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visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Today
we're going to talk about this topic of
vibration isolation, which is a very practical
use of knowing a little bit about vibration. So imagine a
situation you've been in where there's an air
conditioner in a window, and it's causing
your table to shake, or where you're trying to
work, or your bed rattles, or something like that. How many of you have
ever had an experience like that, something's
kind of annoying, messing up a lab
experiment, or whatever? Yeah, you've all experienced
these things, right? So as clever
engineers, are there simple solutions sometimes
to fixing these problems? And that's what we're
going to talk about today. I have a little quick demo
that I'm going to show you. You saw this the
other day, where this is my little squiggle pen,
and it's got a rotating mass inside. And we've looked at
rotating masses a lot. It has some unbalanced,
statically unbalanced rotating mass. Could be a fan blade with a
hunk of chewing gum or something stuck on a blade or
broken piece out of it. Puts in a force me omega squared
cosine omega t, basically an F0 cosine omega t
kind of excitation. And if it happens to be a
flexibly-mounted, mass-spring dashpot system, it'll vibrate. And so I showed you
that the other day. We'll do this, and we'll need
to lower the lights a little. But today, I've set your bed. This is your microscope
here, this little one. Think you can see it
in the foreground. And this is the air
conditioner, or the water pump, or whatever is
causing the trouble. So this is running
at about 28 hertz. If I set the strobe
just right, I can absolutely stop the motion. It doesn't look like
it's moving at all. That's because the strobe
is at exactly the same rate as the squiggle pen. Now I'm going to de-tune the
strobe a little bit so you can see the motion. There is the motion
of this main system. That's causing the problem. But it actually puts
vibration into the tabletop. And next door over here,
I have a little beam. And you can see
that little piece of white moving up and down. It's just a little flat
piece of spring steel with a magnet on the end as a
mass and a piece of white tape so you can see it. But notice it's going
up and down in synchrony with the other one. So this is your microscope
sitting on a lab bench some distance away. The problem vibrations
are being created by the original
imbalance in something, travels through the
floor, gets to your table with the microscope on it. Now your microscope shakes. So the issue is, what
can we do about it? OK. Oh, I do-- yeah, no, I'll
leave this for a second. I'll turn it off. And then I'm going
to have you consider. So I want you to get in
pairs and talk about this. I want, as a group,
we're going to come up with at least three
ways to reduce the vibration of
your microscope-- relatively simple
ways to fix it. How would you do it? So think about. Talk about it. And come up with three ways
of fixing this problem. I'm going to do one more
demo on this in a minute and then put up the transfer
function for the force, the one we had the other day, OK? The picture. All right. Let's have some suggestions. How would you go
about fixing this? All right. You had your hand up first. AUDIENCE: We have it
on suspension systems, and springs. PROFESSOR: Yeah. So put some springs on what? AUDIENCE: Like, have a table
where your microscope is, and then have [INAUDIBLE]. PROFESSOR: So springs
support the microscope. All right. So I had this little magnet
here sitting on this beam. This is that
spring-supported microscope. And I have done a
heck of a lousy job. This thing shakes like crazy. So what do you mean? Not like that. How might you-- OK. So you think you could change
the properties of this system so that it might do better? Let's think about that. OK, what's another idea? AUDIENCE: Change the
length of that spring could change its
natural frequency. PROFESSOR: Yeah. Are you talking about the
microscope one, the receiving one? If you change the length
of it to make it longer, it makes it softer, actually. Make it shorter, it
makes it stiffer. So you would change
its natural frequency. Now, the two ideas together, if
you set the natural frequency correctly, the system
on the receiving end, you can reduce its vibration. And I will demonstrate
that in a second. What's another idea? AUDIENCE: [INAUDIBLE]. PROFESSOR: Acid damping. OK. Well, that's a very
interesting suggestion. Damping, it helps
under one circumstance, but not under some others. And we're going to
explore that today. This is all basically--
we've come up with one of the three ideas. What else? AUDIENCE: Attach
something that vibrates 180 degrees [INAUDIBLE]. PROFESSOR: Oh,
that's interesting. That's the fourth one. I mean, that can be
a little expensive, but these are like
noise-canceling headsets, right? Could we put in something
else somewhere else on the table that
cancels the vibration out where your microscope is? Yeah, you could do that. A little expensive. What else? Yeah. AUDIENCE: You could
cushion [INAUDIBLE]. PROFESSOR: Where
would we put that? AUDIENCE: You could do it
underneath [INAUDIBLE]. PROFESSOR: Yeah. And so that's generally
the same idea. So you're all treating
the microscope. Can you treat something
else in the system? So yeah, we'll fix
the microscope end. But that might knock it
down by a factor of 10. I want to knock it down by
at least a factor of 100, if not a factor of 1,000. AUDIENCE: Fix the
air conditioner. PROFESSOR: Ah-ha-ha-ha-ha! You know, put another piece
of gum on the other blade. Or clean it up, or balance
the rotor, in effect. Rotors are not manufactured
defective usually. They get rejected at QC
before it goes out the door. So fix the rotor. Can be a little expensive
sometimes, but worth it. They do maintenance checks. Actually, they
have accelerometers built in to all expensive
rotating equipment these days. And it's called
condition monitoring. And when they get outside
of certain limits, they shut the thing
down and rebuild it. In electric generator
sets, gas turbines, jet engines on all aircraft
are all incredibly carefully balanced in a way. They start getting
out of balance, they stop and fix them before
the things blow up on them and you have a $20
million problem instead of maybe a $20,000 tear-down. All right? So yeah, you fix the rotor. Third idea. Well, could you-- this
thing's shaking like crazy. What if you change the
length of this beam? Could you stop this thing
from shaking so much? And if you stop this
from shaking so much, would it put so much
excitation into the table? No. So fix the rotor, isolate the
source, isolate the receiver. Now you get at least three ways. And the gentleman up here
came up with the fourth way-- active cancellation. OK. Great. I have a demo of one of these. Sometimes if you're
desperate and you're trying to get some
sleep-- it's your bed that's rattling
or something-- you might want to try the following
that you can do very quickly. So we'll dim the lights again. We'll turn it back on. [VIBRATING] Both are vibrating like crazy. All right. So one way to
detune the receiver is to put a big weight on it
and change its natural frequency by changing the mass. And now the little thing
is hardly shaken at all, because I've detuned just
by changing the mass. Accomplished the same thing
as switching the length. Instead of messing
with the stiffness, I've changed the
mass of the system. This is still
shaking like crazy. OK. So now it's back
to vibrating again. But over here is my source. I don't know if it'll
work so well in this one, because the source is
a lot more massive. But I'm going to put
the same mass on it. Ooh, it made it worse. Ah. And that's another
good demonstration. Excellent. So we could come back
up with the lights. You got to be careful of that. You go to mess with one of these
systems, if you do it wrong, you make the matters worse. First consulting job I ever
had back in 1977 or something like that, they had a
vibration problem on a ship. And the first consultant
in said, stiffen up. It was actually the exhaust
stacks on these 5,000 horsepower diesel engines,
and they were 30 feet tall and shaking like crazy. And the first guy said, stiffen
up those exhaust stacks. And he did exactly
the wrong thing. And it just shook
worse than ever. OK. So now what I'm
going to show you-- what we'll put on
the board today is a little bit of
mathematics to back up how you go about doing the two things. One is isolating the
receiver, or the other one's isolating the source. I'm going to start with
isolating the receiver. But we're going to start
with a little bit of math, a little math tool
that we need that will make life a lot easier for us. Yeah, I'll work here. If you recall last time, now
would be the time to do it, we derived the transfer function
for essentially this system, where we had an
F0 cosine omega t. And we computed the
response, x of t, as some x0 cosine omega
t minus the phase angle. And we worked it
all down to where we could plot it like that. But quite frankly, it was kind
of a lot of lines of math. And it was sort of painful. I actually hated
doing it on the board. But it was easy,
because it was familiar. It's just trig stuff. It was all trig and a
little bit of calculus. So we do that first, because
it all makes sense to you mathematically. But there's a vastly
easier and quicker way to do this, which we'll
address right now. And that's using
complex numbers. So we need a couple bits
of information here. One is Euler's formula. So if you have e raised
to the power i theta, you can show that that is the
same thing as cosine theta plus i sine of theta. That breaks into a real
part and an imaginary part. So if we wanted to express this
excitation, F0 cosine omega t, in complex notation,
we would say it is the real part,
which I'll denote as Re, the real part of
F0 e to the i omega t. And I'm going to just specify
that F0 itself here, this is real and positive. So this is real and positive. So the real part's
going to be F0 times-- and if you break down co ee
to the i omega t into its-- by Euler's formula, it
gives you cosine omega t plus i sine omega t. And the real part is the cosine
part just according to this. OK. Now, another little fact
that I want to show you is if you have a complex
number, a plus bi, I want to express it as
some ce to the i theta. So you want to use
Euler's formula to express a complex number. And if we draw it, the answer
becomes pretty obvious. This is a point up here, a, bi. And this is the imaginary
axis here and the real. And this point, this side is
a, and this side here is b. And this side here, the
length of this triangle, is c. And the angle in here is theta. So now if I ask you,
what's c, well, you say, oh, well, c is obviously
the square root of a squared plus b squared. And theta is a tangent
inverse of b/a, which is the imaginary
part over the real part. If you want to express a
complex number this way, well, the magnitude is
square root of a squared plus b squared. And the phase angle
that you put up here is tangent inverse of the
imaginary part over the real. OK. So now we have
the basic tools we need to take on the
vibration problem. And so we have that
system up there. And our output, from the
way we derived it last time, the output is some x0
cosine omega t minus phi. And one of the reasons it was
so painful doing it this way last time is you
have to-- this is cosine is a function
of both time and phase. And to break it apart
takes a lot of work. So we want to do the same thing,
but with complex variables this time. So I want to express this
then as the real part of-- I could say it's the real
part of x0 e to the i omega t minus phi. And despite Euler's formula if
this is real, just the number. Than this breaks down into
cosine omega t minus phi and i sine omega t minus phi. But here's the beauty of
using complex notation and exponentials. This now becomes the real part
of x0 e to the minus i phi times e to the i omega t. If I could separate these two. And this is what
makes it so much easier to use this approach. And I'm going to call this
part of it just some capital X. And it is a complex number. For sure. If I break this up into cosine
of minus phi and minus i sine phi, it's got an i sine phi. So this is a complex number--
a plus bi kind of form. So in general, this
thing is complex. So this whole thing
is the real part of some X, which I don't
know now, e to the i omega t. So now we can quite
quickly do the derivation we did last time. We're talking about representing
linear systems by some kind of black box-- has a transfer
function in it, which we call H, in this case, x/F. Response x
per unit input F. And remember, we're talking about steady
state response only. And we have, as our input here,
some F0 e to the i omega t. And we have, as an output,
some X e to the i omega t. And we know that we're going
to use the convention that we care about we have to
have real number answers. So we'll be eventually
actually using the real part of the input and
the real part of the output. But to get there, we're going
to use complex notation first and then separate out the real
and imaginary parts at the end. So for our system, we know
the equation of motion, so now it's some F0
e to the i omega t. There's our equation of motion. And I'm going to let x here
be this unknown capital X e to the i omega t. And I'm going to plug it in. And the exponentials
are particularly easy to deal with when
you're taking derivatives. So upon doing that, we
immediately get minus omega squared M plus i omega c plus k. All of that times
Xe to the i omega t equals F0 e to the i omega t. Immediately, I can get rid
of the time-dependent parts. And I can solve for x/F,
which is what we set out to do the other day to find this
transfer function between input and output. So if I solve for
x divided by F, I'm going to get all this
stuff and the denominator on one side. And I'll write it out here. It simply looks like that. And now remember, I can
substitute in some things. I remember k/m is
omega n squared. And zeta in c over 2 omega n M. And I plug those
things in and just rearrange it a little tiny bit. We should come up
with something like we found before, so that x/F,
1/k, and the denominator, 1 minus omega squared over
omega n squared-- not quite yet here-- plus 2 i zeta
omega over omega n. That's what it looks like. So you still have a
complex denominator. And this basically looks like
a number 1 over k times 1 over some a plus bi. There's your a term. Here's your bi term. And the way you deal with
something like this-- you have an a plus bi in the
denominator-- you multiply the numerator and denominator
by the complex conjugate in order to get this into
actually standard a plus bi form. If I do that
symbolically here, it comes out looking like a
minus bi over a squared plus b squared. And that's 1/k e
to the minus i phi over square root of a
squared plus b squared. Because now, see,
the denominator's just a real number. So this whole thing is,
in some form, c plus a di. You could break this into
a real part, complex part. We could say that's
equal to some magnitude times e to the i phi. To get the magnitude,
you take a squared plus b squared square root. It cancels. This is squared the
denominator, so you end up with this part, square
root, in the denominator. This is what the-- we need
to know what phi looks like. Well, phi had better
come out like before, where now phi is minus tangent
inverse of the imaginary part over the real part. And the imaginary
part has a minus here. That's why a minus pops up here. Imaginary part comes from this. The real part comes from there. The common denominator
stuff all cancels out when you take the ratio. So this is tangent inverse of
two zeta omega over omega n all over 1 minus omega
squared over omega n squared, as before. I've skipped a couple of steps,
but we cranked this whole thing out before. This is the same steps that you
would go through to do that. We're just doing this to
get to the phase angle. But this now is exactly the
same thing we got before, which we have plotted up there. We work with magnitude
and phase angle. So the magnitude of
x/F is the same thing as saying the magnitude
of the transfer function. And that transfer function
looks like 1/k, the magnitude, all divided by 1 minus omega
squared over omega n squared squared plus 2 zeta omega over
omega n squared square root. That is the same transfer
function magnitude that we derived last time,
with a lot more work. And this approach,
using complex variables, you can use for any single
input, single output linear system. And we're going to do it to
derive right away the transfer function for the response of
this to motion of the base. So if you follow how we used
this complex variables in e to the i omega t's
to get here, we can now apply the same tools
to do other transfer functions to be a lot more
efficient about it. Before I jump to
this one, remind you how, in practice, we use this. So if the statement
magnitude of x/F equals everything
on the right there. Then in the way
we would normally use this is to say, well,
if you want the magnitude of the response, you take the
magnitude of the input force, multiply it by the magnitude
of the transfer function, evaluate it at the
correct frequency. That would give
you the magnitude. If you want the time
dependence, x of t, well, that's the
magnitude of the force, magnitude of the
transfer function times the real part of e
to the i omega t minus phi. And this gets us back to
when you work this out, this is your x0. And this is your cosine
omega t minus phi. So once you know what
the excitation force is and its frequency, you
put the force in here. You evaluate that thing on the
left at the correct frequency. And you write out
the answer directly. In one of the
homeworks for today, the question just had you
go through the exercise of figuring this out at
three different frequency ratios, like 1/2, 1, and
3, or something like that, would put you to the left of the
peak at 1/2, on the peak at 1, and way off to the right
out at the right edge at 3. And you'll get three
different response amplitudes and three different
phase angles that go with it. All right. So that's how you review. Did the same thing
a different way. And I'm going to move
on to base motion. But any questions
about this now? Yeah. AUDIENCE: Was the e to
the negative ib included in your F of x? PROFESSOR: Yes. AUDIENCE: OK, so
why did the negative b appear again after
your final [INAUDIBLE]? PROFESSOR: The
very top expression up there, it says
x/F. It says we're trying to cast it in the
e to the minus i phi form. That's my goal. And I did that, because
we started over here with the problem that
we had done before, where that's the way we
decided to write the answer. And it turns out that it's
just a convention in vibration engineering that authors and
people have adopted to express the phase angle as minus phi. They could have
done it as plus phi. The plots like this are phi. AUDIENCE: Right. But I guess what I'm wondering,
isn't [INAUDIBLE] x/F. PROFESSOR: Oh, I
see what you mean. It's in there before
you take its magnitude. So the Hx/F, when
it is-- this here is left in complex notation. And this is Hx/F of omega. And it is complex. We take its magnitude. Then the magnitude is
not complex, right? And so we take its magnitude. We get that expression. But when we take
its magnitude, we've thrown away the
phase information. So we have to keep it
and put it somewhere. And so we put it in the
e to the i phi form. And I guess what I
should have done here is now this is--
I've taken-- this is Hx/F, same thing as
x/F, in complex form. And I've said, OK, if
I write it this way, I have just said
it is a magnitude. Times its phase information. I've separated its phase
information from its magnitude by writing it this way. OK. And the phase then is that. And its magnitude is that. Good question. All right. So now let's see if we
can kind of pretty quickly do the same problem
for base motion. So this is our microscope now,
idealizes a mass spring system. So this is our microscope. Has some mass stiffness
damping motion, x of t. And how do you suppose--
where would you measure that motion x of t from? Like, to define
your coordinate here is a major point in
the last homework. Is gravity involved? But only as a constant term,
mg in the equation of motion. It's only there
depending on if you write the equation of motion
in the less desirable way. Where is this measured
from do you guess? Equilibrium position? Static equilibrium position? Or 0 spring force position? How many suggest 0
spring force position? How many suggest
static equilibrium? OK. You got the message. This is from equilibrium,
because you don't have to deal with the mg term. So this is measured
from equilibrium. That's the deflection of
the microscope support. This is the deflection of
the floor that's driving it. Then we know we've got that
table shaking like crazy. That's what's causing
this to vibrate. And we need a free-body diagram. And we approach free-body
diagrams just like before. You imagine positive
motions of x and x dot, positive motions of y and y
dot, and deduce their forces. So positive x gives
you a kx opposing. A positive x dot gives
you a cx dot opposing. A positive y gives you what? A force that results on this. Positive motion of the floor. Positive or negative force? How many think positive? How many think negative? How many aren't sure? How many aren't awake? OK. Look. If I push up on
this-- and now this is fixed when you do
this mental experiment. You fix this momentarily. You cause a positive
deflection here. It compresses the spring. Does the spring
push back or not? So if I'm moving upwards, which
way is the spring pushing? All right. But if I'm pushing
upwards, which way is the spring pushing on the mass? Up. So this one gives me a ky up. And the dashpot does a
similar thing-- cy dot up. And there's also an mg here. But there's also a kx
static, if you will, up. And they cancel. We know that. So we don't have to
deal with the mg terms. So now we can write
our equation of motion. And the equation of
motion for this system is the mass times
the acceleration. That's got to equal
to the sum of all the external forces-- one,
two, three, four of them. And I'll just save a little
time and board space. I'll put them on the correct
sides of the equation. So these are the x-- put the
x terms on the left side. cx dot plus kx. And on the right-hand
side, I get ky plus cy dot. This is my excitation. That's the floor motion. And this is my response
on the left-hand side. So I'm going to let
y of t, the input, be some y0 real positive
times e to the i omega t. And I'm going to assume that
the response is some x, probably complex, e to the i omega t. So this is x of t here. Equals some x I don't
know e to the i omega t. And I'm going to plug those
two into this equation. If I just do that directly,
x is on the left side. y is on the right side. Then I find minus omega squared
m plus i omega c plus k, just like before, xe to the i omega
t equals k plus i omega c y0 e to the i omega t. And nicely, I can for now get
rid of the time-dependent part. And I can solve for
the response that I'm looking for-- x over the
input is real and positive, amplitude of vibration
of the floor. And that I will call
Hx/y of omega, a transfer function, probably
complex, that I can then deal with like I did above. And when I finish manipulating
things, substituting in zetas and omega n squareds
and that kind of thing, this becomes-- well, first,
I'll write it this way. I can write this as a magnitude
times an e to the minus i phi again. That's where I want to go. And when I do that, 1 plus--
a little messier-- 2 zeta omega over omega n
squared square root. This is just the numerator. And the denominator is just the
same as the other single degree of freedom things. 1 minus omega
squared over omega n squared squared
plus 2 zeta omega over omega n squared square
root e to the minus i phi. So now it's the
transfer function as before except the
denominator's a little messy. And there's no 1/k. And I am going to have a
messier expression for phi here. So there is something wrong with
one of the boards this morning. Kind of messy, complicated. Do I ever use it? Rarely. What's important in these
things and what isn't-- really what's important when you're
just trying to get a quick solution to vibration
isolate something, you really want to know what
this is going to come out looking like. You're trying to make
the response x small compared to the input. That's the whole objective. Right now the table might be
moving a half a millimeter or something like
that, but this thing's moving out here five or
six or seven millimeters, 5 or 10 times that. And what we'd really like
is if the table's moving a millimeter, you'd
like this thing out here moving 1/10th
of a millimeter. So the real objective here
is to make this small. It's the magnitude
you care about. Phase you rarely even want
to know or need to know. So we're going to do
a sample calculation. Let's give an example here. So the source is at 20 hertz. So your unbalanced pump,
your unbalanced rotor. Yeah. AUDIENCE: How do we know
in the previous thing that the frequency of
oscillation has to be the same? Like, why wouldn't
it be twice that? PROFESSOR: OK. That's a great question. And I haven't mentioned this
before, and I intended to. These systems that
we're looking at are linear systems, which is
where we started the other day. Linear systems have some
interesting and very useful properties that we depend upon. One was, I said, force
one gives you output one, force two gives you output two. Force one plus two gives
you the sum of the outputs. The other feature
of a linear system is steady state response after
the transients have died away. If the frequency of the
input is at 21.5 Hertz, the frequency of the output
is at 21.5 Hertz, period. Linear systems, the
frequency of the input is equal to the
frequency of the output. That's a really important
little factoid to remember. So I turn on the pump, the
pump's running at 20 Hertz. 20 Hertz times 60 is 1,200
RPM, very common motor speed. So the pump's
running at 20 Hertz. So that fan, it's
got an imbalance. So that means you're
putting excitation into the floor at 20 Hertz. And I want to reduce the
vibration at the microscope by 90%. What that really
means is that my goal is that the magnitude
of x/y is 0.1. And that's the magnitude of
this transfer function, Hx/y. So I want this transfer
function to be 0.1. So just look at the picture. Can I get that answer
to the left of the peak? And what this plot shows
you is this magnitude of the transfer function, for
a variety of values, a damping. And of course, the
lower the damping, the higher the peak
gets at resonance. Right? So no matter what
the damping is, what is the curves all go
to in the left-hand side? They go to 1. And that's really saying the
static response of this system is if you deflect the floor an
inch, the table moves with it. Everything has to move
together when you get down to 0 frequency input. So everything goes
to 1 on the left. You go through
resonance at omega equals a natural frequency. But out to the right, as
the excitation frequency gets higher than the
natural frequency, the response drops off below 1. Which one drops the fastest? As you increase omega
over omega n beyond 1, there's a whole mess of
curves to the right that blend together. And they differ only in damping. Can you tell which
one is the-- let's say if you go to-- at three,
there, the response is at 0.1 for the lowest
curve on that curve, right? And that's the one
with no damping. It's a little
counter-intuitive, right? All right. Well, let's come back to it. Damping does help,
but not at this point. So we need to find
a value of omega over omega n which is greater
than 1 that satisfies this. That's what we're after. And this is kind of
messy to work with. And since I know
the one that works the best is the one
with no damping, we'll solve the no
damping one first, because it makes the
algebra really easy. And then we can go
back and say, now, what happens if you
add some damping? So for the case
there's no damping, the numerator goes to 1. The denominator goes
to just 1 over 1 minus omega squared
over omega n squared. So it becomes that. That simple. And because I want to work
with this ratio bigger than 1, I don't want this
to be negative. And I want to mess with-- keep
carrying along absolute value signs. This is the same thing
as 1 over omega squared over omega n squared minus 1. I just reverse
this, because I know we're going to deal only with
the ones greater than 1 here. And I need this to
be equal to 0.1. And that's just algebra. You could solve that. This implies that omega
over omega n equals root 11, I recall. And that is 3.31. So this is saying on
that curve, if you go out to omega over omega n equals
3.31 right about where that arrow is, the curve for
zero damping drops down to 0.1. And now if, at that
frequency-- ah. So that means we have to
design the spring support such that omega n is equal
to omega over 3.31. But omega-- where'd we start? So F equals 20 Hertz. Omega equals 2 pi f. Do I have that number here? No, but-- so this tells me that
I need a natural frequency that is omega over 3.31, or I need
an fn that is f over 3.31 is 20 Hertz over 3.31. And that number I do have. 6.04 Hertz. So I need a support
whose natural frequency is 20 Hertz divided by 3.31. I need a support whose natural
frequency is 6.04 Hertz. And that's how you go about
designing a flexible base to isolate something
from vibration of whatever it's sitting on. All right. So my f here, 20 Hertz. But my fn needs
to be 6.04 Hertz. That implies multiply by 2 pi. I'm looking for 37.96
radians per second. And that's equal to square root
of k/M. So now what's the M? Well, it's whatever the mass of
the microscope plus its base. Whatever is being
supported by the springs will have that mass. You have to choose the k. So let's say that M total for
this system is 20 kilograms. Solve this equation for k. And that implies that k is
28,827 Newtons per meter. OK? So if we were to
design this system-- and it really mounts up
to in the case of this. Let's see. Beams. The stiffness of a
beam-- ah, that's a good. We'll do this. We have a cantilever here. And we've got a mass on the end. But most of you have
been taking 2001. If you put a force
out here, P, what's the deflection at the
end of a cantilever? AUDIENCE: [INAUDIBLE]. PROFESSOR: OK. So delta is PL cubed over 3EI. And the load, this force, is
equal to some k equivalent times delta, right? This is just a spring. And k times the displacement
is the force it takes to do it. So P's my force. The spring constant
is somehow associated with the rest of this stuff. So if I solve for P over
delta, I get 3EI over L cubed. OK? So if I'm running right at
the natural frequency here and I want to reduce this
to a 1/10th of its motion, I need to change the spring
constant of this cantilever by a factor of-- well, I need
to change the natural frequency by a factor of 3.31. So my k equivalent here
is 3EI over L cubed. And that's what would
go into this equation. But I know that I have a
natural frequency right now. I want it to go down
by a factor of 3.31. So that means I
need to decrease k such that the square root of k
goes down by the factor 3.31. So how much do I have
to change the length? Probably something like
the square root of 3.31. Roughly 2. So if I double the
length of this thing, do you think it's going to work? If I double the length of this
thing and turn it back on, then we shouldn't see
much motion out of this. [VIBRATING] That's moving a lot. It's moving a tiny, tiny bit. So it works. So that's one step of
vibration isolation. Now I'm going to show you a
vibration engineer trick, which is a very handy thing to know. Where's my strong magnet here? So I've got another
beam just like this one. I've got a pretty
massive magnet on it. So it makes another
cantilever beam just like I got over there. OK? So I claim that
with just a ruler, if I clamp this
down at some length, I claim, with just a
ruler, I can predict the natural frequency of that. Take a couple of
minutes and see if you could figure out how to do it. Think about that. Just a ruler. Measurements that I can make. I don't know how
long it actually is. I don't know how thick it is. I know it's steel, but you just
don't have enough information to compute 3EI over L cubed. But simply with a ruler, I'm
going to be able to do this. Talk about it. Think about that while
I set up the experiment. OK. Who's got it figured out? Anybody want to
take a shot at this? So there's my beam. I put the weight on it. What does the beam
do statically? Bends a little, right? kx static equals Mg, right? Has to. So x static is what
I'm calling delta here. So k delta equals Mg. k equals Mg over delta. Natural frequency equals
square root of k/M. Incredibly simple, huh? So what's the experiment
that I would-- what measurement would I make? Delta, right? Put my ruler up there. I measure its static
position like that. Then I put my mass on it, and
I measure the static position again. I measure the delta. And I get a prediction. And I did this in my office. And the delta that I
measured-- I actually set it at a particular length. It was 18 centimeters. Delta measured, I think, 0.5
centimeters, or 0.005 meters. And if you compute omega n
then equals the square root of 9.81 over 0.005. And I want this in Hertz. So I can divide by 2 pi. This comes out as 7.05 Hertz. And Fn measured was 6.57. Pretty good but not perfect. And it's because I've made
an approximation that I glossed over pretty quickly. What has been left out
of this system that would cause the measured
natural frequency to be lower than the predicted? What's been ignored? Yes. AUDIENCE: Damping. PROFESSOR: Damping. Ah. Maybe. How much damping do we
have in this system? Probably at least 10 cycles
to the k halfway, right? Certainly less than 1%. The damped natural frequency is
equal to the natural frequency of the square root of
1 minus theta squared. So this is something
like way less than half a percent difference. So that wouldn't account for it. That's considerably more
than half a percent. So damping couldn't do it. Yeah. AUDIENCE: [INAUDIBLE]. PROFESSOR: Ah, the
mass of the bar. Does this flexure have mass? Yeah. It's probably on the order
of if you stack them all up and compared to that, it
might even be as much as half the mass of the end. And as it vibrates
back and forth, does it have kinetic energy? Yeah. We've ignored the kinetic
energy of the mass. And in fact, that's the
principal error here. We've left out the mass. There's actually a
pretty simple way to-- using energy and just
thinking in Lagrange terms, you can account for
the energy of the mass in this single degree
of freedom system and get a very
accurate prediction. We won't do that today. But I think we'll do that
before the term's out. OK. This applies to any
simple mass spring system in the presence of gravity. So here's a mass. And actually, we're
doing the problem today where I'm moving the base. So here's its base. So this is the table moving. And if I do this, it
clearly makes that move. If I do this really fast,
it doesn't move very much. If I do it close to the natural
frequency, it moves a lot. If I move it very slowly,
as I go up one unit, this follows me exactly. That's why that plot goes to 1. At very, very low frequency,
the support and the mass move exactly together. At very high frequency-- if
I can stop the transient-- I can't do it very well. The mass doesn't move much. The base moves a lot. And at resonance, it goes nuts. OK. The unstretched length of this
spring is about seven inches. The square root
of g over delta, I ought to be able
to predict this. So I did a quick
calculation on that. It was like 1% error. I measured it at 7.36
radians a second. And I predicted it at
7.43 measured 7.36. Same kind of thing--
ignoring the mass of the spring a little bit. So g over delta is a great
little thing to remember. OK. So we have done all but one. Everything we've
started out with today, we've said there's
three ways to fix this, and came up with a fourth way. So in this case, soften
the spring support a lot, so that the natural frequency
is way less than the excitation. We said, what about spring
supporting, softening, flexibly amounting this source, so
that it doesn't put vibration on the table? That's the piece we
haven't addressed. So let's look into
that problem now. So here's our source, some
rotating mass eccentricity causing an excitation. So this has a force
F0 e to the i omega t, which is coming
from the rotating mass. And it applies to the
floor, through the dashpot in the springs, some
FT, I'll call it, F transmitted to the
floor, e to the i omega t. And I want to know-- I need the
H force transmitted per unit force input transfer function. That's what I'm looking for. So now free-body diagram. Now we're going to
make an assumption. We're going to assume that
the motion of the floor, which we'll call y of t,
assume that y is much, much less than x. It's generally true. Whatever's shaking
like crazy, the table's not moving much underneath it. So I'm going to assume, for the
purposes of calculating forces, that this is 0. So for the motion x, what is
the force applied to the floor? So F of t. If you have a positive
displacement x, the force is kx. You have a positive velocity
x, the force pulling up on the floor through
the dashpot is cx dot. So the other way of
saying that is here's our free-body diagram. Here's our F0 e to the
i omega t pulling up. It responds at some x. And the resulting forces
through the spring and the dashpot we
know are kx and cx dot opposing the motion x. Well, by third law, if
these are the forces on the spring and the dashpot,
then down here on the floor, you better have some equal and
opposite forces, kx and cx dot. So this force
produces a motion x. The motion x produces forces in
the mass in the spring, which make the force on
the floor, the spring force, and the dashpot force. OK. So this Ft is-- I
want to write it here. That's all that is-- positive. And I'm going to
assume a solution that we know to work for x,
which is xe to the i omega t. We've plugged it in before. So I plug that in here,
I get a k plus i omega c, xe to the i omega t. So I can just express
my force on the floor in terms of the motion x. And I'm looking for a transfer
function for force transmitted over force in. But force transmitted
is my k plus i omega c, xe to the i omega t. And the force in is
F0 e to the i omega t. Cancel out the
time-dependent part. And it says the transmitted
force over the input force is this little complex
expression times the response x over F. But we
know what that is. That's the transfer
function Hx/F. So this is k plus i omega
c times Hx/F of omega. So this gives us a slightly
different transfer function. Ooh, look at this. Before, when we did
x/y, we ended up with k plus i omega c
y e to the i omega t. And when we did then x/y, we
got the same ratio as this. Exactly the same thing. So I could write all this
out, but-- and let's say I'll do this. Hx/y-- no, no, I won't do that. What I'm going to tell you-- if
you just work through this now, you will find that H force
transmitted over force in is exactly the same as Hx/y. And that what we really care
about is what the magnitude is. So the magnitude of these
two things are the same. And in fact, just work out to
that same expression as before, the 1 plus 2 zeta
omega over omega n squared square root all over
the usual big denominator. So conveniently, for
vibration isolation, the solution to the two
problems are exactly the same. So if you have that one, you
have the transfer function xy that was projected on
the screen a minute ago, it is also the force transmitted
to force in transfer function. So you just have
to remember one. And if you now want
to-- we said, let's say, doubling the length of this just
about accomplished the reducing the vibration of the microscope
by this factor of 10. So if I doubled the
length of this one, I would roughly
do the same thing. I would change the
natural-- this thing is right on the natural
frequency of this beam. That's why it shakes so much. And so it is this system. It's shaking like crazy,
putting force into the table. The table is vibrating,
causing the other one to move. So now if I change this one,
then the same kind of idea. Maybe roughly double its length. Natural frequency diminishes
by a factor of 3 or so. The vibration of this
ought to go way down. And actually, our
little beam out here is picking up more
than the other one. Shh. So this thing is hardly
moving at all now. So by doing that, we've
essentially detuned it. This is no longer running at the
natural frequency of this base. So it's no longer resonant. You're way out on the
curve to the right. So the response of
this isn't very much. That means it doesn't transmit
much force to the base, maybe down by a factor of 8 or 10. That means the table vibration
amplitude drops by that factor. Means that the base motion
over here is now a factor of 10 smaller than it
was to begin with, so that we get a
reduction of 10 here. And we get another reduction of
10 here, because we detuned it. So you might get a factor of 100
reduction by working on both, you see. You've treated the source and
you've treated the receiver. But fortunately, they
use the same curve. So damping. When you do vibration
isolation, you're trying to get well out on
this curve to the right. So there's a couple of
practical engineering things that limit how far you can go. To get further out on
the curve to the right, what do you have to do to the
spring in the system to get stronger or softer? You're trying to make the
natural frequency-- see, the excitation frequency
doesn't change. In order to get omega over
omega n to go bigger and bigger, the excitation's
staying the same. You're having to reduce
the natural frequency. And so what do you have to
do to the spring constant? Decrease it. What is the practical
limit of decreasing the spring that
supports your pump, or your washing machine,
or your air conditioner? AUDIENCE: [INAUDIBLE]. PROFESSOR: Pretty
soon it's just going to-- if it's too heavy--
you put it on there, it's just going to squash
the springs, right? So you can't-- there's limits
to how soft you can make springs to support heavy machines. So there is a practical
limit to how far to the right you can go. But normally, you get out
there as far as you can. And then if the real
system has damping, does it improve or
degrade the performance of your vibration
isolation system? Well, the more the damping
you have, the higher up you are on those curves. So the damping decreases
the performance. But every system has to-- when
you first turn on that motor, the system has to spin up. And you're going to have to
go through that resonance, so that you want some damping. Because if you've got
your scanning electron microscope or your laser
interferometry system set up on a spring-supported table,
if that table has no damping and you walk in the
door and bump it, it is going to sit there
and vibrate all afternoon at its natural frequency due
to the initial conditions. So you need some damping
to prevent problems, either response to
initial conditions, or bumping it, or whatever. Or even as the system
turns on and speeds up, it'll have to go
through that resonance. And it'll vibrate like crazy
as it does, and then finally settle down at the
higher frequency. So you need some damping. But damping does degrade the
steady state performance. And I'm out of time. And we'll see you in recitation. Thanks.
