Module 12 – Lecture 1 – Systems with two degree of freedom


Till now we have been discussing the vibration
problems involving systems which could be described with only one coordinate that means
which possess single degree of freedom. However we encounter larger number of situations where
the number of independent coordinates required describing the systems configuration is more
than one. So therefore, it is necessary to discuss problems where systems have more than
a single degree of freedom. In this direction we will encounter some new concepts where
not present in single degree of freedom systems so before we take up discussions in detail.
First let me present certain basic ideas about these new concepts which will be involved
in place of vibration of systems with more than one degree of freedom.
So we start with the simplest next possible case that means vibration of systems with
two degrees of freedom. Of course, we should also keep in mind that we are discussing undamped
vibration. As just now mentioned that, this system will require two independent coordinates
to describe the systems configuration, it can be in many ways. Maybe, it can be in the
form of two blocks having vibration. It can be seen easily that this system requires two
independent quantities x1 and x2, the displacement of the two masses forms their respective equilibrium
position. Similarly, in case of angular motion we can
have double and in another case of two degree freedom as you can see the displacements from
the equilibrium position. They are represented by two angular rotations of the 2 rods of
length l 1 and l 2, the displacement coordinates are theta1 and theta2. Sometimes, a system
with a two degree freedom system may not involve more than 1 lump of body.
For example, you take a system with a single body m. So, let us consider this body is supported
by three strings. Mass of the body is m and let the body be confined to the plane of the
o. So in that case, again a complete description of the location of this mass will require
two coordinates x and y and therefore again the system will have two degrees of freedom.
Similarly, in engineering we encounter situations where a body is resiliently supported. So, in this case the body�s motion again
is confined to be black board. Maybe, body is the center of mass of the body can have
only vertical motion, but again it can have a tilt. So x and theta will be the two coordinates
here, x and y were the two coordinates. Here, theta1 and theta2 are the two coordinates
required here x1 and x2 are the other two coordinates. So these are some examples of
two degree of freedom systems. Now there are some fundamental points. Let us understand
at the beginning, one thing that we are considering random system to begin with, and I will refer
to situation when damping exists briefly, after we discuss this.
Now one thing we have seen a single degree freedom system, if we disturb the system from
its equilibrium position, it executes simple harmonic motion invariably, without any problem.
In case two degree freedom system suppose, this one I disturb somehow within the bodies
execute simple harmonic motion. The answer is no that is a very fundamental difference
with the single degree freedom of system that means the oscillation, free oscillation of
this will not be necessarily a simple harmonic motion, if the displacement is given in an
arbitrary manner. On the other hand, it is possible to give
the initial condition to start the motion of a two degree freedom system, in such a
way that both the bodies will execute simple harmonic motion and we will see in the case
when one body is executing simple harmonic motion, both the bodies are executing simple
harmonic motion the frequency of that oscillation for both the bodies are same. Phase difference
can be there that means, they can be either in phase or out of phase that we will see
later. But the frequency of oscillations will be same and both will execute simple harmonic
motion only for some specific cases where the initial conditions are adjusted properly.
So therefore, there is a special situation which exists in cases, where more than one
degree of freedom exist that, the vibration is not necessarily simple harmonic if it is
disturbed from the equilibrium position in an arbitrary manner.
On the other hand, just now as I mentioned if we disturb it in a particular way satisfying
certain conditions then the system will execute simple harmonic motion both the bodies at
the same frequency. Now this type of motion that means, where the system is executing
simple harmonic motion with the same frequency all the parts of the body are called either
natural mode of vibration or normal mode of vibration, both terms are used.
So this type of vibration where the whole system executes simple harmonic motion with
the same frequency, that particular type of vibration is called natural mode vibration
or normal mode vibration and the corresponding frequencies are called natural frequency.
Next question will be that how many such natural mode vibration will exist and how many corresponding
natural frequencies will exist. Now that will be exactly equal to number of degrees freedom.
So natural frequency omegan will have number of natural frequency which will be the number
of degrees of freedom. In this case, when you are considering two
degrees of freedom system will have two natural frequencies and two natural mode of vibration.
Now let us start investigating one, two degree freedom system of this lump parameter. We
need not put the other this is also a two degree freedom system and really speaking
we have not lost much in the generality of the case, if we want we can put, only it will
complicate the expressions more we did all the time. This is a two degree of freedom
system and the equations of motion, you can write down, this body in displaced condition
will be here, this will be here, the forces on this, there will be only one force. Here
there will be a force in this direction. Same magnitude and
these are the few diagrams and so the corresponding equations will be, and reorganizing it will
be m1. Now from this, we can express x2 in terms of x1 and x1 2 dot. So let us write
x2 in terms of x1 because this is case of simultaneous frequency equations two unknown
variables and, let us put in a solvable manner by eliminating 1m 1 x2 and x2 2 dot will be
the obviously. Now, if you substitute x2 and x2 dot in this equation say this equation
so then, what we will get finally. You can see now the actually this fourth order
equation. Next one if we take the solution of form. The characteristic equation will
become s to the power 4 plus, and the solution to this will provide the 2 omega squared values
and those two will be the natural frequencies. So if we want to solve these, it can be solved.
The two solutions, two natural frequencies, give the two natural frequencies omegan1 square,
omega n2 square and expressions will be omega n1 the whole square equal to and omega n2
the whole square. Now the solution will require the initial
condition, the initial conditions what we can do we can provide just generalized equation
that means, at t is equal to 0. x1 is x1 0 and the velocity is x1 dot 0. x2 is x2 0 and
the velocity x2 dot is 0. These are the four initial conditions then the complete solution
can be written, I think it is more of an algebra so, what we will do, I will give the general
equation in the final form. Now the reason that I am writing this long expression is
to emphasize one point which will be clear at a later time. So this is the solution for
x1 and we will get a similar solution for x2, it is required for us to write here, otherwise
the point will not be clear. So, these are the general solution given for a simple case
like two degree of freedom system undamped free vibration.
Now it is very clear that they are all mixed up that means two frequencies are involved
omega n1 omega n2 and obviously x1 or x2. None of them are simple harmonic. However,
we can see that it is possible to give initial conditions that is x1 0 x2 0, x1 dot 0 x2
dot 0 in such a manner that the whole system oscillates in a harmonic way. That is all quantities vary in a simple harmonic
way. What is that condition let us see. Before example let both x1 dot 0 and x2 dot 0 be
equal to 0 that is there was no initial velocity given immediately the sine terms go, next
we give an x1 0 by x2 0. So all these quantities are known omega n2 we have already found out,
k2 k1 m1 everything is known. So this quantity is obviously a known quantity. If we keep
our initial displacements in this ratio, then what is going to happen, we will find that
this term x1 by x we will find that this term is going to be 0 and similarly if we substitute
this here we can easily show I am avoiding the algebra that this will be all clear. So
therefore the solutions will be then, so both x1 and x2 are simple harmonic functions of
time with frequency omega n1 that means it is in natural mode oscillation. But this mode of oscillations we have to give
the initial disturbance like this and we can show that all along, this x1 by x2. This ratio
will remain same as this and that is minus k2 by m1. So, whatever ratio I gave that ratio
remains. So this is the first block, this is
the second block and displacement is in this direction so we will find. They vibrate like
this that means, always the ratio of the two will be quantities that means, the shape is
given by the two ratio x1. So this is the first mode and so this is nothing but x1 in
the first mode by x2 of the first and this is the first mode shape, the frequency is
omega n. In a similar way, we could also excite the second mode. How? There if x1 0 by x2
0 we select as, then you will find that this term is 0. We can substitute and verify. Then
again, we are getting same that means, the vibration of the whole system x1 and x2 as
functions which will be simply proportional to cosine omega n2 t and that will be second
mode. They will be very similar expressions where
this will be cosine omega n2 t cosine omega n2 t and that mode will be again something
like this. So we will see that one goes in this direction, at the same time other goes
in this direction, so mode shape will be something like this, and the extreme positions will
be something like this all the time it will vibrate in the opposite. So the ratio x1 0
y x2 0 will be a quantities. Now, why we are studying the normal mode oscillation we should
understand first of all this is something which is mathematically practicable, because
all the variations are simple harmonic functions of time with same frequency.
But we should not also forget the fact that, it is always possible to write any kind of
displacement pattern of a system, in terms of its natural mode just like as an arbitrary
function can be expressed in the form of a series of simple harmonic functions, like
a series, here also it is quite easy to understand that if we give that any kind of oscillation
can be always represented by a combination of its natural mode oscillation.
So, since a general oscillation can be expressed in terms of the simple harmonic natural mode
oscillations, we will investigate only natural mode oscillations and again as we have done
in case of single degree freedom system. Our primary objective will be to find out the
natural frequencies and one extra change in such cases we have to also find out the mode
shapes. Mode shapes means the ratio of the various amplitudes which will be same as ratio
of the various instantaneous values of the coordinates and as many natural frequencies
will have so many different ratio combinations will have.
Let us solve a particular example, of the case as we discussed freedom without any damping.
What happens when damping is present then various modes which are there in the vibration
of the system? The higher modes damped oscillates because, they execute larger number of cycles
at the same time corresponding to the lower frequency mode.
Therefore, different modes dissipate or decayed at different planes that way the concept of
mode itself become a very difficult thing to sustain in such situations and we will
ignore damping. So, it is a system with two blocks and three springs, they are all identical.
This is the first body, this is the second body and their displacements are represented
with the help of x1 and x2. The equations of motion will be derived from the free body
diagram, this being the free body diagram, both is identical masses. The equations of
motion can be written as, for the second body, so we can rewrite this little bit k x2 minus
k x1 minus k x1 so it will be plus 2 k x1 minus k x2 and this one will be 2 k x2 minus
k x1. Now the technique of solving this set of homogeneous equations will be that since
we are studying only natural mode oscillations will assume that both x1 and x2 to be simple
harmonic functions of time with the same frequency so let x1 equal to X1 cosine omega t and x2
equal to X2 cosine omega. Now, as we know in undamped situations their
motion can be either in face or out of face and both can be taken care of by algebraic
sign of x1 and x2, if they are of same sign the motion of the whole body are in place,
if they are opposite signs or 100 to the power 0. Now if we substitute this in these equations what we get simply minus omega square m k
x1 plus 2 k x1 minus k x2 equal to 0. Because omega t everywhere is there and I can take
it out substituting x2 and x1 in the second equation or we can write it in this fashion,
2 k minus m k omega squared x1 minus k x2 equal to 0 minus k x1. So we get in place
of differential equations now what we have done. We have got a set of algebraic equations
both are homogeneous. So therefore, a solution will exist only when the determinant of the coefficients
is 0 and this will give us a characteristic equation. Now, as you can see here, we can
solve the whole equation and we can find out the mode shape so directly I can give the
solution also. So this equation we can definitely solve like this, or if we use this cross value
we will get a m omega squared if we use the minus value then obviously we get the two
natural frequencies omega n1 is equal to k by m and second natural frequency will be
3 k by m. So natural frequencies we have got. Now after we get the natural frequencies for
each natural frequency what will be the mode shape?
If we take this equation for example, if we substitute the value of omega we will get
the ratio that means for the omega n equal to k by m let that be the first mode. So for
this equation we can get x1 in the first mode by x2 will be equal to k divided by 2 k minus
m omega squared. Now this omega squared will be k by m. When omega n2 is equal to 3 k by
m that is the second mode, then we substitute it here x1 of the second mode by x2 equal
to k by 2 k minus m to 3 k because omega n2. So this is simply minus 1 so second mode shape
is the ratio is equal to minus 1 that means they are just equal and opposite here they
are same. Now if we plot the situation how it will look?
Like this is the first body, this is the second body. Let us plot the longitudinal oscillation
in this direction or representation. So x is not in this direction but I am plotting
the displacement of the mass in this direction. So for the first mode what will be the amplitude
and also that means the x1 by x2 is always equal. In the second mode obviously we find
if x1 is here, x2 will be equal and opposite, that will be the mode shape. Remember again
that, it is only these two bodies are moving like this because movement in this direction.
So let us try to, this being a very rather fairly simple case, whether it is possible
to physical explain the results. One possible way let us consider its oscillation,
if they oscillate the same frequency same phase and same amplitude what will happen
in that case? This spring will have no distortion, because the movement of the two masses are
identical at every instant, so distortion being 0 there will be no force.
So then it is as good as this spring not being there and so the system will be equivalent
to and the natural frequency obviously is square root of k by m this is nothing but
2 simple schema systems of identical length. The other mode is that, they are just equal
and opposite, what happens if they are equal and opposite? That is, they are moving like
this, if they are moving like this then midpoint of the spring will have no and so therefore
we can consider the spring is hot and this there is a vault here this point is fixed.
So it becomes again two systems, with this mass one spring of thickness k and this half
of this spring’s thickness is obviously the double because if we make the length of a
spring as half then its thickness is because two halves of springs, if this is k prime
and this is k prime together in series they should give us k and you know when things
are in series. So, k will be equal to as you can see k prime,
k prime by 2. So when k prime will be therefore, if this spring length is half, each half will
have stiffness which is double the stiffness of the original, so this system is equivalent
to now, that means if you total resultant stiffness with mass, it has to overcome to
move is 3 k and therefore omega n2 is going to be natural frequency of this system. There
is another subsystem this side, that we also have seen we will see that physically also
we can explain that there are two situations when the two masses will have the same frequency
harmonic motion and how the physical expansion can be given in this particular case. However
this kind of physical insight may not be possible in more complicated situations and we have
to depend on the mathematical results. Now if there is but in practice one message
that there is no currently that will be motion which will be handling will be simple natural
mode oscillation. But as I mentioned that a general motion can be represented in terms
of natural modes, and we just look into the matter natural or normal mode whatever you
may say. So if a two degree freedom system there will
be two natural modes and general solution. x1 t can be written as we know of course that
x1 1 by x2 1 is a quantity lambda1 and mode shape x1 in second mode will be quantity,
then of course we can write this in terms of, and these equations are there. So how
many unknown quantities will be handling will be because these things will be known after
solving the natural mode oscillations is lambda1 and lambda2 will be known to us so there are
not unknown quantities. Unknown quantities will be x2 1 x2 2 phi1 and phi2. So therefore
x2 and x1 can be written in terms of this four unknown quantities and this four unknown
quantities x2 1 x2 2 phi1 and phi2 can be from initial conditions. There will be four
initial conditions.

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