Problem
Calculate the solution to with IC
and plot the
solution.
Answer
Hence and
, hence
, hence this is underdamped system.
Since , then
We have 2 particular solutions. The first is due to
and the second
is due to
. When the
forcing function is
, we guess
and when the forcing function is the response is
From we find
and
and plug these into
to find
and
, next we find
by
using the IC, and then at the end we add the solution
. Notice that
do not enter into the calculation of
since the impulse
is not effective at
.
Hence
Hence and
. This results in
Hence
Therefore
Now we use IC's to find . At
we obtain
And
At we have
But , Hence
then the above becomes
Hence the final solution is
Substitute values for the parameters above we obtain
This is a plot of the solution superimposed on the forcing functions
The magnitude of the impulse resulting when the mass hits the ground is given by the change of momentum that occurs. Hence
But assuming the mass is dropped from rest, hence , and
where
where
is the
height that mass falls. Hence
Hence the equation of motion is
Since underdamped, , hence the solution is
Taking as time of impact.
Problem
Compute response of the system with IC
and
. Plot the response.
Answer
Where , hence
and
, hence
. This is an underdamped system.
, Hence
The response due to the forcing function is given by
The response due to the other forcing function is given by
Now we determine from IC's
Hence Now to find
But , hence from the above, and noting that
Hence
Therefore
Now we can combine the above solution to obtain the final solution
Substitute numerical values for the above parameters, we obtain
This is a plot of the response
Let the response by . Hence
, where
is the particular solution, which is the response
due the the above forcing function. Using convolution
Where is the unit impulse response of a second order underdamped system which is
hence
Using then
Then the integral becomes
Consider the first integral where
Integrate by parts, where , Let
and let
, hence
Integrate by parts again the last integral above, where , Let
and
let
, hence
Substitute (2) into (1) we obtain
Hence
Now consider the second integral where
Integrate by parts, where , Let
and let
, hence
Integrate by parts again the last integral above, where , Let
and
let
, hence
Substitute (4) into (3) we obtain
Hence
Using the above expressions for , we find (and multiplying the solution by
since
the force is only active from
to
, we obtain
Hence
But
and
Hence can now be written as
And
Hence the overall solution is
The above solution is a bit long due to integration by parts. I will not solve the same problem using Laplace transformation method. The differential equation is
Take Laplace transform, we obtain (assuming and
)
Now we find Laplace transform of
Integration by parts gives
![]() | (8) |
Substitute (8) into (7) we obtain
Hence
Now we can use inverse Laplace transform on the above. It is easier to do partial fraction decomposition
and use tables. I used CAS to do this and this is the result. I plot the solution . I used the following
values to be able to obtain a plot
The acceleration of the mass is measured w.r.t. to the inertial frame, but the spring length is measured relative
to the ground which is moving with displacement
, hence the equation of motion of the mass
is given
by
Therefore
![]() | (1) |
Where is given as
The solution to (1) is given by where
can be found using convolution, and
is
as usual given by
Let us first find . Note that the impulse response
to undamped system is given by
Hence for ,
Integration by parts, where
,
, hence
, therefore
(2) becomes
For
For the first integral in (3), we obtain
For the second integral in (3) we obtain
For the third integral in (3) we obtain
Integration by parts gives
Hence
For
For the first integral in (4), we obtain
For the second integral in (4) we obtain
For the third integral in (4) we obtain
Integration by parts gives
Hence
Hence, the overall response is, assuming zero initial conditions, is given by
Noting that the above becomes
This is a plot of the solution superimposed on top of the forcing function
Let be the function shown above. Let
be its approximation using Fourier series. Hence
Where is the period of
and
For we see that
and
for
, hence
And
And
Hence
These are few terms in the series
This is a plot of the above for increasing number of
Problem
Solve the following system using Laplace transform where the units are in
Newtons and the initial conditions are both zero.
Answer
Divide the equation by we obtain
Let hence the equation becomes
Applying Laplace transform
But due to zero initial conditions, the above simplifies to
From tables, the inverse Laplace transform of is
, but
Hence, letting we see that inverse laplace transform of
is the same as the inverse laplace
transform of
which is
But , hence
or
Problem
Calculate the response spectrum of an undamped system to the forcing function
assuming zero initial conditions.
Answer
Solution sketch: Find the response of the system to the above input. Then find
where this response
is maximum, call this
, then plot
vs.
The system EQM is
For ,
Guess , hence
and
,
hence substitute these into the EQM and compare, we obtain
The input is half sin where , hence
hence the above becomes
Hence and
or
Then the solution becomes
And since then
and take derivative we obtain
And since then the above results in
Hence the solution becomes
Hence
![]() | (1) |
Now we need to find where the maximum is. Take derivative, and set it to zero, we obtain
For , we need to solve
Using , then the above becomes
Hence, either or
for
or the time
which makes the maximum
is one of the following
We now need to find which one of the above 2 solution gives a larger maximum. Using the first solution
, then (1) becomes
And at then (1) becomes
Need now to find which of the above is larger. Let us take the difference and see if the result is positive or negative (is there an easier way?)
Not sure how to continue. Now let us look at The solution here is
But with IC given by and
hence from (1)
and
Hence
![]() | (3) |
And
![]() | (4) |
We need to solve (3) and (4) for and
Combining (3) and (4) we obtain
This is in the form , solve for
we obtain
Hence
And
Ask about the above, why can't I get the answer shown in notes?
To find the response use convolution. Since this is an undamped system, then the impulse response
is
Hence, for
Using , then
Hence the convolution integral becomes
And for
As was done earlier, perform integration by parts, we obtain
But and
, hence the above becomes
Therefore, the final solution is
![]() | (1) |
We can simplify the above more as follows
![]() | (1) |
Hence
![]() | (1) |
Hence
![]() | (1) |
or
![]() | (1) |
Hence
![]() | (1) |
To find where is, we need to find
. Take the derivative, we obtain
Now let for
to find
.
But
Substitute the above into (2) we obtain
Divide by
Hence, the hypotenuse is and so
and
and using these into (1) we find
when
as
But , hence
Hence
Hence
Where
A plot of vs.
gives the response spectrum
Problem Calculate the compliance transfer function for a system described by
where
is the input and
is the displacement.
Answer
Take Laplace transform (assuming zero IC) we obtain
Hence
Hence
Problem
Calculate the frequency response function for the system of problem 3.49 for
Answer
Let
Hence
and
This is a plot of the magnitude and phase