CHAPTER 10.02: EIGENVALUES AND
EIGENVECTORS: A physical example of application of eigenvalues and
eigenvectors 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 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(ωt
- Ø) let suppose so each of the displacement x1 and x2 so i
's go to 1 and 2 will be of the sinusoidal from where Ai is the amplitude of
mass i , and ω is the frequency of vibration
the angle of frequency of vibration and then let’s suppose Ø 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 –Aiω2sin (ωt-Ø)
that is what I’m 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 A1ω2 sin(ωt-Ø) -15(-2A1 sin (ωt-
Ø) + A2 sin (ωt-Ø))=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 -20A2ω2 sin(ωt-Ø)-15(A1sin (ωt-Ø) + A2sin(ωt-Ø))=0
and that will be equal to zero and now you are finding out the sin (ωt-Ø) is a common term so I am going to take that
out because if sin(ωt-Ø)
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(ωt-Ø) here, sin(ωt-Ø)
here, sin(ωt-Ø) 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: -10A1ω2 -15(-2A1+A2)=0, -20A2ω2-15(A1-A2)=0 so this is first equation
this is second equation this is what I get by taking out the sin(ωt-Ø) term out. Ok,
if I collect now the A1 term gives me (-10ω2+30A1)-15A2=0 that's the
first equation and the second equation will be -15A1+(-20ω2+15)A2=0 so
we are basically two (?? 12:06) two unknowns if we consider A1 and A2 to be
the two unknowns and we don’t know what ω 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 (–ω2- 3)A1 - 1.5A2=0 and this one here I suppose I’m
going to divide by 15 I get -0.75A1+-ω2+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 ω2 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 –ω2+3 here, it is -1.5
here, -0.75 here, and –ω2+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 ω2 term here and ω2 term in this matrix here I’ll put 3
and -1.5 here, -0.75 here and 0.75 here and I’ll put ω2 here. So once
that is done I can now see that hey let’s suppose if I assume ω2 =λ
let’s suppose I want to call this another minus but the ω2 that’s
supposed to be λ 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 x λ
is ω2 I’ll call this the X-matrix so what do I get I get [A][X]
–λ[X] is equal to zero vector and that [A][X]=[λ][X] so what is
going to happen is that this is the eigenvalue problem because λ 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] than λ is considered the
eigenvalue of the problem so if I’m able to find λ I'd be able to find
ω2 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. |