Forced Vibrations of a Single Degree of Freedom System (SDOF) & Dynamic Instability Video Lecture | Theory of Machines (TOM) - Mechanical Engineering

in this video we'll examine forced
vibration of the single degree of
freedom system and we will look at
dynamic instabilities that account that
can occur once again we're looking at
our mass spring system we assume that
there's no damping X is the displacement
of the mass positive downwards and the
mass is being forced by some external
force which is harmonic in nature so we
call it f zero times cosine Omega T now
we know from the previous time we looked
at this harmonic oscillator and the link
should be above if you want to go back
and review it but the equations of
motion looks something like MX double
dot of T plus KX dot excuse-me plus KX
of T naught X dot is equal to the force
F of T which in this case is f zero
times cosine Omega T call that equation
one and we also know that the solution
of such an equation is a combination of
the homogeneous and the particular part
so X of T is equal to X homogeneous of T
plus Giller of T in the context of
vibrations the homogeneous solution is
also known as the transient solution and
the particular solution is known as the
steady-state solution
so for the homogeneous solution we
assume a free vibration problem in other
words we make this part equal to zero
okay so a homogeneous solution
and that assumes that M x double-dot I'm
going to leave out the T dependency just
for shorthand but it should be implied
that X is a function of time and the
homogeneous problem the free vibration
problem looks like that MX double dot
plus KX equals zero we'll call this
equation two and we've seen that the
solution to this we'll call X sub H of T
is equal to c1 times cosine of Omega sub
n T where Omega sub n is the natural
frequency plus c2 sine Omega n of T and
we can write that Omega n is equal to
the square root of K divided by m spring
stiffness divided by the mass we'll call
this equation 3 and equation 4 all right
so to find the particular solution
and I've shown this in a previous video
the link to which appears above but to
find the particular solution we use
something called the method of
undetermined coefficients I'm not going
to go through it in huge detail here
because I've done it in this previous
video but fundamentally when we have a
forcing function that sinusoidal either
a sine or a cosine we assume a
particular solution of the form and I
use the constants D here D times cosine
Omega T plus d2 sine Omega T and this
Omega in general is different from the
Omega ends those two omegas in general
are different ok we're going to show
what happens when they're the same and
that's when we get dynamic instability
but for now we're assuming in general
they're different so this would be
equation 5 taking the derivative of
equation 5 with respect to time
X dot sub P of T is equal to minus Omega
D 1 times sine Omega T plus Omega D 2
times cosine Omega T equation 6 ok and
then the acceleration which is the
second derivative X double dot sub T of
T is equal to minus Omega squared D 1
this becomes a cosine of Omega t minus
Omega squared D 2 times sine of Omega T
call that equation 7
all right I'm going to make some space
here to continue on the same page and
now the idea is to substitute seven back
into equation 1 and then we are quite
the because efficients of the similar
terms so what does that mean
mathematically that means let me write
this out substituting 5 to 7 into 1
again from that we get minus Omega
squared let's see how I want to group
this ok so D 1 times K minus Omega M
Omega squared M is times cosine Omega T
plus D 2 into K minus Omega squared M
times sine of Omega T is equal to
f zero cosine of Omega T so all I've
done is I've substituted this into
equation 1 and then I've grouped it I've
grouped everything that's multiplying D
1 and everything that's multiplying D 2
okay and now when we look at the method
of undetermined coefficients on the
left-hand side we have a sine term on
the right hand side we have no sine
terms so the coefficient on the right is
0 so what does that mean if I look at
the sine Omega T term on the left hand
side of 2 K minus Omega squared M is
equal to 0 well this just implies that D
2 is 0 that is equation 8 and similarly
with the cosine Omega T term we find
that D 1 K minus Omega squared M is
equal to F sub 0 since we have a term
with the cosine on the right coefficient
is F sub 0 therefore the coefficients of
the corresponding terms on the left and
the right must be the same that's what
that's telling us all right so this
implies that D 1 is equal to F 0 F sub 0
divided by K minus Omega squared M
that's equation 9 so the particular
solution X P of T is equal to just F 0
divided by K minus Omega squared M times
cosine of Omega T version 10 so all
that's remaining is I should have
probably numbered this but this equation
here which says that the total the total
response X is equal to the homogeneous
response press the particular response
this gives me X of T is equal to c1
cosine Omega in T plus c2 sine Omega NT
plus I do it on the next one plus F zero
divided by K minus Omega squared M times
cosine Omega T now one thing to notice
is what happens mathematically if you
cost your mind back to your your
calculus courses differential equation
class what happens if this Omega was
actually Omega in in this case this
would not be a unique solution since
these would just be the same terms we
would combine these coefficients in
other words and would say cosine Omega
NT what we would need to do is if Omega
is equal to n then of course this would
become an Omega sub N and then we would
have to actually multiply this whole
term back T and that makes it unique and
also solution so in other words if Omega
were equal to Omega n if we were to
force this simple harmonic oscillator at
its natural frequency amplitude would
grow linearly with time and that is the
nature of the failure of the dynamic
instability all right I'm going to show
it try and draw that on the following
page do it straight and then this line
is approximately x equals T gorge
reflection below on this axis we have
time and on the vertical axis we have X
we change color for this we expect the
function to look sinusoidal and growing
the amplitudes of growing linearly
something like that
my drawing is not the best I apologized
but the idea is is that this is a linear
growth phenomenon unlike the exponential
growth we saw in the case of the static
instability dynamic instability is a
linear growth phenomenon well that
brings us to the end of this video I
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the comment section below thank you very
much for watching and I'll catch up with
you in the next video