Damped Harmonic Oscillation

Because of some data technical problems between internet servers it is
suddenly not possible for me to answer the existing thread " Damped harmonic
oscillation" which is the reason to I start a new one. I'm sorry but this
was the only possible way...

"Jan C. Hoffmann" <[Only registered users see links. ]> skrev i en meddelelse
news:bpcrrv$33m$[Only registered users see links. ].aol.com...

The problem is that the one cycle is completely different from another
cycle, but an example could be a function of t^2 followed by another
function t^1 followed by a function of -(t^2).

t [s] x_u [mm]

0/20 0
1/20 5
2/20 15
3/20 30
4/20 50
5/20 75
6/20 100
7/20 125
8/20 150
9/20 175
10/20 200
11/20 220
12/20 235
13/20 245
14/20 240
15/20 245

But maybe you should wait a little to send me your reply, because I have
take a look at the explanation from Anselm in order to find out about you
and Anselm are talking about the same - or maybe you can answer me on that
question ?

Best regards
Torben W. Hansen

"Anselm Proschniewski" <AnselmPro@mercedes-benz-portal.invalid> skrev i en
meddelelse news:bpd100$e6t$[Only registered users see links. ]-felb.debis.de...

Dear Anselm,

Maybe it's possible to answer this thread again - I have tried many times
today without success, due to internet server problems.

You wrote:



If you remember you wrote in a previous message:

M*x'' + D*x' + K*(x-y) = 0 =>
M*x'' + D*x' + K*x = K*y(t) where y(t) is the excitation distance.

I understand this representation of force involving M, D and K combined with
Newtons law F=m*a.

Then you say that the solution is:

x(t) = A + B*t + C*cos(wt) + D*sin(wt) (without damping)

which has to be recalculated for each time sample.

I have three questions as follows:

1) What does A,B,C and D represent ? (A is maybe the initial distance and B
initial velocity ?)
2) Where is y(t), which is the excitation I measure ? ( is y(t) = A+B*t ?)
3) And if the solution upon is without damping, how does it looks like if
the damping is included ?

Best regards
Torben W. Hansen

"Torben W. Hansen" wrote:


A, B, C and D are the four independent coefficients of the solution.
(sorry, I used "D" for the damping matrice too)


Yes. y(t) = A+B*t (linear in parts). For each time step you need new A
and B.

Let's take your excitation for the 1st step. y(t) is a linear function,
y0 = 0 for the beginning and y1 at the end of the step when t=t1.

A = 0 , for y(0) = 0
B = y1/t1 , for x(t) will follow y(t)

The speed is: x'(t) = B - w*C*sin(wt) + w*D*cos(wt)
with w called "omega": w = sqrt(k/m), sqrt = "square root"

Now you have to fullfill both initial conditions:

C = 0 , for x(0) = 0
D = -B/w , for the speed at the beginnig is 0

Then you can compute motion and speed at the end of the first step:

x(t1) = B * t1 + D * sin(wt1)
x'(t1) = B + wD * cos(wt1)

Then you start the next step with new A, B, C and D.

---

With damping the "general" solution is:

x(t) = A + B*t + C*exp(-"delta"t)*cos(wt) + D*exp(-"delta"t)*sin(wt)

with delta = d/(2*m) in 1/sec, d = damping coefficient in N*sec/m
Here w = sqrt(k/m - delta²)

Good Luck.

Anselm aus Stuttgart/Esslingen

"Anselm Proschniewski" <AnselmPro@mercedes-benz-portal.invalid> skrev i en
meddelelse news:bpfgvf$g0s$[Only registered users see links. ]-felb.debis.de...


Thank you very much Anselm,

In this moment I do not have the full overview, but I will work with it and
exercise, in order to obtain better skills.
I have written your explanation into a math tools, which can be seen at
[Only registered users see links. ]
can I ask you to look at the weblink and check it for me ?

I have a noticed that you wrote: w = sqrt(k/m - delta^2) - In another
book I found w = sqrt(delta^2 - k/m) is this a mistake in my book ?

Your explanation looks like an explanation I have seen in literature
concerning Laplace
Transformation. From this literature I understood , Laplace can only be used
to
solve Linear Homogenious Differential Equation, which lead me to this
question:

Can my problem be solved by Laplace ( Then it must be a Linear Homogenious
Differential Equation) ?
- if yes then I will try to dig deep into the Laplace methode
- if no then I have to learn more about solving ODE's in general

Another question is:
I have a limited knowledge about Infinite Series, which can be used to break
down certain kind signal shapes as square, sawtooth, triangle signals etc,
etc... and maybe also my excitation function.
Is it correct understood that the method presented by Jan Hoffmann builds on
the theorem of transforming the excitation function into Fourier Series ?

Now I will intensively read you explanation and and hopefully be able to
program it to C-code in order to build my simulator.

It has not been easy for any of you to help a person with insuffient
knowledge like me, but I think you succeeded - Bravo !
I will continue reading math in order to learn more...

Many thanks to you, Jan and everyone which have been helping me...

Best regards
Torben W. Hansen

"Jan C. Hoffmann" <[Only registered users see links. ]> skrev i en meddelelse
news:bpgieu$4f2$[Only registered users see links. ].aol.com...

Hello Jan - and thanks,

I will look at it very soon, as you see I try to collect as much information
as possible. In this moment I work on understanding the model of Anselm,
because it has been difficult for me to figure out about you and him are
working on the same application or not. Until now I was convinced that you
and Anselm presented two alternative solutions to the same application -
yours based Fourier/Infinite Series or similar - Anselm's based on
Euler-Cauchy polygon method, but now I feel uncertain again. For me it
sounds like both Anselm and you understand my application correctly - is it
possible that you could discuss with Anselm why you think him and you are
talking about a different application ?

Do you think he has studied your drawing Type C ?

Best regards
Torben W. Hansen

"Jan C. Hoffmann" <[Only registered users see links. ]> skrev i en meddelelse
news:bphpl9$2lj$[Only registered users see links. ].aol.com...

Hello Jan,


Yes, but he also wrote:
C*exp(-"delta"t)*cos(wt) + D*exp(-"delta"t)*sin(wt)
You can see the formula on the web here :
[Only registered users see links. ] .
Is this not correct - can this formula together with the "polygon methode"
not solve my problem ?


I do not doubt anything you say. I have to ask Anselm directly and inform
him what you told me,
and then I hope that him and you together could kill one of the methodes, so
I can concentrate understanding the right one - I hope that you understand
my confusion.

I printed out every message that you and Anselm wrote. I dare not taking the
risc to loose any of the recommandations until the methode is clarified.

Now I will try to contact Anselm...

Best regards
Torben W. Hansen

"Anselm Proschniewski" <AnselmPro@mercedes-benz-portal.invalid> skrev i en
meddelelse news:bpfgvf$g0s$[Only registered users see links. ]-felb.debis.de...

Hello again Anselm,

The confusion continues, because Jan wrote:

<Cited from Anselm>
</Cited from Anselm>


I dare not taking the risc to loose any of the recommandations until the
methode is clarified.

I hope that him and you together could kill one of the methodes, so
I can concentrate understanding the right one, it will make me very happy -
Thanks in forward...

Best regards
Torben W. Hansen

"Jan C. Hoffmann" wrote:

[...] Shit OE !

No. I wrote:


The speed x'(t) gets a little complicated, I am too lazy to derivate it
now. Torben, do you know how to derivate this formula exactly ?

x'(t) = B - wC*exp(-"delta"t)*sin(wt) + wD*exp(-"delta"t)*cos(wt)
should be a good approximation if delta is very small.

Anselm aus Stuttgart/Esslingen

"Anselm Proschniewski" <AnselmPro@mercedes-benz-portal.invalid> skrev i en
meddelelse news:bphung$jit$[Only registered users see links. ]-felb.debis.de...

Yes, I know - and I answered to Jan:
C*exp(-"delta"t)*cos(wt) + D*exp(-"delta"t)*sin(wt)
Therfore it would be nice if you and him could clarify the disagreements if
any, which I also suggested to Jan.


now.
No not in this moment.
Before this dialog I had some basic basic knowledge of polynomias, limits,
derivatives and integrals, complex numbers, vectors, infinite series and
more.
I have already in conjunction with our dialog during the last 3 week been
reading more about Laplace transformation, vectors, determinants, linear
combination, parametric representation, matrices, homogenious linear ODE's
and therefore there is a little bit caos in my head. I am not afraid of
spending the necessary time to learn it, but I must know the model to work
at before I can know the sort of math to learn and use. I'm not familiar
with the math I just have learned and therefore it will take me much time to
work out the these equations. Another problem is that I have no teacher -
I'm self studying.


By the way, I put your explanation into formula and graphs at the link
below, will you check it and correct me ?
[Only registered users see links. ] .

Best regards
Torben W. Hansen

"Anselm Proschniewski" <AnselmPro@mercedes-benz-portal.invalid> skrev i en
meddelelse news:bpfgvf$g0s$[Only registered users see links. ]-felb.debis.de...

Hello Anselm,

In order to resume you wrote:

delta = d/(2*m)
w = sqrt(k/m - delta²)
y(t) = A+B*t
A = 0 , for y(0) = 0
B = y1/t1
C = 0, for x(0) = 0
D = -B/w
x(t) = A + B*t + C*exp(-"delta"t)*cos(wt) + D*exp(-"delta"t)*sin(wt)
x'(t) = B - w*C*exp(-"delta"t)*sin(wt) + w*D*exp(-"delta"t)*cos(wt)
should be a good approximation if delta is very small.

How should C be calculated for the next time step ?

---

As I understood :
A: initial distance
B: initial speed

you wrote:

Are C and D vector composants for the harmonic part of the equation - or
what ?

Best regards
Torben W. Hansen