00:12:27.485
So let's think about
a simple pendulum.
00:12:43.050 --> 00:12:49.770
So our common massless
string and a bob on the end,
00:12:49.770 --> 00:12:58.215
some length L,
equation of motion.
00:13:03.460 --> 00:13:05.810
And this is point A up here.
00:13:05.810 --> 00:13:20.930
IZZ with respect to A. Theta
double dot plus MgL sine theta
00:13:20.930 --> 00:13:22.580
equals 0.
00:13:22.580 --> 00:13:23.830
That's the equation of motion.
00:13:23.830 --> 00:13:25.329
With no damping,
that's the equation
00:13:25.329 --> 00:13:27.080
of motion in this system.
00:13:27.080 --> 00:13:30.060
Is it a linear
differential equation?
00:13:33.500 --> 00:13:35.790
And to do the
things that we want
00:13:35.790 --> 00:13:40.210
to be able to do in this
course, like vibration
00:13:40.210 --> 00:13:43.240
with harmonic inputs
and so forth, we
00:13:43.240 --> 00:13:46.040
want to deal with
linear equations.
00:13:46.040 --> 00:13:48.940
So one of the topics for
today is linearization.
00:13:48.940 --> 00:13:52.550
So this is one of the simplest
examples of linearization.
00:13:52.550 --> 00:13:57.470
We need a linearized
equation, and we
00:13:57.470 --> 00:14:01.790
need to remember in a
couple of approximations.
00:14:01.790 --> 00:14:06.410
So sine of theta, you can
do Taylor series expansion.
00:14:06.410 --> 00:14:13.780
It's theta minus theta cubed
over 3 factorial plus theta
00:14:13.780 --> 00:14:19.220
to the fifth over 5 factorial
plus minus and so forth.
00:14:19.220 --> 00:14:26.970
And cosine of theta
is 1 minus theta
00:14:26.970 --> 00:14:32.030
squared over 2 factorial plus
theta to the fourth over 4
00:14:32.030 --> 00:14:35.620
factorial, and so forth.
00:14:39.110 --> 00:14:42.480
So what we say, what do we mean
when we linearize something?
00:14:42.480 --> 00:14:44.410
So linearization means
that we're essentially
00:14:44.410 --> 00:14:51.940
assuming the variable that we're
working with is small enough
00:14:51.940 --> 00:14:55.710
that the right hand side,
an adequate approximation
00:14:55.710 --> 00:15:00.200
of this function, is to keep
only up to the linear terms
00:15:00.200 --> 00:15:03.220
on the right hand side.
00:15:03.220 --> 00:15:08.580
For sine, the term
raised to the 1 power,
00:15:08.580 --> 00:15:09.850
that's the linear term.
00:15:09.850 --> 00:15:12.720
This is a cubic term,
a fifth order term.
00:15:12.720 --> 00:15:15.790
We're going to throw those away
and say, this is close enough.
00:15:15.790 --> 00:15:18.320
For cosine, it's 1
minus theta squared.
00:15:18.320 --> 00:15:19.810
We throw these away.
00:15:19.810 --> 00:15:25.440
The small angle approximation
for cosine as it's equal to 1.
00:15:25.440 --> 00:15:27.940
That's the simplest
example of linearization,
00:15:27.940 --> 00:15:32.350
of a non-linear term.
00:15:32.350 --> 00:15:37.120
So when you linearize
this equation of motion,
00:15:37.120 --> 00:15:47.150
we end up with IZZ with respect
to A theta double dot plus MGL
00:15:47.150 --> 00:15:51.660
theta equals 0, and we get
our familiar natural frequency
00:15:51.660 --> 00:15:54.820
for Bob as square
root of g over L.
00:15:54.820 --> 00:15:57.745
So we need linearization to be
able to do pendulum problems.
00:16:04.865 --> 00:16:05.365
Hmm.
00:16:08.790 --> 00:16:10.690
OK.
00:16:10.690 --> 00:16:12.530
Or maybe let's do
an example here.
00:16:12.530 --> 00:16:15.060
I've got a pendulum that
we'll do an experiment
00:16:15.060 --> 00:16:16.230
with this morning.
00:16:16.230 --> 00:16:20.820
But 30 degrees is
about like that.
00:16:20.820 --> 00:16:23.220
17 degrees is about like that.
00:16:23.220 --> 00:16:25.380
That's quite a bit of angle.
00:16:25.380 --> 00:16:32.720
Is that small in the sense that
I'm linearizing this equation?
00:16:32.720 --> 00:16:39.520
So 17 degrees happens to be--
I'll have to use this here.
00:16:47.918 --> 00:16:54.550
That's actually 17.2
degrees equals 0.3 radians.
00:16:58.750 --> 00:17:10.140
Sine of 0.3 is 0.2955.
00:17:10.140 --> 00:17:14.240
And if we fill out and
look at these terms,
00:17:14.240 --> 00:17:22.220
the lead term here is 0.3--
so plug in the 0.3-- minus--
00:17:22.220 --> 00:17:29.530
and the second term when you
cube 0.3 and divide it by 6,
00:17:29.530 --> 00:17:40.750
the second term is minus-- what
is my number here-- 0.0045.
00:17:40.750 --> 00:17:44.130
And if you subtract that from
this, you get exactly this.
00:17:44.130 --> 00:17:48.110
So to four decimal
places, you only
00:17:48.110 --> 00:17:51.210
need two terms in this series
to get exactly the right answer.
00:17:51.210 --> 00:17:54.750
This thing out here, this fifth
order term, is really tiny.
00:17:54.750 --> 00:17:58.180
But the approximation, if
we say, OK, let's skip this,
00:17:58.180 --> 00:18:02.620
we're saying that 0.3
is approximately 0.2955.
00:18:02.620 --> 00:18:05.670
Pretty close.
00:18:05.670 --> 00:18:08.872
So up to 17 degrees,
0.3 radians,
00:18:08.872 --> 00:18:10.080
that's a great approximation.
00:18:14.060 --> 00:18:17.871
So it's a little high by about,
I think it's about 1 and 1/2%
00:18:17.871 --> 00:18:18.370
high.
00:18:26.250 --> 00:18:30.070
So for pretty large
angles for pendula,
00:18:30.070 --> 00:18:32.040
that simple linearization
works just fine.
00:18:34.700 --> 00:18:40.370
OK, once we get it linearized,
that equation of motion
00:18:40.370 --> 00:18:45.420
is of exactly the same
form as the one up there.
00:18:45.420 --> 00:18:47.380
We don't have any damping in it.
00:18:47.380 --> 00:18:50.510
We could add some damping.
00:18:50.510 --> 00:18:54.510
We can put a damping in here
with a torsional damper-- ct
00:18:54.510 --> 00:18:56.110
theta dot.
00:18:56.110 --> 00:18:59.570
And now that equation is
of exactly the same form
00:18:59.570 --> 00:19:04.690
as the linear oscillator,
linear meaning translational
00:19:04.690 --> 00:19:06.270
oscillator.
00:19:06.270 --> 00:19:09.480
Have that inertia term, a
damping term, a stiffness term.
00:19:09.480 --> 00:19:12.700
It's a second order linear
differential equation,
00:19:12.700 --> 00:19:14.499
homogeneous linear
differential equation,
00:19:14.499 --> 00:19:15.790
nothing on the right hand side.
00:19:22.730 --> 00:19:25.400
Because they're
exactly the same form,
00:19:25.400 --> 00:19:31.940
then the solution for
decay, transient decay
00:19:31.940 --> 00:19:35.670
from initial conditions,
takes on exactly the same form
00:19:35.670 --> 00:19:40.590
except that it has an
initial angle, theta 0,
00:19:40.590 --> 00:19:42.430
and I use the
approximation here.
00:19:42.430 --> 00:19:50.940
Cosine omega dt plus theta
0 dot, the initial velocity,
00:19:50.940 --> 00:20:01.070
over omega d sine omega dt all
times e to the minus zeta omega
00:20:01.070 --> 00:20:02.960
nt.
00:20:02.960 --> 00:20:06.030
So that's the exact same
transient decay equation,
00:20:06.030 --> 00:20:08.850
but now cast in angular terms.
00:20:15.310 --> 00:20:20.460
And if you wanted to express
it as a cosine omega t minus v,
00:20:20.460 --> 00:20:24.080
then A would be this squared
plus this squared square root
00:20:24.080 --> 00:20:30.570
and the phi would be a
similar calculation as we
00:20:30.570 --> 00:20:32.690
have up there someone here.
00:20:35.330 --> 00:20:38.900
Just the B term over the
A term, tangent numbers.
00:20:38.900 --> 00:20:41.790
All right.
00:20:41.790 --> 00:20:42.610
What's it good for?
00:20:42.610 --> 00:20:45.350
So I use this, this equation
gets used quite a lot.
00:20:45.350 --> 00:20:46.695
It has some practical uses.
00:20:51.140 --> 00:20:57.000
Let's do an example, a little
more complicated pendulum.
00:20:57.000 --> 00:20:57.840
Draw a stick maybe.
00:21:01.470 --> 00:21:02.580
Center of mass there.
00:21:07.295 --> 00:21:15.020
A, IZZ with respect
to A. We'll call it
00:21:15.020 --> 00:21:18.155
ML cubed over 3
for a slender rod.
00:21:23.310 --> 00:21:26.340
And now, what I
want to do is I have
00:21:26.340 --> 00:21:34.490
coming along here a mass,
a bullet, that has mass m.
00:21:34.490 --> 00:21:39.940
Has velocity vi
for initial here,
00:21:39.940 --> 00:21:46.990
and that's its linear
momentum, p initial.
00:21:49.540 --> 00:21:52.532
And it's going to hit
this stick and bed in it.
00:21:52.532 --> 00:21:53.990
So you've done this
problem before.
00:21:53.990 --> 00:21:54.490
Yeah.
00:21:54.490 --> 00:21:56.720
AUDIENCE: [INAUDIBLE].
00:21:56.720 --> 00:21:57.470
PROFESSOR: Pardon?
00:21:57.470 --> 00:21:59.730
AUDIENCE: ML cubed [INAUDIBLE].
00:21:59.730 --> 00:22:01.450
PROFESSOR: ML cubed over 3.
00:22:05.850 --> 00:22:06.470
Good.
00:22:06.470 --> 00:22:08.620
I don't know why I was
thinking cubed this morning.
00:22:08.620 --> 00:22:10.090
ML squared over 3.
00:22:10.090 --> 00:22:10.990
Good catch there.
00:22:10.990 --> 00:22:12.360
OK, so we have it.
00:22:12.360 --> 00:22:14.250
This is mass moment of inertia.
00:22:14.250 --> 00:22:15.764
This is a pendulum.
00:22:15.764 --> 00:22:17.180
This bullet's going
to come along.
00:22:20.870 --> 00:22:22.550
So this is exactly
what I've got here.
00:22:22.550 --> 00:22:24.100
I'll get it in the picture.
00:22:24.100 --> 00:22:25.240
Yeah.
00:22:25.240 --> 00:22:27.140
So it's initially at rest.
00:22:27.140 --> 00:22:29.690
Coming along, this
paper, this clip here,
00:22:29.690 --> 00:22:31.005
it represents the bullet.
00:22:31.005 --> 00:22:33.530
So it's swimming along.
00:22:33.530 --> 00:22:35.800
Hits this thing, sticks to
it, and when it hits it,
00:22:35.800 --> 00:22:37.870
it does that.
00:22:37.870 --> 00:22:40.900
So then this thing after it
hits swings back and forth.
00:22:40.900 --> 00:22:43.420
So what's the response
of this system
00:22:43.420 --> 00:22:44.915
to being hit by the bullet?
00:22:44.915 --> 00:22:49.330
Well, I claim you can do it
entirely by evaluating response
00:22:49.330 --> 00:22:51.620
to initial conditions.
00:22:51.620 --> 00:22:55.625
But we need to use one
conservation law to get there.
00:22:59.790 --> 00:23:03.260
So what's conserved on impact?
00:23:03.260 --> 00:23:05.640
Is linear momentum
conserved on impact?
00:23:09.930 --> 00:23:12.580
How many think yes?
00:23:12.580 --> 00:23:14.270
Linear momentum conserved.
00:23:14.270 --> 00:23:16.330
How many think angular
momentum's conserved?
00:23:16.330 --> 00:23:16.830
Hm.
00:23:16.830 --> 00:23:19.780
Good. you guys have learned
something this year.
00:23:19.780 --> 00:23:20.600
That's great.
00:23:20.600 --> 00:23:22.255
Why is linear momentum
not conserved?
00:23:27.560 --> 00:23:31.130
Because are there any
possible other external forces
00:23:31.130 --> 00:23:33.320
on the system?
00:23:33.320 --> 00:23:34.270
At the pin joint.
00:23:34.270 --> 00:23:36.860
You can have reaction
forces here and there.
00:23:36.860 --> 00:23:38.910
You have no control of them.
00:23:38.910 --> 00:23:41.350
But the moments
about this point,
00:23:41.350 --> 00:23:44.640
are there any external
moments about that point
00:23:44.640 --> 00:23:45.780
during the impact?
00:23:45.780 --> 00:23:46.280
No.
00:23:46.280 --> 00:23:48.405
They're reaction forces,
but there's no moment arm.
00:23:48.405 --> 00:23:49.830
So there's no moments.
00:23:49.830 --> 00:23:52.780
So you can use conservation
of angular momentum.
00:23:52.780 --> 00:23:56.620
So H1 I'll call it
here with respect to A
00:23:56.620 --> 00:24:01.410
is just R cross Pi.
00:24:01.410 --> 00:24:07.030
And the R is the length
in the I direction.
00:24:07.030 --> 00:24:10.730
P is in the j direction, so
the momentum is in the k.
00:24:10.730 --> 00:24:17.970
So this should be
mv initial times L,
00:24:17.970 --> 00:24:20.830
and its direction is
in the k hat direction.
00:24:20.830 --> 00:24:23.960
So that's the initial
angular momentum
00:24:23.960 --> 00:24:25.760
of the system with
respect to this.
00:24:25.760 --> 00:24:27.520
This has no initial
angular momentum
00:24:27.520 --> 00:24:29.550
because it's motionless.
00:24:29.550 --> 00:24:32.010
And since angular
momentum is conserved,
00:24:32.010 --> 00:24:35.785
that H2 we'll call
it with respect to A
00:24:35.785 --> 00:24:40.910
has got to be equal to
H1 with respect to A,
00:24:40.910 --> 00:24:49.880
and that will then be IZZ
with respect to A theta dot.
00:24:49.880 --> 00:24:54.040
But I need to account
for the mass, this thing.
00:24:54.040 --> 00:24:58.215
So the total mass moment of
inertia with respect to A
00:24:58.215 --> 00:25:01.780
is IZZ with respect
to A plus M-- what?
00:25:06.580 --> 00:25:09.250
Now I've got the total mass
moment of inertia with respect
00:25:09.250 --> 00:25:13.335
to this point, that of the stick
plus that of the initial mass
00:25:13.335 --> 00:25:15.140
that I've stuck on there.
00:25:15.140 --> 00:25:18.380
And this must be
equal to theta dot.
00:25:18.380 --> 00:25:19.980
And I put a not down
here because I'm
00:25:19.980 --> 00:25:23.540
looking for my equivalent
initial condition.
00:25:23.540 --> 00:25:31.810
And this then is
mv initial times
00:25:31.810 --> 00:25:37.730
L. Then I can solve
for theta dot 0,
00:25:37.730 --> 00:25:50.520
and that looks like mv initial
L over IZZ A plus mL squared.
00:25:50.520 --> 00:25:55.081
And everything on the
right hand side you know.
00:25:55.081 --> 00:25:57.330
You know the initial velocity,
the mass of the bullet,
00:25:57.330 --> 00:26:01.470
the length of the distance
from the pivot, mass moment
00:26:01.470 --> 00:26:05.240
of inertia, and the additional
mass moment of inertia.
00:26:05.240 --> 00:26:06.892
These are all
numbers you plug in,
00:26:06.892 --> 00:26:08.100
and you get a value for this.
00:26:08.100 --> 00:26:11.970
And once you have a value
for this, you can use that.
00:26:14.750 --> 00:26:17.110
In this problem, what's theta 0?
00:26:21.320 --> 00:26:25.460
The initial angular
deflection at time t0
00:26:25.460 --> 00:26:29.550
plus right after
the bullets hit it.
00:26:29.550 --> 00:26:35.600
And it hasn't moved because it
hasn't had time to move yet.
00:26:35.600 --> 00:26:37.710
At some velocity,
it takes finite time
00:26:37.710 --> 00:26:39.500
to get a deflection.
00:26:39.500 --> 00:26:42.700
So there's zero initial
angular deflection,
00:26:42.700 --> 00:26:48.020
but you get a step up in
initial angular velocity.
00:26:48.020 --> 00:26:51.180
And so the response
of this system
00:26:51.180 --> 00:27:05.210
is theta t is theta 0 dot
over omega d sine omega dt.
00:27:05.210 --> 00:27:06.060
So what's omega d?
00:27:15.950 --> 00:27:18.930
Remember, I'll
define a few things.
00:27:18.930 --> 00:27:25.880
In this case, this
is ct over 2 IZZ
00:27:25.880 --> 00:27:32.675
z A plus little ml
squared-- we have
00:27:32.675 --> 00:27:37.120
to deal with all the quantities
after the collision-- 2 times
00:27:37.120 --> 00:27:38.690
omega n.
00:27:38.690 --> 00:27:47.080
That's the damping ratio
for this torsional pendulum,
00:27:47.080 --> 00:27:48.170
with this pendulum.
00:27:48.170 --> 00:27:49.960
It's the damping constant.
00:27:49.960 --> 00:27:53.170
2 times the mass, the
inertial quantity,
00:27:53.170 --> 00:27:57.300
times omega n for a
translational system
00:27:57.300 --> 00:28:04.260
at c over 2 m omega n.
00:28:04.260 --> 00:28:07.840
For a pendulum system,
it's the torsional
00:28:07.840 --> 00:28:13.040
damping over 2 times the mass
moment of inertia times omega
00:28:13.040 --> 00:28:14.490
n.
00:28:14.490 --> 00:28:22.630
And omega n, well, it
is going to calculate
00:28:22.630 --> 00:28:23.930
the natural frequency.
00:28:23.930 --> 00:28:27.730
It's just MgL divided
by IZZ plus this.
00:28:27.730 --> 00:28:29.540
Maybe you ought to
write that down.
00:28:35.790 --> 00:28:38.630
So always for a sample singular
[INAUDIBLE] oscillator,
00:28:38.630 --> 00:28:40.340
you want the undamped
natural frequency.
00:28:40.340 --> 00:28:42.050
Ignore the damping term.
00:28:42.050 --> 00:28:45.880
Take the stiffness term
coefficient here and divide it
00:28:45.880 --> 00:28:47.570
by the inertial coefficient.
00:28:47.570 --> 00:28:51.510
But we care about the natural
frequency after the impact,
00:28:51.510 --> 00:28:58.055
so this is going to be-- ah.
00:28:58.055 --> 00:29:01.740
The trouble is here I
don't know for this system,
00:29:01.740 --> 00:29:03.910
I haven't worked out yet,
what this term looks like.
00:29:03.910 --> 00:29:04.698
What is it?
00:29:07.710 --> 00:29:11.920
This result right here
is for the simple Bob.
00:29:11.920 --> 00:29:18.880
For this stick, it's MgL over
2 plus the little m times l.
00:29:18.880 --> 00:29:20.690
Little more messy.
00:29:20.690 --> 00:29:27.060
So MgL over 2 plus little ml.
00:29:27.060 --> 00:29:30.880
That'll be the-- come from the
potential energy in this system
00:29:30.880 --> 00:29:37.490
all over IZZ A plus mL squared.
00:29:37.490 --> 00:29:39.230
So you get your
natural frequency out
00:29:39.230 --> 00:29:42.146
of that expression.
00:29:42.146 --> 00:29:42.645
OK.
00:29:49.450 --> 00:29:53.970
So you do this problem
sometimes before when
00:29:53.970 --> 00:29:56.660
you do, say, somebody asks
you how high does it swing.
00:29:56.660 --> 00:29:57.250
AND so forth.
00:29:57.250 --> 00:30:00.030
Well, you can do
it by conservation
00:30:00.030 --> 00:30:02.050
of energy, et cetera.
00:30:02.050 --> 00:30:04.520
But now, you have
actually exact expression
00:30:04.520 --> 00:30:10.130
for the time history of
the thing after the impact,
00:30:10.130 --> 00:30:12.420
including the
effects of damping.
00:30:12.420 --> 00:30:19.870
And if you were to draw
the result of this function
00:30:19.870 --> 00:30:23.970
of theta as a function
of time, this one
00:30:23.970 --> 00:30:30.020
starts with no initial
displacement but a velocity
00:30:30.020 --> 00:30:30.685
and does this.
00:30:37.600 --> 00:30:39.830
And that's your
exponential decay envelope,
00:30:39.830 --> 00:30:42.150
and this is time.
00:30:45.910 --> 00:30:47.970
Now, what-- yeah.
00:30:47.970 --> 00:30:48.470
Excuse me.
00:30:51.970 --> 00:30:52.970
AUDIENCE: [INAUDIBLE].
00:30:52.970 --> 00:30:54.465
PROFESSOR: Why this one?
00:30:54.465 --> 00:30:57.860
AUDIENCE: Yes. [INAUDIBLE].
00:30:57.860 --> 00:30:59.170
PROFESSOR: Excuse me.
00:30:59.170 --> 00:30:59.870
Forgot the g.
00:30:59.870 --> 00:31:02.996
I mean, it accounts for
the restoring moment,
00:31:02.996 --> 00:31:04.870
the additional little
bit of restoring moment
00:31:04.870 --> 00:31:08.040
that you get from having
added the mass of this thing
00:31:08.040 --> 00:31:09.250
to it, right.
00:31:09.250 --> 00:31:12.810
So it has by itself
MgL sine theta,
00:31:12.810 --> 00:31:13.970
and we linearize that too.
00:31:13.970 --> 00:31:18.650
So it's MgL theta, and those
two terms would add together.
00:31:18.650 --> 00:31:22.380
So you just have
a second term here
00:31:22.380 --> 00:31:27.030
that has MgL like behavior.
00:31:27.030 --> 00:31:31.470
How I most often personally
make use of expressions
00:31:31.470 --> 00:31:34.560
like this, or the
one for translation,
00:31:34.560 --> 00:31:39.140
is because
experimentally, if I'm
00:31:39.140 --> 00:31:42.440
trying to predict the
vibration behavior of a system,
00:31:42.440 --> 00:31:46.360
one of the things you want
to know is the damping.
00:31:46.360 --> 00:31:48.790
And one of the simplest
ways to measure damping
00:31:48.790 --> 00:31:53.460
is to give a system an initial
deflection or initial velocity
00:31:53.460 --> 00:31:56.420
and measure its decay,
and from its decay,
00:31:56.420 --> 00:31:58.240
calculate the damping.
00:31:58.240 --> 00:32:06.470
So the last time I gave you
an expression for doing that,
00:32:06.470 --> 00:32:13.720
and that was a damping
ratio 1 over 2 pi
00:32:13.720 --> 00:32:18.110
times the number of cycles that
you count, that you watch it,
00:32:18.110 --> 00:32:27.750
times the natural log of
x of t over x of t plus n
00:32:27.750 --> 00:32:31.620
periods of vibration.
00:32:31.620 --> 00:32:32.550
This has a name.
00:32:32.550 --> 00:32:35.940
It's called the logarithmic
decrement, this thing.
00:32:35.940 --> 00:32:38.250
So if somebody
says log decrement,
00:32:38.250 --> 00:32:40.990
that's where they're
referring to this expression.
00:32:51.265 --> 00:32:54.340
A comment about this.
00:32:54.340 --> 00:33:07.380
In this expression, x
of t must be zero means
00:33:07.380 --> 00:33:12.080
if your measurement-- we have
x of t here, or theta of t--
00:33:12.080 --> 00:33:14.100
they must be zero mean.
00:33:14.100 --> 00:33:16.670
There must be
oscillations around zero
00:33:16.670 --> 00:33:19.460
or you have to have subtracted
the mean to get it there
00:33:19.460 --> 00:33:21.730
because if this is
displaced and is oscillating
00:33:21.730 --> 00:33:25.480
around some offset, then this
calculates and will get really
00:33:25.480 --> 00:33:25.980
messed up.
00:33:25.980 --> 00:33:28.250
It's got an offset
plus an offset here
00:33:28.250 --> 00:33:29.160
plus an offset there.
00:33:29.160 --> 00:33:31.500
It means it's
totally meaningless.
00:33:31.500 --> 00:33:36.970
So you must remove the mean
value from any time history
00:33:36.970 --> 00:33:39.360
that you go to do this.
00:33:39.360 --> 00:33:42.790
So there's an easier way the
same expression-- and this is,
00:33:42.790 --> 00:33:45.130
in fact, the way I use this.
00:33:45.130 --> 00:33:48.030
A plot out like just
your data acquisition
00:33:48.030 --> 00:33:50.420
grabs it, plots it for you.
00:33:50.420 --> 00:33:53.410
I take this value
from here to here,
00:33:53.410 --> 00:33:57.510
and this is my peak
to peak amplitude.
00:33:57.510 --> 00:34:01.640
And then I go out n cycles
later and find the peak
00:34:01.640 --> 00:34:03.940
to peak amplitude.
00:34:03.940 --> 00:34:06.920
And so this is perfectly,
this is just the same as 1
00:34:06.920 --> 00:34:15.530
over 2 pi n, but now you
do natural log of x peak
00:34:15.530 --> 00:34:22.969
to peak t over x peak to
peak at t plus n periods.
00:34:25.670 --> 00:34:27.844
And that now, peak
to peak measurement,
00:34:27.844 --> 00:34:29.010
you totally ignore the mean.
00:34:29.010 --> 00:34:30.870
Doesn't matter where you are.
00:34:30.870 --> 00:34:35.010
You want the here to here,
here to here, plug it in there,
00:34:35.010 --> 00:34:37.320
and you're done.
00:34:37.320 --> 00:34:49.219
OK, so let's-- I got 1, 2, 3, 4.
00:34:49.219 --> 00:35:01.060
Let's let n equal 4, and let's
assume this expression here--
00:35:01.060 --> 00:35:05.830
n is 1 over 2 pi times 4.
00:35:05.830 --> 00:35:10.090
And let's assume that
in these four periods
00:35:10.090 --> 00:35:15.230
from-- that's 1 period,
2, 3, 4 getting out here
00:35:15.230 --> 00:35:21.710
to this fourth, four periods
away, that this is one fifth
00:35:21.710 --> 00:35:23.610
the initial.
00:35:23.610 --> 00:35:25.945
So this would be the
natural log of 5.
00:35:28.980 --> 00:35:32.290
So 1 over 2 pi times
4, natural log of 5,
00:35:32.290 --> 00:35:36.895
and you run the
numbers, you get 0.064,
00:35:36.895 --> 00:35:42.575
or what we call 6.4% damping.
00:35:42.575 --> 00:35:43.450
That's how you do it.
00:35:43.450 --> 00:35:46.630
That's the way you do a
calculation like that.
00:35:46.630 --> 00:35:52.150
Now, I gave you a quick rule of
thumb for estimating damping,
00:35:52.150 --> 00:35:54.910
and this is what
I-- I can't work.
00:35:54.910 --> 00:35:58.220
I don't do logs
in my head, but I
00:35:58.220 --> 00:36:01.050
can do damping
estimates without that
00:36:01.050 --> 00:36:06.400
because I know
that zeta is also--
00:36:06.400 --> 00:36:09.040
if I just plug in
some numbers here
00:36:09.040 --> 00:36:13.680
and run them all
in advance is 0.11
00:36:13.680 --> 00:36:16.695
divided by the number
of cycles to decay 50%.
00:36:20.132 --> 00:36:21.590
So we're going to
do an experiment.
00:36:26.990 --> 00:36:29.620
And I guess it can be
seen with the camera.
00:36:29.620 --> 00:36:33.000
So here's my pendulum.
00:36:33.000 --> 00:36:38.030
This is my initial amplitude,
and this is about half.
00:36:38.030 --> 00:36:45.190
So if I take this thing over
here, like that, then let go,
00:36:45.190 --> 00:36:49.780
and count the cycles that
it takes to decay halfway,
00:36:49.780 --> 00:36:50.889
we can do this experiment.
00:36:50.889 --> 00:36:51.930
So let's do it carefully.
00:36:51.930 --> 00:36:54.160
So line it up like
that, and you're
00:36:54.160 --> 00:36:57.480
going to help me tell-- you
count how many cycles it
00:36:57.480 --> 00:36:59.250
takes till it gets to here.
00:36:59.250 --> 00:37:15.330
So 1, 2, 3, 4, 5, 6, 7, 8.
00:37:15.330 --> 00:37:16.920
About eight cycles.
00:37:16.920 --> 00:37:20.170
So it decayed halfway
in eight cycles.
00:37:20.170 --> 00:37:27.016
So zeta is approximately 0.11/8.
00:37:27.016 --> 00:37:30.240
It's 1 and 1/2%, 1.4%,
something like that.
00:37:33.042 --> 00:37:34.500
Perfectly good
estimate of damping.
00:37:34.500 --> 00:37:44.780
Now, the stopwatch here, we
can do this experiment again.
00:37:47.660 --> 00:37:49.100
I want you to count.
00:37:49.100 --> 00:37:52.000
You're doing the counting.
00:37:52.000 --> 00:37:57.390
And I'm going to say start,
and I want you to count cycles
00:37:57.390 --> 00:37:59.570
until I say stop.
00:37:59.570 --> 00:38:04.040
Now, I'll probably stop on 10
to make the calculation easy,
00:38:04.040 --> 00:38:08.090
so quietly to yourself
count the number of cycles
00:38:08.090 --> 00:38:11.950
from the time I release
it until the time I stop.
00:38:11.950 --> 00:38:13.060
Come back here.
00:38:19.110 --> 00:38:22.150
So this time, the
backdrop doesn't matter.
00:38:22.150 --> 00:38:23.870
I just want you to count cycles.
00:38:27.200 --> 00:38:29.220
And I'll start-- I'll
let it get going,
00:38:29.220 --> 00:38:30.730
and when it comes
back to me is when
00:38:30.730 --> 00:38:33.705
I'm going to start the
stopwatch because I
00:38:33.705 --> 00:38:35.580
have a hard time doing
both at the same time.
00:38:35.580 --> 00:38:38.190
So start.
00:38:57.810 --> 00:38:59.320
How many cycles?
00:38:59.320 --> 00:39:01.320
AUDIENCE: [INAUDIBLE].
00:39:01.320 --> 00:39:11.390
PROFESSOR: So I
got 17.84 for 10.
00:39:11.390 --> 00:39:25.010
So 10 divided by 10 is
1.784 seconds per cycle.
00:39:25.010 --> 00:39:26.960
Can't write like that.
00:39:26.960 --> 00:39:29.450
1.784 seconds per cycle.
00:39:29.450 --> 00:39:33.159
The frequency would
be 1 over that, right.
00:39:33.159 --> 00:39:34.950
The thing you have to
be careful about when
00:39:34.950 --> 00:39:44.400
you're counting cycles is if
I start here, that's 0, 1, 2.
00:39:44.400 --> 00:39:47.160
A very common human mistake is
when you're counting something
00:39:47.160 --> 00:39:49.380
like this is to say
one when you start,
00:39:49.380 --> 00:39:52.610
and then you're going
to be off by one count.
00:39:52.610 --> 00:39:54.020
Follow me?
00:39:54.020 --> 00:39:59.320
If I start 0, 1, 2.
00:39:59.320 --> 00:40:01.960
So I start the clock on
zero, but the first cycle
00:40:01.960 --> 00:40:04.030
isn't completed till
one whole cycle later.
00:40:04.030 --> 00:40:06.985
So be careful how you count.
00:40:06.985 --> 00:40:07.485
OK.
00:40:23.670 --> 00:40:35.380
Now we're going to shift
gears and take on a new topic,
00:40:35.380 --> 00:40:40.450
and that's the response
to a harmonic input,
00:40:40.450 --> 00:40:43.823
some cosine omega t excitation.
00:40:43.823 --> 00:40:44.769
Yeah.
00:40:44.769 --> 00:40:46.065
AUDIENCE: [INAUDIBLE].
00:40:46.065 --> 00:40:47.190
PROFESSOR: What is omega d?
00:41:02.340 --> 00:41:05.224
So it's the damped
natural frequency.
00:41:05.224 --> 00:41:06.390
That's how it's referred to.
00:41:06.390 --> 00:41:09.560
It is the frequency
you observe when you do
00:41:09.560 --> 00:41:11.782
an experiment like we just did.
00:41:11.782 --> 00:41:14.430
The actual oscillation
frequency when
00:41:14.430 --> 00:41:16.760
it's responding to
initial conditions
00:41:16.760 --> 00:41:18.930
is slightly different
from the undamped,
00:41:18.930 --> 00:41:23.590
but if you have light damping,
if you have even 10% damping,
00:41:23.590 --> 00:41:27.550
0.1 squared is 0.01.
00:41:27.550 --> 00:41:29.370
That's 0.99 square root.
00:41:29.370 --> 00:41:32.090
It's 0.995.
00:41:32.090 --> 00:41:33.660
So you're only off by half.
00:41:33.660 --> 00:41:35.600
They're only half a
percent difference.
00:41:35.600 --> 00:41:38.450
So for lightly damped systems,
for all intents and purposes,
00:41:38.450 --> 00:41:41.250
mega n and mega d are
almost exactly the same.
00:41:43.920 --> 00:41:44.420
OK.
00:41:47.020 --> 00:41:53.890
We now want to think about--
we have a linear system putting
00:41:53.890 --> 00:41:56.360
a force into it.
00:41:56.360 --> 00:42:01.080
It looks like some
F0 cosine omega t.
00:42:01.080 --> 00:42:05.430
And out of that system, we
measure a response, x of t.
00:42:09.380 --> 00:42:14.300
And inside this box here is
my system transfer function.
00:42:14.300 --> 00:42:17.545
It's the mathematics that
tells me I can take F of t
00:42:17.545 --> 00:42:21.687
in and predict
what x of t out is.
00:42:21.687 --> 00:42:23.270
So I need to know
the information that
00:42:23.270 --> 00:42:25.720
goes into this box, and of
course, the real system-- this
00:42:25.720 --> 00:42:27.280
is just the mechanical system.
00:42:27.280 --> 00:42:29.880
Force in, measured output out.
00:42:29.880 --> 00:42:33.660
This is what we call a
single input single output
00:42:33.660 --> 00:42:37.820
system, SISO, single input
single output linear system.
00:42:42.515 --> 00:42:44.140
And there's all sorts
of linear systems
00:42:44.140 --> 00:42:46.265
that you're going to study
as mechanical engineers,
00:42:46.265 --> 00:42:49.380
and you've already begun, I'm
sure, studying some of them.
00:42:49.380 --> 00:42:51.720
One of the properties
of a linear system
00:42:51.720 --> 00:42:57.220
is that you put a force in, F1,
and measure a response out, x1.
00:42:57.220 --> 00:42:58.970
And then you try a
different force, F2,
00:42:58.970 --> 00:43:01.540
and you measure a
response out, x2.
00:43:01.540 --> 00:43:03.600
What's the response if
you put them both in
00:43:03.600 --> 00:43:05.300
at the same time, F1 and F2?
00:43:09.550 --> 00:43:13.050
You just add the responses
to them individually.
00:43:13.050 --> 00:43:15.360
So F1 gives you x1.
00:43:15.360 --> 00:43:16.690
F2 give you x2.
00:43:16.690 --> 00:43:20.010
F1 plus F2 gives you
x1 plus x2, and that's
00:43:20.010 --> 00:43:25.020
one of the characteristics
of a linear system.
00:43:25.020 --> 00:43:30.240
We use that concept to be
able to separate the response.
00:43:30.240 --> 00:43:33.330
Our calculation's about
the response of a system,
00:43:33.330 --> 00:43:37.100
like our oscillator here,
separate its response
00:43:37.100 --> 00:43:42.070
to transient effects, transience
being initial conditions.
00:43:42.070 --> 00:43:45.750
They die out over time-- that's
why we call them transients--
00:43:45.750 --> 00:43:48.690
and steady state effects.
00:43:48.690 --> 00:43:50.450
So cosine omega t,
you can leave it
00:43:50.450 --> 00:43:52.950
running for a long, long time,
and pretty soon, the system
00:43:52.950 --> 00:43:58.930
will settle down to responding
just to that cosine omega t.
00:43:58.930 --> 00:44:00.900
And that we call steady state.
00:44:00.900 --> 00:44:02.800
And we use them separately.
00:44:02.800 --> 00:44:04.640
So we've done
initial conditions.
00:44:04.640 --> 00:44:06.920
Now we're going to look
at the steady state
00:44:06.920 --> 00:44:13.490
response of a-- say our
oscillator, our mass spring
00:44:13.490 --> 00:44:17.920
dashpot, to a harmonic
input, F0 cosine omega t.
00:44:20.850 --> 00:44:22.260
Another brief word.
00:44:22.260 --> 00:44:33.080
If I have a force, F0 cosine
omega t would look like that.
00:44:35.810 --> 00:44:42.770
And the response that I
measure to start off with my--
00:44:42.770 --> 00:44:47.090
it's sitting here at zero
when you turn this on.
00:44:47.090 --> 00:44:52.780
And it's going to do some odd
things initially, and then
00:44:52.780 --> 00:45:02.850
eventually settle down to some
long term steady response.
00:45:02.850 --> 00:45:04.900
The amplitude stays constant.
00:45:04.900 --> 00:45:08.360
It stays angle with
respect to the input
00:45:08.360 --> 00:45:09.840
isn't necessarily the same.
00:45:09.840 --> 00:45:18.547
There's some
possibly phase shift.
00:45:18.547 --> 00:45:20.880
And that's so the two, if
you're plotting them together,
00:45:20.880 --> 00:45:21.671
they won't line up.
00:45:21.671 --> 00:45:24.090
But see this messy
stuff at the beginning?
00:45:24.090 --> 00:45:27.600
When you first turn this
on, it jumps from here
00:45:27.600 --> 00:45:31.740
to here, that force, and it
gives it a kick to begin with.
00:45:31.740 --> 00:45:36.390
And this will have some
response initially due
00:45:36.390 --> 00:45:38.750
to that transient start up.
00:45:38.750 --> 00:45:42.910
And this response is all
modeled by the response
00:45:42.910 --> 00:45:44.770
to initial conditions.
00:45:44.770 --> 00:45:47.310
And it'll die out after a
while, this messy stuff.
00:45:47.310 --> 00:45:49.280
What's the frequency?
00:45:49.280 --> 00:45:54.950
What frequency do you expect
this initial, erratic looking
00:45:54.950 --> 00:45:55.817
stuff to be at?
00:45:59.400 --> 00:46:03.670
Its response to
initial conditions.
00:46:03.670 --> 00:46:08.250
What is the model for a
response to initial conditions?
00:46:08.250 --> 00:46:09.962
What's the frequency
of the response
00:46:09.962 --> 00:46:12.170
to initial conditions of
the single degree of freedom
00:46:12.170 --> 00:46:14.470
system?
00:46:14.470 --> 00:46:16.580
We have an equation
over here, right?
00:46:20.300 --> 00:46:22.640
The top has a cosine
term and a sine term.
00:46:22.640 --> 00:46:24.690
Part of it's a response
to initial displacement.
00:46:24.690 --> 00:46:27.560
Part of it's a response
to the initial velocity.
00:46:27.560 --> 00:46:31.880
Any of this start up stuff can
be cast as initial conditions,
00:46:31.880 --> 00:46:33.620
and the response to
initial conditions
00:46:33.620 --> 00:46:37.420
is always at the natural
frequency period.
00:46:37.420 --> 00:46:41.040
No other frequencies for same
degree of freedom systems.
00:46:41.040 --> 00:46:43.270
So you get a behavior
that's oscillating
00:46:43.270 --> 00:46:45.010
at its natural frequency.
00:46:45.010 --> 00:46:49.640
Mixed in there is a response
at the excitation frequency.
00:46:49.640 --> 00:46:52.090
And after a long
time, the response
00:46:52.090 --> 00:46:54.180
is only excitation frequency.
00:46:54.180 --> 00:46:55.150
This is now out here.
00:46:55.150 --> 00:46:56.590
This is omega.
00:46:56.590 --> 00:47:04.280
In here, you have omega
and omega d going on.
00:47:04.280 --> 00:47:06.430
So this is messy.
00:47:06.430 --> 00:47:08.330
Usually isn't
important, but it is.
00:47:08.330 --> 00:47:11.690
There are ways of getting the
exact solution, but mostly,
00:47:11.690 --> 00:47:15.910
vibration engineers, you're
interested in the long term
00:47:15.910 --> 00:47:21.905
steady state response to what
we call a harmonic input.
00:47:21.905 --> 00:47:22.405
OK.
00:47:40.510 --> 00:47:44.980
So we'll work a
classic single degree
00:47:44.980 --> 00:47:52.360
of freedom oscillator problem--
excited by F0 cosine omega t.
00:47:52.360 --> 00:47:55.800
You've done this in
1803, but now we'll
00:47:55.800 --> 00:48:00.230
do it using engineering
terminology.
00:48:00.230 --> 00:48:04.150
We'll look at it the way a
person studying vibration
00:48:04.150 --> 00:48:06.160
would think about this.
00:48:06.160 --> 00:48:07.535
We know the equation of motion.
00:48:21.070 --> 00:48:24.640
And I'm interested in the
steady state response.
00:48:24.640 --> 00:48:28.470
So this is x, and I'll do-- you
just write it once like this.
00:48:28.470 --> 00:48:29.650
SS, steady state.
00:48:35.750 --> 00:48:38.520
I'm only interested in
its-- after those transients
00:48:38.520 --> 00:48:39.400
have died out.
00:48:43.680 --> 00:48:46.130
And that steady
state response I know
00:48:46.130 --> 00:48:51.100
is going to be some
amplitude X0 cosine omega
00:48:51.100 --> 00:48:54.320
t minus some phase angle
that I don't necessarily
00:48:54.320 --> 00:48:55.810
know to begin with.
00:48:55.810 --> 00:48:58.280
But that's my input.
00:48:58.280 --> 00:48:59.810
This is my output.
00:48:59.810 --> 00:49:04.820
I plug it into here and turn the
crank and see what falls out.
00:49:11.510 --> 00:49:16.230
So you plug both of
those in, and you
00:49:16.230 --> 00:49:20.480
get two-- you
get-- this is going
00:49:20.480 --> 00:49:23.360
to be a little writing
intensive for a few minutes.
00:49:38.220 --> 00:49:46.760
So you plug the X0 cosine omega
t into all of these terms.
00:49:46.760 --> 00:49:50.070
The m term gives you
minus m omega squared,
00:49:50.070 --> 00:49:55.920
the k term gives you a
k, and the damping term,
00:49:55.920 --> 00:50:11.950
minus c omega sine omega
t minus v. All of that
00:50:11.950 --> 00:50:18.620
equals the right hand
side-- F0 cosine omega t.
00:50:18.620 --> 00:50:20.441
So this just purely
from substitution
00:50:20.441 --> 00:50:22.065
and then gathering
some terms together.
00:50:36.070 --> 00:50:40.450
I'm going to divide
through by k, by k.
00:50:40.450 --> 00:50:42.347
If I divide through by
k, k divided by-- this
00:50:42.347 --> 00:50:43.700
gives me a one.
00:50:43.700 --> 00:50:46.100
This gives me an
m over k, which is
00:50:46.100 --> 00:50:49.425
1 over the natural frequency
squared, for example.
00:50:49.425 --> 00:50:54.400
And I'm going to put
this into a form that
00:50:54.400 --> 00:50:58.470
is the standard form for
discussing vibration problems.
00:50:58.470 --> 00:51:04.770
So this equation can be
rewritten in this form.
00:51:04.770 --> 00:51:15.320
1 minus omega squared over
omega n squared cosine omega
00:51:15.320 --> 00:51:26.990
t minus v minus 2 zeta omega
over omega n sine omega t
00:51:26.990 --> 00:51:34.740
minus v. All that's still
equal to F0 cosine omega t.
00:51:34.740 --> 00:51:36.910
So this is getting into
kind of more standard form.
00:51:36.910 --> 00:51:38.270
So there's 1 minus omega.
00:51:38.270 --> 00:51:40.100
This now, this
omega over omega n,
00:51:40.100 --> 00:51:45.210
is called the frequency ratio,
and you see a lot of that.
00:51:45.210 --> 00:51:48.670
And I've substituted
n here. c omega over k
00:51:48.670 --> 00:51:53.230
turns out to be 2 zeta
omega over omega n.
00:51:53.230 --> 00:51:55.640
So this frequency
ratio appears a lot.
00:51:55.640 --> 00:52:02.000
in our-- let's see here.
00:52:10.220 --> 00:52:16.590
You need a couple of trig
identities-- cosine omega t
00:52:16.590 --> 00:52:31.510
minus v. Cosine omega t cosine
phi plus sine omega t sine phi,
00:52:31.510 --> 00:52:44.880
and sine omega t minus phi
gives you [INAUDIBLE] sine.
00:52:44.880 --> 00:52:57.310
Sine omega t cosine phi minus
cosine omega t sine phi.
00:52:57.310 --> 00:52:59.820
So that's a trig identity
you actually use quite a bit
00:52:59.820 --> 00:53:01.990
doing vibration problems.
00:53:01.990 --> 00:53:07.290
We need them, so we
take these, plug them
00:53:07.290 --> 00:53:11.980
in in all these places, and
do quite a bit of cranking.
00:53:11.980 --> 00:53:13.190
Yep.
00:53:13.190 --> 00:53:15.690
AUDIENCE: [INAUDIBLE].
00:53:15.690 --> 00:53:17.280
PROFESSOR: Yeah.
00:53:17.280 --> 00:53:19.850
Thank you.
00:53:19.850 --> 00:53:22.670
And that's called,
that f over k,
00:53:22.670 --> 00:53:26.442
is how much the spring
would move statically,
00:53:26.442 --> 00:53:28.750
at which the point
would move statically.
00:53:28.750 --> 00:53:31.570
We'll need that term also.
00:53:31.570 --> 00:53:32.850
OK.
00:53:32.850 --> 00:53:36.120
You do all of this.
00:53:36.120 --> 00:53:37.375
Here, I'll call these--
00:53:59.678 --> 00:54:00.680
OK.
00:54:00.680 --> 00:54:12.188
So this is C.
That's expression C.
00:54:12.188 --> 00:54:13.470
I can't see that probably.
00:54:13.470 --> 00:54:24.960
Call this D, this E. So
you plug D and E into C
00:54:24.960 --> 00:54:27.105
and work it through,
you get two equations.
00:55:03.650 --> 00:55:05.750
You break it into
two parts because one
00:55:05.750 --> 00:55:08.350
is a function of cosine
omega t, and then
00:55:08.350 --> 00:55:10.890
you have another part after
this substitution that's
00:55:10.890 --> 00:55:14.880
a function of sine omega t,
and you can separate them.
00:55:38.230 --> 00:55:40.120
But there's no
sine omega t force.
00:55:40.120 --> 00:55:41.710
On the right hand
side, you get zero.
00:55:41.710 --> 00:55:43.410
There are two equations here.
00:55:43.410 --> 00:55:47.800
How many unknowns do we have?
00:55:47.800 --> 00:55:55.110
All we know when we start
this thing is the input,
00:55:55.110 --> 00:56:04.700
and we have unknown
response amplitude,
00:56:04.700 --> 00:56:07.850
and we have an unknown phase
that we're looking for.
00:56:07.850 --> 00:56:08.730
How many equations?
00:56:08.730 --> 00:56:09.500
How many unknowns?
00:56:09.500 --> 00:56:11.440
Two and two.
00:56:11.440 --> 00:56:13.170
So you can do a lot
of cranking, which
00:56:13.170 --> 00:56:14.600
I have no intention
of doing here,
00:56:14.600 --> 00:56:20.885
and solve for the amplitude
of the response and the phase.
00:56:40.690 --> 00:56:45.480
And every textbook-- the
Williams textbook does this.
00:56:45.480 --> 00:56:50.920
There are two readings
posted on Stellar
00:56:50.920 --> 00:56:54.790
by [? Row. ?] Every textbook
goes through these derivations
00:56:54.790 --> 00:56:56.090
that I've just done.
00:56:56.090 --> 00:56:57.443
Nick, you've got a question.
00:56:57.443 --> 00:56:58.359
AUDIENCE: [INAUDIBLE].
00:57:00.900 --> 00:57:01.740
PROFESSOR: Pardon?
00:57:01.740 --> 00:57:03.370
AUDIENCE: [INAUDIBLE].
00:57:03.370 --> 00:57:06.215
PROFESSOR: Yeah, I
keep forgetting it.
00:57:06.215 --> 00:57:07.050
You're right.
00:57:07.050 --> 00:57:08.650
So we got a k here.
00:57:08.650 --> 00:57:13.480
And notice, this equation,
we throw away that for now.
00:57:13.480 --> 00:57:15.270
We get rid of this for now.
00:57:15.270 --> 00:57:17.420
We have these two
equations and two unknowns
00:57:17.420 --> 00:57:22.400
are just algebraic equations
There's not time dependent.
00:57:22.400 --> 00:57:24.210
We can get rid of that part.
00:57:24.210 --> 00:57:26.940
So we've now reduced
this to algebra,
00:57:26.940 --> 00:57:31.330
and the answer is
plotted up there.
00:57:31.330 --> 00:57:33.290
You've probably seen it before.
00:57:33.290 --> 00:57:40.780
It says that x0 is F0 over k--
I can get it right this time
00:57:40.780 --> 00:57:45.030
from the get go-- over
a denominator, which
00:57:45.030 --> 00:57:49.420
appears again and again
and again in vibration.
00:57:49.420 --> 00:57:54.530
Omega squared over
omega n squared
00:57:54.530 --> 00:58:02.570
squared plus 2 zeta
omega over omega n
00:58:02.570 --> 00:58:11.161
squared square root, the
whole thing, and an expression
00:58:11.161 --> 00:58:11.660
for phi.
00:58:16.050 --> 00:58:23.630
Tangent inverse of 2 zeta
omega over omega n, 1
00:58:23.630 --> 00:58:30.000
minus omega squared
over omega n squared.
00:58:30.000 --> 00:58:32.310
So you can solve all
that-- this mess over here
00:58:32.310 --> 00:58:34.490
for these two quantities.
00:58:34.490 --> 00:58:36.690
Do you need to remember this?
00:58:36.690 --> 00:58:39.770
You ever going to be
asked this on a quiz?
00:58:39.770 --> 00:58:42.550
Not by me.
00:58:42.550 --> 00:58:46.320
You ever going to have to use
this on a quiz and in homework?
00:58:46.320 --> 00:58:48.820
Absolutely.
00:58:48.820 --> 00:58:52.470
So the takeaway
is today know how
00:58:52.470 --> 00:58:56.200
to use those response to
initial condition formulas
00:58:56.200 --> 00:58:59.950
and damping and these two.
00:58:59.950 --> 00:59:04.560
So when you plot,
when you plot these,
00:59:04.560 --> 00:59:07.320
you get this picture up there.
00:59:07.320 --> 00:59:09.700
And we need to talk about
the properties of this.
00:59:13.380 --> 00:59:15.125
Remember, this
omega over omega n
00:59:15.125 --> 00:59:16.940
is the same called
the frequency ratio.
00:59:16.940 --> 00:59:19.930
It's just the ratio of
the excitation frequency
00:59:19.930 --> 00:59:22.790
to the natural
frequency of the system.
00:59:22.790 --> 00:59:27.090
And when they're equal, for
example, this ratio is one.
00:59:27.090 --> 00:59:30.820
This whole thing in
parentheses goes to zero.
00:59:30.820 --> 00:59:36.090
This expression over here goes
to 2 zeta, because that's one.
00:59:36.090 --> 00:59:39.210
2 zeta squared square
root is just 2 zeta.
00:59:39.210 --> 00:59:41.440
When omega equals omega
n, this whole expression
00:59:41.440 --> 00:59:45.400
is F0 over k divided
by 2 zeta, for example.
00:59:45.400 --> 00:59:47.570
And that's called
resonance, and that's
00:59:47.570 --> 00:59:50.795
when you're right at where
that peak goes to its maximum.
00:59:55.840 --> 00:59:59.290
Let's talk about this
expression for a moment.
00:59:59.290 --> 01:00:02.790
If we have our cart,
our mass-spring
01:00:02.790 --> 01:00:04.930
dashpot we started here.
01:00:04.930 --> 01:00:06.860
If you apply a
force, a static force
01:00:06.860 --> 01:00:14.640
F0 and stretch the spring
by an amount F0 over k.
01:00:14.640 --> 01:00:21.350
So x-- what we'll call x
static is just F0 over k.
01:00:25.390 --> 01:00:46.710
And if I want to plot, I
want to-- this has a name.
01:00:46.710 --> 01:00:49.210
This is called, this
ratio here, this gives you
01:00:49.210 --> 01:00:52.230
the magnitude of the response.
01:00:52.230 --> 01:00:53.690
It goes by a variety of names.
01:00:53.690 --> 01:00:56.890
Some people call it
a transfer function.
01:00:56.890 --> 01:01:00.190
Some people call it a
frequency response function.
01:01:04.230 --> 01:01:05.930
I write it
intentionally this way.
01:01:05.930 --> 01:01:12.310
This is I put output over input
because this expression has
01:01:12.310 --> 01:01:15.170
units of output over input.
01:01:15.170 --> 01:01:18.440
So I just write it like
this, remind myself
01:01:18.440 --> 01:01:20.560
what this transfer
function is about.
01:01:20.560 --> 01:01:24.850
The input is force, the
output is displacement.
01:01:24.850 --> 01:01:27.510
This expression
has units of force,
01:01:27.510 --> 01:01:29.024
force per unit displacement.
01:01:38.710 --> 01:01:46.900
If I go to here, if I try to
plot this-- let me start over.
01:01:46.900 --> 01:01:49.340
If I try to plot
this, it's going
01:01:49.340 --> 01:01:52.960
to be depending on the exact
value of the spring constant
01:01:52.960 --> 01:01:55.960
and the exact value of
the force every time.
01:01:55.960 --> 01:01:59.840
I have to get a unique plot
every time I go to do this.
01:01:59.840 --> 01:02:02.750
So textbooks and
engineers, I don't want
01:02:02.750 --> 01:02:04.960
to have to remember this part.
01:02:04.960 --> 01:02:10.140
This is where all of the content
is in is in this denominator,
01:02:10.140 --> 01:02:11.320
and it's dimensionless.
01:02:11.320 --> 01:02:17.000
So what I'd really like to
plot is x0 over x static.
01:02:19.620 --> 01:02:27.040
And if I do that, that is x0
over the quantity F0 over k.
01:02:27.040 --> 01:02:30.090
If I just divide-- this is x
static-- it would bring this
01:02:30.090 --> 01:02:35.800
to this side, then this
expression, this is just 1
01:02:35.800 --> 01:02:40.320
over that denominator.
01:02:40.320 --> 01:02:43.810
And sometimes, I think in
the handout by [? Row ?],
01:02:43.810 --> 01:02:46.155
they just call this h of omega.
01:02:48.700 --> 01:02:52.810
It's dimensionless,
frequency over frequency,
01:02:52.810 --> 01:02:56.690
and that's actually
what's plotted up there.
01:02:56.690 --> 01:03:01.950
And this is called-- has
different names also.
01:03:01.950 --> 01:03:06.670
Magnification factor,
dynamic amplification factor,
01:03:06.670 --> 01:03:10.360
because the ratio
of x to x static
01:03:10.360 --> 01:03:20.760
if this is the dynamic effects
magnify the response compared
01:03:20.760 --> 01:03:21.780
to the static response.
01:03:21.780 --> 01:03:24.370
So it might be this
over this might be 10.
01:03:24.370 --> 01:03:30.020
I mean, the dynamic response is
10 times the static response.
01:03:30.020 --> 01:03:32.140
OK, how do you--
to sum this up--
01:03:32.140 --> 01:03:37.380
and we'll be kind of
getting close to the end.
01:03:37.380 --> 01:03:39.770
We want to talk just about
the properties of this.
01:03:39.770 --> 01:03:40.940
How do we use this?
01:03:52.830 --> 01:03:57.730
So in practical use, you
have an input specified,
01:03:57.730 --> 01:04:00.260
some force cosine omega t.
01:04:00.260 --> 01:04:02.900
You know you have a single
degree of freedom oscillator
01:04:02.900 --> 01:04:09.160
that is governed by
equations like that one,
01:04:09.160 --> 01:04:10.940
and you want to
predict the response.
01:04:10.940 --> 01:04:18.450
Well, you say x of t is
equal to the magnitude
01:04:18.450 --> 01:04:28.437
of the force times the--
and you divide that by-- we
01:04:28.437 --> 01:04:29.311
could do it this way.
01:04:34.230 --> 01:04:36.820
The magnitude of the
force divided by k,
01:04:36.820 --> 01:04:38.180
which is the static response.
01:04:40.710 --> 01:04:43.720
To predict x0, we just have
to predict this quantity,
01:04:43.720 --> 01:04:46.490
multiply it by F0 over k.
01:04:46.490 --> 01:04:48.180
So you know this.
01:04:48.180 --> 01:04:50.110
You better know that
about your system,
01:04:50.110 --> 01:04:55.530
and you multiply it by
this quantity magnitude
01:04:55.530 --> 01:04:57.260
of h of omega.
01:04:57.260 --> 01:05:05.400
And the time dependent part
is times cosine omega t
01:05:05.400 --> 01:05:07.110
minus the phase angle.
01:05:07.110 --> 01:05:10.610
And the phase angle, you
get either off the plot
01:05:10.610 --> 01:05:15.214
or from the-- have
I written it down?
01:05:20.540 --> 01:05:22.573
I haven't written the
phase angle down yet.
01:05:34.180 --> 01:05:35.770
It's kind of a messy
expression too.
01:05:35.770 --> 01:05:39.200
That's why we plot it.
01:05:39.200 --> 01:05:45.330
2 zeta omega over omega n
over 1 minus omega squared
01:05:45.330 --> 01:05:48.210
over omega n squared.
01:05:48.210 --> 01:05:52.150
But by knowing just
this plot, what you just
01:05:52.150 --> 01:05:54.500
put in every textbook about
vibration in the world,
01:05:54.500 --> 01:05:57.770
by knowing this
magnification factor,
01:05:57.770 --> 01:06:01.960
calculating the static response,
multiplying the two together,
01:06:01.960 --> 01:06:04.820
you have the amplitude
of the response,
01:06:04.820 --> 01:06:08.490
and its time dependence is
cosine omega t minus the phase
01:06:08.490 --> 01:06:09.880
angle.
01:06:09.880 --> 01:06:11.785
And I've got a
little example here.
01:06:21.460 --> 01:06:23.110
Actually, rather
than the example,
01:06:23.110 --> 01:06:25.550
I've gone to all the
trouble of setting this up.
01:06:43.110 --> 01:06:44.460
All right.
01:06:44.460 --> 01:06:48.210
This is just a beam.
01:06:48.210 --> 01:06:49.860
Where's my other little beam?
01:06:56.910 --> 01:06:59.225
And a beam is just a spring.
01:07:02.420 --> 01:07:04.230
Put a mass on the and.
01:07:04.230 --> 01:07:07.720
This is basically a single
degree of freedom system.
01:07:07.720 --> 01:07:09.500
It has a natural frequency.
01:07:09.500 --> 01:07:13.130
The beam has a
certain stiffness.
01:07:13.130 --> 01:07:16.520
And now, in this case,
we're interested in response
01:07:16.520 --> 01:07:18.750
to some harmonic input.
01:07:18.750 --> 01:07:23.230
So any of you know
what a squiggle pen is.
01:07:23.230 --> 01:07:25.045
This is a kid's toy.
01:07:25.045 --> 01:07:26.035
AUDIENCE: Excuse me.
01:07:26.035 --> 01:07:29.005
Can you move the camera a tad
to the left so the [INAUDIBLE]?
01:07:32.794 --> 01:07:33.460
PROFESSOR: Yeah.
01:07:36.450 --> 01:07:38.920
So all throughout
the term, we've
01:07:38.920 --> 01:07:41.920
studied rotating masses
quite a bit, right.
01:07:41.920 --> 01:07:44.730
This thing has a rotating mass
that you can see in the end.
01:07:44.730 --> 01:07:46.420
I mean, when you leave it, you
can come down and take a look.
01:07:46.420 --> 01:07:47.544
It has a low rotating mass.
01:07:47.544 --> 01:07:50.030
It's actually a pen,
but it's a kid's toy.
01:07:50.030 --> 01:07:50.980
Shakes like crazy.
01:07:55.860 --> 01:08:00.599
And now we need the lights down.
01:08:06.830 --> 01:08:14.550
And it happens that the--
I've got a strobe light here,
01:08:14.550 --> 01:08:17.149
and I've kind of
preset the frequency
01:08:17.149 --> 01:08:21.870
so it's very close to the
frequency of vibration
01:08:21.870 --> 01:08:23.090
of this beam.
01:08:23.090 --> 01:08:28.939
So there's a rotating mass in
this pen going round and round,
01:08:28.939 --> 01:08:30.810
and it puts a force
into the system
01:08:30.810 --> 01:08:36.000
that looks like F0 cosine omega
t in the vertical direction.
01:08:36.000 --> 01:08:38.000
Also does it in the
horizontal, but vertical
01:08:38.000 --> 01:08:40.899
is our response direction.
01:08:40.899 --> 01:08:43.129
So it's putting
in a force, and I
01:08:43.129 --> 01:08:48.220
have set the length of this beam
so that the natural frequency
01:08:48.220 --> 01:08:50.180
of this beam with
this mass on the end
01:08:50.180 --> 01:08:54.090
is exactly very close
to being the frequency
01:08:54.090 --> 01:08:54.840
of the excitation.
01:08:57.540 --> 01:09:01.430
And the flash rate is slightly
different than the vibration
01:09:01.430 --> 01:09:04.542
rate, so you see it
illuminated at many positions
01:09:04.542 --> 01:09:05.750
as it goes through the cycle.
01:09:05.750 --> 01:09:09.720
So you see it going up and down.
01:09:09.720 --> 01:09:13.090
So if I mismatch it quite a
bit, then you see it going.
01:09:13.090 --> 01:09:15.200
And actually, if you look
at the very right end,
01:09:15.200 --> 01:09:18.446
you can see a white
thing going up and down.
01:09:18.446 --> 01:09:20.029
That's the mass where
you can actually
01:09:20.029 --> 01:09:24.950
see the mass in the very
end of the-- right there.
01:09:24.950 --> 01:09:28.220
You can see something
going around and round.
01:09:28.220 --> 01:09:31.529
There, that's the rotating mass.
01:09:31.529 --> 01:09:33.580
So the beam is
going up and down,
01:09:33.580 --> 01:09:37.710
and I've got this, the vibration
frequency of that mass going
01:09:37.710 --> 01:09:40.771
round and round equal to the
natural frequency of the beam
01:09:40.771 --> 01:09:41.729
of the mass in the end.
01:09:47.920 --> 01:09:56.480
And it's moving quite a bit,
and I'll loosen my clamp,
01:09:56.480 --> 01:09:59.300
and I'm going to change
the length of the beam.
01:09:59.300 --> 01:10:00.399
I've shortened it.
01:10:08.890 --> 01:10:11.540
And now it's still
moving up and down
01:10:11.540 --> 01:10:17.160
but not as much because the
frequency of the rotation
01:10:17.160 --> 01:10:19.250
of the eccentric
mass is no longer
01:10:19.250 --> 01:10:21.260
close to the natural
frequency of the system.
01:10:24.210 --> 01:10:27.300
In fact, I've made
the natural frequency
01:10:27.300 --> 01:10:34.510
of the system-- you can
bring the lights back up--
01:10:34.510 --> 01:10:37.030
I've made the natural
frequency of the system.
01:10:37.030 --> 01:10:40.310
By making the beam shorter,
I've made it stiffer.
01:10:40.310 --> 01:10:43.235
So the natural
frequency has gone up.
01:10:43.235 --> 01:10:46.290
The frequency of the rotation
of the eccentric mass
01:10:46.290 --> 01:10:47.920
has stayed about the same.
01:10:47.920 --> 01:10:49.530
So what's happened
to that frequency
01:10:49.530 --> 01:10:52.970
ratio, omega over omega n?
01:10:52.970 --> 01:10:54.620
So less than one or
greater than one.
01:10:57.230 --> 01:10:59.630
So the omega n has gone up.
01:10:59.630 --> 01:11:00.585
Omega stayed the same.
01:11:03.690 --> 01:11:06.740
The frequency ratio when
you shorten this beam
01:11:06.740 --> 01:11:14.570
is less than one, and the
properties of this transfer
01:11:14.570 --> 01:11:17.400
function, we call it--
this magnification factor
01:11:17.400 --> 01:11:19.050
looks like this.
01:11:19.050 --> 01:11:22.900
When we're exciting
it right at one-- this
01:11:22.900 --> 01:11:27.580
is omega over omega n--
you write it resonance.
01:11:27.580 --> 01:11:31.800
When you excite it at a
frequency ratio less than one,
01:11:31.800 --> 01:11:34.750
you start dropping
off this backside,
01:11:34.750 --> 01:11:38.700
and the response goes down.
01:11:38.700 --> 01:11:43.370
And if you excite it at
frequencies much greater
01:11:43.370 --> 01:11:46.420
than the natural frequency,
you end up way out here.
01:11:46.420 --> 01:11:47.350
I can do that too.
01:11:59.820 --> 01:12:05.390
So how much you think
it will vibrate now?
01:12:05.390 --> 01:12:07.690
A lot?
01:12:07.690 --> 01:12:08.456
A little?
01:12:16.560 --> 01:12:17.289
Hardly-- oops.
01:12:17.289 --> 01:12:19.330
Oh, I've brought it out
so much you can't see it.
01:12:24.780 --> 01:12:32.450
Hardly moving at all, and
that's because in terms
01:12:32.450 --> 01:12:36.640
of this terminology of
magnification factors, transfer
01:12:36.640 --> 01:12:41.720
functions, this is a
plot of x over x static.
01:12:41.720 --> 01:12:43.530
It goes right here.
01:12:43.530 --> 01:12:46.300
When you go to zero
frequency, you are at static,
01:12:46.300 --> 01:12:51.052
so the response at
very, very low frequency
01:12:51.052 --> 01:12:53.010
goes to being the same
as the static frequency.
01:12:53.010 --> 01:12:55.530
So in this plot, it goes to one.
01:12:55.530 --> 01:12:59.650
At resonance, you
put in one here.
01:12:59.650 --> 01:13:00.910
This goes to zero.
01:13:00.910 --> 01:13:04.060
That becomes a one
2 zeta squared.
01:13:04.060 --> 01:13:12.350
This height here
is 1 over 2 zeta.
01:13:12.350 --> 01:13:14.770
So the dynamic
amplification at resonance
01:13:14.770 --> 01:13:16.665
is just 1 over 2 times
the damping ratio.
01:13:16.665 --> 01:13:19.280
You have 1% damping.
01:13:19.280 --> 01:13:21.322
Twice that is o2.
01:13:21.322 --> 01:13:24.720
1 over o2 is 50.
01:13:24.720 --> 01:13:29.100
So if you only had 1% damping,
the dynamic amplification,
01:13:29.100 --> 01:13:31.600
the amount that this
vibrates greater
01:13:31.600 --> 01:13:35.190
than its static response,
is a factor of 50 greater.
01:13:35.190 --> 01:13:39.420
But then as you go higher
in this omega over omega n,
01:13:39.420 --> 01:13:42.460
and you get way out here, and
you get almost no vibration
01:13:42.460 --> 01:13:43.530
at all.
01:13:43.530 --> 01:13:46.296
And that's what's happened
when I've lengthened this.
01:13:50.180 --> 01:13:50.680
OK.
01:13:55.930 --> 01:14:02.090
So there's your introduction
to linear systems.
01:14:02.090 --> 01:14:05.090
In this case, a single
degree of freedom system
01:14:05.090 --> 01:14:10.570
that vibrates an
oscillator, we're
01:14:10.570 --> 01:14:15.230
talking about steady state
response, not the part
01:14:15.230 --> 01:14:18.440
of the solution, the
mathematical solution,
01:14:18.440 --> 01:14:20.850
to the initial conditions.
01:14:20.850 --> 01:14:24.740
So all of this has been
about steady state response
01:14:24.740 --> 01:14:27.240
of our simple oscillator
to what we call
01:14:27.240 --> 01:14:32.880
a harmonic input,
F0 cosine omega t.
01:14:32.880 --> 01:14:34.720
So what's important
that you need
01:14:34.720 --> 01:14:36.990
to remember and be able to use?
01:14:39.560 --> 01:14:43.520
This concept here, this
idea of a transfer function.
01:14:43.520 --> 01:14:44.720
That's really important.
01:14:44.720 --> 01:14:46.570
You might want to
remember it though
01:14:46.570 --> 01:14:51.960
as this dimensionless
quantity x over x static.
01:14:51.960 --> 01:14:55.010
Just remember the shape
of this transfer function.
01:14:55.010 --> 01:14:59.170
Magnitude of the amplification,
1 over 2 zeta at resonance.
01:14:59.170 --> 01:15:03.380
And it goes to one
at low frequency.
01:15:03.380 --> 01:15:06.770
The high frequency,
it drops away off.
01:15:06.770 --> 01:15:10.540
Next time, we're
going to pick up
01:15:10.540 --> 01:15:14.120
the topic of what we
call vibration isolation.
01:15:14.120 --> 01:15:16.540
The practical thing
to know as engineers
01:15:16.540 --> 01:15:20.784
is when you have a
significant vibration
01:15:20.784 --> 01:15:22.200
problem, like in
a lab, and you're
01:15:22.200 --> 01:15:24.750
looking through your microscope,
and the floor vibration is
01:15:24.750 --> 01:15:28.510
causing trouble with
your microscope,
01:15:28.510 --> 01:15:32.820
and you can't move the
subway, what can you
01:15:32.820 --> 01:15:35.989
do to solve that problem?
01:15:35.989 --> 01:15:37.530
Well, you might be
able-- what if you
01:15:37.530 --> 01:15:40.620
put some kind of a flexible
pad under the microscope?
01:15:40.620 --> 01:15:43.120
You might be able to reduce the
vibration of the microscope.
01:15:45.860 --> 01:15:47.110
Things like that.
01:15:47.110 --> 01:15:49.750
That is the topic of
vibration isolation,
01:15:49.750 --> 01:15:52.690
so we're going to get
into that next time.
01:15:52.690 --> 01:15:54.024
Thanks.