In this segment, we will take a physical example of what is the application of eigenvalues
and eigenvectors. There are many physical applications of eigenvalues and eigenvectors,
but we will just concentrate on a simple one which you can follow based on your knowledge
of physics. So let's suppose somebody gives you says that Hey, I'm going to give you a
spring here, I'm going to put a mass next to it and I'm going to put another spring
here and another mass right here . Now what I want to do is let's suppose the spring stiffness
is k here the spring stiffness is k here we're assuming the two springs are the same stiffness.
Let's suppose we want to say this is mass m1 and this is mass m2; that s the mass of
the first mass and mass of the second mass. What we want to do is we want to be able to
measure what x1 and x2 are so we are trying to figure out, hey, how does this mass m1
and m2 move when you are either forcing it to move by maybe let s suppose you might pull
it or you might keep on adding some forcing function to it, but let's suppose we take
a simple case where you simply pull this mass m2, m1 also gets pulled with it, and then
let's suppose you leave it alone what s going to happen is that the m1 mass is going to
oscillate and m2 is also going to oscillate, or maybe not oscillate, depending on the values
of k. In this case, they are going to oscillate because there's no damping; we're not assuming
any damping. We want to be able to figure out how does x1 and x2 vary as a function
of time and then we will see hey how does it relate to our application of eigenvalues
and eigenvectors so if you look at the free-body diagram of the mass m1 so you have m1 mass
right here let's suppose if I assume x1 to be greater than zero and then I assume x2
to be greater than x1 so I m not saying that this is what's really happening when the system
is moving but let's suppose we are saying that at a particular time x1 is greater than
zero and x2 is greater than x1 so if that s the case now what s going to happen is that
if x1 is greater than zero this spring is going to stretch so if you look at the k1
spring it is going to stretch so if it is stretching that means that is has to have
a tensile force in it. So if that's the tensile force in the spring k1 that means the reaction
of that force will be like this on the mass itself so that will be kx1 so because this
going to stretch by a distance of x1 and the amount of force should be stored in the spring
the force should be experienced by the spring will be kx1 so since this direction is like
this that means the equal and opposite force has to act on the mass will be kx1 when we
are assuming x2 is greater than x1 than if we take the spring two again if x2 is greater
than x1 that means that this spring has stretched because its moving x2 more than x1 so this
is stretched out because x1 is moving a little less than what the x2 is moving so it'll be
stretched out like this so the amount of force should be existing in the spring would be
k(x2-x1) so if that is the case then we know that hey that is on this side right here so
that means what we'll have k(x2-x1) going this way. So it's just the opposite force
should be existing so your force is k(x2-x1) going this way on the spring on this side
so the reaction from the mass would be that one now these are the only forces acting on
mass m1 and what we are going to do is we are going to sum the forces on mass m1 so
I m going to get -kx1 + (kx2 - x1) and what this is going to be equal to so this is minus
kx1 because let s suppose we say negative is to the left side and positive is to the
right side so we get minus kx1 from here plus k(x2-x1) from here and then we equal to the
inertia force in the mass and that will be M1(d2x1 /dt2) so that s what I m going to
get from there so if I m going to simplify this I don t need to simplify it will be like
that so that is what our first equation is going to be for mass m1 . Let's look at m2.
On m2 you can very well see that the only force that should be existing will be this
way (kx2 - x1) because that I can note from here that hey since the force in m1 is going
to be to the right than the equal and opposite force has to act on m2 in the opposite direction
or I can look at it this way that hey the force is going this way in the spring so that's
the opposite of that is being experience by the mass by Newton s 3rd law of motion so
with M2 right here so what I get is k(x2 - x1) is equal to that is the force going in the
negative direction there or in the negative x2 direction so the total inertia force will
be M2(d2x2 /dt2) so this is first differential equation and this is the second differential
equation so what we are getting is that the system is governed by two separate differential
equation which are simultaneous in nature because we are finding out the two dependent
variables x1 and x2 are in both of these equations right here. So let s see if we put some numbers
in here what happens so let's choose M1=10 so I m not using any units on that so I m
keeping it unit-less, M2=20 and let s suppose k=15 so what does that do to our equation?
So the first equation which I write here will turn out to be that I get (10 d2x1 /dt2) = -15x1+15(x2-x1)
and if I take all of these to be left inside what do I get from there so I have (15x2 - 15x1)
so I get (10d2x1/dt2) + 15(2x1-x2) =0 so if I take this to the left hand side this is
what I will be able to get that equation one rewritten in terms of the numbers and everything
taken to the left hand side similarly the second equation which I ll write here if I
do the same kind of substitution I will get 20 d2x / dt2 -15(x1-x2)=0 and that will be
my second equation done simply by substituting M2=20 here, k=15 here and taking that to the
left hand side that is what I m going to get as my two equations now so these are the two
equations which are going to govern my system coupled with some initial conditions but let
s go and see how this is related to eigenvalues and eigenvectors.
So from vibration theory if you have taken the vibrations class you will be quite familiar
with this but if you have not I'm just giving it to you that your displacements will be
of this form Aisin(wt - "phi") let suppose so each of the displacement x1 and x2 so i
's go to 1 and 2 will be of the sinusoidal form where Ai is the amplitude of mass i , and
w is the frequency of vibration the angle of frequency of vibration and then let s suppose
"phi" is given as the phase shift. So if we assume that is the case then let s go and
see that if we substitute back into our differential equation what happens we have (d2xi /dt2)
in both differential equations so if i take the second derivative of this I will get Ai
w^2 sin (wt-"phi") that is what im going to get from the second derivative of this xi
and now I m going to substitute this into my differential equation so if I look at the
first differential equation I m going to get -10 A1 w^2 sin(wt-"phi") -15(-2A1 sin (wt-
"phi") + A2 sin (wt-"phi"))=0 that s what I m going to get from the first differential
equation and the second differential equation if I make these substitutions of xi and d2xi/dt2
in this case being x2 I will get -20A2 w^2 sin(wt-"phi")-15(A1sin (wt-"phi") + A2sin(wt-"phi"))=0
and that will be equal to zero and now you are finding out the sin (�t-�) is a common term so I am
going to take that out 'cause if sin(wt-"phi") is zero everywhere, which is not really the
case, then our previous solution for my displacement, which is not true, so I can very well get
rid of this sin(wt-"phi") here, sin(wt-"phi") here, sin(wt-"phi") here, same thing here,
same thing here and same thing here. What does that give me? That basically gives me
the following thing, it gives me: -10A1 w^2 -15(-2A1+A2)=0, -20A2 w^2-15(A1-A2)=0 so this
is first equation this is second equation this is what I get by taking out the sin(wt-"phi")
term out.
Ok, if I collect now the A1 term gives me (-10w^2+30A1)-15A2=0 that's the first equation
and the second equation will be -15A1+(-20w^2+15)A2=0 so we are basically two equations two unknowns
if we consider A1 and A2 to be the two unknowns and we don t know what w is but let s suppose
we consider it to be two unknowns then in this case now what we are going to find out
is that if I simplify this a little bit further and I suppose I divide it by 15 throughout
I might be able to so if I divided it by 15 throughout I get (-w^2- 3)A1 - 1.5A2=0 and
this one here I suppose I m going to divide by 15 I get -0.75A1+-w^2+0.75A2 =0 and the
reason why I did this because here I divided by ten and I got this here I divided by twenty
and the only reason I m doing this is because I want to make the coefficient of w^2 in
both cases of course with A1 I only have one there so -1( A1) here and -1(A2)here so if
I now write it like in the matrix form let s see what happens if I write it in the matrix
form I get A1 here and I get A2 here and of course I'll get 0 and 0 here because that's
the right hand side so if I look at the quotient of A1 it is w^2+3 here, it is -1.5 here,
-0.75 here, and w^2+0.75 here. I can break it down further I can write it like this I
can say hey let me only put known quantities here in this matrix and then I'll have something
else here an unknown quantities associated with this so what I mean by saying that is
if I put A1 and A2 here and I don t want to put w^2 term here and w^2 term in this matrix
here I ll put 3 and -1.5 here, -0.75 here and 0.75 here and I ll put w^2 here. So once
that is done I can now see that hey let s suppose if I assume w^2 ="lambda" let s suppose
I want to call this another minus but the w^2 that s supposed to be "lambda" what do I get
from here and I call this I suppose I ll call this to be the A-matrix I ll call this to
be the X- matrix and that's "lambda" is w^2 I ll call this the X-matrix so what do I get I get [A][X]
"lambda"[X] is equal to zero vector and that [A][X]=["lambda"][X] so what is going to happen is that this is
the eigenvalue problem because "lambda" how is the eigenvalue defined, eigenvalue is defined
as if hey if there is a non-zero vector [X] for which [A] times the non-zero vector [X]
is equal to some scalar times the non-zero vector [X] then "lambda" is considered the eigenvalue
of the problem so if I m able to find "lambda" I'd be able to find w^2 which will give me the
angle frequency within which the system is running so that's why the eigenvalues show
us that hey how we can use for the physics this one is called the eigenvalue of course
and what is the corresponding vector [X] is called the eigenvector and you can very well
see that the eigenvector is a measure of A1 and A2 which is simply a measure of the amplitudes
which the spring mass system is having so that's how the physics of the problem and
the eigenvalue eigenvector concept are involved. That's the end of this segment.