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So today, we are going to start the lecture
on uncoupled motions. See in the last class
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I derived or showed that equation of motion
for an oscillating system for ship turns out
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to be of having a formed something like mass
plus some part into well, I just put an simple
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this thing mode of motion, b into z dot plus
c z equal to F, where what I said is that
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this part this was what was called as radiation
forces.
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This part is m z dot dot was mass inertia
force, this was hydrostatic restoring forces
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and this was wave exciting forces
and in specifically, this could be called
added mass force and this could be called
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damping. But I think we will specify it to
be radiation damping or potential damping
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force because it is arising out of only radiation
damping as as we have derived earlier.
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Say let us call it damping, because other
wall also might look similar. Now having said
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that so this was single degree. If actually
I consider a 6 degree of freedom that is a
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ship in waves, then every motion will affect
every motion. And all this force, get coupled
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means, you will have some radiation force
arising in the heave direction because of
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roll motion etcetera etcetera. But today we
are going to talk about that to start with
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a simple heave equation of motion single heave
equation of motion.
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So now, that will look like the equation for
that will have then m plus let me call this
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a z now. Now
exciting force, we know the f. But I explained
to you that in waves see everything is periodic
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with a periodicity of omega e. So therefore,
this can be written as some amplitude of exciting
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force, let us say I call it this thing.
And I can write it as cos omega e t well we
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can also put a phase here but that depends
on let me put a phase here also, let us say
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we can also put a phase here as beta. Because
the phase you can also make it 0 depending
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on where you start from. Now this is a single
degree equation of motion for heave, where
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I will call this to be heave added mass. We
can put it also z here and this is heave damping
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and I also let me put it z also that is heave
restoring force. Now before I discuss about
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these terms, we will first let us talk about
the solution of this equation motion how does
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it look like.
You know, this is a equation of motion if
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the bodies undergoing only heave and nothing
else. And we are assuming that as it goes
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heave up and down, that is body is undergoing
this mode of motion. It is not exciting or
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causing any other mode of motion just uncoupled,
you know we are making this assumption. Now
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you see this equation of motion, we will start
now supposing we all know this it looks like
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and vibration equation. If I made this 0 and
if I remove damping what I end up getting
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is what is known as undamped natural motion
or equation for undamped free undamped motion.
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So, let us look at that so, if I just consider
the inertia force and c z equal to 0 then
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this is what is my so, called undamped free
motion right
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Now this, undamped free motion has a solution
which will look like z equal to A sin omega
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z t plus b cos actually either you can call
it, this way or I can call it let me call
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it A 1 b 1. So that, I can write it as a sin
or cos omega z t minus beta z, as you knows
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it is the same way. Either you can write in
as a 2 sinusoidal term or one sinusoidal term
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with a phase gap.
This is how the solution is and the most significant
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thing here is w z, this is given by square
root of c z by m plus a z which is what is
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called natural period natural heave period
sorry sorry natural heave frequency. So that,
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natural period is simply 2 pi is going to
be 2 pi by natural heave period of course,
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I mean this implied here by omega z.
Now, natural period is very very very very
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important concept, this is why we have to
insist on that afterwards. But this we know
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this is what is called well, we can also specify
it is natural heave period for undamped motion
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natural free period, if you want to add the
word undamped you can also add that. Because
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I have reduced the damping right I do not
have damping here.
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How does this solution looks like well it
will look like a sign curve. The beta is going
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to be a measure of this that is a phase that
is the phase, this beta measures in some sense
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this how much later with respect to well at
t equal to 0. It is having a motion, essentially
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if I know A and if I know beta z, then this
solution is fully complete which means that
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in this equation you of course, I am assuming
we know m A z c z. So, here if I know well
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if I know that if I know omega z So, omega
z is known the only unknown is A and beta
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z and of course, I can always solve it. If
I had two initial condition but that is a
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trivial thing not very important.
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Now coming to the next level, that is my damped
natural period, free damped motion. If I consider,
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that means here I have a damping but no forcing.
So, here the equation will look like m plus
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a z double dot plus so, this as a solution
this solution we all know this. So, we need
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not go through how mu t, we can call it d
omega d this is d t plus A 2 sin omega d t
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or which is equivalently e power of minus
d t I can call it this way. A sin omega d
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t minus eta, let us put something eta z d
let me just call it some phase angle. See
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remember that any signal with a phase gap
can also be written as sum sum of 2 sin.
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Because you know sin a plus b is sin a cos
b plus cos a sin b, if you actually apply
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you will end up getting this relation. This
is very straight forward, I mean what I mean
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is that two sinusoidal curve is essentially
one sin curve with a phase curve if you add
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this up if you add this and a cos curve. You
will find out that summation is nothing but
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another sin curve starting with a phase curve.
So, one can use either this expression or
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this expression I mean, so this is interchangeable.
We need to know, that there are two unknowns
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in a signal A 1 A 2 or A and eta z 2 unknowns
which ever form you write so this is known.
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In this case, this signal looks like that
this part is a free vibration but there is
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a decay part. Now, this nu is given by
b by b z by 2 m plus a z, this is called the
decaying constant.
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In fact this solutions that I wrote, we had
made an assumption here that b z is small
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decaying constant is small. See if the decaying
was very very high, then it could have just
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dived down a periodic motion but in a it is
presumed. Water is a very less viscous water
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as such even if you take viscosity damping.
Of course, here we are not considering even
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viscosity, then the damping is very low.
As a result b z is always very smaller b z
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by this this mu is very small so, it is always
looking like that. So, this is a relevant
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solution for this as far as heave equation
of motion is concerned. Of course, the other
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thing is w d this is given by, that this is
a damp period frequency z square minus mu
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square root over. Now here or therefore, I
have got T d equal to two pi by omega d at
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this point so, this is how this is the solution
how does it look like.
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It was it will look like decaying, this is
my z and this is my p and this decay this
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decay is e power of minus mu t depending on
this mu, it was large as decay faster longer
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this part this is decay decay parameter right.
What is this as we have found out this is
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actually b by b z by 2 I just wrote it in
last slide.
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So, this is a decaying parameter once again,
if I want the solution very simple I have
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two unknowns because the solution that I looking
back at that two unknowns. One was you can
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call it either this and this or this and this
if I know this, this and this the solution
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is like this the solution is fully known.
If I could determine this to unknowns I can
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determine this to unknown if I have two condition
such as somebody says let us say that, I have
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got in initial condition.
Say, let me put it this way that is A 1 A
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2 or A and eta z t as this one as I mentioned
and for that you need two condition. So, we
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can put in form of initial condition. For
example, I can say that for example, let me
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see I can say that at t equal to 0 my z equal
to z a and also t equal to 0 my z dot equal
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to 0. That is very actually, very you know
valid point what I mean is that it is something
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like you are pushing the body down. So, I
know at t equal to 0 what is my see I pushed
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it down to an extent z a. So, I know what
is my initial displacement that is what is
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at t 0 z is how much I pushed it down no imposed
motion left my hand. So, at that time an initial
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velocity z dot is 0. So, I know this for example,
so if I know this I can find out c 1 c 2
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I just I give an example of that see, if I
look back here again I will write it in this
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form. In c 1 c 2 form, let us say if I wrote
z as again e power of minus nu t A 1 cos omega
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d t plus A 2 sin omega d t, I have z dot given
by minus nu e power of minus nu t is becomes
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minus A 1 sin omega d t plus A 2.
So, I know that see at t equal to 0, z is
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equal to from this equation c 1, from one
I will get z is equal to
simply A 1 right. You see because at t 0,
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this is 0, this is also 1 so, z becomes A
1 cos you know 0 that is A 1. And which of
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course is equal to my z a, because this my
initial condition z a becomes a.
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[FL] omega d will come right that is what
you are saying thank you, see here your omega
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d will come right sorry sorry you are absolutely
sorry absolutely, we are going to get actually
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you are right this was actually for this differentiation.
Then I am going to get one more term that
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is e power of minus mu t right right right
sorry about that. We are going to get here,
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well well let me let actually deduct that
wait, let me write it now you’re you are
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absolutely right. So, this is I am sorry this
is going to be first minus mu e power of minus
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mu t, then A 1 cos omega d t plus A 2 sin
omega d t right plus e power of minus mu t.
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Here I m going to get this differentiation
that is what you are saying, right that is
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minus, we are going to get A 1 omega d into
sin omega d t right plus here, we are going
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to get A 2 omega d into cos omega d t that
is right right you are absolutely through.
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So, in any case what therefore, is happening
from this one. Obviously, at t 0 my z dot
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becomes this this goes off, then my z dot
will become minus this A 1 right plus I will
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end up getting omega d into A 2.
So this becomes my one equation and this becomes
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one. So, from there this is actually equal
my initial condition is 0, so I end up getting
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0 equal to this. So my point, you say here
is that I have found A 1 is z 1 and here I
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have got 0 equal to minus mu z a plus omega
d A 2. So, this gives me A 2 equal to mu z
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a by omega d right.
Really speaking, this I just went through
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this, it is very trivial my full point of
saying here was that we had two equations
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sorry one equation with two unknowns. So,
to solve that I need two equations, so to
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get the two equations I need two conditions.
It can be any two condition I can say any
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two condition. So that, I can get two independent
equations, so this is once you know that two
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equations you can solve for it that is all.
A really it is also not very important it
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just to prove the point, but what is most
important for us, now is the the forced part
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the full solution.
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That is now you’ll look at forced damped
equation of motion the final one. So, here
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I have got let me call it here F, let us call
it F 0 here, this e. Now, you know this has
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everybody knows this it has two solution,
its solution can be written as a sum of the
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few so, called complimentary solution as the
full solution. So, it essentially the solution
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can be written as z equal to e power of minus
mu. One is solution for the homogeneous that
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is going to make it 0 and 1 is for the full
power. So, this is something like or A 1 let
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me call it.
Now here will come an interesting part, plus
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there’ll be sinusoidal solution forced solution
e t minus say epsilon z. Now you see how does
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it look like the first one is this one. This
one is going down like that and a second one
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is actually a forced solution of omega t.
You are actually adding the two together solution
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now the question is that what is our interest.
Our interest is obviously in only in this
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part, why because you see given sufficient
time our this thing will decay because of
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mu factor. So, essentially if I were looking
at a steady step solution. That means, if
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I am holding for a long time what would happen
the body will begin to heave only with this
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oscillation, forced oscillation because this
one will decay down.
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In fact, the full solution may look like,
if you were to draw experiment something like
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that and ultimately it will become a straight
line. So, this is what I am looking at this
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is what is my interest fully, so we are looking
at only the steady part of the solution. Again
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here my unknowns are this and this and what
you notice here is that this is oscillating
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at that exciting frequency omega e yes. No
no no no not at all, all structures you see
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the question is this decay the we are not
talk of q or see or the question. It is nothing
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to structures here it is that we are looking
at a this is a decay factor minus v by A,
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when minus v by a would make e power omega
t small if t, in fact in real life just 1
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minute will become 0 actually.
Because see you are having e power of minus
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b by 2, A this factor whatever way it is with
t if you give t 1 second, 2 second etcetera
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shortly it becomes 0 in practical. In fact,
in reality it may become actually, you would
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not even notice it will be actually 0 in almost
like 3, 4 periods 10 second 15 second or so.
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Remember it is nothing to do with structure
it is connected to the decay part of it.
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And we are only looking at the steady part
of the solution, when we are saying forget
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about few more structures, even structure.
When the wave comes I am looking at for 1
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hour, 2 hour, 3 hour kind of solution not
in 10 seconds that goes off. And second thing
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is that wave always existing, you cannot actually
fix a time when it started you know, when
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you are looking at that.
See suppose you today go on boat well, it
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does not oscillating yesterday onwards not
that it has started 0, cannot see and suddenly
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started. So, what you see is always a steady
solution the decay part is just not there
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and simulation also shows. In fact, we will
see later on by numbers b by 2 A and that
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how it is how fast it decays. Now, we will
look at steady part solution and that part
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of solution of course, we know it.
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So it turns out that the solution for this
steady part, again from vibration is given
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by like this z a is given as z static into
this solution. When z static is known as static
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heave amplitude, well we are calling heave
amplitude but what I am saying is that typical
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solution for a vibration forced vibration
equation. This will be F 0 by c z is this
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a solution mu z this this all we know it,
that is why I will not going to go through
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this in the the solution part.
What we what we need to know in this course
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is to tell the typicality of heave motion.
You know like how does it heave, how much
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it does it etcetera, etcetera. Because this
part anybody was studied, you know any vibration
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system mechanically you’ll we will find
out
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standard things and the other things is the
epsilon right, what we wrote epsilon z. Now
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we had to write the kappa part and of course
I had missed out here one. That is this this
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part this is given by
it is called non dimensional damping factor.
See what I mean here is that solution for
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this equation is fully known to everybody.
It is you can write it this way and let us
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look at the diagram that is more revealing.
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So, it will look something like this, if I
were to look here, mu this is here, mu equal
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to omega e by this is always easier actually,
what we are plotting is z verses omega. Now,
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you can write this in z was omega e, if you
want to write instead of omega e. We are writing
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a non dimensional form of omega, that is omega
e by omega z. And instead of writing z we
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are writing mu because z is mu into z s t,
z s t is a constant so, it is a same thing.
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This is why it is easier this becomes something
like that it goes
to etcetera, where this is for damping 0 and
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00:28:21,240 --> 00:28:28,240
this is increasing, that is damping.
This is actually one magnification factor,
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00:28:34,850 --> 00:28:41,850
see the thing is that again let me put it
this way. This is actually one undamp one
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00:28:43,870 --> 00:28:49,840
and this is actually the damp one. See what
is happening understand this all of you have
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00:28:49,840 --> 00:28:54,860
seen this, what what does it show it shows
that this equation that if I have the body
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00:28:54,860 --> 00:29:01,860
here heave. Number one is that at low frequency,
very low frequency my we need to see this
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00:29:09,760 --> 00:29:16,760
very very important you know. At low frequency
at the limit of omega tends to 0, my magnificent
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00:29:19,070 --> 00:29:26,070
factor is 1 which means it simply goes up
as F 0 by F 0 by what c
at this low frequency. At this region first
196
00:29:38,730 --> 00:29:45,200
of all, if I had no damping, I would find
out that at tuning factor omega a by omega
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00:29:45,200 --> 00:29:50,530
z my magnification is theoretically infinity
but if you give a slight damping it will come
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00:29:50,530 --> 00:29:53,809
down to finite value.
As you keep giving more damping, it actually
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00:29:53,809 --> 00:30:00,809
the peak ships towards this side lower side
because this correspond to omega d the damping
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00:30:02,040 --> 00:30:09,040
natural damping period. You go down this side,
here it will eventually come down ascend particularly
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00:30:09,860 --> 00:30:16,860
to 0. Now the important point of this we all
know this you see this part is known. But
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00:30:17,110 --> 00:30:22,480
we want to know is not the solution part say,
importance of this solution with respect to
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00:30:22,480 --> 00:30:29,480
heave motion. Remember this equation m x dot
dot plus b x dot plus c x equal to F. It is
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00:30:32,429 --> 00:30:38,500
actually this vibration equation, everybody
have studied it is ubiquitously occurring
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00:30:38,500 --> 00:30:45,440
in nature any oscillating system, this is
you know phenomena string vibration, any vibrating
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00:30:45,440 --> 00:30:49,799
system, even atom vibration whatever.
The question is not in this solution all I
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00:30:49,799 --> 00:30:55,390
showed here is a typical solution which also
all of you know very well. What is important
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00:30:55,390 --> 00:31:01,110
is that the characteristic of that with respect
to heave mode of motion that is very important
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00:31:01,110 --> 00:31:05,820
in heave. See heave is not going to be like
atom oscillation, it is not going to be like
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00:31:05,820 --> 00:31:10,040
a string oscillation, what is the special
like that characterizes the motion.
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00:31:10,040 --> 00:31:16,720
So, here I want to tell you this see look
at this low end, what the low end means, now
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00:31:16,720 --> 00:31:23,720
let me look another picture, then now omega
e tends 2 0. What does it mean, it means very
213
00:31:24,750 --> 00:31:31,750
long waves infinite period. So, you have this
very long wave wave body here, now we will
214
00:31:39,190 --> 00:31:46,190
again write down the equation of motion well.
I just now you know z, you can write z as
215
00:31:55,600 --> 00:32:02,600
z a, where I will now this time I am going
to write the sort of you know e power of minus
216
00:32:02,830 --> 00:32:09,830
I omega e t, let me write it this way.
But there is also e phi may be e power of
217
00:32:10,730 --> 00:32:17,730
minus I beta or no I what we call epsilon
in the real e power of minus i. I used the
218
00:32:20,250 --> 00:32:27,250
term earlier, let us see the phase phase was
used as epsilon phase. So, we will use as
219
00:32:27,390 --> 00:32:31,919
epsilon z there is not important really speaking
but I mean I want to add this now. You see
220
00:32:31,919 --> 00:32:38,510
z dot is going to be proportional to minus
I omega e into this thing and z double dot
221
00:32:38,510 --> 00:32:45,510
is proportional to omega square e into this.
What I want to say velocity has a proportionality
222
00:32:47,080 --> 00:32:53,549
of omega e multiplied by displacemen,t acceleration
has proportionality omega e square. Now when
223
00:32:53,549 --> 00:32:59,940
omega e is very small omega e is very small
what would happen obviously so no. So, this
224
00:32:59,940 --> 00:33:06,940
equation will look like minus omega e square
into something z a, minus I omega e into something
225
00:33:08,710 --> 00:33:15,710
plus you can say c z into z equal to F, now
it look something like that.
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00:33:16,049 --> 00:33:23,049
Let us pay attention on that see the first
term the force, the added mass force is in
227
00:33:24,650 --> 00:33:30,630
proportion to omega e square. Second term
damping is in proportion, of course forget
228
00:33:30,630 --> 00:33:36,799
the phase gaping proportion to omega e. Third
one is not it is a constant. Now when omega
229
00:33:36,799 --> 00:33:42,510
e tends to 0, this will actually have no influence
very less value this also will have very less
230
00:33:42,510 --> 00:33:46,669
value. So, my F z becomes simply F by c z
that is what is happened.
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00:33:46,669 --> 00:33:50,650
Now, what is c z we will come to that but
it means is that it is happening so slow.
232
00:33:50,650 --> 00:33:54,660
In fact, we will find that the phase, also
phase gap become zero with respect to exciting
233
00:33:54,660 --> 00:34:00,299
force, this will simply go up and down with
the water. Means if water goes up body goes
234
00:34:00,299 --> 00:34:05,780
the water comes on body comes down it will
simply the wave with hydrostatic force.
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00:34:05,780 --> 00:34:12,780
So that means, every time if you stand here
you are going to find your water surface a
236
00:34:13,869 --> 00:34:18,059
free boat, same the free boat is going to
remain the same. You do not do not even feel
237
00:34:18,059 --> 00:34:21,740
that you are moving, because first of all
it is moving very fast. So, that is what is
238
00:34:21,740 --> 00:34:25,919
happened in tsunami type of wave very long
wave, you do not even know that there is wave
239
00:34:25,919 --> 00:34:30,409
coming. Because you have actually you have
gone up and come down with the water.
240
00:34:30,409 --> 00:34:34,970
So, as you stand you always look at the water,
you you do not feel you are moving. First
241
00:34:34,970 --> 00:34:41,049
of all omega is small means t e is very large.
So, it takes all the time just image it takes
242
00:34:41,049 --> 00:34:46,489
10 minutes to go up 1 meter, so how would
you feel it. So this is one extreme of this
243
00:34:46,489 --> 00:34:52,729
graph, you know this part and we can call
this region to be hydrostatic motion. That
244
00:34:52,729 --> 00:34:59,019
is this side of motion is dominated by hydrostatic
force, it is a hydrostatic force that dominate.
245
00:34:59,019 --> 00:35:04,269
You can make all kind of mistake in a z, b
z does not matter, do not even bother very
246
00:35:04,269 --> 00:35:10,190
low frequency the body is going to go up and
down. So, I can make 100 percent a z in estimation
247
00:35:10,190 --> 00:35:15,069
does not matter I still get a good estimation.
This is why the small boat in large waves
248
00:35:15,069 --> 00:35:21,239
the simply ride the wave, so this is what
we call hydrostatic dominated motion. Now
249
00:35:21,239 --> 00:35:25,789
coming to this other side, this is this side
now also we have see, now let us look at this
250
00:35:25,789 --> 00:35:28,130
side, high frequency side.
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00:35:28,130 --> 00:35:33,369
Now here again if I look at this equation,
see it was omega e square into something that
252
00:35:33,369 --> 00:35:40,369
was my inertia force plus I omega, say it
is minus here into something plus c z into
253
00:35:41,380 --> 00:35:47,729
something equal to force. Now you see when
omega is very high, high frequency motion
254
00:35:47,729 --> 00:35:54,729
is very fast. What happens this becomes the
much more dominating force this also becomes
255
00:35:56,529 --> 00:36:02,200
fairly dominating force this becomes compatibly
less important. So, what happen is that it
256
00:36:02,200 --> 00:36:06,630
become inertia dominated and you’ll find
out that, the force of this and force of this
257
00:36:06,630 --> 00:36:10,999
is opposite in sign. As a result there’s
a phase gap, I did not put the phase gap.
258
00:36:10,999 --> 00:36:14,910
But if you are to put a phase gap between
them theoretically, there will be a change
259
00:36:14,910 --> 00:36:21,910
of ninety degree here the phase gap. That
means, look at this earlier I had got c z
260
00:36:22,430 --> 00:36:26,440
into something is F, now I have got minus
omega square minus something into z equal
261
00:36:26,440 --> 00:36:31,729
to f. Obviously, the z a in this case if it
was like that in the other case it is going
262
00:36:31,729 --> 00:36:36,640
to be just opposite.
So, what is happening this is very important,
263
00:36:36,640 --> 00:36:43,640
in this case in the first case my boat was
riding this. But now in the lower end my boat
264
00:36:44,670 --> 00:36:50,479
is out of phase with that that means, when
the wave is rising, it is sinking it is more
265
00:36:50,479 --> 00:36:55,319
dangerous. You see that is what I wanted to
tell you from physical point of view. That
266
00:36:55,319 --> 00:37:00,329
means, if I were to draw this here now in
this case when the water is rising my boat
267
00:37:00,329 --> 00:37:04,279
is sinking.
Here when water is this thing my boat is rising.
268
00:37:04,279 --> 00:37:11,279
So, you see I end up getting an oscillation
related to the water surface much more pronounce
269
00:37:11,460 --> 00:37:16,869
which is of course more important. Because
what happens just imagine this case, the free
270
00:37:16,869 --> 00:37:20,579
surface is rising the way this crest but my
ship is going down. So, there is a chance
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00:37:20,579 --> 00:37:27,579
of water my my free body is reducing.
So inertia dominated motion typically have
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00:37:28,329 --> 00:37:35,329
a responses which is out of phase with the
forcing of the waves typically. Now the third
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00:37:38,700 --> 00:37:45,299
thing that is important is the other part
that is damping part. Now this this part now
274
00:37:45,299 --> 00:37:52,299
this part this small part here this values
peak values depends purely on b or mu k value,
275
00:37:53,150 --> 00:37:58,049
k is nothing but damping. You will find out
that if you get k 0 is infinity, if you get
276
00:37:58,049 --> 00:38:01,799
k with 0.001 it will come down to final value
drastically
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00:38:01,799 --> 00:38:08,799
So, this is extremely sensitive on damping.
So, this part inside part what we can call
278
00:38:09,489 --> 00:38:16,309
damping dominated motion but it is very small
part only. Because really speaking estimating
279
00:38:16,309 --> 00:38:21,339
this amplitude requires a good estimate of
damping b very good estimate of damping b
280
00:38:21,339 --> 00:38:26,809
and damping is always very difficult to estimate.
But we can call it damping dominate and in
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00:38:26,809 --> 00:38:32,660
fact that theoretically this is where there
was a change of phase of you knows like that
282
00:38:32,660 --> 00:38:38,440
e z value, e z value actually undergoes a
phase change in this case. So, this is the
283
00:38:38,440 --> 00:38:41,970
kind of modes of motion that we find.
284
00:38:41,970 --> 00:38:48,430
So, we will find out therefore, that you see
three cases, one is high frequency wave there
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00:38:48,430 --> 00:38:55,430
is a it is going like that it
is in this case sinking more and here rising
more. You know this gap is changing go to
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00:39:06,960 --> 00:39:13,960
this long wave it is always riding the wave,
it is always having the same. That means,
287
00:39:18,460 --> 00:39:24,569
this free boat part remains constant more
or less constant so and it is very easy to
288
00:39:24,569 --> 00:39:31,009
study this motion. Because I can I do not
have to actually determine a z, b z etcetera,
289
00:39:31,009 --> 00:39:38,009
again if I were to look at this equation of
motion. In fact, what would have happened
290
00:39:41,259 --> 00:39:45,599
is that you know that even this part this
has got two part. Actually this was if you
291
00:39:45,599 --> 00:39:52,599
know the inertia and I said that in the first
class it is incident with and scattered with
292
00:39:53,749 --> 00:39:59,479
together combines and gives this F e, this
F e in principal.
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00:39:59,479 --> 00:40:06,479
Now, at very low frequency the body is going
to disturb the waves very less, so scattering
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00:40:09,789 --> 00:40:16,789
also is very small. Because very long wave,
the body simply goes up and down since it
295
00:40:16,849 --> 00:40:21,039
goes up and down, it does not you know like
dissect or it is not going to disrupt the
296
00:40:21,039 --> 00:40:24,049
wave field. Even if the body was not there
wave will looks exactly the same. So, this
297
00:40:24,049 --> 00:40:28,670
becomes very small as such because these are
also proportionate to omega e actually.
298
00:40:28,670 --> 00:40:33,700
So, you end up getting only this part very
easy to find out, very easy to figure out,
299
00:40:33,700 --> 00:40:40,700
no problem at all. And let us say you made
a you are doing a solution, you made a mistake
300
00:40:41,180 --> 00:40:46,829
in this estimate which is of course, is very
usual to do. Because hydrodynamics is on this
301
00:40:46,829 --> 00:40:51,739
and on this, this, this and this. So, in static
dominated force there is no hydrodynamics
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00:40:51,739 --> 00:40:57,049
practically, you can make all mistake in hydrodynamics
and you end up getting a good estimate.
303
00:40:57,049 --> 00:41:02,160
In inertia dominated force which happens to
be more critical in some sense because there
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00:41:02,160 --> 00:41:07,380
is a out of phase. You unfortunately require
to have a good estimate of this force and
305
00:41:07,380 --> 00:41:13,809
this force, and this force. Because it is
the hydrodynamic forces that decide inertia
306
00:41:13,809 --> 00:41:20,789
dominated motion. And it is inertia dominated
motion, where typically I have a tendency
307
00:41:20,789 --> 00:41:25,180
over free boat to change.
So so therefore, this is a situation where
308
00:41:25,180 --> 00:41:32,180
things that is critical I might demand on
calculating a z, b z and F a s dynamics correctly
309
00:41:33,390 --> 00:41:40,390
is also become more important. See however,
opposite that the critical scenario, I have
310
00:41:40,440 --> 00:41:45,279
less hydrodynamics, I could have afforded
to have bad or you know or poor estimate of
311
00:41:45,279 --> 00:41:49,460
hydrodynamics. But this is a situation here
that you cannot ignore hydrodynamics. Therefore
312
00:41:49,460 --> 00:41:55,499
it is the hydrodynamic forces that are more
important for inertia dominated this thing
313
00:41:55,499 --> 00:42:02,499
and also for damping, if you look at that
for that damping part. So, this part is extremely
314
00:42:02,920 --> 00:42:05,019
important for us to understand.
315
00:42:05,019 --> 00:42:12,019
So, again if I were to look back this, I just
draw it here again a typical this. So, this
316
00:42:13,920 --> 00:42:20,650
once again I call this z a equal to something
right, you know like z a equal to mu into
317
00:42:20,650 --> 00:42:27,650
well z s t and I plotted only mu here. Whichever
you plot mu becomes 1 means it is hydrostatic
318
00:42:27,789 --> 00:42:34,789
force etcetera. So, this edge lower frequency
range is once again, I tell you hydrostatic
319
00:42:37,739 --> 00:42:44,739
dominated, this side inertia dominated, this
is damping dominated but this can be narrowed
320
00:42:52,690 --> 00:42:59,690
down to a small part, this damping. Now you
tell me we will try to tell from physical
321
00:43:01,680 --> 00:43:07,009
point of view, what is the importance now
look at that. Typically, you know you’ll
322
00:43:07,009 --> 00:43:13,739
find out that or we will find out afterwards
from the point of view of wave length.
323
00:43:13,739 --> 00:43:19,759
What is the wave length in open ocean, we
have seen it may be something like may be
324
00:43:19,759 --> 00:43:26,710
50 meter going to say 500 meter something
like that. If you if you look at that it will
325
00:43:26,710 --> 00:43:33,489
be of this range in fact, lambda by L if L
is ship length may range some value depending
326
00:43:33,489 --> 00:43:40,489
on the L of course. Now what happen is, if
I take a boat small boat of 30 meter or 40
327
00:43:42,710 --> 00:43:48,279
meter let us say 50 meter boat. Then all the
waves are existing which appeared to be long
328
00:43:48,279 --> 00:43:53,219
with respect to the boat because lambda by
small h is 1 and goes to 10.
329
00:43:53,219 --> 00:43:58,569
So, relatively what would happen the boat
that have a tendency to follow the wave. So,
330
00:43:58,569 --> 00:44:04,380
mostly if I have a small boat, then we are
looking at always relatively high frequency
331
00:44:04,380 --> 00:44:10,729
motion, why I say relatively because again
this response will remain remember omega e
332
00:44:10,729 --> 00:44:17,170
by omega z. So, omega z comes in picture,
omega z will of course become low this I will
333
00:44:17,170 --> 00:44:22,700
come later on. But let us say if the shape
is this and if the waves are much longer with
334
00:44:22,700 --> 00:44:24,930
respect to that then you are afraid to go
up and down.
335
00:44:24,930 --> 00:44:29,880
But if the shape is much longer with respect
to this. Then obviously, what happens is that
336
00:44:29,880 --> 00:44:36,880
you have a situation, where you have got essentially
a relatively high frequency wave. Now I can
337
00:44:38,900 --> 00:44:44,660
relate omega is lambda, remember omega e can
be relate to omega a into omega g k lambda
338
00:44:44,660 --> 00:44:51,660
omega e, can be related to omega wave, can
be relate to lambda wave. So, I have omega
339
00:44:52,589 --> 00:44:59,589
is lambda wave by L ship. So, I can support
some of this make it in order of this, I can
340
00:44:59,940 --> 00:45:05,160
relate that in some sense it is order of wave
length by ship length. Now what is happening
341
00:45:05,160 --> 00:45:12,160
obviously, when wave length by ship length,
lambda by L is actually lambda is small, it
342
00:45:12,969 --> 00:45:19,969
is small means omega e is high. When lambda
lambda is larger long wave means omega e is
343
00:45:25,749 --> 00:45:31,319
of course small, peroid is large.
You know that is what typically is the question
344
00:45:31,319 --> 00:45:37,390
of lambda by e, now you see for a given lambda
if L is small then obviously, what will happen
345
00:45:37,390 --> 00:45:43,319
for a given lambda L is small, this value
becomes large. That means, small shapes feel
346
00:45:43,319 --> 00:45:50,319
the waves to be large, large ships will fill
the waves small. Now going by the open ocean
347
00:45:50,599 --> 00:45:57,599
52 meter wave to 500 wave length obviously,
what happen if a 50 meter one ship fills 500
348
00:45:57,809 --> 00:46:02,789
long ship as 10 times its length. But a 500
meter long ship or a typical 200 meter long
349
00:46:02,789 --> 00:46:08,940
ship a ship typically we have got say 100
to say 300 meter. L ocean goings are something
350
00:46:08,940 --> 00:46:14,339
like that a typical ocean going ship is something
like 100 meter, 200 meter, 300 meter. You
351
00:46:14,339 --> 00:46:19,859
know that a v L c may of this length, a typical
small cargo ship also of this length.
352
00:46:19,859 --> 00:46:24,940
So for this once obviously, what happen say
300 meter long ship, there are waves which
353
00:46:24,940 --> 00:46:31,579
is 200 meter that is going to be small. And
therefore for actual practical shape, we are
354
00:46:31,579 --> 00:46:38,229
in this region for large ocean going ships,
you see I cannot avoid studying hydrodynamics.
355
00:46:38,229 --> 00:46:42,710
I am trying to emphasize the importance of
that. That if you look at large ocean going
356
00:46:42,710 --> 00:46:49,710
ships, then large ocean going ships would
encounter waves where the encounter frequency
357
00:46:50,200 --> 00:46:57,200
is moderately high and therefore, inertia
force dominates as an example you will find
358
00:46:57,749 --> 00:46:58,089
out
359
00:46:58,089 --> 00:47:03,380
These example if you ever go to port you will
find out there is a large ship. If you see
360
00:47:03,380 --> 00:47:10,380
there are some small ripples and I you used
to always wonder personally myself, that in
361
00:47:10,420 --> 00:47:15,700
such cases the boat does not move up and down.
You you think that and there is a small boat
362
00:47:15,700 --> 00:47:20,190
next to it, you’ll find out that as if the
boat is grounded. You know I used to feel
363
00:47:20,190 --> 00:47:24,569
wondering that this boat since grounded, you
know these waves are flashing by and ship
364
00:47:24,569 --> 00:47:31,019
does not move a small boat here is slashing
up and down. So that means, what in height
365
00:47:31,019 --> 00:47:36,469
this is, so small it is actually so high frequency
the inertia. In fact, it is something in this
366
00:47:36,469 --> 00:47:41,690
region for this boat that it does not have
any force to move up and down at all.
367
00:47:41,690 --> 00:47:48,690
But slightly more and something like that
this will now begin to up and down. And of
368
00:47:49,660 --> 00:47:54,999
course, in open ocean these things always
happens and in this case these motions are
369
00:47:54,999 --> 00:47:59,630
this region. We are shifting to this region
remember for 300 meter long ship to go to
370
00:47:59,630 --> 00:48:06,630
this region, I need 1 kilometer long like
wave. That means, lambda by L to be 3, if
371
00:48:07,140 --> 00:48:14,140
L is 300 meter lambda is 900 meter but most
lambdas in open ocean are 300, 400 meter itself.
372
00:48:14,140 --> 00:48:19,369
So, I end up getting lambda by L to be around
0.75, 1, 1.5 time
373
00:48:19,369 --> 00:48:25,420
And in such cases I have inertia domination.
So, my point of saying is that large ocean
374
00:48:25,420 --> 00:48:32,420
going waves, you cannot avoid studying dynamics.
And in order to get dynamics, now I have this
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problem of having to estimate the dynamic
part of this. This part I have to estimate
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to some moderately you know acceptable degree.
In the other case I could ignore it completely
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I can even make it 0 also some very rough
estimate. If I met 100 percent in that I might
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have made 5 percent estimate in of motion.
In the in the low frequency range but here
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if I make 100 percent error, I might make
50 percent error in my motion, that is very
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important. That is that tells us that add
and mass damping and exciting force becomes
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an important parameter for us to study. You
me just cannot avoid it, now just last last
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few minutes of this class.
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Let me talk about this natural period, c z
well omega z equal to root over of c z by
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m plus a z to do that I first need to have
an estimate of c z. How what is c z, I will
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00:49:42,759 --> 00:49:47,569
talk about that but before that let me tell
you the difference between omega z and omega
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d. See omega d is omega z square minus mu
square mu is b by m plus a z 2. This mu is
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a very small number, typically for water it
is not you know water is not thick oil, water
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is one of the least viscous thing. Of course,
there is no viscosity also and this is actually
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damp radiation damping.
What is origin of that I will tell you later
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on in next classes when we discuss b and a.
But what I want to say here is this is small.
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So this is small means, this square is even
smaller. So, suppose this is point 1, this
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00:50:29,509 --> 00:50:34,989
I 0.01 numerically then what happened omega
d is approximately same as omega z. So in
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00:50:34,989 --> 00:50:41,989
all practical purpose, when we study natural
period and we say resonance as you know that,
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when omega e is omega z it is resonating what
is why we call that is this peak peak value.
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So, we can estimate the resonance simply taking
omega d to be s omega z. So, our importance
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become omega z rather omega d, which is also
very useful for me. Because if I had to do
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omega d, I had to get mu mu is to know b z,
b z is more difficult to estimate. So therefore,
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let me talk about this is very important because
if my omega e happens to be this, I am going
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to be very large heave motion right.
So, natural period becomes a very very very
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00:51:21,299 --> 00:51:28,299
important point, especially for off shore
structure and off shore structure. It is digging
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00:51:34,479 --> 00:51:41,479
oil from somewhere always stationary, if it
begins to heave very large you might stand
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00:51:42,479 --> 00:51:48,249
here and not feel but this oil, this line
is going to have this tension. So, you actually
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00:51:48,249 --> 00:51:55,119
cannot allow or cannot have oil explosion
going on if the heave exceeds some value say
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00:51:55,119 --> 00:51:59,609
3 meter or 2 meter or something other
So, you know if it is heaving very largely
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00:51:59,609 --> 00:52:03,259
you actually end up having these problems.
So heave becomes very important, when it will
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00:52:03,259 --> 00:52:10,259
have large if the natural period meets the
exciting period remembers. So, natural period
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00:52:10,479 --> 00:52:16,769
becomes very important very very important
from the point of view of study. Now for for
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this I need this now, what is c z we will
now try to tell about that you see the c z.
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00:52:23,160 --> 00:52:30,160
And I will just I think that, we talk about
this c z tomorrow’s class because we do
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00:52:30,900 --> 00:52:36,569
not have time but we understand here one thing
is that c z is hydrostatic restoring force.
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00:52:36,569 --> 00:52:42,680
So it purely comes from hydrostatic, there
is no hydrodynamics having said that hydrostatic
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00:52:42,680 --> 00:52:46,450
it is not exactly the boyancy force which
is what we will try to find out it is the
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00:52:46,450 --> 00:52:50,999
unbalanced hydrostatic force. And remember
that we have always called hydrostatic force
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00:52:50,999 --> 00:52:57,289
F s as something into z and this something
is this c z.
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00:52:57,289 --> 00:53:02,950
So, we will derive that c z expression tomorrow.
So, that we can estimate this value for practical
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structures and figure out which range they
are. Then we will find out very ironically
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00:53:08,479 --> 00:53:13,180
that, it turns out what typical shapes omega
z is actually in the range in which oceans
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exist, waves exist. So, we will get back to
that tomorrow