Applications of Laplace Transformation-I Video Lecture - Computer Science Engineering (CSE)

dear viewers the applicant the title of
my lecture is applications of Laplace
transformation in order to show the real
power of Laplace transformation in its
applications to various problems of
engineering mathematics we shall derive
some more properties of the Laplace
transformation two very important
properties of the Laplace transformation
concerned the sifting on the a sexist
and the shifting on the T axis the
shifting on the S axis we had covered in
our previous lectures on Laplace
transformation where we had shown that
the replacement of s by s minus a in F s
which is the Laplace transform of the
function f T corresponds to the
multiplication of the original function
f T by the exponential function e to the
power a T in the seconds of thing shift
in the second shifting theorem which
concerns the shifting on the T axis we
shall show that the shifting if we
replace T by T minus a in the function f
T then it corresponds roughly to the
multiplication of the Laplace transform
FS of F T by e to the power minus s the
precise formulation of the second
shifting theorem is as follows
it says that if LF T is equal to FS
where s is greater than gamma then e to
the power minus s into FS a greater than
or equal to 0 is the Laplace transform
of ft minus a into u ad where u at E is
the unit step function defined as u at E
is equal to 0 when T is less than a and
u at E is equal to 1 when T is greater
than or equal to a this is the graph of
the function U at T in the interval 0 to
L that is when T lies between 0 and a UA
T is defined as 0 and when T is greater
than or equal to a u it is defined as
equal to 1
let us look at the proof of the second
shifting theorem the Laplace transform
of ft minus f into u a T by definition
can be written as integral over 0 to
infinity e to the power minus st into f
t minus a into u a T DT which is equal
to integral over a to infinity e to the
power minus st into f t minus a DT
because u a T is equal to 0 over the
interval 0 to a and over the interval a
to infinity it is defined as 1 so the
integral over 0 to infinity reduces to
the integral over a to infinity and UAT
becomes 1 now let us make a substitution
here let us put tau is equal to t minus
a so when we put tau is equal to t minus
here the limits of integration change
from a to infinity to 0 to infinity and
DT becomes equal to D tau and hence we
get a laugh ft minus f ua t equal to
integral over 0 to infinity e to the
power minus tau plus a into F tau D tau
which can be written as e to the power
minus s integral over 0 to infinity e to
the power minus s tau into F tau D tau
now integral over 0 to infinity e to the
power minus s tau into F tau D tau is
the Laplace transform of the function f
tau so we can say it is equal to F s and
thus we get the Laplace transform of ft
minus au at e equal to e to the power
minus s into F s now if we take a
particular case here let us assume that
ft is equal to 1 for all T greater than
or equal to 0 then we will have the
Laplace transform of ft minus a into u
ax t equal to Laplace transform of u T
because ft is equal to 1 for all T
greater than or equal to 0 so we get
Laplace transform of U at T equal to e
to the power minus s into Laplace
transform of F T that is Laplace
transform of 1 we know that Laplace
transform of 1 is 1 by s whenever s is
greater than zero this we have shown
earlier in our previous lecture on
Laplace transformation
so it is so the Laplace transform of U
at e will be equal to e to the power
minus s over s whenever s is greater
than zero now let us let us study a
remark quite often we have to find the
Laplace transform of ft into UAT where
the function ft lacks the shifted from
ft minus a if we have ft minus here here
in place of ft we can directly apply the
second shifting theorem but quite often
it is not so instead of ft minus a we
have ft and we have to find the Laplace
transform of ft into u ax t
so in such a case we do the following ft
into ua t may be written as ft minus a
plus a into u ax t now ft minus a plus a
is a function of t minus f which we have
denoted by capital f t minus a so we get
F T into u a t equal to f t minus a into
u ax t and then by the second shifting
theorem that is theorem number one the
Laplace transform of F T into u at T
will be equal to Laplace transform of ft
minus a into u ad now Laplace transform
of ft minus a into u a T by sit by
theorem one is e to the power minus s
into Laplace transform of the function
capital F T which is equal to e to the
power minus s into Laplace transform of
F T F T is equal to f of T plus a
because we have assumed that capital F
of t minus a is equal to small F T so f
of F T is equal to f of T plus a so we
get the Laplace transform of F T into u
ax T equal to e to the power minus s
into Laplace transform of F T plus a so
whenever we want to find the Laplace
transform of the function f t into u ax
t what we will do is we will write it as
e to the power minus s into Laplace
transform of F T plus a in the second
shifting theorem we had F of t minus a
here and we had F
here now the changes that we have ft
here and here we have f of t plus s so
we get the Laplace transform of the
function f of T plus a in the function
of T we replace T by T plus a and then
find its Laplace transform in order to
get the Laplace transform of the
function f T into u ax T now let us
study an example there's done the second
shifting theorem let us find the Laplace
transform of the function f T where F T
is defined as 0 for T less than PI by 2
F T is defined as sine T 4 PI by 2 less
than or equal to T less than 3 PI by 2
and F T is defined as 0 for T greater
than or equal to 3 PI by 2 so we can
write the function f T in terms of unit
step functions as f T equal to sine T
into u PI by 2 t minus u 3 PI by 2 T
because when T is less than PI by 2 u PI
by 2 is T 0 u 3 PI by 2 is T is also 0
so f T is equal to 0 and when T is equal
to PI by 2 are more than PI by 2 but
less than 3 PI by 2 then you PI by 2 T
will be equal to 1 by u 3 PI by 2 T will
be 0 so f T will be equal to sine T and
when T is equal to 3 PI by 2 are more
than 3 PI by 2 u PI by 2 T will be equal
to 1 u 3 PI by 2 T min will also be
equal to 1 so f T will be equal to 0
hence F T the given function f T can be
described in terms of unit step
functions u PI by 2 T and u 3 PI by 2 TS
F T equal to 7 T times u PI by 2 t minus
u 3 PI by 2 T now let us apply the
second shifting theorem so when you take
the Laplace transform of this violently
property of the Laplace transform LF T
will be equal to L of cyant into u PI by
2 T minus L of sine T into u 3 PI by 2 T
now we can see here that this is Aleph
sine T into the u pi by 2 T is not
actually in the form of the second
shifting theorem rather it is in the
form of the remark which follows the
second shifting
where we had discussed that how to find
the Laplace transform of ft into u ax t
so using that remark you will have the
following Laplace transform of the
function f t will be e to the power
minus s is PI by 2 here so e to power
minus PI s by 2 into Laplace transform
of F of T plus a F T sine T is PI by 2
so Laplace transform of sine T plus PI
by 2 we have minus e to the power minus
s a is three primary two here so we got
it / - three PI s by 2 into the Laplace
transform of sine T plus 3 PI by 2 now
sine T plus PI by 2 is cos T so Laplace
transform of cos T we have here and then
la plage transform of sine T plus 3 PI
or sine T plus 3 PI by 2 is equal to
minus cos T so we have minus - plus here
e to the power minus 3 PI s by 2 into
Laplace transform of cos T and Laplace
transform cos T we know it is s over s
square plus 1 so we get the right-hand
side as it 4 minus PI s by 2 in to s
over s square plus 1 plus each power
minus 3 PI s by 2 in to s over s square
plus 1 which is further equal to s into
e to the power minus PI s by 2 plus e to
power minus 3 PI s by 2 over s square
plus 1 let's take another example on
this down second shifting theorem here
we have to find the inverse Laplace
transform of s into e to the power minus
s by 2 plus PI into e to the power minus
s over s square plus PI square we can
express the inverse Laplace transform of
s e to power minus s by 2 plus PI to
power minus s over s square plus v
square as the sum of the inverse Laplace
transforms of s into each power minus s
by 2 over s square plus PI square and PI
into e to power minus s over s square
plus PI square and we know that s over s
square plus PI square is the Laplace
transform of cos PI T and here we have
if you identify with e to power minus s
here s into H power minus s by 2 tells
us that AZ
equal to half here so we have in we have
to find inverse Laplace transform of e
to the power minus s into FS where a is
1/2 and FS is s over s square plus PI
square whose inverse Laplace transform
is cos PI T and so by second shifting
theorem inverse Laplace transform of s
into e to power minus s by 2 over s
square plus PI square will be you eighty
you get e becomes you half T into F T
minus AF t minus M means F T is cos T
cos PI T so f t minus a will be T aft -
Emily f t minus half that will be giving
us cos pi t minus 1/2 and then we have
inverse Laplace transform here of pi e
to the power minus s over s square plus
v square PI over N square plus PI square
is the Laplace transform of sine PI T
function and e to power minus s tells us
that it is equal to 1 here so using
second sifting theorem we get u 1 T that
is ua t becomes u 1 T and then F of t
minus 1 so f T is sine PI T so f t minus
1 becomes sine pi t minus 1 so we get
the inverse Laplace transform s cos pi
in into t minus half into u 1/2 T plus
sine PI T minus 1 into u 1 T which we
can write in the simplified form as cos
pi t minus 1/2 become sine PI T into u
1/2 T and sine PI T minus 1 becomes sine
minus sine PI T and so we get minus sine
PI T into u 1 T and we can simplify it
further we can write it as sine PI T
times you have t minus u 1 T now let's
take another function f T which is
defined as 0 when T is greater than R
greater than or equal to 0 but less than
PI by 2 and equal to cos T when T is
greater than or equal to PI by 2 and let
us find the Laplace transform of this
function so again we can express this
function as in terms of unitaster
functions we can write f TS cos T into u
PI by 2 T
20 is less than PI by 2 you PI by 2 is 0
you PI by 2 t is 0 so ft becomes 0 and
when T is equal to PI by 2 or more than
that then you PI by 2 T is equal to 1 so
we get ft equal to cos T and then
Laplace transform of cos T into u PI by
2 T will be equal to e to the power
minus s a is equal to PI by 2 here so 8
4 minus PI by 2 into s and then Laplace
transform off ft plus a ft is cos T here
is PI by 2 so we get the Laplace
transform of cos T plus PI by 2 and cos
PI by 2 plus theta we know this is equal
to minus sine theta so we get minus L
sine T inside L of sine T here so we get
minus e to power minus PI by 2 interest
I love science I'm L of sine T is when
over s square plus 1 so we get minus e
to the power minus PI by 2 s over s
square plus 1 now we are going to study
periodic functions periodic functions
occur in many practical problems and in
most of the cases they are more
complicated than the single sine or
cosine functions we are going to show
that if the function f T is periodic
with period t and it is piecewise
continuous over a length T then the
Laplace transform of the function exists
so let us define first a periodic
function a periodic function f is
defined if a periodic function defined
for all T greater than 0 is said to be
periodic with period T greater than 0 if
f of T plus T is equal to F T for all T
greater than 0 and if f is periodic with
period t then we can easily see by
induction on n that f of T plus NT is
equal to F T where n takes values 1 2 3
and so on now for such a function we
have L of F T equal to 0 to infinity e
to the power minus s T into f T DT we
shall see that for the existence of the
integral of a periodic function with
period t we just need the piecewise
continuity of the function f so let us
see how
when this Laplace transform of the
function of T edges so far for the
periodic function L of F T can be
written as integral 0 to infinity e to
the power minus s T into F T DT and now
using periodicity we can we'll what we
will do we will plug the interval 0 to
infinity in two parts
0 to t t2 to t2 t2 3t and so on and make
use of the periodicity of the function f
T so what we will do in the second
integral onwards on the right side that
is integral t to 2t integral to 2t - 3 T
and so on in the we shall make
substitutions T equal to tau plus TT
equal to tau plus 2 PT equal to tau plus
3 T and so on and use periodicity of the
function f to obtain LF T is equal to
integral 0 to te to the power minus s T
as tau F tau D tau plus integral over 0
to T e to the power minus s tau plus T
into F tau D tau here when you
substitute T equal to tau plus T using
periodicity of the function f f tau plus
T will be f tau e to power minus s T
will become e to the power minus s tau
plus T and DT will become D tau the
limits of integration which are TN to T
will change to 0 to T here they will
become 0 n T and in the integrand we
will have e to the power minus s tau
plus 2t into f tau D tau again using
periodicity here this way we shall then
have LF T equal to 1 plus e to the power
minus s T plus e to the power minus 2 s
T and so on multiplied by integral 0 to
te to the power minus s tau into F tau D
tau now this is a geometric series the
ratio is e to the power minus s T so we
can write the sum of the series as 1
over 1 minus e to the power minus s T
and then multiplied by integral 0 to te
to the power minus s tau F tau D tau
this integral
occurs here exists FFT is a piecewise
continuous function and of course
periodic with period t so thus we have
the following theorem for a periodic
function if F T is a piecewise
continuous function having a period T
greater than 0 then the Laplace
transform of the function f T edges and
is given by 1 over 1 minus e to the
power minus s T into integral 0 to t e
to the power minus s tau F tau D tau
just take an example on this hard kill
let us find the Laplace transform of the
square wave function which is shown here
in this figure you can see here that the
function f given by this figure is
periodic with period 2 a here you can
see the graph of the function over the
interval 0 to a it takes the value a and
then over the interval a to 2 a it takes
the value minus a and then the graph of
the function f over the length 2 a is
repeated over from 2 a 2 4 8 and so on
so it is a periodic function with period
to it the given function is periodic
with period P that we have denoted by
deep so T is equal to 2ei here hence F T
and F T is clearly a piecewise
continuous function on the length T
equal to 2 a so Laplace transform F T of
the function given function exists and
will be obtained from this result 1 over
1 minus e to the power minus s T
integral over 0 to te to the power minus
s tau F tau D tau so T is equal to 2 a
gives us a Laplace transform of the
given function of T s 1 over 1 minus e
to the power minus 2 EI s and then
integral over 0 to 2 a is broken up into
two parts integral over 0 to a and then
integral over a 2 to a integral over 0
to a we have in the integral over D Rho
2 a we use the definition of F F
equal to a in this part of the interval
and in the interval a two to a function
f takes the value minus a so we have put
those values of f and then we integrate
e to the power minus s tau it integral
of it four minus s tau with respect to
tau gives us into e to the power minus
tau over minus s and so when you put the
limits 0 a here and in my 2a here what
we get is the following a times 1 minus
2 times e to the power minus s plus e to
the power minus 2 s over s into 1 minus
e to the power minus 2 s now 1 minus 2 e
to the power minus s plus e to the power
minus 2 s is the square of 1 minus e to
the power minus s so we replace this
numerator by a times 1 minus e to the
power minus s whole square in the
denominator we have s into 1 minus e to
power minus 2 s 1 minus e to power minus
2 s can be factorized as 1 minus e to
the power minus s into 1 plus e to the
power minus s and then 1 minus e to
power minus s can be cancelled from the
numerator and denominator and we will be
getting a times 1 minus e to the power
minus s over s times 1 plus e to the
power minus s now if you multiply in the
numerator and denominator by e to the
power s by 2 then the then what we will
get a over s times e to the power s by 2
minus e to the power minus s by 2
divided by e to the power s by 2 plus e
to the power minus s by 2 which we know
that denotes the tan hyperbolic of s by
2 so we have the Laplace transform of
the given square wave function as a over
s into tan hyperbolic yes by 2 now we
are going to discuss the Dirac Delta
function the Dirac Delta function is
also called the impulse function and
there has applications in these problems
where a large force occurs is applied
over a very small
for a large force is applied for a very
short time
say for example in the case of bending
of beams when we when we used the point
loads the the point load means we are
applying a very large pressure over a
very small area so in such cases we use
the Dirac Delta function the Dirac Delta
function is also called as unit impulse
function our impulse function and it was
introduced by British theoretical
physicist Paul Dirac so it is applied
are used in problems there are large
forces applied for a very short time or
a large force x over a very small area
for example in the loading of a beam and
because of its nature it is defined as
Delta t minus a is equal to infinity at
t equal to a and 0 and T is not equal to
it so and further that integral of delta
t minus a over the interval minus
infinity to infinity is defined as equal
to 1 we are going to show that the
Laplace transform of Delta t minus a
exists and is equal to e to the power
minus s for all a greater than or equal
to zero so let us consider the function
Delta Epsilon t minus a which is defined
as 1 over epsilon for T greater vigor
than a and less than a plus Epsilon and
0 elsewhere let us look at the graph of
this Delta Epsilon
t minus a function Delta Epsilon t minus
a function by its definition can be
drawn like this this DX is this delta
epsilon t minus ada presenting that
vertical axis and then we have over the
interval a to a plus epsilon Delta
Epsilon
t minus a taking the value 1 by Epsilon
and 0 elsewhere so that this Delta
Epsilon t minus a is then the derivative
of the function u epsilon T minus a
which is defined as 0 for T less than a
and 1 by epsilon x t minus a for a less
than T less than a plus Epsilon and 1
40 greater than a plus or not clearly
you can see that the derivative of you
epsilon t minus a gives us Delta t minus
Delta Epsilon t minus a so this the
graph of you epsilon t minus a is like
this over the interval and for values of
T less than a it is taking value 0 over
the interval a to a plus Epsilon it is
given by this line segment 1 by epsilon
t minus a having slope 1 by Epsilon and
then over the values of T greater than a
plus Epsilon it is taking value 1
further we note that as epsilon tends to
0 you epsilon t minus a tends to you a
t1 that is the unit step function which
is 0 for T less than a and equals 1 for
T bigger than or equal to a and that's
why we can regard Delta t minus a as the
derivative of U at T Delta t minus a was
the limit of u epsilon at Delta t minus
a was the limit of Delta Epsilon
t minus a delta epsilon t minus a was
the derivative of u epsilon t minus a so
and the limit of you epsilon t minus a
as epsilon tends to 0 is UAT so we say
that delta t minus a is the derivative
of u ax t and since it is the derivative
of u t you can write it as u a t equal
to integral over minus infinity to t
delta t minus f TT now if f is assumed
to be continuous at t equal to a then
from the mean value theorem of integral
calculus we will be getting integral
over minus infinity to infinity f T into
Delta Epsilon t minus a DT equal to F T
into 1 by Epsilon duty because Delta
Epsilon
t minus a assumes value 1 by epsilon
over the interval a to epsilon a 2 a
plus epsilon and elsewhere it is defined
as 0 so this integral is equal to
integral over a to a plus epsilon F T
into 1 by epsilon into DT now
let us use the mean value theorem here
if we assume that F is continuous at T
equal to a we can apply the mean value
theorem of integral calculus and then
this will be equal to 1 by epsilon is a
constant we have put the constant out
here and then a plus epsilon minus a a
plus epsilon minus a into the value of
the function at some intermediate point
theta is an intermediate point of the
interval a to a plus Epsilon so where
theta is some point in the interval a to
a plus epsilon and then let us now take
let epsilon go to 0 when epsilon goes to
0
we shall have Delta Epsilon t minus a
tending to delta t minus a so we will
have the left hand side tends to
integral over minus infinity to infinity
f t into delta t minus a DT while the
right hand side epsilon will cancel with
this epsilon F theta will tend to FA s
epsilon tends to 0 so the value of the
integral of f t into delta t minus a
over the interval minus infinity to
infinity gives us f a the value of the
function f at the point a now let us
take a particular case let us assume f T
to be equal to e to the power minus s T
then it follows that Elif Delta t minus
is equal to e to the power minus s so if
you take F T equal to e to the power
minus s in the integral then Laplace
transform Delta t minus a Laplace
transform Delta t minus a will be
integral 0 to infinity e to the power
minus s T into Delta t minus a DT so
that will give us e to the power minus s
because we have shown that integral over
minus infinity to infinity of T Delta t
minus a DT is equal to F a so from that
F T equal to e to power minus s T will
transform to FA that is e to the power
minus s Laplace transform Delta t - is
it / - yes now if you take a equal to
zero here then we get that Laplace
transform delta T is e to the power 0
that is 1 so Laplace
transform of delta T function is equal
to 1 R we can say that inverse Laplace
transform of 1 is equal to Delta T now
which is very useful result let us take
an example on this Dirac Delta function
let us find the integral of F T into
Delta dash t minus a over the interval
minus infinity to infinity we shall see
that it is equal to minus F dash a now
making use of the integration by parts
integral over minus infinity to infinity
F T Delta dash t minus a DT can be
written as f T into Delta t minus a and
evaluated at Madison Finity n plus
infinity and then minus integral over
minus infinity to infinity F dash T into
Delta t minus a DT now Delta t minus is
defined as 0 everywhere except at T
equal to a very text value infinity so
this will change go to 0 as T goes to
infinity or minus infinity and the right
hand side will then become so we get the
integral over minus infinity to infinity
F T at a dash t minus a DT equal to
minus F dash a because this becomes 0 as
T goes to plus infinity or minus
infinity and this we have earlier seen
that if F T is continuous at T equal to
a then integral over minus infinity to
infinity F T Delta t minus a DT is equal
to F F so from that we get the value of
this integral s f dash a so we get the
value of the derived integral s minus F
dash a now let us find the Laplace
transform of T cube into Delta t minus 4
so Laplace transform of T cube into
Delta t minus 4 is equal to integral 0
to infinity e to the power minus s T
into T cube into Delta t minus 4 DT now
so what we have now we have before this
this can be identified with the integral
over minus infinity to infinity of T
Delta t minus a DT so here a is equal to
4 and we that integral over minus
infinity
infinity F T Delta t minus a DT will
then reduce to zero to infinity because
over the interval minus infinity to
infinity to zero Delta t minus a will be
equal to zero so that integral if you
identify this integral with that this
will be equal to the value of this
integral will be equal to e to the power
minus 4 s into 4 cubed if this is we can
regard this as ft then it will become FA
is equal to 4 here so the value of the
integral will be 4 cubed e to the power
minus here's a similar cases here if you
take integral or minus infinity to
infinity sine 2t Delta t minus PI by 4
DT then the value of this integral again
by this property of the Dirac Delta
function that integral over minus
infinity to infinity f T Delta t minus a
DT is FA will get the value of the given
integral that is integral over minus
infinity to infinity sine 2t Delta t
minus PI by 4 DT s sine 2 times 5 by 4
this is sine 2a is equal to PI by 4 so
sine 2 pi by 4 and which is sine PI by 2
and we know that sine PI by 2 is equal
to 1 now we shall study the example of a
differential equation where the Dirac
Delta function is used let us solve the
differential equation by double dash
plus by equal to Delta t minus a where
we are given that by V equal to y 0
equal to 0 a and B are some real
constants so let's take the Laplace
transform of the given differential
equation we will get Laplace transform
of using linearity of the Laplace
transformation we get the Laplace the
Laplace transform of the left hand side
s blahblah transform of Y Double Dash
plus the Laplace transform of by so
Laplace transform of Y double dash we
know is given by s square by BA
minus s by 0 minus PI - 0 and Laplace
transform of bi is by bar Laplace
transform of the right hand side that is
Laplace transform Delta t minus a we
have seen it is given by e to the power
minus years
now we can just take the because we are
not they are given no value of ID s 0 so
let us assume that by - 0 is equal to K
then we can find the value of BI bar
solve this equation for B bar and we get
Y bar equal to e to the power minus AS
plus bi - 0 that is it for minus es plus
K divided by y 0 is given equal to 0 so
we get Y bar equal to e to power minus s
plus K over s square plus 1 now let us
take the inverse Laplace transform of
the Laplace transform of Y bar we shall
have y equal to sine t minus f into UT
minus a plus K sine T we can then use
the initial condition that is by at V is
equal to 0 so when you put t equal to V
here we get 0 equal to sine B minus a
into u B minus a plus K sine B and
therefore the value of K is equal to
minus sine B minus a into u B minus a
over sine B and hence Y is equal to sine
t minus a into u t minus a minus K K
value of K s minus sine B minus a u B
minus a over sine B so this value we are
substituting here so by is equal to sine
t minus a into u t minus a minus sine v
minus a into u b minus a over sine B
into scientists study some applications
of the Laplace transformation first be
study the applications of the Laplace
transformation to the problems in
dynamics let us consider the case of a
particle of mass M which can perform
small oscillations about a position of
equilibrium under a restoring force M
and square times the displacement it is
started from rest by a constant force F
which X for a time T and then ceases
show that the amplitude of subsequent
oscillations is given by 2 F
over em and square into sign and T y2 so
now say this is the position of
equilibrium at the particle was here at
time T the particle is here the palm
mass of the particle is M and at time T
it is at a distance X from the position
of the equilibrium that is o acted upon
by forces there is a restoring force a
minus square X and this is the force F
which is acting for a time T so the
force F which is acting for the time T
only can be represented as and then it
seizes afterwards can be represented in
terms of unit step functions as f into 1
minus u times 1 minus UT minus T and
therefore the question of the motion of
the mass M will be equal to M X double
dot that is M into d square X by DT
square equal to resultant of the force F
which is acting for the time T only and
then the restoring force so we get MD
square X by DT square equal to minus ml
square into X plus F times 1 minus u t
minus T now dividing by M we get d
square X by DT square plus n square X
equal to F by M into 1 minus u t minus T
let us now take the Laplace transform of
both sides Laplace transform of X double
dot R X double dash will give us s
square X bar minus SX 0 minus X - 0 X 0
is equal to 0 because at the time T
equal to 0 the particle was at the point
O and it was at the at rest so X - 0 is
also 0 so making use of those initial
conditions we get the Laplace transform
of the equation of motion as s square
plus n square into X bar equal to F over
m s into 1 minus e to the power minus
STD Laplace transform on the right hand
side we get the Laplace transform of
Banaras 1 over s and laplace transform
of u t minus TS e to the power minus s T
over s so the Laplace transform of the
right hand side gives us
F / M s into 1 minus e to the power
minus s T after we make use of the given
initial conditions that is X at T equal
to 0 is 0 and X dash at T equal to 0 is
0 now solving for X bar we get X bar
equal to F into 1 minus e to the power
minus s T over M s into S square plus n
square let us break it now into its
partial fractions we can write X bar
further s F into 1 minus e to the power
minus s T over m and then burn over s
into s 2 square plus n square when we
break into partial fractions we get 1
over n square times 1 over s minus s
over s square plus n square now in order
to find the displacement of the particle
M from the equation of equilibrium X in
order to find X let's take the inverse
Laplace transform of this equation so
then we will get X equal to now F over m
n square F over m square is a constant
so we have kept it like that only F over
m square then this one multiplied by 1
over s minus s over s square plus n
square when we take the inverse Laplace
transform of that we get inverse Laplace
transform 1 over s s 1 inverse Laplace
transform of s over s square plus n
square s cost NT so we get 1 minus cos
NT as the inverse Laplace transform of
this expression so 1 into this whole
thing when we take the inverse Laplace
transform that we get when minus cos
empty and then minus this minus is here
then e to the power minus s T multiplied
by this then we take in restaurant
inverse Laplace transform of e to the
power minus n e to power minus s T into
this expression by second shifting
theorem will be equal to 1 minus cos n t
minus T into UT minus T now we can study
the various cases here when T will be
more than 0 but less than capital T UT
minus T the unit is step function UT
minus T will take value 0 so X will be
equal to F over a minus square into
- cause NT and when T will take value
more than capital T UT minus T will be
equal to one so the displacement X will
be equal to F over a - square into 1
minus cos n t minus 1 - cos and t minus
t and this can be simplified further as
F over a - square into cos + t - t - cos
sin t now making use of the formula cos
a minus cos B as 2 sine a plus B by 2
into sine B minus a by 2 we get this
expression further equal to F over a -
square into 2 sine + T by 2 into sine +
t minus T by 2 and thus the amplitude of
the oscillations of the mass M about the
position of equilibrium will be equal to
2 F by M + square into sine and T by 2
now let's study another example on
dynamics where a body falls from rest in
a liquid the density of the liquid is
1/4 of that of the body if the liquid
offers a resistance proportional to the
velocity of the body and the velocity
approaches the limiting value of 9
meters per second then let us find the
distance which the body falls in 5
seconds so this is the liquid and this
is the body of mass M after it has
fallen in the liquid mg is the weight of
the body acting downwards 1 by 4 times
mg 1 by 4 times mg body first whose
density is 1 photon density of the
liquid is 1/4 of the density of the body
so we get 1 by 4 MV acting upwards and
then MK b MK b m 1 by 4 mg is the up
thrust which is acting upwards and then
MK b we are taking the resistance is
proportional to the velocity for
convenience they are writing
distances M K into V V the velocity the
constants of proportionality we are
writing as M into K so that B maybe we
may cancel this M from the equation we
can divide that equation by M so why for
convenience we are writing the
resistance equal to MK V so MK V is
acting upwards 1 by 4 mg is acting
upwards which is the up thrust and then
mg which is the weight of the body is
acting downwards so they are these are
the forces that are acting on the body
at time T after it has fallen a distance
X from the surface of the water the X is
measured from the surface of the sorry
from the surface of the liquid so this
is the situation after and at an instant
T and thus the equation of motion will
be given by M into DV by DT equal to mg
minus 1 by 4 mg minus m kb this is the
resultant force acting on the body and
the M into DV by DT is therefore equal
to the resultant force which which gives
us the question as DV by DT plus KB
equal to 3 by 4 G after we divide the
equation by M and simplify it the
initial condition is that at T equal to
0 the body was at rest that is give us
equal to 0 and let us now take the
Laplace transform of this equation when
we take the laplace transform here the
Laplace transform of DV by DT will give
us s into V Bar minus b0 b0 is equal to
0 and then K times V Bar and then 3 by 4
into G is a constant then n of 1 l of 1
is 1 by yes so this what we get after
taking the Laplace transform of this
equation and we can then find V Bar from
here V Bar give will give us 3 by 4 into
G by s into S Plus K let us bracket into
its partial fractions V Bar will be 3
gee by 4 K into 1 by s minus 1 by s plus
K and when
take the inverse Laplace transform of
this equation we shall have the value of
V which is 3 gee by 4 K into 1 minus e
to the power minus KT inverse Laplace
transform of 1 by s is 1 and inverse
Laplace transform of 1 over s plus case
e to the power minus KT now when T tends
to infinity e to power minus KT will go
to 0 so we will tend to three G by 4 K
the limiting velocity of the body is
therefore 3G by 4 K but we are given
that the limiting velocity is 9 meters
per second so so thus we have three G by
4 K equal to 9 R we have k equal to G by
12 and hence V equal to 3 gee by 4 K
into 1 minus e to the power minus K T
implies that DX by DT is equal to 9 into
1 minus e to the power minus K T V we
know V is equal to DX by DT and 3G by 4
K is equal to 9 so we get the
differential equation again DX by DT
equal to 9 times 1 minus e to the power
minus K T let us call this as question
number 1 and the initial condition is
that at t equal to 0 the body was at the
surface of the liquid that is X is equal
to 0 so now Laplace transform of if you
take the Laplace transform of the
equation 1 we are going to solve it for
X so then we will have SX bar minus X 0
at 0 is equal to 0 and the right hand
side will give us 9 times Laplace
transform of minus 1 over s - Laplace
transform of e to the power minus KT s 1
over s plus K R we can say that X bar is
equal to 9 into 1 over s square minus 1
by s into 1 by s minus 1 over s plus K
after we've read it into partial
fractions so the inverse Laplace
transform of this equation then gives us
X equal to 9 times inverse Laplace
transform of 1 over s square is T and
then we have my
one over k 1 over k is equal to 12 over
G this is the value of 1 over K and then
1 over s inverse Laplace transform 1
over s is 1 inverse Laplace transform of
1 over s plus K is e to the power minus
KT and K is equal to G by 12 so we get
it 4 minus KT as e to the power minus GT
bihter
now let us put T equal to 5 and G equal
to nine point eight meters per second
square in this expression to find the
value of x which is which will give us
the distance the body by trouble by the
body in five seconds so we get X equal
to 9 into five that is 45 then we
calculated the value of this 1 minus e
to the power minus G is 9.8 ps5 divided
by 12 the value of this expression comes
out to be 1 minus point zero point zero
1 7 and then 9 times 12 by G 9 times 12
by G gives us 11 point zero 2 so we get
the value of XS 34.1 7 meters now let us
study some problems on the simple
electric circuit a simple electric
circuit consists of a resistance given
by our inductance L and capacitance C
which are connected in series and a
switch is provided to connect our
disconnect the circuit and then a source
of electro-motive power is also there so
then if we then we know that the if Q is
the charge on the condenser at time T
and I denotes the current in the circuit
at time T then Q is equal to then DQ by
DT is equal to I and the voltages
developed across the resistance
inductance and capacitance are given by
L di by DT L di by DT is the voltage
developed
the inductance RI is the voltage
developed across the resistance and Q by
C is the voltage developed across the
capacitance which is equal to the Volta
the electro-motive source which is
provided in the given circuit here it is
in this problem it is e delta T so by
kirchoff's law the L di by DT plus RI
plus Q by C will be equal to e delta T
and DQ by DT will be equal to I now we
are further given that and that it is
here it is also assumed that the
connecting wires have neck syllable
resistance so if I is the current at any
time subsequent time T we have to find
the limit of I as T just to 0 now let us
take just all these equations using the
Laplace transformation method so when we
take the Laplace transform of this
equation we'll get this L is the L this
L is a constant so L times Laplace
transform of di by DT and Laplace
transform where di by DT will give us s
into Ivar minus I I 0 I 0 is equal to 0
because we have given initial condition
so at T equal to 0 the charge on the
condenser is 0 and at T equal to G of
the current in the circuit is also 0 so
we get L into S into Ivar plus R times I
bar plus Q Bar over C C is the
capacitance which is constant the
Laplace transform of Q will be Q R equal
to e times Laplace transform of delta T
we have seen earlier that it is equal to
1 so we get laplace transform of e delta
T as e and the Laplace transform of DQ
by DT will be s into Q Bar minus Q 0 Q 0
is equal to 0 and which will be equal to
I bar so this equation when we take the
Laplace transform of this equation will
get s Q Bar equal to AI bar and
hence we get the equations for life
equations LS plus R into I bar plus Q
Bar by c equal to e & SQ r equal to Iowa
now substituting the value of Q bar from
here as I bar over s in this equation
will have the following LS square plus
RS plus 1 by C into I bar equal two
years which can be expressed as s plus M
whole square plus n square into I bar
equal to e by L into s after dividing by
L where m is equal to R by 2 L and n
square is equal to 1 by Cl minus R by 12
whole square and we can then express
this equation as I've R equal to e by L
s we can write as s plus M minus M over
s plus M whole square plus n square we
have added and subtracted I am here in
order to take the inverse Laplace
transform of this equation so when you
take the inverse Laplace transform of
this equation in this Laplace transform
of 5r will give us ie IVA L is a
constant and then inverse Laplace
transform of this side expression will
be inverse Laplace transform of s plus M
/ S Plus M whole square plus M square -
inverse Laplace transform of M / S Plus
M whole square plus n square inverse
Laplace transform of S Plus M / S Plus M
whole square plus n square will be e to
the power minus M T into cos NT because
Laplace transform of cosine T is s over
s square plus n square and here s is
being replaced by s plus M so by first
shifting theorem inverse Laplace
transform of this will be e to the power
minus M T into cos sin T and similarly
inverse Laplace transform of M over s
plus M whole square plus na square will
be e to the power minus MT into sine NT
so after taking inverse Laplace
transform we get I equal to e by L into
cos NT minus M by n into sine NT into e
to the power minus empty and we then
have as T tends to 0 the limit of
is ebuy el thus we have the following
remark that that even though at
initially the current was zero in the
circuit hi zero is equal to zero a large
current develops intense instantaneously
in the circuit due to the impulsive
voltage which is applied at T equal to
zero and we are finding the limit of
limit that is e yl d by L is the limit
of this current which develops
instantaneously in the circuit due to
the impulsive voltage so Eva L is the
limit of this current as T tends to zero
now let's take one more example on this
RLC circuit the flow of current I in the
arizona circuit when the initial
electromotive force is ET is governed by
the differential equations di by DQ by
DT equal to I and L di by DT plus RI
plus q by c equal to ET and when we take
the Laplace transforms of the above
equations we get s Q Bar equal to alpha
I bar plus RI bar plus Q Bar by C equal
to e bar so younger studying a general
case where ET denotes the electromotive
source of voltage i mean ii teachers the
voltage of the electro-motive source so
here we again we are assuming that i en
q both are 0 at t equal to 0 and when we
eliminate q bar from these two equations
are we put the value of q bar as AI bar
over s from this equation into this
equation we will get s square plus R by
L into s plus 1 by LC into I equal to e
bar s over m so
ivar is the Laplace transform of ET so
this can be written as s plus M whole
square plus n square into y bar equal to
e bar s by L and where m is R by 2 L and
square is 1 by LC minus R square by 4 L
square as we had seen earlier and
therefore I bar is equal to Y bar s
4 s plus M whole square plus n square
into and after solving for IVA and when
we take the inverse Laplace transform of
this equation we can then find the value
of the current I if at t equal to 0 now
let us study some particular cases of
the electro-motive source voltage ET if
we assume that at t equal to 0 at
constant voltage is applied that is e t
is equal to a constant e then we will
get is after taking inverse Laplace
transform et is equal to e so e bar will
become e by s and therefore e bar s will
be equal to e so we will have e here and
then inverse Laplace transform of so 1
over s plus M whole square plus n it's
curve will be e to the power minus MT
into sine NT by and and so I will be
equal to e upon n L into e to the power
minus M T into sine NT now if if bun by
LC minus R square by 4 L square is equal
to zero that is if RL and C have values
such that 1 by LC becomes equal to R
square by 4 L square we will have the
inverse Laplace transform of this
equation as because in that case and a
square will be zero so we will have I've
R equal to e VAR s over s plus M whole
square into L so when we will take the
inverse Laplace transform the the
inverse Laplace transform Y bar will be
ie and we will get Eve RS e VAR s is
equal to e so we will get e here L is a
constant will come like that and then 1
over s plus M whole square inverse
Laplace transform inverse Laplace
transform of 1 over s whole is F 1 over
s square is equal to T so by first
lifting theorem inverse Laplace
transform of 1 over s plus M whole
square will be equal to PE to the power
minus M T and if it so happens that RL
and C have values such that 1 by LC
becomes less than R square by 4 L square
then the N square which is occurring
here will be replaced by minus K square
and we
then we have Ivar equal to e bar s over
s plus M whole square minus K square
into and so then when we will take the
inverse Laplace transform we shall get I
equal to e upon K L into e to the power
minus MT into sine hyperbolic KT because
we know that the Laplace transform of
sine hyperbolic KT is K over s square
minus K square so when we multiply by e
to the power minus empty and take the
Laplace transform we will get K over s
plus M whole square minus K square and
so inverse Laplace transform here we
will get will give give us I equal to e
by K L into it / - empty into sine
hyperbolic K T if n square is equal to
minus K Square and where we are assuming
that 1 by LC minus R square by 4 L
square is less than 0 now in our next
lecture we shall discuss some more
applications of the Laplace
transformation like the application of
Laplace transformation to bending of
beams and the application of Laplace
transform to the boundary value problems
which occur in the in engineering
mathematics they'll find the L and the
solution of the heat conduction equation
by using the Laplace transformation
method thank you