# 27. Vibration of Continuous Structures: Strings, Beams, Rods, etc.

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: Today we are going

to talk about the vibration of continuous systems. Not covered on

the quiz, but it’s a really important part

of real-world vibration and the most– one of the

easiest ones to demonstrate, I’ve shown you this one

before, is the taut string. But I want to show you something

unusual about– something you may not know about strings. Wait until it calms

down here a little bit. OK, so this is your guitar

string or a piano string. It’s under tension. We’ve already seen

that it exhibits natural frequencies

in mode shape, so there’s the first mode. Looks like half a sine wave. Has a particular frequency

associated with it. Get it to stop doing

that– but if I excite it at twice the frequency– I

don’t know if I can do this. There we go. That turns out to

be exactly twice the frequency of the first. The mode shaped

one full sine wave. The mode shapes for a taut

string are sine n pi x over L. But strings can do

something else kind of neat. And that is if I

hit this thing– I’m going to wait till

it calms down here. If I give this

thing just a pulse, what do you expect to see? Are you going to see vibration? Tell me what you see. What do you see happening? Something running

back and forth. Right? What you’re seeing

is wave propagation. It’s not really vibration. Vibration we see of its modes

and standing waves and things like that. Right? So the taut string satisfies

an equation of motion that’s called the wave equation. We’re going to talk quite a

bit about that this morning. And the wave equation has

its name give something away. The wave equation describes

continuous systems of a particular kind that

support travelling waves. And so the string

will both support– I can give it a little pluck. I’ll try to just place it in

a particular shape and let go. There it is. And that little pluck just goes

back and forth back and forth at a particular speed. So is there a relationship

between the speed at which things can

travel in a string and the natural

frequencies of the string? Well, we’ll get into that today. And I’m going to start

by just showing you a little something that

comes from my research and– let’s see. Let me do this. I think this will work. Hear that? As I go slower, does

frequency go up or down? It’s kind of slow, and

I’m going to speed up. Right? Goes up as the speed goes up. So that’s the result

of the phenomenon called flow-induced vibration. And I’ll give you

a very brief intro to flow-induced vibration. You have a cylinder sitting

still, flow coming by it– water or air. The cylinder is diameter D,

velocity U, for the flow. What happens in the wake of that

cylinder, vortices are formed. And just like if you’re

paddling a canoe or something and stick a paddle

in the water, you’ll see vortices shed off the side. First you get one

that’s positive and then one that’s negative

And so one full cycle of this is from here to here. There’s a frequency

to this shedding. And the shedding

frequency, FS, in hertz, can be predicted by a simple

dimensionless parameter called the Strouhal number, St U over

D. And that’s approximately 0.2 U/D for stationary cylinders. You can predict the frequency at

which these vortices are shed. Now, associated with

the shedding of vortices is a lift force. I’ll call it some

FL cosine omega t, which is 2 pi FS, times t. So at this frequency

of vortex shedding there is a transverse force. There’s actually an

inline force also, which I’ll call FD for drag. And it goes like cosine

2 omega s times t. It’s twice the

frequency of that. And so you’ll get some

inline oscillatory excitation and what we call cross-flow

oscillatory excitation. And this is the cause

of lots of things that the people who work on it

call flow-induced vibration. Now, an amazing thing happens is

if this cylinder is elastically mounted or is flexible, and

that force starts to act on it, it will begin to vibrate. And the amazing thing,

as it begins to vibrate, it correlates the

shedding of these vortices all along the cylinder. So it’s like soldiers

marching in step going across the bridge. If everybody’s walking

randomly, then the bridge doesn’t respond too much. But if everybody

marches together, you can put a pretty

good excitation into it. Well, the motion of

the cylinder itself organizes these vortex shedding

all along the cylinder, so they’re all marching in step. And that means the force is

all correlated on the length. And you can get some pretty

substantial response. So that’s the subject called

flow-induced vibration. And with that, I’m going

to show you a few slides. Let’s dim the lights a little

bit, if you could, to see this. There’s some pictures I

just want you to see better. All right. So I do flow-induced vibration. I’ve been doing

this– working on this for all my professional career. And it’s applied, primarily,

to big, flexible cylinders in the ocean. Particularly associated with the

things that the US Navy does. Long cables and things and

also the offshore oil industry. Next slide. Can we dim the lights? Can we dim the lights? I want you to be able to see. This is a tension leg platform. It’s one of the

structures that’s used in the offshore

industry to produce oil. And one of these might be

moored in 3,000 feet of water, 1,000 meters of water. Might weigh 20,000 tons. And what’s connecting

it– what holds in place– are steel cylinders

a half a meter in diameter, 3,000 feet

long, going vertically down from each of those three

pontoon legs sticking out. And they’re under

a lot of tension. And in fact, it pulls the

thing down into the water so the buoyancy

of the whole thing puts tension on these cylinders. But now, what happens

if an ocean current comes by those cylinders? Vortex shedding, and

the cylinders vibrate. And if they vibrate, over time

they will fatigue and fail. OK. Next slide. There’s a picture of a real one. That’s a bigger one

called Marco Polo. It’s on a launch ship that’ll

take it out to the site that it is. And the ship will lower

and it will slide off. So these are big. Next slide. This is a diagram of

the Gulf of Mexico. South America is at the bottom. The Yucatan Peninsula is

sticking up there right in the middle of the bottom. This is a picture of

satellite imagery of currents in the Gulf of Mexico. And there’s a current that

flows up off of South America into the Gulf of Mexico,

goes around in a loop, and then comes out. You can see Florida sticking

down in there on the right. That current comes

out of the Gulf, goes around the tip of Florida,

and goes up the Atlantic Coast, and is known as the Gulf Stream. But it starts as

a big current that comes into the Gulf of Mexic. And, every now and then, that

current pinches off an eddy. And that’s what that red

circle is in the middle. And it’s an eddy that’s many,

many kilometers in diameter with surface currents on the

order of a meter per second or more. And those are the biggest

threat for causing flow-induced vibration

failures of long members from hanging off of

offshore structures. Next. So I’ve been doing research

in this area for a long time. This is a picture taken

in the summer of 1981. It is a piece of steel

pipe about 2 inches in diameter and 75 feet long. It’s under 750

pounds of tension, and it’s pinned at each end. It behaves almost exactly

like my rubber cord here. It has natural frequencies,

and it will vibrate if a current comes by it. So this is actually a sandbar. And at low tide, we’d do

all the work putting it up. Then, as the tide

comes in, the flow is perpendicular

to the cylinder, and vortices start shedding. And as the pipe

begins to move, they get organized all

along the length. And a typical

response mode was when the vortex shedding frequency,

therefore the lift force frequency, coincided with

the natural frequency. Then you’d expect it to give

quite a bit of response. The diagram on the left

is if you cut the cylinder and looked down its axis,

this is the trajectory that you’d see

the cylinder make. It would sit there and just

make big figure eights. So up and down vertical

is its vertical motion. Flow’s coming from,

say, left to right. Its vertical motion

is up and down. In-line motion’s like this. And exactly such a phase it

just makes big beautiful figure eights. That’s the kind of

motion you’d see. OK? So then, very much what I was

talking about a minute ago, very much behavior

dominated by vibration. Vibration in the third mode,

cross flow, was a typical one. And fifth mode,

inline, was typical. But as cylinders

go in the ocean, that one’s kind of short. Third mode vibration

is sort of low. So as years have

gone by and oil is being produced in deeper

and deeper and deeper water, the cylinders we’re

putting out there get longer and longer

and longer and longer. And the modes that are

excited by currents coming by get quite high. So this is an experiment we did. It was roughly a

1/10 scale model. Model is almost 2 inches

in diameter, 500 feet long. Scale that up by a

factor of 10, you’re up around 20 inches in

diameter and 5,000 feet long, which is exactly the

size of the drilling riser that BP had hung

off the drilling ship when the blowout occurred. It’s a piece of steel pipe,

21 inches in diameter, 3/4 of an inch wall

thickness, 5,000 feet long, under a lot of tension. And when ocean

currents come by, it behaves just like this string. And so we’re out– this

is a 1/10 scale model. So we put a big weight on

the bottom of the cylinder, put it behind a boat, and

towed it in the Gulf Stream. Next picture. So there’s the boat. It’s an oceanographic vessel. It’s actually a catamaran. Next. This is a spool that had

our test cylinder on it. There’s a reddish object

down on the bottom which is– that’s a 750-pound

piece of railroad wheel, and it’s the weight

on the bottom. And so you’d spool this

thing off, lower it down, and then do your tests. Next. Top, we measured

tension inclination. And then we also had– it’s

a pin joint at the top, so it would vibrate freely. Inside, though,

was fiber optics. Next. We had eight optical fibers. And in those optical

fibers were what we call optical strain gauges. So we had 280 optical strain

gauges instrumented up and down that pipe so we could

measure its vibration. And so you’re looking at a

cross section of the pipe. There were two optical

fibers in each quadrant, and each one of those

fibers had 35 sensors on it. Next. This is typical

experimental case. This is the surface. This is 500 feet down. This is the current profile. So the flow velocity

is about 2 feet per second near the

surface, up to 4 feet per second down on the bottom. And this is the region

where most of the excitation was coming from that would drive

the flow-induced vibration. This is measured RMS strain

caused by the bending vibration in the cylinders. And peak– the maximum

strain– is about right there. Next. Typical response spectrum. Basically, the frequency content

at three different locations. Down deep, in the

middle, near the top. This is frequency. So this would be the

peak that describes the principal cross-flow

vibration at the vortex shedding frequency. Next. This is position, bottom to top. This is time, and these

are strain records from all of those strain sensors. There’s a strain sensor about

every 2 meters along here. But what you’re seeing

is– this is evidence. The red is the

amplitude and red– let’s say red is positive strain

and blue is negative strain. And so at any location on the

pipe where it’s vibrating, it’s going to go from red to

blue, red to blue, red to blue. But it’s showing you that

they’re highly correlated all along the

length, that there’s a red streak all lined up, but

it’s not parallel to the pipe. It’s inclined. This is showing you

wave propagation. The behavior of the

pipe is completely dominated by wave propagation,

not by standing wave vibration. So totally different than

that short pipe in 1981. The wave equation. Let’s imagine we have a long

pipe or a string like that, and it can carry waves

traveling along it. The position at any location on

here– here’s a coordinate x. We describe the

motion at a point by a coordinate w of x and t. So it’s a function of

where it is and time. What describes the

motion of something which obeys the wave equation

is the following equation. Partial squared w with

respect to x squared equals 1/c squared partial

squared w with respect to t squared. That’s what’s known as the

one-dimensional wave equation. And the one-dimensional

wave equation governs an incredibly broad

category of physical phenomena. Light behaves according

to the wave equation. Sound propagating

across the room to you is governed by

the wave equation. Longitudinal vibration of rods,

torsional vibration of rods– all governed by

the wave equation. So it’s worthwhile to know

a little bit about the wave equation. And what I showed

you this morning, it has this kind

of duality to it. You can have things that vibrate

with standing waves and mode shapes, but the same

system can support waves that travel along it. So let’s figure out why that is. So I’m going to do the

derivation for you of the wave equation for a

string, just so you know where it comes

from because then that general derivation applies

to all these different things. So imagine you’ve got now–

we’re interested in eventually getting to vibration. So I’m going to make this

a finite length string. And it has this position we’ll

describe as a w of x and t. It has a tension, T, a

mass per unit length, m. So this is like

kilograms per meter is the mass per unit length of

this thing which can vibrate. So tension. Mass per unit length. L, the length of it. What other parameters

do we need? That’ll do for the moment. Now– so let’s draw

it again without. In some displaced position

and what’s exciting it may be my vortex

shedding, and so I’m going to draw that excitation here. And that we’ll describe

as F of x and t, some force per unit length. So this has units of

newtons per meter. Now, in that little–

there may also be drag forces,

the fluid damping. So I’m going to cut out a

little piece of this cylinder and do a force balance on

that piece of cylinder. So basically, F equals ma. We’re just applying Newton

to this piece of cylinder. And I’ll draw it right here. A little section

of it is curved. Here’s horizontal. There’s horizontal. We need to evaluate

all the forces on it. So the tension on this

end– so like that. And the tension on this end

is some different angle. This we’ll call theta 1. This we’ll call theta 2. And along here are my

excitation forces, F of x and t. There may be some resistance–

drag forces, damping. That’ll be a damping

constant, R of x, which is force per unit

length per unit velocity, times– the force

on this would have to be multiplied

by the velocity, so the derivative of this

displacement with respect to time. That’s the force along here,

and it can vary with position. Have we accounted

for everything? Ah, well, this is position

x, and this is at x plus dx. So this little element

is dx in length. And this is all

for small motions. And if you assume

small motions, then you can say theta 1 is approximately

equal to sine theta 1. That’s also approximately

equal to tan theta 1. And that’s equal to the

derivative of w with respect to x, just the slope. We’re going to take

advantage of that. Theta 2, same thing. It’s approximately equal to

tan theta 2 here, and sin and all those things. But that, then– the slope

has changed a little bit when you go through dx. And this is equal to the

slope on the left-hand side plus the rate of change

of the slope times dx. So the slope on the left,

this is now the slope on the right-hand side. And so now, all

that’s left to do is to write a force balance

for that little piece on the element dx. So if positive, upward. We have a T sine theta. But because sine theta

is approximately tan theta is equal to

dw dx, then there’s an upward force on the

right-hand side, which is T. And this turns into partial

squared w with respect to x squared dx. So on the right-hand

side– positive upwards– you have

T times the partial of w with respect to x, plus

partial square w with respect to x squared dx. That’s the upward force

on the right-hand side. On the left-hand side, we have a

downward force, minus T partial of w with respect to x. And you notice that this one’s

going to cancel that one. We have minus R of x partial

w with respect to t– that’s the velocity– dx long. Because that’s force

per unit length. And have we missed anything? So that’s the sum of

the external forces on this little slice. And that has to be equal

to– what did Newton say? The mass, which is the mass

per unit length, times dx, is the total mass,

times the acceleration, partial squared w with

respect to t squared. So this cancels this term. And then you notice I’m

left with everything as just something dx,

something dx, something dx. Get rid of the dx’s,

and I can write– oh, I left out something. I left out my distributed

force, F of x and t dx. It’s positive as it’s drawn. It’s over here also. So this, and I

cancel out that dx. So I put them all together

now and assemble them. I can write down the equation

that governs this motion. So T partial square

w with respect to x squared minus r of x times

velocity plus f of x and t equals m partial square w

with respect to t squared. And that just says that the

sum of the forces on the object equals its mass times

its acceleration. Now, if we’re interested in

natural frequencies and mode shapes, when we’ve been doing

one and two degree of freedom systems, and we want to get

the natural frequencies in mode shapes, we temporarily let

the damping be 0 and the force be 0, right? So we want to do

the same thing now. We’re interested in how do you

find the omega n’s and what I call the psi n’s. Because now the mode

shapes are functions. And so this is a

natural frequency and the mode shape for mode n. We know there’s lots of modes. So we let r of x and

f of x and t be 0. And when we do that,

this term goes away. This term goes away. I’m just left with T partial

squared w with respect to x squared equals this. And I’m going to

divide through by t. So I get partial squared w with

respect to x squared equals 1 over T over m partial squared

w with respect to t squared. And this T/m quantity

turns out to be the speed of wave

propagation in the medium. And that is the wave equation. So we’ve just found the

wave equation for the string just by applying Newton’s law

to a little section of string. You can do that for the vibrate. You’re going to

do the same thing, cut out a little

section of a beam, do the force balance on it,

set it equal to the mass times acceleration. And for a beam, you’ll get

a fourth order differential equation. And it’s not the wave equation. It still vibrates, but

it’s not governed by what we call the wave equation. OK, so this is the one

dimensional wave equation. This quantity T/m is

the phase velocity. It’s called phase velocity. You know, that’s a

good one to remember. For a simple string, the speed

of phenomena running down the string is the square

root of the tension divided by the mass per unit length. And if you had a long string,

I put that little pluck in it, and you can see that pluck

running back and forth on it. That’s the speed it’s going at. Basically, it’s called–

well, so if I have my string, and I put a little bump

on it, and that bump goes zipping along,

your eye will see this thing propagating at c. So to get natural

frequencies in mode shapes, we basically need to

solve this equation. And it’s quite

straightforward to do. And a technique

known as separation of variables works, which

means that all you’re doing is saying, I believe

that I’m going to be able to write the

solution as some function of x only times some function of

time only, product of two terms. And that in fact– because

we’re interested in vibration. You can tell me what

the function of time is. You’re going to tell me half the

solution just from observation. What is it? Just the time dependent part. It’s the same as anything

else that vibrates. So a single degree

of freedom system, what is the time dependent

function that we substitute in to find the natural frequency? AUDIENCE: [INAUDIBLE] PROFESSOR: Say again? AUDIENCE: e to the i omega t. PROFESSOR: e to the i

omega t would be just fine. Cosine omega t works. Sine omega t works. But e to the i omega t

is pretty easy to use. Because it’s so simple

to take the derivatives. So we can guess

that this is going to be some W of x times

Ae to the i omega t. And plug it in. Plug it into our wave

equation over here. So I’ll make sure I

write it consistently. So we plug this

into the first term. It’s two derivatives

with respect to x. So this is just– and the

time-dependent part just stays outside. And on the right-hand side, when

we plug it in here, 1 over c squared, two derivatives

with respect to time, it’s going to give me

minus omega squared, so minus omega squared

over c squared. And then it gives me back

W of x Ae to the i omega t. And now I can get rid of

the Ae to the i omega t’s. And I’m left with just an

equation involving x only. And it’s an ordinary

differential equation in w of x. So it turns into d2W dx

squared plus omega squared over c squared W equals 0. And you’ve seen this

equation before. Does this not look like,

have some similarity to, Mx double dot plus kx equals 0? They’re basically

the same equation. This one’s a function of x. That one’s a function of time. And we know the solution

to this one is some x of t is some amplitude

e to the i omega t. So therefore, we can guess

that the solution to this one is W of x is going to be–

I’ll write it as some B. Now I need a function of x. But it can be just like this–

e to the i, and I’ll say kx. I know that’s going

to be a solution. So let’s plug it in. If I plug that in, I

get minus k squared Be to the ikx plus omega squared

over c squared Be to the ikx equals 0. Well, now I get rid of these. And what I found out

is that k squared is omega squared over c squared. And this has a name– k. It’s called the wave number. And it also happens to

be 2 pi over lambda. We’ll come back to that. Lambda is the wavelength. You have sinusoidal waves

running through the medium. 2 pi over lambda is the

same as omega over c. And this is called

the wave number– really important

quantity if you’re trying to understand wave

propagation in systems. And actually, this

one, this definition applies to all wave

bearing systems, whether or not they

obey the wave equation. It’ll apply to waves

traveling down a beam as well. So the definition of wave number

is frequency divided by speed, or 2 pi over the wavelength. Well, let’s see. We can’t go much further with

just the wave equation itself. In order to get the

natural frequencies, we have to invoke

other information that we know in the problem. In particular, we

know that in order to get natural frequencies,

we had to create conditions where this could vibrate. In particular, I fix that

end, and I fix this end, and I put some tension on it. And now it’ll vibrate. But it clearly has

something to do with its ends and its length. And so this is a

boundary value problem. And we have to invoke the

boundary conditions to actually finish finding the natural

frequencies and mode shapes. Apply the boundary

conditions– so I assumed here that my W of x is going to

look something like that. In order to get a little

more information out of this, I’m going to write now W of x

in an alternative form that’s equally valid. And I’ll call it B1 cosine

kx plus a B2 sine kx. And I could relate that to

e to the ikx, B to the ikx, by real and imaginary

parts, and so forth. This is a real part. I’m saying in general it could

have a cosine part and also a sine part. But now I know my boundary

conditions are W at x equals 0. W of 0 is what? What’s the displacement

at x equals 0? AUDIENCE: [INAUDIBLE] PROFESSOR: 0. That’s the pin. That’s the end

where it’s fixed at. And we started out here

with a second order partial differential equation. And a second order equation

requires two boundary conditions. A fourth order

equation for the beam will require four

boundary conditions. We only have to find two. One of them is it has

no motion on the left. So you plug in 0 for x. Cosine of 0 is 1. Sine of 0 is 0. So we find out that

this is B1 times 1. But it has to be 0 as

the boundary condition. So that implies B1 is 0. There’s no cosines

in this answer. And W at L is 0. And so that says B2

sine kL equals 0. And that’s true. That’s only true

if kL equals n pi. So now I’ve found

out that there’s, just for vibration of

a finite length string, only particular

values of k that work. So that says that there are

special values of k which I’ll call k sub n which

are equal to n pi over L. And from that, we now

have our mode shapes. Because we can say,

ah, well, there’s special solution

for this W of x that applies only when we satisfy

the boundary conditions. And that will be some

undetermined amplitude. B2 came from the sine term. And those are our mode shapes. And now the natural

frequencies– once you know mode shapes,

natural frequencies actually become pretty trivial to find. In this case, if we know

that’s the mode shape, then how do we get the

natural frequencies? Well, we know that–

what’s the definition of k? Therefore, the

particular values of k that were allowed

solutions here are going to correspond to

particular values of omega n. And therefore, omega n squared

is just kn squared c squared. And that’s n pi

over L squared T/m. That’s omega n squared. So the natural

frequencies of a string are n pi over L root T/m. And this is in

radians per second. And I like to work

in hertz sometimes. So the natural frequencies

in hertz– omega n over 2 pi. And that becomes n

over 2L root T/m. So the first natural frequency,

f1, is 1 over 2L root T/m. Now, let’s draw. What’s the mode shape

for the first mode? Well, it’s half a sine

wave, vibrates like that. It’s full wavelength. I didn’t leave myself

quite enough room. That’s half a wavelength

of a sine wave. So the full wavelength

would be like that. This is of length L. And

so is this piece over here. So the lambda is 2L for

this particular problem. Let’s see, how do I want

to pose this question? So how long does it take

for a wave or disturbance to travel the length

of this finite string? How long does it take it

to go down there and back? How would you calculate that? Distance equals rate times time. What’s the distance? 2L. What’s the speed? c. So the length of time ought

to be 2L over c, right? So the time required– and

2L divided by T over m. But f1 is T/m divided by 2L. Hmm. So the period– so there’s

a direct connection between propagation speed,

frequencies, wavelengths. They’re very closely related. So the natural frequency of

the first mode of this string, that frequency, is exactly

1 over the length of time it takes for a disturbance

to travel down and back. So with that depth

of understanding of how the wave

equation behaves, you can guess the behavior

of lots of other things that behave like that,

like my rod here. I’ll do a little demo

with it in a second. So for example, the

longitudinal vibration, stress waves running

up and down this thing, obey the wave equation. So if I take this thing

and drop it on the floor, it’ll bounce off the floor. How long does it take

to bounce off the floor? So what do you think actually–

what physics has to happen? What’s required to make this

thing bounce off the floor? So we’re going to consider

the floor infinitely rigid. It hits the floor. It actually stays there for

some finite length of time, and then it leaves. So physically, when I was

holding up my string, if I smacked the end, what happened? A pulse took off, ran down

the end, reflected, came back. And that was one round trip. What do you suppose

happens here? I put a pulse into the end. Is it a tension or compression,

the strain that’s felt? AUDIENCE: Compression PROFESSOR: Compression. So a little compression

pulse is put into the end. That compression pulse

then, when it first hits, the compression and the speed

of propagation is finite. So that compression wave

starts traveling up here. Behind the compression wave,

this rod has come to a stop. In front of the

compression wave, the rod doesn’t know

it hit the ground yet. It’s still moving down. So that compression

wave travels up, and it is decelerating

each little slice of mass as it passes through. It brings it to a stop. And so the compression

reaches the top end. The cylinder has come to a stop. The end is free. It can’t take any strain. So an equal and

opposite tension wave has to start to make the sum

of them go to 0 at the end. The boundary condition

at the end is no strain. So it reflects as

a tension wave. Now you have a tension

wave going down. And what it does is it

accelerates every atom as it goes by, as it goes past it. So everything is stopped now. Now it starts down,

and this thing starts rebounding from– the

top rebounds from the floor before the bottom does. The top starts going up. All of it– more

and more goes up. And one hits the bottom. The tension wave hits the

floor, and it jumps off. So how long does it take? Right? And what do you guess

the natural frequency of a free-free rod is? Now, it has a funny mode shape. The mode shape is not half

a sine wave like this. The displacement of the

rod, it has free ends. The ends are moving a lot. But I’ll give you a clue. [ROD RINGING] I can hold it in the

center and not damp it. What do you think the

mode shape looks like? Half a wavelength long,

ends are free– cosine, maximum displacement,

goes to zero, maximum negative displacement. So it’s half a

wavelength long, but it’s a cosine half a wavelength. And the full wavelength is 2L. So this has mode shapes. The mode shapes– I’ve applied

different boundary conditions. These are free-free

boundary conditions. The mode shapes are

cosine n pi x over L. But they have to obey

a certain other law that we know about,

conservation of momentum. Because I’ve got

gravity to deal with, I have to hang on to this thing. But I’ve picked a place to hang

onto it that you can hear it. I’m not affecting the motion. There’s no motion

where I’m holding it. So if I were out in

space, I could do this– [ROD RINGING] –and just let it hang

there in space, right? And it would sit there and ring. What is happening to the

center of mass of this system as it vibrates? AUDIENCE: [INAUDIBLE] PROFESSOR: Stationary. So half of the mass

of this thing’s got to be moving that way. And half of the mass

has to be moving that way so that the total

center of mass doesn’t move. Well, cosine mode

shape, positive here, negative there, perfectly

symmetric, center of mass doesn’t move. So there’s all sorts

of neat little problems that you can solve just by

knowing the wave equation and figuring out

boundary conditions. How many of you stand

in the shower at home and sing, and

every now and then, you hit a note, man, you

just sound great, right? And it’s just all

this reverberation. How many of you have done that? OK, right, what’s going on? AUDIENCE: [INAUDIBLE]

Natural frequency? PROFESSOR: You’ve

hit a– somebody said natural frequency. Of what? AUDIENCE: [INAUDIBLE] PROFESSOR: Huh? AUDIENCE: [INAUDIBLE] PROFESSOR: You’ve hit the

natural frequency of the shower stall itself. If the shower stall

is a meter across, pressure waves– and

you plot pressure inside of the shower,

the lowest mode if you’re plotting pressure. Well, let’s plot actually

molecular movement. What’s the boundary

condition at the wall, the molecules at the wall? They can’t move, right? 0. So the molecular

motion at resonance in the shower stall, the

molecules, the pressures making them move back and forth, looks

like back to the string again. This is L. The first

natural frequency of sound waves bouncing

off the walls in the stall is 1 over 2L root

times c, whatever c is. And c is the speed

of sound in air, which is 340 meters per second. So 340 meters per

second divided by 2L– so if it’s 1 meter

across here, it’s 340 divided by 2, 170 hertz. So that first note you

can hit in the 1 meter across shower stall is about

170 hertz– pretty low. But you can hit second mode. It’d be twice

that, and so forth. OK, what about an organ pipe? This is an organ pipe, wood. It’s got a stoppered end. Actually, let’s do it

without the stopper. Now it’s an open organ pipe. [ORGAN NOTE] Basic wave equation–

how would you model its boundary conditions? So you can talk about maybe

particle molecular motion. This is, now again, just

sound waves, so air particles. And this is now longitudinal. Things are moving inside. So what’s the boundary condition

at this end, free or fixed? Free. And here it’s quite open, so

the boundary condition on here is free. So for the molecular motion

in a free-free organ pipe, you have to get back to that

half a wavelength cosine thing. And if you wanted to

plot pressure instead, you can write the wave

equation in terms of pressure. Pressure is– this is

pressure relief here and pressure relief there. So in fact, if is

displacement of the molecules, pressure would plot like that. You’d have what’s called

a pressure relief boundary condition. But again, it’s a

half wavelength long. What do you think the

first natural frequency of this organ pipe is? The period would be 2L over c. The frequency

would be c over 2L. So the frequency for the organ

pipe open end f1 is c over 2L. [ORGAN NOTE] Check your intuition. I’m going to close the

end– still an organ pipe. Is the frequency now going

to be higher or lower? Take a vote. How many think the

frequency is going to go up? Raise your hands, commit. All right, down. We’ve got a lot of uncertainty. All right, let’s

do the experiment. [ORGAN NOTE] [LOWER ORGAN NOTE] How come? I find that actually

kind of counterintuitive. Until I learned this, I would

have guessed the opposite way. What’s going on with

pressure in a closed pipe? Well, here at the orifice

where the sound is actually generated, it’s the pressure. If we wanted to plot

pressure at the opening, that’s a pressure relief place. So it’s 0. But at the other end where

the stopper is, it’s maximum. How many wavelengths is that? A quarter. And so the length of time

it takes for the thing to go through one

complete period is going to be 4L over c, half

the frequency of the open pipe. OK, so the wave equation

is really quite powerful, governs lots of things. I’ve got 10, 15

minutes left here. I don’t want you to go away

thinking that the whole world behaves like the wave equation. Because there are some

important other physical systems that we care about. And I’m going to

show you just one. And that’s the

vibration of the beam. So here’s the cantilever beam. The whole table is moving. And you can see it

up on the screen. OK, so its first mode

vibration, tip moves maximum. It kind of looks like

a quarter wavelength. It roughly is, but not exactly. So let’s draw a cantilever. And most of you have had 2.001. So if you put a load

P out here– bends, goes through a

displacement delta. So you know that delta equals

Pl cubed over 3EI, right? And what’s this I? AUDIENCE: [INAUDIBLE] PROFESSOR: Area

moment of inertia. Now that you’ve been

doing dynamics all term, we talk about mass

moments of inertia. There’s also area

moments of inertia. So this is the area moment

of inertia of a beam. In this case, our beam is

a little rectangular cross section. And the neutral axis is

here, a little variable y at displacement. I is the integral

of y squared dA. And dA is just a little

slice of area here, dA. And the integral of y squared

dA is your cross sectional area moment of inertia in the

direction of bending. So that is I. You can also

write it as a kappa squared A. And we ran into

this in dynamics. We called it the

radius of gyration. You had the same thing with

area moments of inertia, the radius of gyration. This is going to be really

helpful in a second. So if you solve the

force balance for a beam like I did for the string,

take a little slice, do force balance for

transverse motions– I’m not going to grind it out. And temporarily neglect

external forces and damping. I want to get to the natural

frequencies and mode shapes. So the free vibration,

no damping, equation looks like EI partial

4 w with respect to x to the fourth plus rho A

partial squared w with respect to t squared equals 0. And now this is

density, mass density. And the A, this A, is the

area of the cross section. So it’s just some bh with

thickness times the width. So rho times A is a

mass per unit length. And so mass per

unit length times dx would be the

little mass associated with the element times

the acceleration should be the forces on the element. So that’s the fourth

order partial differential equation that describes

the vibration of a beam. And you have to apply

the boundary conditions. And for the string, it was

just B1 cosine B2 sine. For the beam, it’s B1 cosine

plus B2 sine plus C2 cosh plus D2 sinh x. And then you have to apply

four boundary conditions and solve for B1, B2, and

so forth, all four of those. I won’t do it. But that’s how you do it. Separation– and separation

of variables works again. So we solve this, apply

the boundary conditions. What are the

boundary conditions? Just so you understand

what I mean by the boundary conditions, what are they

for a free-free beam, zero motion at the wall? No strain at the end, no

bending moment at the end, no sheer force at

the end– so there’s no second derivative,

no third derivative. And at the wall, the slope

is 0, the first derivative. No slope comes into the wall,

but the slope is 0 there. So those are the different

kind of boundaries. So if you have a

free-free beam, you have no bending at either end

and no strain at either end. Fixed-fixed beam–

no displacement, zero slopes at both at ends,

and all different combinations. And every different

combination gives you different natural frequencies. So you apply the boundary

conditions, and for a beam, you find out that

for all beams omega n can be written as some

beta n, a parameter, squared, I’ll call it, times the

square root of EI over rho A. And this thing varies according

to the boundary conditions. Now that’s what you get shown

in every textbook in the world. And I have a very hard

time visualizing this, getting physical

intuition by that. So something you never

see in a textbook but I often do is let’s

replace I with kappa squared A. And you get a square root of E

over rho and a square root of I over A. But I is kappa

squared A. The A’s cancel. It’s the square root

of kappa squared. So this, you know

what E over rho is? E over rho, square

root of E over rho, is the sound speed

in a solid material. So the speed of stress

waves traveling up and down this thing is the square

root of E over rho. [ROD RINGING] This is aluminum. It’s about 4,000

meters a second. So if you know just the

properties of the material, you have that. And that says then

omega n for beams is some beta n squared a

parameter times kappa CL. And this thing, this

is often written CL. It’s the longitudinal

sound speed. This is sound speed for waves

traveling through the medium. So this tells you if you

make the beam twice as thick, what do you do to its

natural frequencies? Doubles– instantly

you know that. So bending properties depend a

lot on the radius of gyration. And I’ll give you a

few natural frequencies for different

boundary conditions just so you see what

they behave like. So a pin-pin beam

looks like that. So you put a plank across the

stream, rocks on both sides, you’ve got a pin-pin

beam, basically. It’s set there in rock. So some length L

has properties EI. So the natural frequencies

for a pin-pin beam, the beta n’s, are

just n pi over L. And so your natural

frequencies– omega n looks like n pi over L

quantity squared kappa CL. And for the cantilever,

the natural frequencies look like omega n pi squared

over 4L squared, is the beta n. And I’ll write it this

way again– EI over rho A. You can always

go back and do that. Or you can call it kappa CL. This is also kappa CL. But then there are

some numbers you’ve got to use here– 1.194 squared. That’s the first mode. Second mode– 2.988 squared. And then after that– 5

squared, 7 squared, 9 squared. So this is the natural

frequency of a cantilever. Pi squared over 4L

squared times 1.194 squared kappa CL, that’s

this natural frequency. And one final case,

because I can show it to you– the free-free case. So that’s a beam bending

that vibrates like that. And I happen to know on a

beam for the first mode– this is the first mode of a beam. Where these nodes are,

where there’s no motion, I should be able to hold

it there and not damp it. And that turns out to be at

about the quarter points. So whack it like that. [ROD RINGING] And do it again. [ROD RINGING] All right, so I want you to

hold it about right there. Nope, you can’t hold it

like that, though– just got to balance it. Because you’ve got to be

right where the node is. [ROD RINGING] You can hear that

little bit lower tone. That’s that free-free

bending mode. And it’s just sitting. You can feel it vibrating

a little bit but not much. When you’re right

in the right spot, you’re right on the mode shape. You can almost see it if

you hit it hard enough. So that’s the free-free beam. And the free-free beam has

natural frequencies omega n, again, pi squared over 4L

squared kappa CL 3.0112 squared, 5 squared, 7 squared,

9 squared, so as you go up in n. So those are the

natural frequencies of a free-free beam. Oh, one last fact about

beams– so this is now a steel beam under no tension. It can support its own

weight, long though. So can a beam support

waves traveling down the beam, transverse waves

traveling down the beam? What do you think? Well, if it can support

this, it can probably support waves, right? So waves will

propagate in a beam even though this is fourth order

partial differential equation. But how fast do they go? That’s the question. So this is a beam. And I want to know about

waves traveling down it. And I’m not going

to go through– this would take another hour or so to

show you where this comes from. But here’s my beam. Here’s a disturbance

traveling along it with some speed that

I’m going to call CT. It’s transverse wave speed. It’s the speed you’d see

a crest of a wave moving at running down that beam. CT for a beam– square

root of omega kappa CL. And CL, again, is the

square root of E over rho. That’s the speed of

sound in the material. That just turns up in here. So what does this tell you

about the frequency dependence of the speed? Does the speed change

with frequency? Omega kappa CL– it’s

proportional to frequency. High frequency waves go faster

than low frequency waves in a beam. I didn’t emphasize it when

we were talking about it. But the wave equation,

what was c for the string? For the wave equation, the

speed of wave propagation was square root of T/m. Was it frequency dependent? Always traveled

at the same speed. And so there’s an

important consequence. So for anything that

obeys the wave equation, the speed of propagation is

a constant and independent to frequency. So I can make any initial

shape that I make in this thing and let it go. Its initial disturbance,

that little shape will stay that shape and run

up and down the thing forever. And that shape– you could

imagine a little pluck like this to start with. You could imagine

doing a Fourier series to approximate that. It would be made up of a

bunch of different Fourier components. And yet for something that

bears the wave equation, that little pluck will just

stay the shape of that pluck and run around forever. But not so in a beam. If you did that in a beam,

if you come up and put an impulse into a

beam, all that energy would start out together. But in very brief time, the

high frequency information would get out in front of the

low frequency information. And if you were way down this

beam, and somebody up a mile away whacks one end, and

you’re down further along, you’ll see high

frequency waves past you, and then lower frequency,

and finally really slow ones coming by, the

really long waves. So that’s called dispersion. So beam waves are dispersive. Things that obey the wave

equation are non-dispersive. The energy all travels at

the same speed independent of frequency. All right, so that’s

it for the term. I’ll see you guys

on next Wednesday.

thaaaaaaaaaaaaaaaaaaaaaaaaaaaaanks MIT

That's an outstanding instructor. Thinking of my school now makes me vomit.

thank you sir

thank you sir

πππ

THIS IS BRIEF RATHERTHAT WE LEARN,KEEP UP IT

this lecturer is the best.

really great lecture, thanks Dr j.Kim Vandiver

thank you MIT ^^

Need a clap emoji for times like this.

8.003 had me thinking I knew physics but no mention of reference frames, almost no angular momentum!

Great lecture, thanks!

Great lesson

That's a beautiful lecture in every sense. MIT provides deeply insightful lectures. Makes the reading literature very comprehensible.

Thank u sir and thanks MIT

I'm a master student and I could say this the best and highest standard lecture I've had during my study life. Please please put other courses of MIT for mechanical engineering that all student around the world enjoy course

This content should not be restricted! Please, release it all over the planet!

Great Professor…

Amazing way to demonstrate this complex topic. Anyone know the book, professor is following for these lectures?

amazing professor. clear, engaging. no words, simply amazing

Amazing professor I m from Algeria I liked this video very much I hope that you have more videos for civil engineering

Have a question regarding the organ pipe experiment. The frequencies played are approximately 370 Hz and 210 Hz respectively. So it's not exactly the half but something like 0.57 times the first frequency. What's the reason of this?

Anyone from india studying mechanical engineering? πππ