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J. KIM VANDIVER: Today's lecture
is not mathematically hard,
but it's really important to
establish vocabulary today.
We're going to talk
about vibration
for the rest of the term.
And vibration is essentially
applied dynamics.
So up until now, we've been
finding equations of motion,
but not solving them.
Did you notice that?
I've almost never asked you to
solve the equation of motion
that you've just discovered
using Lagrange or whatever.
The rest of the
term, we're actually
going to be talking mostly
about the resulting motion.
The equations of motion
are pretty easy to find.
You have all the techniques that
you need to know for finding.
And now, we're going to talk
about how things vibrate.
So why do we choose vibration?
Vibration, one, is an
incredibly common phenomenon.
We wouldn't have speech
without vibration.
You wouldn't have musical
instruments without vibration.
It's a positive thing when
it's making good music.
It's a negative thing when
it's keeping you awake at night
because the air conditioner
in the next room
is causing something
to rattle in the room
and it's driving you nuts.
So you can want it,
it can be desirable,
and you cannot want it.
And you need to know ways
of getting rid of it.
And so we're going to talk
about vibration, about making
vibration, about
suppressing vibration,
about isolating sensitive
instruments from the vibration
of the floor, things like that.
So that's the topic of
the rest of the term.
And today, we're going to talk
about single degree of freedom
systems.
And you might think that we're
spending an awful lot of time
on single degree
of freedom systems.
But actually, there's
a reason for that.
Lots of things in
real life, like-- this
is just an aluminum rod.
This will vibrate.
And continuous
systems, which this is,
have a theoretically infinite
number of degrees of freedom.
Yet when it comes to
talking about its vibration,
it is conceptually easy to
think about the vibration
of an object like this,
one natural frequency,
one natural mode at a time.
And in fact, you can
model that natural mode
with its single degree
of freedom equivalent.
And that's the way I
approach vibration.
So if you can isolate
one particular mode,
you can literally model it
as a Mass-Spring-Dashpot.
So you need to understand the
Mass-Spring-Dashpot behavior
inside and out, because it's
the vocabulary we use to do
much more complicated things.
So a single degree
of freedom system,
like the simple pendulum,
has a natural frequency.
In this case, it has mode shape.
Here's another one, kind of
fun, single degree of freedom.
This obviously
involves rotation.
And you can figure
that out using
Lagrange or whatever, single
degree of freedom systems.
But now, I'm going to excite
one mode of vibration of this.
[CLANG]
[HIGH-PITCHED TONE]
Hear the real high pitch?
I'll get it down here by the
mic so that people at home
can hear it-- about a
kilohertz, way up there.
And that's one natural
mode of this thing
in longitudinal vibration.
When I thump it sideways--
[CLANG]
[LOWER TONE]
you hear a lower tone.
Hear that?
[HUMS LOW]
rather than--
[HUMS HIGH]
[CLANG]
That's bending
vibration of this thing.
But each mode of
vibration I can think
of in terms of its equivalent
single degree of freedom
oscillate.
So we'll get to talking
about these things
a little bit-- continuous
systems-- in the last couple
lectures of the term.
But for today then,
we're really going
to develop this vocabulary
around the vibration
of single degree
of freedom systems.
So let's start.
All right.
So to keep it from
being totally boring,
I'm going to start with a
little Mass-Spring-Dashpot that
has two springs.
And they're of such a
length that unstretched,
they just meet in the middle.
And then, I'm going
to take a mass
and I'm going to squeeze it
in between these two springs--
I can't draw a spring very
well today-- and this is k1
and this is k2 and here's m.
And we'll put it on a roller
so it's obviously constrained
to motion in one direction.
And I'll pick this
point here as the place
I'm going to put my
inertial coordinate.
So my inertial coordinate's
just measured from
or happens to be where the
endpoints of these two springs
were.
Now, to squeeze
the spring in here,
I have this clearly
pre-compression
in these springs.
So we are no longer in a
zero-force state, right?
So and I want to get the
equations of motion in this.
And moreover, I
want to predict--
I want to find out what's
the natural frequency
of this spring.
So let's check your intuition.
So write down on
your piece of paper
whether or not the natural
frequency will be different
because there's pre-compression,
or whether or not
that pre-compression
in the springs
has nothing to do with
the natural frequency.
So write down on your
paper "natural frequency
is different" or "natural
frequency is the same."
Let's have a prediction here.
And then, we'll set
about figuring this out
and in the course
of doing it, we'll
develop a little vocabulary.
All through the
course so far, when
we've done equations
of motion, we've
usually picked the
zero-spring-force position.
And we sort of led you down
this rosy path that suggests
that's the way we do it.
But there are other
ways that you're
going to find that are
preferable to that, sometimes.
So that's one of the reasons
I'm doing this example.
So let's do a free body diagram.
And if I held this
mass, for example,
right at the center when
I put the springs in,
it's obvious that this
spring gets compressed
by half of the
length of the mass
and this spring gets
compressed by half
of the length of
the mass, right?
So this is going to be L long.
So if I held it
right in the middle,
it would compress L/2 and L/2.
But then, when I release
it, if these springs
are a different
spring constant, it's
going to move a little bit.
So the force on this
side pushing back
is sum k1 times L/2
minus the distance
that I move in that direction,
which would relieve it.
And the force on this
side also pushes back.
It's k2 times L/2 over
2 plus x, because when
I go in that direction, I'm
compressing it even further.
And those are the total
forces in the x-direction
on this body.
There's an N and an
mg, which we know
we don't have to deal
with because we're only
interested in motion
left and right.
All right?
So we can say sum of the
forces in the x-direction,
mass times the acceleration.
And those forces are k1 L/2
minus x minus k2 L/2 plus x.
And that's the complete equation
of motion for this problem.
And rearrange it so
that I get the functions
of x together here.
So mx double dot plus
k1 plus k2 times x
equals L/2 times k1 minus k2.
And that's your
equation of motion.
It's non-homogeneous.
This is all constants
on the right-hand side.
And on the left-hand side are
the functions of x, right?
So what's the natural
frequency of the system?
AUDIENCE: Square root
of k1 plus k2 over m.
J. KIM VANDIVER: I hear a
square root of the quantity k1
plus k2, the stiffness,
divided by m, k over m,
a usual
Mass-Spring-Dashpot system.
Did the pre-compression
have anything
to do with the
natural frequency?
I won't ask you to embarrass
yourselves, but a few of you
probably got that
wrong, all right?
So there's a lesson in this
that I want you to go away with.
and I'll say it once.
And that is when
an external force
has nothing to do with
the motion coordinates
in the problem.
It doesn't affect the
natural frequency.
These come from external forces.
These are these
pre-compressions, right?
And I can separate them out and
they are not functions of x.
The stuff on the right-hand
side of the equation,
that's not a function
of the motion variable--
cannot affect the
natural frequency.
So I'll give you another one.
This is our common thing
hanging from a stick.
I've taken my system
I built the other day
for a different purpose,
but now, it's just a mass
hanging from a spring.
And it's right now at
its equilibrium position
or there's non-zero
force in the spring.
It clearly has a
natural frequency.
And is that natural frequency
a function of gravity?
And so if you go to write
the equation to motion
of this system, you would find
mx double dot plus kx equals mg
g.
But the mg is not
a function of x.
The natural frequency's
again, the square root of k/m.
Now, we want to talk about
solving this differential
equation.
And because it's got
this constant term
in the right-hand side,
it's non-homogeneous,
which is kind of a nuisance
term in terms of dealing
with a differential equation.
It'd be a lot nicer if the
right-hand side were 0.
So I want to make the
right-hand side of this one 0.
And draw a use of a
conclusion from that.
First thing I need
to know is I'd
like to know what is the static
equilibrium position of this.
And when you go to compute
static equilibrium,
you look at the
equation of motion
and you say, make all
motion variables things
that are functions of time 0.
So no acceleration--
you're left with this.
So you just solve this for
whatever the value of x is
and I'll call it x
of s for x-static.
And you'll find
that, oh, well, it's
that term divided
by k1 plus k2, k1
minus k2 all over k1 plus k2.
And that's the static position.
So now, let's say, ah,
well, we started off
with this motion variable that
wasn't arbitrarily defined
at the middle.
And let's say that,
well, it's made up
of a static component,
which is a constant, just
a value, plus a
dynamic component I'll
call x of d, which moves.
This is the function of time.
This is a constant.
It's not a function of time.
And that means if I
take its derivative,
I might need a value for x dot.
That goes away.
It's just xd dot.
And x double dot
is xd double dot.
And let's substitute this
into my equation of motion.
So it becomes m xd double
dot plus k1 plus k2 times--
and now, this term has got two
pieces now-- times xd plus k1
plus k2 times xs
equals L/2 k1 minus k2.
All right?
Now if I say, well, let's
examine the static case,
then this goes away.
For the static equilibrium
case, this term is 0.
This term is 0 because
the dynamic motion
is 0 in the static case.
That xd is motion about the
static equilibrium position.
So for static case,
these two terms go away
and we know that
this equals that.
But if that's true, we
can get rid of these.
They cancel one another.
These terms cancel and I'm
left with m xd double dot
plus k equivalent, I'll
call it, xd equals 0.
So the k equivalent's
just the total stiffnesses
in the system, whatever
works out, right?
In this case, it's k1 plus
k2 and the natural frequency,
omega n, is the square root
of k equivalent divided by m.
So most often, if you're
interested in vibration,
you're interested in
natural frequencies,
you're interested in solving
the differential equation,
you will find it advantageous to
write your equations of motion
around the static
equilibrium position.
So I could have started
this problem by saying,
whatever the static
equilibrium position is
of this thing, that's
what I'm measuring x from.
And then, I would have come
to this equation eventually.
You'd have to figure out what is
the static equilibrium position
and know what you're doing,
but once you know it,
then you have the answer.
Now, the same thing is
true of that problem.
That's a non-homogeneous
differential equation
for the hanging mass.
And we derive the
equations of motion things
for this many different
ways this term, all right?
But we usually said,
zero-spring force.
But now, if you started
from here and said,
this is the static
equilibrium position,
what's the motion
about this position,
then you'd get the equation
with 0 on the right-hand side--
lots of advantages
there to using that.
All single degree of
freedom oscillators
will boil down to this equation.
This is one involving
translation,
but for a simple pendulum.
This object, for example, is a
pendulum, but it's rotational.
So it's a pendulum, but
it's one degree of freedom.
All pendulum problems,
if you do them
about equilibrium
positions, boil
down to some I with
respect to the point
that they're rocking
about, theta double
dot plus some Kt, torsional
spring constant theta,
equals 0.
They take the same form.
So all translational
single degree
of freedom systems, all
rotational single degree
of freedom systems, it's the
same differential equation--
just this involves mass
and linear acceleration.
This involves mass moment
of inertia and rotational
acceleration.
So everything that I
say about the solution
to single degree
of freedom systems
applies to both
types of problems.
So let's look into the solution
of this equation briefly.
Mostly, I'm doing this to
establish some terminology.
So a solution I know or
I can show that xd of t,
the solution to this
problem-- notice,
are there any external forces,
by the way, excitations,
f of t's or anything?
No.
So this thing has no
external excitation
that's going to make it move.
So it's only source of
vibration or motion is what?
Comes from-- I hear
initial conditions, right?
You have to do
something to perturb it
and then it will vibrate.
So here it is.
It's about its
equilibrium position.
I give it an initial
deflection and let go.
Or it's around its
initial condition
and I give it an
initial velocity.
It also responds to some
combination of the two.
So initial conditions
are the only things
that account for
motion of something
without external excitation.
And that motion, I can
write that solution
as A cosine omega t.
You'll find this is
a possible solution.
B sine omega t is another
possible solution.
Sum A cosine omega t minus
phase angle's also a solution.
And sum A e to the i omega t
you'll find is also a solution.
Any of those things
you could throw in
and the precise values of these
things, the A's, the B's, the
phi's, and so forth depend on--
AUDIENCE: the
initial conditions.
J. KIM VANDIVER: The
initial conditions.
So let's do this one quickly.
All right.
And I'll choose And I'm going to
stop writing the x sub d here.
This is now my position
from the equilibrium point.
So x of t-- I'm
going to say, let
it be an A1 cosine omega
t plus a B1 sine omega t
and plug it in.
When I plug it into
the equation of motion,
x double dot requires you to
take two derivatives of each
of these terms.
Two derivatives of cosine
gives you minus omega cosine.
Two derivative sine minus omega
squared cosine minus omega
squared sine.
So the answer comes
out minus m omega
squared plus k equivalent
here times A1 cosine
plus B1 sine-- omega t's
obviously in them-- equals 0.
So I just plugged in
that equation of motion.
I get this back.
This is what I started with.
That's x.
In general, it is not
equal to 0, can take on all
sorts of values.
So that's not generally 0
and that means this must be.
And from this, then,
when we solve this,
we find that omega what we
call n squared is k over m.
And that's, of course, where our
natural frequency comes from.
This is called the
undamped natural frequency,
because there's no dampening
in this problem yet.
We get the square
root of k over m
is the natural
frequency of the system.
Let's find out
what are A1 and B1.
Well, let's let x0 be
x at t equals 0 here.
And if we just plug that in
here, put t equals 0 here,
cosine goes to 1.
This term goes away.
So this implies
that A1 equals x0.
So we find out right away
that the A1 cosine omega
t takes care of the response
to an initial deflection.
And we need a x dot here minus
A1 omega sine omega t plus B1
omega cosine omega t.
That's the derivative of x.
You know the solution's that,
so its first derivative,
the velocity, must
look like this.
And let's let v0 equals
x dot at t equals 0.
When we plug that in,
this term goes away
and we get B1 omega
and cosine is 1.
So therefore, B1
is v0 over omega.
But in fact, the
omega's omega n,
because we already found that,
that the only frequency that
satisfies the equation
of motion when
you have only initial
conditions in the system,
the only frequency that
is allowed in the answer
is the natural frequency.
So we now know B1 is v0
over omega n and A1 is x0.
So if I give you any combination
of initial displacement
and initial velocity, you can
write out for me the exact time
history of the motion.
X0 to cosine omega t plus
v0 over omega n sine omega t
is the complete solution for a
response to initial conditions.
So any translational
oscillator one degree
of freedom where you have
a translational coordinate
measured from its
equilibrium position
has the equation of
motion-- actually,
you've done this enough.
But if we added a force here
and we added some damping
and I wanted the equation
of motion of this,
you know that it's m
x double dot plus b x
dot plus kx equals F of t.
And so you're going to be
confronted with problems--
find the equation of
motion in a system.
It comes up looking
like that and they say,
what's the natural frequency?
And I've been a little sloppy.
I really mean, what's the
undamped natural frequency?
And so to find the
undamped-- when
one says that, what's the
undamped natural frequency,
you just temporarily let b
and F be 0, just temporarily,
and solve then for omega n
equals square root of k/n.
It's what you do.
And then, so we know this
is a parameter that tells us
about the behavior of the
system, which we always
want to know for the single
degree of freedom systems.
What is the natural
frequency of the system?
And we know for b
equals 0 and F of 0,
then the response can be only
due to initial conditions.
So we have x of t.
We know it's going to be
some x0 cosine omega n
t plus v0 over omega
n sine omega n t.
And every simple vibration
system in the world
behaves basically like this
from initial conditions.
It'll be some part responding
to the initial displacement,
some part to the
initial velocity.
And damping is going to make
it a little bit more complex,
but not actually by much.
The same basic terms appear even
when you have damping in it.
This can be expressed as sum
A cosine omega, in this case,
n t minus the phase angle.
And it's useful to know
this trigonometric identity
to be able to put
things together
into an expression like that.
And you'll find out that A is
just the square root of the two
pieces.
It's a sine and cosine term.
So you have an x0 squared
plus a v0 over omega n squared
square root.
Remember, this is any A and
B. It's just a square root
of A squared plus B squared.
That's what we're doing here.
And the phase angle,
the tangent inverse
of this-- we've been calling
this like an A and this
is the B quantity.
So tangent inverse of--
get my signs right--
B over A, which
in this case then
is tangent inverse of
v0 over x0 omega n.
That's all there is to it.
And finally, another
trig thing that you
need to know-- we're going
to use it quite a bit--
is that if you have an
expression A cosine omega
t minus phi, that's equal to the
real part of A e to the i omega
t.
And if A is real and-- I don't
want to write it that way--
when A is real, it's A times
e to the i omega t minus phi,
because Euler's formula
says e to the i theta
equals cosine theta
plus i sine of theta.
So if you have an
i omega t minus phi
here, you get back
a cosine omega
t minus phi and another term,
an i sine omega t minus phi.
So you can always express
that as the real part of that.
So we're going to need
that little trig identity
as we go through the term.
Now, I've found in many
years of teaching vibration
that something
that many students
find a little confusing is
this notion of phase angle.
What does "phase
angle" really mean?
So I'll try to explain it to
you in a couple different ways.
So let's look at
what this vibration
that we're talking about
here, x0 cosine omega
t plus v0 over omega n
sine-- what's it look like?
So that's-- we've just got our--
and we see what it looks like.
But if you plot the motion of
this thing just versus time,
what's it look like and where
does phase angle come into it?
So this is now x
of t and this is
t equals 0 and this
undamped system
is essentially going
to look like that.
And this is the value
x0, the amplitude,
the initial condition on
x that you began with.
And right here, the slope-- v0
is the slope, the initial slope
of this curve, right, because
the time derivative is F x dot.
If we were plotting x dot, the
initial velocity is omega x0.
And so it's just the
slope is v0 here.
So this is your
initial velocity.
This is the-- and I
didn't-- yeah, that's right.
This is the initial
displacement.
The total written
out mathematically,
it looks like that.
And I'm plotting this function,
A cosine omega t minus phi.
Yeah?
Did I see a hand up?
AUDIENCE: Does x0 at t equals
0 or is it a little bit after?
J. KIM VANDIVER:
Well, I was just
looking at it myself and
said, this can't be right.
This has got to be the initial
condition on x and this
has to be the initial
condition on v.
Now, whatever this
turns out to be is
whatever it turns out to be.
You have some initial velocity.
You have some
initial displacement.
The system can actually
peak out sometime later
at a maximum value, right?
And that maximum value is that.
So this over here
is the square root
of x0 squared plus v0 over
omega n squared square root.
That's what the peak value is.
And this system's undamped,
so it just goes on forever.
So the question is, though,
what is this gap here
between when it starts and
when it meets its maximum?
Well, when we use an
expression like-- we
said we can express this as
some A cosine omega t minus phi.
It's just the point at
which the cosine then
reaches its maximum.
So if this axis here is omega
t, if we plot this actually
versus omega t,
then one full cycle
here is 2 pi or 360 degrees.
So if you plot it versus
omega t, then this gap in here
is just phi.
That's the delay in
angle, if you will,
that the system goes
through between getting
from the initial
conditions to getting
to the peak of the cosine.
And phi must also then
be equal to some omega
n times a delta tau, I'll
call it, some time delay.
So if this is plotted--
if this axis is time--
not omega t, but time-- then x
the same plot, this delay here,
this is a time delay.
And when you plot
it against time,
it's a delay in time
to get to the peak.
And omega n delta
tau, this delay,
must be equal to
the phase angle.
So the delta tau, this time
delay, is phi over omega n.
So you can think about
this as a delay in time
or as a shift in phase angle,
depending on whether or not
you want to plot this thing
as a function of omega t
or as a function of time.
But you're going to need
this concept of phase angle
the rest of the term.
Want to ask any
questions about phase?
Because we're doing vibration
for the remainder of the term,
this is an introduction to a
topic called linear systems.
And so this is basically the
fundamental stuff in which you
then, when you go
on to 2004, which
is controls and
that sort of thing,
this is the basic intro to it.
And we'll talk more about linear
system behavior as we go along.
Now, we're going to do something
that you've-- much of this
stuff I know you've seen before.
Some of the new parts is
just vocabulary and ways
of thinking about
vibration that engineers
do that mathematicians
tend not to.
So you have seen most of
this stuff before where?
AUDIENCE: 1.803.
J. KIM VANDIVER: 1.803, right?
You've done all this.
And a year ago last May, in
May, I taught the 1803 lecture
with Professor Haynes Miller.
Now, if you had
1.803 last spring,
I think you had
somebody different.
But he invited me to come here.
It was in the same
classroom and we
taught the second-order ordinary
differential equation together.
It was really a lot of fun.
He said, well,
here's what we do.
And then, I said,
oh, well, engineers
look at it the following way.
So what I'm going
to show you is what
he and I did in class that day.
You can go back and
watch that on video.
It's kind of fun.
But I'll give you
my take on it today.
So this is the
engineer's view of what
you've already seen in 1.803.
So we have that system and we
have that equation of motion.
And the engineers
and mathematicians
would more or less agree to
that m x double dot plus bx.
But I went and looked at
the web page last night.
Last spring, the person
used c instead of b.
Haynes Miller the
year before used b.
So you can't depend on
any absolute consistency.
So let's start off with our
homogeneous equation here.
And I'm looking now for the
response to initial conditions
with damping.
You've done this in 1.803.
You know that you can solve
this by assuming a solution
of a form Ae to the st.
Plugging it in gives
you a quadratic equation
that looks like s squared
plus sb plus k equals 0.
This has roots.
I left out my m here, so it
starts off looking like that.
You divide through
by the m. s squared
plus b/m s plus k/m equals 0.
And that's where
Haynes would leave it.
And he'd give you the entire
answer in terms of b/m and k/m
and that kind of thing.
Engineers, we like to call that
the natural frequency squared.
And this term, we'd
modify to put it
in a terminology that is more
convenient to engineering.
So I'll show you
how that works out.
When you solve this
quadratic just using
the quadratic equation,
you get the following.
You get that the roots,
there's two of them.
I'll call them S1 and 2.
The roots to this
equation look like
minus b over 2m plus or minus
square root of b squared
over 4m squared minus k/m.
And that's what you'd
get to do in 1.803.
And an engineer would say, well,
let's change that a little bit.
So my roots that I
would use for S1 and 2,
I just factor out--
that's omega n squared.
I can factor that out and it
becomes omega n on the outside.
And I put an omega n in the
numerator and denominator here,
as well.
So I get roots that
look like-- so I've just
manipulated that a little bit.
I have a name for this term.
I use the Greek letter
zeta is b over 2 omega n
m is the way I remember
it in my brain.
It's called the damping ratio.
And if I say that, then the
roots, S1 and 2 for this,
look like minus zeta
omega n plus or minus
omega n times the square
root of zeta squared minus 1.
And those are the roots
that a vibration engineer
would use to describe
this second-order linear
differential equation
solution homogeneous solution.
Those are the roots
of the equation.
And when you have no damping,
then this term goes away
and you're left with-- and I
left an i out of here, I think.
No, I'm fine.
The i comes out of here.
So for one thing to absolutely
take away from today
is to remember this.
That's our definition of damping
called the damping ratio.
When that's 1, it's a number
we call critical damping.
I'll show you what
that means in a second.
And when it's greater than
1, the system won't vibrate.
It just has exponential decay.
If it's less than 1,
you get vibration.
And that's why we like to use
it this way as it's meaningful.
Its value, you
instantly know if it's
greater than or
less than 1, it's
going to change the
behavior of the system
from vibrating to not vibrating.
So now, there's four
possible solutions to this.
I'm not going to elaborate on
all of them, but zeta equals 0,
we've already done.
We know the answer to that.
Response to initial
conditions-- simple.
We know that one.
We have another solution
when zeta's greater than 1.
When zeta's greater than
1, this quantity here
is the inside is greater than 1,
so it's a real positive number.
And both the roots of this
thing are completely real.
And you know that the--
remember the response,
we hypothesize in the
beginning that response
looks like some Ae to the st.
So now, we just plug back in.
This is our st value.
We can plug them back
in and we will get
the motion of the system back.
So for zeta greater
than 1, st comes out
looking like minus zeta
omega n t plus or minus
square root of zeta
squared minus 1 times t.
And you just plug this in,
and x is just e to the st.
But these are just
pure real values.
And you'll find out that the
system from initial conditions
on velocity and
displacement just--
[WINDS DOWN]
and dies out.
Zeta equals to 1.
Then, st is just minus-- you get
a double root-- minus omega nt,
twice.
And the solution for this, I
can write out the whole thing.
x of t here is just
some A1 plus t A2 e
to the minus zeta omega n t.
And again, it looks--
it's just some kind
of damp, not very interesting
response, no oscillations.
And then finally,
zeta less than 1.
And this is the only one--
this one produces oscillation.
And the solution for st is plus
or minus-- minus zeta omega n
t, a real part, plus
or minus i omega n t
times the square root
of 1 minus zeta squared.
Now, I've turned around
this zeta squared minus 1.
This is now a negative number.
A square root of a
negative number gives me i.
And now, I turn this
around, so this is just
a real positive number.
So when you get i
into this answer,
what does it tell you that
the solution looks like?
AUDIENCE: Sines and cosines.
J. KIM VANDIVER: Sines
and cosines, right?
So now, this gives you sines
and cosines with a decay.
This is an exponential to
e to the minus zeta omega n
t multiplied by a
sine and a cosine.
And so this is the
interesting part.
So most of the work of
the rest of this term,
we're only interested
in this final solution.
And what it looks like for this
one-- so for zeta less than 1,
x of t is some Ae to
the minus zeta omega n
t times a cosine omega
d t-- make it d times
t minus a phase angle--
come out looking like that.
And if you draw it, depends on
initial conditions, so again,
a positive velocity and
a positive displacement.
It does this, but
then it dies out.
It's very similar to
the undamped case,
except that it has
this damping that
causes it to die out with time.
But this right here, this is
still the initial slope is v0
and the initial
displacement here is x0.
And I'm now going to give you
the exact expressions for this
and we'll talk about it.
Another way of writing
this then in terms
of the initial conditions is
this looks like x0 cosine omega
d t plus v0 over omega d.
So expanding this out,
this result clearly
has to depend on the
initial displacement
and on the initial velocity.
Now, what's this?
I keep writing this omega d.
So notice in here
in the solution,
it's omega n times the square
root of 1 minus zeta squared.
So the frequency that's in
here isn't exactly omega n.
It's omega n altered by a bit.
Omega sub d is called the
damped natural frequency.
And it's equal to omega
n times the square root
of 1 minus theta squared.
The system actually oscillates
at a slightly different
frequency.
And for most systems
that vibrate at all,
this damping term
is quite small.
And when you square it,
it gets even smaller.
So this is usually a number
that's 0.99, oftentimes,
or even bigger than that.
This is very close to 1 for
all small amounts of damping.
But being really
careful about this
in including it
everywhere, that's
what this result looks like.
And this little thing, psi,
this little phase angle here,
is tangent inverse of theta
over the square root of 1
minus theta squared.
And this number--
when damping is small,
this is a very small number.
And most of the time of
problems that we deal with,
the damping will be small.
So let's say, for
small damping--
and by that, I mean
zeta, say, less than 10%,
what we call 10%, 0.1.
And if you have a little
more-- you don't care too much
about the precision,
it might even be 20%.
Actually, if it were 0.2,
squared is 0.04, right?
1 minus 0.04-- 0.96
square root, 0.98.
So even with 20%
damping, the difference
between the undamped
natural frequency
and the damped natural
frequency's 2%.
So for most cases with any
kind of small damping at all,
we can write an approximation
which is easier to remember.
And it's all I carry
around in my head.
I can't remember
this, quite frankly.
Don't try to and I
would instead express
the answer to this as
just x0 cosine omega
d t plus v0 over omega d
sine omega damped times time
times e to the minus
zeta omega n t.
So why do I bother to
carry the omega d's along
if I just said that they're
almost exactly the same.
For light damping, then omega
n's approximately omega d.
Well, you need to
keep this one in here
because even though it's only
2% difference at 20% damping,
if you say the solution is omega
n when it's really omega d,
this thing will accumulate
a phase error over time.
So it's gets bigger and
bigger, this error here,
because you haven't taken
care of that little 2%.
That 2% can bite you after
you go through enough cycles.
So I keep omega d in
the expression here.
But other than that, it's almost
exactly the same expression
that we just came up to
for the simple response
of an undamped system
to initial conditions,
x0 cosine plus v0
over omega n sine.
And now, all we've added to
it is put the transient decay
and the fact that it
decays into the expression
and changed the frequency
it oscillates at to omega
d instead of omega n.
So I'm going to try to
impress something on you.
If I took this pendulum
and my stopwatch,
measured the natural
frequency of this thing,
I could get a very accurate
value if I do it carefully.
Then, I take the same object
and I dunk it in water
and it goes back and forth.
And it conspicuously
goes back and forth
but dies down now after
a while, because it's
got that water damping it.
But I measure that
frequency and it's
10% different, 20% different.
And I have seen people make
this mistake dozens of times.
You say, that's the experiment.
Explain why.
What's the reason that that
measured frequency has changed?
Got any ocean engineers
in the audience?
All right.
So why does-- if you put
the pendulum in water--
and it's still oscillating now.
So it isn't so damp that it's--
[BLOWS]
So it's got some damping.
It's dying out and the
natural frequency's
changed by 15% or 20%.
What's the explanation?
And the answer you always
get from people is, damping.
Why?
Because everybody's been
taught this thing, right?
And they all then assume that
the change in the frequency
is caused by damping.
But damping couldn't
possibly be the reason,
because with 20% damping,
this thing'll die out
in about two swings
and it's done.
That's a lot of
damping, actually,
but it only accounts for 2%
change in natural frequency,
not 15%.
Hmmm.
So what causes the
change in the frequency?
AUDIENCE: Buoyancy
of the pendulum?
J. KIM VANDIVER:
No, not buoyancy.
That could actually
have an effect.
That's actually-- I should
say, yes, you're partly right.
There's another reason.
When the thing is swinging back
and forth there in the water,
it actually carries
some water with it.
Effectively, the
kinetic energy-- you
now know how to do
vibration problems.
Find the equations
of motion accounting
for the potential energy
and the kinetic energy.
The kinetic energy
changes, because some water
moves with the object and
it's called added mass.
It literally-- there is water
moving with the object that
has kinetic energy
associated with the motion
and it acts like
it's more massive.
It is dynamically more massive.
There's water moving with it.
So trying to impress
on you that damping
doesn't cause much of
a change in systems
that actually vibrate.
Really observe the vibration.
If you can observe
the vibration,
damping cannot possibly
account for a very large shift
in frequency.
What's the motion look like?
Let's move on a little bit here.
So that's what this
solution looks like.
We know it depends on
initial conditions.
The distance from here to here
will make this a time axis.
This is one period.
So this is tau d.
That's the damped
period of vibration.
And we know that x of t is some
Ae to the minus theta omega n t
cosine omega d t
minus a phase angle.
We can write that
expression like this.
And this term, this
is just a cosine.
This term repeats
every period, right?
If it's at maximum value here,
exactly one period later,
it's again at its maximum.
So the cosine term
goes to 1 every 2 pi
or every period
of motion, right?
So I want to take-- I'm going
to define this as the value
at x at some time t.
I'll call it t0.
And out here is x at t0 plus
n tau d, n periods later.
So this is the period,
defined as period.
Remember, omega d
is the same thing
as 2 pi times the
frequency in hertz.
And frequency is 1 over
period, 2 pi over the period.
So remember, there's
a relationship
that you need to remember
now that relates radian
frequency to frequency in
cycles per second in hertz
to frequency
expressed in period.
All right?
This would be tau d here
and this would be an f d.
For any frequency,
you can say that.
At omega is 2 pi f
is 2 pi over tau.
So you've got to
be good with that.
But now, so here we are, two
peaks separated by n periods.
And I want to take the ratio of
x of t to x of t plus n tau d
here.
And that's just
going to be then my--
when I take that ratio, x
of t has cosine omega d t
minus phi in it.
And n periods later, exactly
the same thing appears, right?
So the cosine term
just cancels out.
This just is e-- and
the A's cancel out.
That's the initial conditions.
It's e to the minus
zeta omega n t--
and I guess I called
it t0-- over e
to the minus zeta omega n
t0 plus n damped periods.
And if I bring this
into the numerator,
the exponent becomes positive.
The t0 terms, minus zeta omega
and t0 plus, those cancel.
And this expression is just
e to the plus zeta omega
n times n td.
And the last step that I want to
do to this, what I'm coming up
with is a way of
estimating-- purposely
doing this-- is this transient
curve we know is controlled
by a damping, by zeta.
I want to have an experimental
way to determine what is zeta.
And I do it by computing
something called
the logarithmic decrement.
So if I take the natural
log of x of t over x of t
plus n periods, it's the
natural log of this expression.
So I just get the exponent back.
This then is n zeta
omega-- I guess
I better to do it
carefully-- omega n n tau d.
The tau d is 2 pi over omega
and I get some nice things
to cancel out here.
So this natural log over the
ratio-- this is n zeta omega n
and this is 2 pi
over omega d, which
is omega n times the square
root of 1 minus zeta squared.
Omega n's go away.
And for zeta small, this term's
approximately 1, in which case
this then becomes n 2 pi zeta.
And zeta equals 1 over
2 pi n natural log
of this ratio of x of
t over x of t plus nt.
So experimentally, if you
just go in and measure your--
if you plot out the response,
you measure a peak value,
you measure the peak
value n periods later,
compute the log of that
ratio, divide by 1 over 2
pi n, the number of
periods, you have
an estimate of the
natural frequency--
estimate of the damping
ratio, excuse me.
And to give you one quick
little rule of thumb here,
so this is an experimental
way that very quickly, you
can estimate the damping
of a pendulum or whatever
by just doing a
quick measurement.
So if it happens
that after n periods,
this value is half
of the initial value,
then this ratio is 2, right?
So x of t-- some n periods
later, this is only half
as big.
This value's 2.
The natural log of 2 is some
number you can calculate.
So there's a little rule.
If you just work
that out, you find
that zeta equals 1 over 2 pi n
50% times the natural log of 2.
And you end up here
was 0-- let me do this
carefully-- 1 over 2 pi,
n 50%, natural log of 2.
And that is 0.11 over n 50%.
That's a really handy
little engineer tool
to carry around in your head.
So if I have an oscillator,
this little end here,
I can do an experiment.
Give it initial deflection
and it starts off
at six inches or three
inches amplitude.
And you let it oscillate
until you see it die down
to half of that value.
So let's say, one,
two, about four cycles
this thing decays by about 50%.
Four cycles-- plug
4 into that formula.
You get about 0.025.
Agree?
2 and a 1/2% damping.
Really very convenient
little thing
to carry around with
you-- measure pendulum,
how much damping does it have?
And now, this is
what I'm saying.
Most things that have
any substantial amount
of vibration, the damping is
going to be way less than 10%.
If it dies, if it takes
one cycle for the amplitude
to decrease, one cycle for the
amplitude to decrease by 50%,
how much damping does it have?
AUDIENCE: 11%.
J. KIM VANDIVER: 11%.
So 11% damping is
a lot of damping.
The thing starts out here and
the next cycle, it's half gone,
and the next cycle after
that, it's half of that.
And so in about three
cycles, it's gone.
So if you see anything that's
vibrating any length of time
at all, its damping
is way less than 10%
and this notion of small
damping is a perfectly good one.
And I'll close by just
saying one other thing.
If something vibrates a
lot, the damping's small.
You need small
damping for things
to actually vibrate very much.
This thing, this is vibrating--
[HIGH TONE]
that high-pitched one,
that's about a kilohertz.
How many cycles do you
think it's gone through
to get down to 50% of
that initial amplitude
that you could hear?
A few thousand?
How much damping do
you think this rod has?
Really tiny, really tiny.
All right.
So even though all we
talked about today was
single degree of
freedom oscillators,
I hope you learned a few
things that we'll carry now
through the rest of the term.
We'll use all these
concepts that we did today
to talk about more
complicated vibration.
Good luck on your 2.001 quiz.
See you on Tuesday.
