The following content is

provided under a Creative Commons license. Your support will help

MIT OpenCourseWare continue to offer high quality

educational resources for free. To make a donation or to

view additional materials from hundreds of MIT courses,

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.

Quem precisa do RHAO agora?

**Any way that we can get the title of the text that was used that semester?

hi,

i am interested to know is there any advantage of dealing in frequency domain while dealing with dynamics of continuous system? specially the catenary type structures (SCR)?

if is there any what are they? any reference ?

please help me in that

شباب في عرب هون عم يحضرو؟

i am sorry but anyone know what is this device he was holding in his hands dealing with frequency? help plz

Thank you sir！

8.03 in 2 lectures minus EM waves

18:30 He ment to say that linearization of pendulum systems with small angles is acceptable.

Thank you! This is great content. The experiment helped visualize the dynamic amplification ratio really well

This is just gold to mech students

Excuse me but why is the k for the bar bullet problem is

(M*g*L/2 + m*L*g)?at 30:56

o_O

56:51 I presume he refereed to the book; S. RAO – MECHANICAL VIBRATIONS

Really nice!