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Last lecture, we studied free vibration problems. Specifically, we studied the behavior of two
systems.
One is a damped natural vibration and another is undamped vibration. In damped one, the
damping is present and in undamped, natural vibration damping is not present. Now, we
specifically took two systems. This is the spring mass system. We have studied that problem
and then another model was that in which we attached a dashpot. We attach a dashpot here
and we put like this. Here, if we make the free body diagram of this, this was having
kx, the force. Therefore, governing equation was naturally, mx double dot is equal to minus
kx. Whereas in this system, this is mass m and this is kx. This will basically be cx
dot. Therefore, governing equation was mx dot is equal to minus kx minus cx dot. No
external force term was present.
In this lecture, today, we are going to study forced vibrations with damping and without
damping. Vibrations that take place under the excitation of external forces are called
forced vibrations. When we put that in this one, when the excitation is oscillatory, the
system is forced to vibrate at the excitation frequency. What happens, when we apply a force
on a spring mass system, suppose we know a mass is there and it is held by a spring then
we displace the mass. Naturally, we have created a disturbance from its equilibrium position.
So, the natural vibrations may start, but the natural vibrations that is free vibrations,
they will decay with time because of the presence of the damping.
If damping is not there, then of course these free vibrations will not decay, but because
there is a damping, these decay with time. In any system, damping is always present.
There is no system in which there is no damping. All systems, because undamped case is just
an idealization, therefore what happens? These vibrations will decay, but because there is
a force present that keeps vibrating, we are mostly interested in the steady state response.
Then the system will vibrate with the excitation frequency. If the frequency of excitation
coincides with one of the natural frequencies of the system then we get a condition of resonance
in which dangerously large oscillations may result. Large amount of oscillations take
place when there is a condition of the resonance. Now, we are going to study the mathematical
treatment of that forced vibration. French mathematician J Fourier, who was born in 1768
and died in 1830, showed that any periodic motion can be represented by a series of sine’s
and cosines that are harmonically related. That means you have a periodic function, suppose
you have a periodic function fx, it can be represented by a0 plus a1 sin omega x plus
a2 sin 2 omega x like that, there may be so many terms upto infinite. Then you have b1
cos omega x plus b2 cos 2omega x like that. So, cosine and sin terms are there. In a general
case, that Fourier, we will not discuss about the Fourier series method.
However, we know that if we have any periodic motion that can be represented by a series
of sine’s and cosine’s. Therefore, the study of forced vibration under the action
of a harmonic excitation force is very important, because suppose if we can find out the expressions
for that what happens? When the force f is equal to a1 sin omega t then we can take another
force f is equal to another a2 sin 2 omega t. Therefore, under the action of combined
terms, the response will also be combined, because this system if we take k and m and
even damping coefficient c as a constant, then this is a linear system. That means,
we can simply find out the displacement by superposition under the two circumstances.
Therefore, let us try to study that. This is mass then this is a spring then this is
dashpot and this is x. There is a force F acting here. Now, this will be F0 sin omega
t. Let us assume that F is equal to F0 sin omega t that means you have got a sinusoidal
forcing function. This is kx then this is cx dot; kx is the restoring force and kx is
this one. Therefore what happens? Now this is the free body diagram of mass m. There
is a force F0 sin omega t. This is kx and this is cx dot.
Then consider the spring-mass-dashpot system. The mass is displaced by distance x and its
free body diagram is as shown. From this figure, we see that the equation of motion is mx dot
dot plus because the equation is basically mx dot is equal to minus kx minus cx dot.
Therefore, it can be written as plus F0 sin omega t. Therefore, this can be written as
mx dot dot plus kx plus cx dot is equal to F0 sin omega t.
Solution to this equation consists of two parts; one is called the complementary function,
which is the solution of the homogeneous equation. That means, if we solve mx double dot plus
cx dot plus kx equal to 0, we will get one solution that is called solution of the homogeneous
equation. This equation mx dot plus cx dot plus kx is equal to 0 is called homogenous,
because if x equal to a, is the solution then x is equal to constant times a, is also a
solution. That means, if one solution is x then naturally cx is also a solution. So,
it is totally homogeneous that it is called complementary function. The other one is that
particular integral that depends upon the term on the right hand side. In this case,
it is F0 sin omega t. So, the combined motion is basically just the combination of complementary
function and particular integral. The physical significance of this is that
when we displace a mass by some force, then combined effect which we get, is the sum of
these two things; that means, natural vibrations, that mass particle starts vibrating in natural
mode and then forcing term vibrations are also present. Therefore, the motion is the
combination of these two. The particular solution in the equation is a steady-state oscillation
of the same frequency omega as that of the excitation.
We can assume particular solution to form this one. We just assume, let us see that
what happens? x is equal to let us say X sin omega t minus
phi, where X is the amplitude of oscillation and phi is the phase of the displacement with
respect to the excitation force. If you substitute this expression in the equation of motion,
you get like this; minus m omega square X sin omega t minus phi, because X sin omega
t minus phi has been differentiated two times. So, you get minus m omega square X sin omega
t minus phi plus c omega X cos omega t minus phi. It has been differentiated one time plus
kx sin omega t minus phi is equal to F0 sin omega t, or k minus m omega square sin omega
t minus phi plus c omega cos omega t minus phi into x is equal to F0 sin omega t, that
expression we got. We have to find out the expression for x.
If we carryout trigonometry here, can I simplify this expression, k minus m omega squared sin
omega t minus phi. If we just see that if I can represent them in trigonometric form,
if this is theta and if this is k minus m omega square and this is c omega and this
is under root of k minus m omega square plus omega square plus c square omega square. Finally,
we can get sin theta type of term.
If we divide this whole thing by k minus m omega square divided by that thing this we
are dividing this thing by this expression that means square of these two terms this
and this then take the under root, we divide by sin omega t. Similarly, here c omega t
and here this is cos omega t minus phi, like that multiplied; since we divided by under
root k minus m omega square plus c square omega square. Therefore, we have to multiply
it also by this thing then this expression remains same. Therefore, the previous expression
and this expression is basically same. This is equal to F0 sin omega t.
We can take this common. Here k minus m omega square whole square plus c square omega square
multiplied by cos theta, because this can be written cos theta and this is sin theta
cos omega x is equal to F0 sin omega t where theta is basically like this.
This is k minus m omega square and this is c omega. Therefore, by Pythagoras theorem,
this is k minus m omega square plus c square omega square. Simplifying this expression,
we may write this as sin omega t plus phi, because this is the formula of sin omega t
plus phi. This one cos omega t minus theta this is sin omega t plus theta minus phi basically,
because this is a b. So, sin omega t minus phi plus theta. Apply the formula of sin a
plus b; so, sin a cos b. So, sin omega t minus phi cos theta sin cos omega t minus phi and
this is theta. So, this is sin omega t minus phi cos theta cos omega t minus phi sin theta.
So this is like this. In this expression this is omega t this is
equal to like that so sin omega t minus phi plus theta is equal to F0 sin omega t. We
have obtained this expression.
If we compare the right and left hand side of this thing, sin omega t, this has to be
equal. So, omega t plus theta minus phi should be equal to 0.
That means we have to say, where X equal to F0 by this one and omega t plus theta minus
phi is equal to omega t. This condition has to be because for all t, if it has to be correct
then this relation has to be satisfied which gives phi is equal to theta; that means, which
is equal to basically, tan inverse c omega k minus m omega square. We can say tan theta
is equal to c omega divided by this thing. So, tan omega theta is equal to tan inverse
c omega k minus m omega square. We can write X and phi. So, X is equal to F0 under root
k minus m omega square plus c omega square.
We can write X as F0 by k divided by k. So, 1 minus m omega square by k whole square plus
c omega by k whole square and c omega by k 1 minus m omega square by k. It is to be noted
that F0 by k is the static deflection of the spring under the action of that force F0 k.
Maximum amplitude of the force is F0. So, if you apply maximum force then you will get
the static deflection as F0 by k.
We know that omegan is equal to under root k by m that is the natural frequency of undamped
oscillation. Now, Cc is equal to 2m times omegan which is called critical damping and
zeta is equal to C by Cc, that is called damping factor. We can write c omega by k as c by
Cc Cc omega by k, that means 2 zeta omega by omegan, because Cc is equal to 2 Cc by
Cc 2g. This Cc is equal to 2 m into omegan and m by k is basically 1 by omegan square.
We just do that simple. So, we will keep C omega by k is 2 zi where zi is damping factor
omega by omegan.
This gives us X divided by F0 by k is equal to 1 under root divided by under root 1 minus
omega by omegan Whole Square whole square plus 2zeta omega by omegan whole square. This
is also zeta. This tan and tan phi is equal to 2zi omega by omegan 1 minus omega by omegan
Whole Square. Why we have written this in this form? As
we already mentioned, F0 by k is the maximum static deflection. X divided by F0 by k gives
the ratio of that dynamic amplitude to static amplitude or maximum static deflection. Similarly,
here we get phi, phi is the phase angle, because we started like this.
We assumed that the force was F0 sin omega t but our displacement was sin omega t minus
phi; that means, there was a phase difference between two. So, this gives the idea about
the phase difference.
In this case, if c is equal to 0 then there is a no phase difference. There is no damping
present, then but if there is a damping present then there is a phase difference between the
applied force and the response we get. So, that is because effect of damping when it
is like that, so that damping also introduces a force. Therefore, this comes X is equal
to F0 by k under root 1 minus m omega square by k whole square plus c omega by k whole
square and phi is equal to C omega by K whole divided by 1 minus m omega square by K. In
this, this is tan phi.
Using this, now notice that in this equation A, F0 by k is the static deflection and X
is the amplitude of vibration. So, amplitude of vibration depends naturally, what is the
frequency ratio? That means omega by omegan. What is the natural frequency? Therefore,
the ratio X by F0 by k is often called the magnification factor. How much times the dynamic
amplitude is more than the static deflection and depends on omega.
Now, plots of magnification factor versus omega by omegan have been plotted, we have
shown. Actually for three damping values that is; tau is equal to 0.01 that is very low
damping, tau is equal to 0.2 and tau is equal to 1. In this case, at omega by omegan equal
to 1. If damping is very low then a resonance type condition occurs. So, we see that the
magnification becomes under almost infinity. Otherwise also, nearby this region, near the
natural frequencies, the magnification factor becomes very high.
However, if we have critical damping then magnification factor is always less than 1;
that means if omega is very high compared to the natural frequency then the magnification
factor approaches zero. This is very interesting, that means at omega is equal to 0. That means
in static case, if you just apply some force then there will be displacement. That will
be basically X by F0 by k. Therefore, you will get the same type of thing, but if we
apply a very high frequency then displacement approaches 0. Why? Because the mass is not
able to respond, because of the dashpot it is not able to move at all. Therefore, there approach is 0.
Similarly, but when tau is equal to 0.2 then at resonance it will be very high, but of
course, it will not be infinite. If there is a damping present in the system, it will
not allow the vibration to become infinite, of course it becomes very high. Similarly,
in the case of the low damping, it is like this.
When the damping is very small, the magnification factor is very large in the vicinity of natural
frequency. Hence, the forcing function should not be closed to natural frequency, if the
vibrations have to be avoided. For very large value of omega by omegan, the magnification
factor becomes very small. Hence, the operating or forcing frequency should be four to five
times the natural frequency, for best results. That means, if you have the natural frequency
of some system is 100 hertz. If you have the forcing function frequency at 400 hertz then
you know that it will not. We might have noticed these type of examples, like a fan makes lot
of noise when it is run at a low rpm, but when the rpm is increased, then it does not
make that much noise because of this phenomena. Similarly, some cars also, when they run at
a low speed then they will show vibrations, but at high speed the vibrations are not seen.
For undamped forced vibration, now undamped, one damping factor zeta is equal to 0. Thus, we
get X is equal to F0 by k 1 minus omega by omega square and we get phi is equal to 0.
There is no phase difference between the forcing function and the response. Then we are getting
this type of expression. Here also, it is seen that if omega is very small then X is
equal to, of course F0 by k, but if omega is very large then this X will approach 0
because of this term F0 by k. So, this term will approach 0; so, F0 by k divided by minus
omega by omegan square. This term is approaching infinity. You get something like, here 0.
So, this is the interesting thing. If you have a spring mass system like this,
there is a very flexible spring. If you apply force of a very high frequency, this will
not be able to move that means it will not carryout any vibration.
Now, complementary solutions known as the transient solution, is of no special interest
since with time it dies out with a small amount of damping which can never be completely eliminated.
Therefore, we have not discussed about the complementary solution which are called transient
solution. This is of no special interest. However, in undamped case, one can study actually
and one can get some insight about the cases in which damping is almost 0.
Let us discuss, the other type of problem that is the problem involving the support
motion. What is this, that if you have kept this on a support, this is a fixed support
like this and this is mass and this is a spring and this is dashpot. Now, support itself is
getting excited. Supporting is being excited and support is undergoing the displacement
y and mass itself is undergoing the displacement x. This is the equilibrium position of the
mass and x is the displacement from this position.
Let y be the harmonic displacement of the support point. If we measure the displacement
x of the mass from an inertial reference, we can make from the free body diagram of
the mass. It is subjected to spring and viscous force. Equation of motion is like this. This
is mass, it is stretched. Suppose, you have pulled the mass by amount x, but the support
by that time also has moved by a distance y. So, net stretching of the spring is k(x-y).
Therefore, the spring will provide a force k(x-y) in this direction.
Similarly, the net velocity of the mass from inertial reference frame, means from outside
observer which is on a fixed frame, which is from that frame he is observing and he
says that net velocity of the mass is x dot minus y dot. This is the velocity that is
the relative velocity, because otherwise the absolute velocity of the particle is x dot;
mass m is x dot. Therefore, support itself is moving with y. Therefore, net velocity
is relative velocity between this thing is x dot y dot. That means dashpot is there.
So, dashpot’s piston is moving with x dot. Cylinder itself is moving with y dot. So,
difference is x dot minus y dot. Therefore, cx dot minus y dot is equal to
k(x-y). Now, put z is equal to x minus y. If we put z is equal to x, we get mx dot first,
by Newton’s law, mx dot is equal to minus k x minus y minus c x dot minus y dot. Now,
we are getting the terms x minus y. Therefore, better to write it like z, the y variable
gets reduced like that. You say z is equal to x minus y, we get mz
double prime. Here, we do not have any y term. So, if we put that here also, mx dot minus
y dot then since we have added here, therefore, from here we can subtract that thing. We can
add here also so it becomes minus my dot. Therefore, it becomes mz double dot plus cz
dot plus kz is equal to minus m y double dot, and this is y is equal to, if we assume that
support is getting excited by y is equal to y sin omega t. Therefore, this is minus m
omega square y sin omega t. So, you are getting this thing.
Its solution can be immediately written as z is equal to Z sin omega t minus phi. Because
this type of thing we have already done. This is like a force. Instead of the force F0 sin
omega t, we are having m omega square Y. So, m omega square Y is like a force. So, z is
equal to Z sin omega and Z is equal to m omega square by Y under root k minus m omega square
whole square plus C omega square and tan phi is equal to C omega k minus m omega square.
In this case, if the frequency ratio omega by omegan is large then Z by Y becomes equal to one for all values
of the damping ratio. Let us discuss this point in detail. We have seen m omega square
tan phi Z is equal to m omega square Y under root k minus m omega square plus C omega square.
Now, tan phi is equal to C omega k minus m omega square.
In this case, Z by Y is equal to one for all values of the damping ratio. How can we prove
this one? Let us write Z is equal to omega by omegan is large; that means, Z is equal
to m omega square Y divided by k minus m. So, we take that. Maybe we can take m square
common, so, m can be taken common here. This becomes k by m. k by m minus omega square
whole square plus C square omega square. This is equal to m omega square Y. This is m and
this will become k by m minus omega square omega square. So, this is under root omegan
square minus omega square whole square plus C square omega square.
Now, omega by omegan is large that means omegan is smaller, much smaller in fact in comparison
to omega square. Therefore, here omegan square minus omega square can be written as omega
square. So, this will become omega to the power 4 and this becomes m omega square Y
divided by m under root omega to the power 4 plus c square omega square. Since omega
is a very large number, omega whole term is very high in comparison to C square omega
square. First of all, damping is always less than
1 and omega. Therefore, what happens? This I can write as m omega square Y divided by
m omega square that is omega square. So, this is equal to Y. Therefore, Z is equal to Y.
Irrespective of damping Z is equal to Y if omega by omegan is large. That means, if the
support is being excited by a very high this one then Z will be equal to Y, but then what
happens, Z is equal to Y but Z is basically X minus Y. That means, we have said Z is equal
to X minus Y and so, that is equal to Y. So, what happens is that Z by Y is almost this
one. That means, the relative displacement is almost same as the support displacement.
Using the characteristic of this type of response, you know that we have got response and understanding
that we can design various types of systems. You know what is our primary interest? Sometimes,
you have to design the system for measuring the acceleration. Sometimes, you have to isolate
the vibrations. So, these types of things are there.
This briefly we have discussed. Let us discuss one or two other interesting things. This
factor, which you see here, F0 by k 1 minus omega by omegan square whole square, if omega
by omegan becomes equal to 1, this will become 0.
This factor is 1 by 1 minus omega by omegan square is called resonance factor, rho. We
can denote it by rho. Therefore, if omega is equal to 0, the resonance factor is 1.
If omega is equal to very large, resonance factor is 0.
If omega is equal to omegan resonance factor is infinity. We have seen that you perhaps
get the idea that at omega is equal to omegan, that amplitude of the vibration becomes very
high. Let us solve the differential equation when there is a resonance. We know that the
differential equation, suppose mx dot plus kx is equal to kt say x, kx is equal to F
cos omega t. Let us say, omega is omegan. So, omegan t and this can be written as x
dot plus k by m x is equal to F by m cos omegan t. Now, k by m is equal to omegan square.
Hence, x dot plus omegan square x is equal to F by m cos omegan t. This is the differential
equation which has to be solved. Here, we can see that, if we just put x is
equal to A sin omega t, that will not satisfy this differential equation in this case, because
this is the thing therefore, in this particular case, when the you are getting that same omegan
here and omegan here like that, therefore, you will get a particular solution. In this
case, is xp is equal to t times A cos omegan t plus B sin omegan t that is it and for this
C a. Now, we can have these are the constants. We can have cases.
Let us see, what happens when x is equal to 0 at time t is equal to 0 and x dot is also
equal to 0, that means displacement and velocity both component are 0. So, what will happen?
This will become, at t is equal to 0. So, let us write the solution again. x is equal
to t and this is A cos omegan t plus B sin omegan t. Now at t is equal to 0, x is equal
to 0. Let us see x dot is equal to A cos omegan t plus B sin omegan t. So, we have to put
that condition and after that, we will get some solution which will be equal to xp is
equal to F0 divided by 2m omega0 t sin omega0 t. That means, at natural frequency the vibrations
will always keep increasing. So, this amplitude of vibration will always keep increasing.
So, this type of phenomena will come. So, with time the magnitude always keeps on increasing.
This is what we observe. This is the type of resonance. So, this condition has to be
avoided.
Let us see that if in case damping is very small and your omega is very near to omega
0, that means omega is almost omegan then x is equal to F0 divided by m omega 0 square.
This is omega omegan square minus omega square cos omega t. This is the particular integral
and this can be written as F0 divided by A1 minus omega by omegan whole square and this
is cos omega t. Therefore we consider that even if there is
no damping, let us take the case of no damping, then there will be a homogeneous solution
also. That will be the complementary part. Complementary function is given by C is a
constant cos times omegan t minus phi, where phi is some angle. Therefore, x is equal to
C cos omegan t minus phi plus F0 divided by m omegan square minus omega square cos omega
t. This term has come like this. In this, we put the condition that at time t is equal
to 0, x is equal to 0 and at time t dot t is equal to 0, x dot is also 0.
That means, we are applying the forcing function at the time when the particle is at rest and
from that equilibrium position. Then we will not show that required algebra. Then C and
phi can be eliminated and you get x is equal to F0 divided by m. This will be equal to
omegan square minus omega square. This is cos omega t minus cos omegan t. So, this can
be written as cos omega t minus cos omegan t. We have the formula for cos C minus cos
t.
This can be written as, x is equal to 2F0 divided by m omegan square minus omega square
sin omega 0 plus omega divided by 2 into t sin omega 0 minus omega divided by 2 into
t. So, if we plot this thing here, plot may look something like this and then again it goes like this. What happens,
because omega 0 minus omega is very small, therefore, this time period will be very large
and we will get the phenomena of beats type of thing. So, here we will be getting the
beats type of thing. Like that, we can actually study different type of problems.
We will conclude here. So, today I have discussed about forced vibration problem in which we
considered damping and this one. Let us consider the cases, you will always not getting a spring
mass system. Sometimes you have got the axial rod, if you have got axial rod and you are
applying a force. Here, there may be a mass and you are applying a force here. Even this
type of problem can be solved by a spring mass system. So, you model it like a spring
and this is the mass. Some portion of the rod mass can also come here and it can be
damped at that point and then it becomes like that one.
Supposing you are applying the displacement; this is the force, this is a rod of length
L and this is the force. If you apply a force F, cross sectional area is A. Therefore, this
will give stress is equal to F by A and strain is equal to F by A divided by A times E. E
is the Young’s modulus of elasticity. Therefore, displacement delta of the end point is equal
to FL divided by AE. F by delta is called stiffness k that is equal to AE by L. In this
case, the spring constant can be written as AE by L and you can actually solve this problem.
So, axial vibrations of this system will be more, having more frequency if A is more.
Similarly, if you have some cantilever beam type of thing here and if you apply some load
delta here, you have F is equal to PL cube by 3 EI, delta is equal to PL cube by 3 EI.
This is known from the strength of material. You will come to know about that thing experimentally.
Otherwise, what happens? Take a cantilever beam. Apply some force and measure its deflection.
Then you know that one can plot P versus delta P is the force and slope of this gives you
stiffness dp by d delta. As the length increases, its stiffness decreases.
Therefore, you can do that one simple experiment. You can make that you take a beam type of
thing and here between two adjustable clamps you put it and clamp it properly from some
screw. You can always see that when the length was large and if you displace it slightly,
you can observe its natural vibrations. It will vibrate with a longer time period that
means less frequency. If the same thing you displace and you make like this thing has
been displaced this side and now you have like this here. If you vibrate then the frequency
of vibration will be very high and this is how it will be doing. Therefore, these types
of simple experiments one can do and one can observe. In the next lecture, we will discuss
the vibration of rigid bodies.