20. Linear System Modeling a Single Degree of Freedom Oscillator

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visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Are we ready
with the concept questions from the homework this week? How do we get
different– there we go. I looked at one,
it was one thing. I looked at the
other, [INAUDIBLE]. Does g enter into
the expression for the undamped natural frequency? And most people said no, but
about a third of you said yes. If you have worked
on that problem now, you have already
discovered the answer. So you’ll find that g does
not come into the expression. When you do a pendulum,
g is in the expression. And there’s a question
on the homework about what’s the difference. How can you predict
when g is going to be involved in a natural
frequency expression and when it is not? I want you to think
about that one a bit, maybe talk about it
at– if there’s still questions about it,
we talk about it in recitation on
Thursday, Friday. OK. So does g enter into
the expression here? I’m sure you know this simple
pendulum, the natural frequency square root of g over l. For a simple
mass-spring dashpot, the natural
frequency is k over m whether or not it’s
affected by gravity. So there’s something
different about these two. OK, let’s go onto the third one. “In an experiment, this system
is given initial velocity observed to decay in
amplitude of vibration by 50% in 10 cycles. You can estimate
the damping ratio to be approximately,” what? Well, it gives a little rule
of thumb I gave you last week. I’ll go over it again today. 0.11 divided by the number
of cycles did decay 50%. So it took 10
cycles, 0.11 divided by 10, 0.01, 1.1% damping. OK, next. “At which of the three
excitation frequencies will the response
magnitude be greatest?” You’ve done oscillators
excited by cosine omega t kind of things before. So at the ratio 1,
most people said it’s where it would
be the largest. Why at 1? Anybody want to
give a shout here? What happens when
you drive the system at its natural frequency? It’s called resonance, and we’re
going to talk about that today. So it’s when the
frequency ratio is one that– and for the system
being lightly damped that you get the largest response. Finally– which
one are we on here? Oh, this one. Can all the kinetic energy be
accounted for by an expression of the form IZ omega squared. By the way, I brought this
system if you haven’t. So there’s a really
simple demo, but it has all sorts of–
so in this case, we’re talking about that motion. It certainly has some I omega
squared kind of kinetic energy, but does the center of gravity
translate as it’s oscillating? What’s the potential
energy in the system? By the way, any time
you get an oscillation, energy flows from potential
to kinetic, potential kinetic. That’s what oscillation is. So there has to be an exchange
going on between kinetic energy and potential energy. And if there’s no
losses in the system, the total energy is constant. So the kinetic energy’s
certainly in the motion, but when it reaches
maximum amplitude, what’s its velocity when
it’s right here? Zero. So all of its energy
must be where? In the potential. And where’s the
potential in this system? Pardon? In the string. That’s not stored in the string. There’s only two sources
of potential energy we talk about in this class. Gravity and– AUDIENCE: Strings. PROFESSOR: Strings. Well, these strings
don’t stretch, so there’s no spring
kinetic energy. We’ve got potential energy. There must be gravitational
potential energy. How is it coming
into this system? So he says when it turns,
the center of gravity has to raise up a
little bit, and that’s the potential energy
in this system. The center of gravity goes
up and down a tiny bit. So is there any velocity
in the vertical direction? Is there any kinetic
energy associated with up and down motion? Yeah, so that doesn’t
entirely capture it, 1/2 I omega squared. Is it an important
amount of energy? I don’t know, but there is
some velocity up and down. My guess is that it
actually isn’t important, that the answer is it
does move up and down. It has to, or you would not
have any potential energy exchange in the system. OK. Is that it? OK. Let’s keep moving here. Got a lot of fun things
to show you today. So last time, we talked about
response to initial conditions. I’m going to finish up
with that and then go on to talking about excitation
of harmonic forces. So last time, we were
considering a system like this– X is measured
from the zero spring force in this case. Give you an equation
of motion of that sort. And we’ve found that
you could express x of T as a– I’ll give you the exact
expression– x0 square root of 1 minus zeta squared. Cosine omega damped times time
minus a little phase angle. And there’s a second term here,
v0 over omega d sine omega dt. And the whole thing times e
to the minus zeta omega n t. So that’s our response
to an initial deflection x0 or an initial velocity v0. That’s the full kind
of messy expression. There’s another way of writing
that, which I’ll show you. Another way of saying is that
it’s in x0 cosine omega dt plus v0 plus zeta omega n
x0, all over omega d sine omega d times t e to
the minus zeta omega nt, the same exponent. This is your
decaying exponential that makes it die out. And so I just rearranged
some of these things. There’s another little
phase angle in here now. So you have just a cosine
term, this proportional x0, and a sine term, which
has both v0 and x0 in it. The x0 term, if
damping is small, this term is pretty small
because it’s x0 times zeta. When you divide by omega
d, which is almost omega n, that goes away. So this term, the
scale of it is zeta x0. So if this is 1% or 2%,
that’s a very small number. I gave you an
approximation, which for almost all practical
examples that you might want to do, make it
approximately sine here. So this is x0 cosine omega dt
plus v0 over omega d sine omega dte to the minus zeta omega nt. And this is the practical one. For any reasonable system that
has relatively low damping, even 10% or 15% damping, you
get part of the transient decay comes from x0 cosine, the other
part v0 over omega d sine. That’s what I can remember in my
head when I’m trying to do it. Now the question, the thing
I want to address today is what’s this useful for? My approach to
teaching you vibration is I want you to go
away with a few simple, practical understandings
so that you can actually solve some vibration
problems, and one of them is just knowing this allows
you to do a couple things, and we’ll do a couple of
examples this morning. By the way, this form, this is
A cosine plus B sine expression. And I label them
A and B. A and B, they’re both of the
form A1 cosine omega t plus B1 sine omega t. And you can always add
a sine and a cosine at the same frequency. If I put just any
frequency here, they just have to be the same. You can always take an
expression like that and rewrite it as some
magnitude cosine omega t minus a phase angle. And the magnitude is
just a square root of the sum of the squares–
A1 squared plus B1 squared. And the phase angle
is the tangent inverse of the sine term
over the cosine term. So you can always
rewrite sine plus cosine as a cosine omega t minus
v. We use that a lot, and that’ll be used
a lot in this course. And then, of course,
if this whole thing is multiplied by an e to the minus
8 omega mt, then so is this. OK. OK, what are these
things useful for? And we’ve derived this all
for a mass spring system. Is that equation
applicable to a pendulum? So this expression is applicable
to any single degree of freedom system that oscillates. You just have to
exchange a couple things. So let’s think about
a simple pendulum. So our common massless
string and a bob on the end, some length L,
equation of motion. And this is point A up here. IZZ with respect to A. Theta
double dot plus MgL sine theta equals 0. That’s the equation of motion. With no damping,
that’s the equation of motion in this system. Is it a linear
differential equation? And to do the
things that we want to be able to do in this
course, like vibration with harmonic inputs
and so forth, we want to deal with
linear equations. So one of the topics for
today is linearization. So this is one of the simplest
examples of linearization. We need a linearized
equation, and we need to remember in a
couple of approximations. So sine of theta, you can
do Taylor series expansion. It’s theta minus theta cubed
over 3 factorial plus theta to the fifth over 5 factorial
plus minus and so forth. And cosine of theta
is 1 minus theta squared over 2 factorial plus
theta to the fourth over 4 factorial, and so forth. So what we say, what do we mean
when we linearize something? So linearization means
that we’re essentially assuming the variable that we’re
working with is small enough that the right hand side,
an adequate approximation of this function, is to keep
only up to the linear terms on the right hand side. For sine, the term
raised to the 1 power, that’s the linear term. This is a cubic term,
a fifth order term. We’re going to throw those away
and say, this is close enough. For cosine, it’s 1
minus theta squared. We throw these away. The small angle approximation
for cosine as it’s equal to 1. That’s the simplest
example of linearization, of a non-linear term. So when you linearize
this equation of motion, we end up with IZZ with respect
to A theta double dot plus MGL theta equals 0, and we get
our familiar natural frequency for Bob as square
root of g over L. So we need linearization to be
able to do pendulum problems. Hmm. OK. Or maybe let’s do
an example here. I’ve got a pendulum that
we’ll do an experiment with this morning. But 30 degrees is
about like that. 17 degrees is about like that. That’s quite a bit of angle. Is that small in the sense that
I’m linearizing this equation? So 17 degrees happens to be–
I’ll have to use this here. That’s actually 17.2
degrees equals 0.3 radians. Sine of 0.3 is 0.2955. And if we fill out and
look at these terms, the lead term here is 0.3–
so plug in the 0.3– minus– and the second term when you
cube 0.3 and divide it by 6, the second term is minus– what
is my number here– 0.0045. And if you subtract that from
this, you get exactly this. So to four decimal
places, you only need two terms in this series
to get exactly the right answer. This thing out here, this fifth
order term, is really tiny. But the approximation, if
we say, OK, let’s skip this, we’re saying that 0.3
is approximately 0.2955. Pretty close. So up to 17 degrees,
0.3 radians, that’s a great approximation. So it’s a little high by about,
I think it’s about 1 and 1/2% high. So for pretty large
angles for pendula, that simple linearization
works just fine. OK, once we get it linearized,
that equation of motion is of exactly the same
form as the one up there. We don’t have any damping in it. We could add some damping. We can put a damping in here
with a torsional damper– ct theta dot. And now that equation is
of exactly the same form as the linear oscillator,
linear meaning translational oscillator. Have that inertia term, a
damping term, a stiffness term. It’s a second order linear
differential equation, homogeneous linear
differential equation, nothing on the right hand side. Because they’re
exactly the same form, then the solution for
decay, transient decay from initial conditions,
takes on exactly the same form except that it has an
initial angle, theta 0, and I use the
approximation here. Cosine omega dt plus theta
0 dot, the initial velocity, over omega d sine omega dt all
times e to the minus zeta omega nt. So that’s the exact same
transient decay equation, but now cast in angular terms. And if you wanted to express
it as a cosine omega t minus v, then A would be this squared
plus this squared square root and the phi would be a
similar calculation as we have up there someone here. Just the B term over the
A term, tangent numbers. All right. What’s it good for? So I use this, this equation
gets used quite a lot. It has some practical uses. Let’s do an example, a little
more complicated pendulum. Draw a stick maybe. Center of mass there. A, IZZ with respect
to A. We’ll call it ML cubed over 3
for a slender rod. And now, what I
want to do is I have coming along here a mass,
a bullet, that has mass m. Has velocity vi
for initial here, and that’s its linear
momentum, p initial. And it’s going to hit
this stick and bed in it. So you’ve done this
problem before. Yeah. AUDIENCE: [INAUDIBLE]. PROFESSOR: Pardon? AUDIENCE: ML cubed [INAUDIBLE]. PROFESSOR: ML cubed over 3. Good. I don’t know why I was
thinking cubed this morning. ML squared over 3. Good catch there. OK, so we have it. This is mass moment of inertia. This is a pendulum. This bullet’s going
to come along. So this is exactly
what I’ve got here. I’ll get it in the picture. Yeah. So it’s initially at rest. Coming along, this
paper, this clip here, it represents the bullet. So it’s swimming along. Hits this thing, sticks to
it, and when it hits it, it does that. So then this thing after it
hits swings back and forth. So what’s the response
of this system to being hit by the bullet? Well, I claim you can do it
entirely by evaluating response to initial conditions. But we need to use one
conservation law to get there. So what’s conserved on impact? Is linear momentum
conserved on impact? How many think yes? Linear momentum conserved. How many think angular
momentum’s conserved? Hm. Good. you guys have learned
something this year. That’s great. Why is linear momentum
not conserved? Because are there any
possible other external forces on the system? At the pin joint. You can have reaction
forces here and there. You have no control of them. But the moments
about this point, are there any external
moments about that point during the impact? No. They’re reaction forces,
but there’s no moment arm. So there’s no moments. So you can use conservation
of angular momentum. So H1 I’ll call it
here with respect to A is just R cross Pi. And the R is the length
in the I direction. P is in the j direction, so
the momentum is in the k. So this should be
mv initial times L, and its direction is
in the k hat direction. So that’s the initial
angular momentum of the system with
respect to this. This has no initial
angular momentum because it’s motionless. And since angular
momentum is conserved, that H2 we’ll call
it with respect to A has got to be equal to
H1 with respect to A, and that will then be IZZ
with respect to A theta dot. But I need to account
for the mass, this thing. So the total mass moment of
inertia with respect to A is IZZ with respect
to A plus M– what? Now I’ve got the total mass
moment of inertia with respect to this point, that of the stick
plus that of the initial mass that I’ve stuck on there. And this must be
equal to theta dot. And I put a not down
here because I’m looking for my equivalent
initial condition. And this then is
mv initial times L. Then I can solve
for theta dot 0, and that looks like mv initial
L over IZZ A plus mL squared. And everything on the
right hand side you know. You know the initial velocity,
the mass of the bullet, the length of the distance
from the pivot, mass moment of inertia, and the additional
mass moment of inertia. These are all
numbers you plug in, and you get a value for this. And once you have a value
for this, you can use that. In this problem, what’s theta 0? The initial angular
deflection at time t0 plus right after
the bullets hit it. And it hasn’t moved because it
hasn’t had time to move yet. At some velocity,
it takes finite time to get a deflection. So there’s zero initial
angular deflection, but you get a step up in
initial angular velocity. And so the response
of this system is theta t is theta 0 dot
over omega d sine omega dt. So what’s omega d? Remember, I’ll
define a few things. In this case, this
is ct over 2 IZZ z A plus little ml
squared– we have to deal with all the quantities
after the collision– 2 times omega n. That’s the damping ratio
for this torsional pendulum, with this pendulum. It’s the damping constant. 2 times the mass, the
inertial quantity, times omega n for a
translational system at c over 2 m omega n. For a pendulum system,
it’s the torsional damping over 2 times the mass
moment of inertia times omega n. And omega n, well, it
is going to calculate the natural frequency. It’s just MgL divided
by IZZ plus this. Maybe you ought to
write that down. So always for a sample singular
[INAUDIBLE] oscillator, you want the undamped
natural frequency. Ignore the damping term. Take the stiffness term
coefficient here and divide it by the inertial coefficient. But we care about the natural
frequency after the impact, so this is going to be– ah. The trouble is here I
don’t know for this system, I haven’t worked out yet,
what this term looks like. What is it? This result right here
is for the simple Bob. For this stick, it’s MgL over
2 plus the little m times l. Little more messy. So MgL over 2 plus little ml. That’ll be the– come from the
potential energy in this system all over IZZ A plus mL squared. So you get your
natural frequency out of that expression. OK. So you do this problem
sometimes before when you do, say, somebody asks
you how high does it swing. AND so forth. Well, you can do
it by conservation of energy, et cetera. But now, you have
actually exact expression for the time history of
the thing after the impact, including the
effects of damping. And if you were to draw
the result of this function of theta as a function
of time, this one starts with no initial
displacement but a velocity and does this. And that’s your
exponential decay envelope, and this is time. Now, what– yeah. Excuse me. AUDIENCE: [INAUDIBLE]. PROFESSOR: Why this one? AUDIENCE: Yes. [INAUDIBLE]. PROFESSOR: Excuse me. Forgot the g. I mean, it accounts for
the restoring moment, the additional little
bit of restoring moment that you get from having
added the mass of this thing to it, right. So it has by itself
MgL sine theta, and we linearize that too. So it’s MgL theta, and those
two terms would add together. So you just have
a second term here that has MgL like behavior. How I most often personally
make use of expressions like this, or the
one for translation, is because
experimentally, if I’m trying to predict the
vibration behavior of a system, one of the things you want
to know is the damping. And one of the simplest
ways to measure damping is to give a system an initial
deflection or initial velocity and measure its decay,
and from its decay, calculate the damping. So the last time I gave you
an expression for doing that, and that was a damping
ratio 1 over 2 pi times the number of cycles that
you count, that you watch it, times the natural log of
x of t over x of t plus n periods of vibration. This has a name. It’s called the logarithmic
decrement, this thing. So if somebody
says log decrement, that’s where they’re
referring to this expression. A comment about this. In this expression, x
of t must be zero means if your measurement– we have
x of t here, or theta of t– they must be zero mean. There must be
oscillations around zero or you have to have subtracted
the mean to get it there because if this is
displaced and is oscillating around some offset, then this
calculates and will get really messed up. It’s got an offset
plus an offset here plus an offset there. It means it’s
totally meaningless. So you must remove the mean
value from any time history that you go to do this. So there’s an easier way the
same expression– and this is, in fact, the way I use this. A plot out like just
your data acquisition grabs it, plots it for you. I take this value
from here to here, and this is my peak
to peak amplitude. And then I go out n cycles
later and find the peak to peak amplitude. And so this is perfectly,
this is just the same as 1 over 2 pi n, but now you
do natural log of x peak to peak t over x peak to
peak at t plus n periods. And that now, peak
to peak measurement, you totally ignore the mean. Doesn’t matter where you are. You want the here to here,
here to here, plug it in there, and you’re done. OK, so let’s– I got 1, 2, 3, 4. Let’s let n equal 4, and let’s
assume this expression here– n is 1 over 2 pi times 4. And let’s assume that
in these four periods from– that’s 1 period,
2, 3, 4 getting out here to this fourth, four periods
away, that this is one fifth the initial. So this would be the
natural log of 5. So 1 over 2 pi times
4, natural log of 5, and you run the
numbers, you get 0.064, or what we call 6.4% damping. That’s how you do it. That’s the way you do a
calculation like that. Now, I gave you a quick rule of
thumb for estimating damping, and this is what
I– I can’t work. I don’t do logs
in my head, but I can do damping
estimates without that because I know
that zeta is also– if I just plug in
some numbers here and run them all
in advance is 0.11 divided by the number
of cycles to decay 50%. So we’re going to
do an experiment. And I guess it can be
seen with the camera. So here’s my pendulum. This is my initial amplitude,
and this is about half. So if I take this thing over
here, like that, then let go, and count the cycles that
it takes to decay halfway, we can do this experiment. So let’s do it carefully. So line it up like
that, and you’re going to help me tell– you
count how many cycles it takes till it gets to here. So 1, 2, 3, 4, 5, 6, 7, 8. About eight cycles. So it decayed halfway
in eight cycles. So zeta is approximately 0.11/8. It’s 1 and 1/2%, 1.4%,
something like that. Perfectly good
estimate of damping. Now, the stopwatch here, we
can do this experiment again. I want you to count. You’re doing the counting. And I’m going to say start,
and I want you to count cycles until I say stop. Now, I’ll probably stop on 10
to make the calculation easy, so quietly to yourself
count the number of cycles from the time I release
it until the time I stop. Come back here. So this time, the
backdrop doesn’t matter. I just want you to count cycles. And I’ll start– I’ll
let it get going, and when it comes
back to me is when I’m going to start the
stopwatch because I have a hard time doing
both at the same time. So start. How many cycles? AUDIENCE: [INAUDIBLE]. PROFESSOR: So I
got 17.84 for 10. So 10 divided by 10 is
1.784 seconds per cycle. Can’t write like that. 1.784 seconds per cycle. The frequency would
be 1 over that, right. The thing you have to
be careful about when you’re counting cycles is if
I start here, that’s 0, 1, 2. A very common human mistake is
when you’re counting something like this is to say
one when you start, and then you’re going
to be off by one count. Follow me? If I start 0, 1, 2. So I start the clock on
zero, but the first cycle isn’t completed till
one whole cycle later. So be careful how you count. OK. Now we’re going to shift
gears and take on a new topic, and that’s the response
to a harmonic input, some cosine omega t excitation. Yeah. AUDIENCE: [INAUDIBLE]. PROFESSOR: What is omega d? So it’s the damped
natural frequency. That’s how it’s referred to. It is the frequency
you observe when you do an experiment like we just did. The actual oscillation
frequency when it’s responding to
initial conditions is slightly different
from the undamped, but if you have light damping,
if you have even 10% damping, 0.1 squared is 0.01. That’s 0.99 square root. It’s 0.995. So you’re only off by half. They’re only half a
percent difference. So for lightly damped systems,
for all intents and purposes, mega n and mega d are
almost exactly the same. OK. We now want to think about–
we have a linear system putting a force into it. It looks like some
F0 cosine omega t. And out of that system, we
measure a response, x of t. And inside this box here is
my system transfer function. It’s the mathematics that
tells me I can take F of t in and predict
what x of t out is. So I need to know
the information that goes into this box, and of
course, the real system– this is just the mechanical system. Force in, measured output out. This is what we call a
single input single output system, SISO, single input
single output linear system. And there’s all sorts
of linear systems that you’re going to study
as mechanical engineers, and you’ve already begun, I’m
sure, studying some of them. One of the properties
of a linear system is that you put a force in, F1,
and measure a response out, x1. And then you try a
different force, F2, and you measure a
response out, x2. What’s the response if
you put them both in at the same time, F1 and F2? You just add the responses
to them individually. So F1 gives you x1. F2 give you x2. F1 plus F2 gives you
x1 plus x2, and that’s one of the characteristics
of a linear system. We use that concept to be
able to separate the response. Our calculation’s about
the response of a system, like our oscillator here,
separate its response to transient effects, transience
being initial conditions. They die out over time– that’s
why we call them transients– and steady state effects. So cosine omega t,
you can leave it running for a long, long time,
and pretty soon, the system will settle down to responding
just to that cosine omega t. And that we call steady state. And we use them separately. So we’ve done
initial conditions. Now we’re going to look
at the steady state response of a– say our
oscillator, our mass spring dashpot, to a harmonic
input, F0 cosine omega t. Another brief word. If I have a force, F0 cosine
omega t would look like that. And the response that I
measure to start off with my– it’s sitting here at zero
when you turn this on. And it’s going to do some odd
things initially, and then eventually settle down to some
long term steady response. The amplitude stays constant. It stays angle with
respect to the input isn’t necessarily the same. There’s some
possibly phase shift. And that’s so the two, if
you’re plotting them together, they won’t line up. But see this messy
stuff at the beginning? When you first turn this
on, it jumps from here to here, that force, and it
gives it a kick to begin with. And this will have some
response initially due to that transient start up. And this response is all
modeled by the response to initial conditions. And it’ll die out after a
while, this messy stuff. What’s the frequency? What frequency do you expect
this initial, erratic looking stuff to be at? Its response to
initial conditions. What is the model for a
response to initial conditions? What’s the frequency
of the response to initial conditions of
the single degree of freedom system? We have an equation
over here, right? The top has a cosine
term and a sine term. Part of it’s a response
to initial displacement. Part of it’s a response
to the initial velocity. Any of this start up stuff can
be cast as initial conditions, and the response to
initial conditions is always at the natural
frequency period. No other frequencies for same
degree of freedom systems. So you get a behavior
that’s oscillating at its natural frequency. Mixed in there is a response
at the excitation frequency. And after a long
time, the response is only excitation frequency. This is now out here. This is omega. In here, you have omega
and omega d going on. So this is messy. Usually isn’t
important, but it is. There are ways of getting the
exact solution, but mostly, vibration engineers, you’re
interested in the long term steady state response to what
we call a harmonic input. OK. So we’ll work a
classic single degree of freedom oscillator problem–
excited by F0 cosine omega t. You’ve done this in
1803, but now we’ll do it using engineering
terminology. We’ll look at it the way a
person studying vibration would think about this. We know the equation of motion. And I’m interested in the
steady state response. So this is x, and I’ll do– you
just write it once like this. SS, steady state. I’m only interested in
its– after those transients have died out. And that steady
state response I know is going to be some
amplitude X0 cosine omega t minus some phase angle
that I don’t necessarily know to begin with. But that’s my input. This is my output. I plug it into here and turn the
crank and see what falls out. So you plug both of
those in, and you get two– you
get– this is going to be a little writing
intensive for a few minutes. So you plug the X0 cosine omega
t into all of these terms. The m term gives you
minus m omega squared, the k term gives you a
k, and the damping term, minus c omega sine omega
t minus v. All of that equals the right hand
side– F0 cosine omega t. So this just purely
from substitution and then gathering
some terms together. I’m going to divide
through by k, by k. If I divide through by
k, k divided by– this gives me a one. This gives me an
m over k, which is 1 over the natural frequency
squared, for example. And I’m going to put
this into a form that is the standard form for
discussing vibration problems. So this equation can be
rewritten in this form. 1 minus omega squared over
omega n squared cosine omega t minus v minus 2 zeta omega
over omega n sine omega t minus v. All that’s still
equal to F0 cosine omega t. So this is getting into
kind of more standard form. So there’s 1 minus omega. This now, this
omega over omega n, is called the frequency ratio,
and you see a lot of that. And I’ve substituted
n here. c omega over k turns out to be 2 zeta
omega over omega n. So this frequency
ratio appears a lot. in our– let’s see here. You need a couple of trig
identities– cosine omega t minus v. Cosine omega t cosine
phi plus sine omega t sine phi, and sine omega t minus phi
gives you [INAUDIBLE] sine. Sine omega t cosine phi minus
cosine omega t sine phi. So that’s a trig identity
you actually use quite a bit doing vibration problems. We need them, so we
take these, plug them in in all these places, and
do quite a bit of cranking. Yep. AUDIENCE: [INAUDIBLE]. PROFESSOR: Yeah. Thank you. And that’s called,
that f over k, is how much the spring
would move statically, at which the point
would move statically. We’ll need that term also. OK. You do all of this. Here, I’ll call these– OK. So this is C.
That’s expression C. I can’t see that probably. Call this D, this E. So
you plug D and E into C and work it through,
you get two equations. You break it into
two parts because one is a function of cosine
omega t, and then you have another part after
this substitution that’s a function of sine omega t,
and you can separate them. But there’s no
sine omega t force. On the right hand
side, you get zero. There are two equations here. How many unknowns do we have? All we know when we start
this thing is the input, and we have unknown
response amplitude, and we have an unknown phase
that we’re looking for. How many equations? How many unknowns? Two and two. So you can do a lot
of cranking, which I have no intention
of doing here, and solve for the amplitude
of the response and the phase. And every textbook– the
Williams textbook does this. There are two readings
posted on Stellar by [? Row. ?] Every textbook
goes through these derivations that I’ve just done. Nick, you’ve got a question. AUDIENCE: [INAUDIBLE]. PROFESSOR: Pardon? AUDIENCE: [INAUDIBLE]. PROFESSOR: Yeah, I
keep forgetting it. You’re right. So we got a k here. And notice, this equation,
we throw away that for now. We get rid of this for now. We have these two
equations and two unknowns are just algebraic equations
There’s not time dependent. We can get rid of that part. So we’ve now reduced
this to algebra, and the answer is
plotted up there. You’ve probably seen it before. It says that x0 is F0 over k–
I can get it right this time from the get go– over
a denominator, which appears again and again
and again in vibration. Omega squared over
omega n squared squared plus 2 zeta
omega over omega n squared square root, the
whole thing, and an expression for phi. Tangent inverse of 2 zeta
omega over omega n, 1 minus omega squared
over omega n squared. So you can solve all
that– this mess over here for these two quantities. Do you need to remember this? You ever going to be
asked this on a quiz? Not by me. You ever going to have to use
this on a quiz and in homework? Absolutely. So the takeaway
is today know how to use those response to
initial condition formulas and damping and these two. So when you plot,
when you plot these, you get this picture up there. And we need to talk about
the properties of this. Remember, this
omega over omega n is the same called
the frequency ratio. It’s just the ratio of
the excitation frequency to the natural
frequency of the system. And when they’re equal, for
example, this ratio is one. This whole thing in
parentheses goes to zero. This expression over here goes
to 2 zeta, because that’s one. 2 zeta squared square
root is just 2 zeta. When omega equals omega
n, this whole expression is F0 over k divided
by 2 zeta, for example. And that’s called
resonance, and that’s when you’re right at where
that peak goes to its maximum. Let’s talk about this
expression for a moment. If we have our cart,
our mass-spring dashpot we started here. If you apply a
force, a static force F0 and stretch the spring
by an amount F0 over k. So x– what we’ll call x
static is just F0 over k. And if I want to plot, I
want to– this has a name. This is called, this
ratio here, this gives you the magnitude of the response. It goes by a variety of names. Some people call it
a transfer function. Some people call it a
frequency response function. I write it
intentionally this way. This is I put output over input
because this expression has units of output over input. So I just write it like
this, remind myself what this transfer
function is about. The input is force, the
output is displacement. This expression
has units of force, force per unit displacement. If I go to here, if I try to
plot this– let me start over. If I try to plot
this, it’s going to be depending on the exact
value of the spring constant and the exact value of
the force every time. I have to get a unique plot
every time I go to do this. So textbooks and
engineers, I don’t want to have to remember this part. This is where all of the content
is in is in this denominator, and it’s dimensionless. So what I’d really like to
plot is x0 over x static. And if I do that, that is x0
over the quantity F0 over k. If I just divide– this is x
static– it would bring this to this side, then this
expression, this is just 1 over that denominator. And sometimes, I think in
the handout by [? Row ?], they just call this h of omega. It’s dimensionless,
frequency over frequency, and that’s actually
what’s plotted up there. And this is called– has
different names also. Magnification factor,
dynamic amplification factor, because the ratio
of x to x static if this is the dynamic effects
magnify the response compared to the static response. So it might be this
over this might be 10. I mean, the dynamic response is
10 times the static response. OK, how do you–
to sum this up– and we’ll be kind of
getting close to the end. We want to talk just about
the properties of this. How do we use this? So in practical use, you
have an input specified, some force cosine omega t. You know you have a single
degree of freedom oscillator that is governed by
equations like that one, and you want to
predict the response. Well, you say x of t is
equal to the magnitude of the force times the–
and you divide that by– we could do it this way. The magnitude of the
force divided by k, which is the static response. To predict x0, we just have
to predict this quantity, multiply it by F0 over k. So you know this. You better know that
about your system, and you multiply it by
this quantity magnitude of h of omega. And the time dependent part
is times cosine omega t minus the phase angle. And the phase angle, you
get either off the plot or from the– have
I written it down? I haven’t written the
phase angle down yet. It’s kind of a messy
expression too. That’s why we plot it. 2 zeta omega over omega n
over 1 minus omega squared over omega n squared. But by knowing just
this plot, what you just put in every textbook about
vibration in the world, by knowing this
magnification factor, calculating the static response,
multiplying the two together, you have the amplitude
of the response, and its time dependence is
cosine omega t minus the phase angle. And I’ve got a
little example here. Actually, rather
than the example, I’ve gone to all the
trouble of setting this up. All right. This is just a beam. Where’s my other little beam? And a beam is just a spring. Put a mass on the and. This is basically a single
degree of freedom system. It has a natural frequency. The beam has a
certain stiffness. And now, in this case,
we’re interested in response to some harmonic input. So any of you know
what a squiggle pen is. This is a kid’s toy. AUDIENCE: Excuse me. Can you move the camera a tad
to the left so the [INAUDIBLE]? PROFESSOR: Yeah. So all throughout
the term, we’ve studied rotating masses
quite a bit, right. This thing has a rotating mass
that you can see in the end. I mean, when you leave it, you
can come down and take a look. It has a low rotating mass. It’s actually a pen,
but it’s a kid’s toy. Shakes like crazy. And now we need the lights down. And it happens that the–
I’ve got a strobe light here, and I’ve kind of
preset the frequency so it’s very close to the
frequency of vibration of this beam. So there’s a rotating mass in
this pen going round and round, and it puts a force
into the system that looks like F0 cosine omega
t in the vertical direction. Also does it in the
horizontal, but vertical is our response direction. So it’s putting
in a force, and I have set the length of this beam
so that the natural frequency of this beam with
this mass on the end is exactly very close
to being the frequency of the excitation. And the flash rate is slightly
different than the vibration rate, so you see it
illuminated at many positions as it goes through the cycle. So you see it going up and down. So if I mismatch it quite a
bit, then you see it going. And actually, if you look
at the very right end, you can see a white
thing going up and down. That’s the mass where
you can actually see the mass in the very
end of the– right there. You can see something
going around and round. There, that’s the rotating mass. So the beam is
going up and down, and I’ve got this, the vibration
frequency of that mass going round and round equal to the
natural frequency of the beam of the mass in the end. And it’s moving quite a bit,
and I’ll loosen my clamp, and I’m going to change
the length of the beam. I’ve shortened it. And now it’s still
moving up and down but not as much because the
frequency of the rotation of the eccentric
mass is no longer close to the natural
frequency of the system. In fact, I’ve made
the natural frequency of the system– you can
bring the lights back up– I’ve made the natural
frequency of the system. By making the beam shorter,
I’ve made it stiffer. So the natural
frequency has gone up. The frequency of the rotation
of the eccentric mass has stayed about the same. So what’s happened
to that frequency ratio, omega over omega n? So less than one or
greater than one. So the omega n has gone up. Omega stayed the same. The frequency ratio when
you shorten this beam is less than one, and the
properties of this transfer function, we call it–
this magnification factor looks like this. When we’re exciting
it right at one– this is omega over omega n–
you write it resonance. When you excite it at a
frequency ratio less than one, you start dropping
off this backside, and the response goes down. And if you excite it at
frequencies much greater than the natural frequency,
you end up way out here. I can do that too. So how much you think
it will vibrate now? A lot? A little? Hardly– oops. Oh, I’ve brought it out
so much you can’t see it. Hardly moving at all, and
that’s because in terms of this terminology of
magnification factors, transfer functions, this is a
plot of x over x static. It goes right here. When you go to zero
frequency, you are at static, so the response at
very, very low frequency goes to being the same
as the static frequency. So in this plot, it goes to one. At resonance, you
put in one here. This goes to zero. That becomes a one
2 zeta squared. This height here
is 1 over 2 zeta. So the dynamic
amplification at resonance is just 1 over 2 times
the damping ratio. You have 1% damping. Twice that is o2. 1 over o2 is 50. So if you only had 1% damping,
the dynamic amplification, the amount that this
vibrates greater than its static response,
is a factor of 50 greater. But then as you go higher
in this omega over omega n, and you get way out here, and
you get almost no vibration at all. And that’s what’s happened
when I’ve lengthened this. OK. So there’s your introduction
to linear systems. In this case, a single
degree of freedom system that vibrates an
oscillator, we’re talking about steady state
response, not the part of the solution, the
mathematical solution, to the initial conditions. So all of this has been
about steady state response of our simple oscillator
to what we call a harmonic input,
F0 cosine omega t. So what’s important
that you need to remember and be able to use? This concept here, this
idea of a transfer function. That’s really important. You might want to
remember it though as this dimensionless
quantity x over x static. Just remember the shape
of this transfer function. Magnitude of the amplification,
1 over 2 zeta at resonance. And it goes to one
at low frequency. The high frequency,
it drops away off. Next time, we’re
going to pick up the topic of what we
call vibration isolation. The practical thing
to know as engineers is when you have a
significant vibration problem, like in
a lab, and you’re looking through your microscope,
and the floor vibration is causing trouble with
your microscope, and you can’t move the
subway, what can you do to solve that problem? Well, you might be
able– what if you put some kind of a flexible
pad under the microscope? You might be able to reduce the
vibration of the microscope. Things like that. That is the topic of
vibration isolation, so we’re going to get
into that next time. Thanks.

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