Question

anyone know how to deal with graphs like the one on this site: http://www.ltcconline.net/greenl/courses/204/appsHigherOrder/forcedVibrations.htm the one right before "Forced vibrations with damping"

Answer 1

What is your goal?

Answer 2

I agree with nowhereman's question. We can deal with graphs like these, but what do you need from these graphs?

Answer 3

Thanks for responding! I am given a graph that looks similar to what I have linked you to but need help finding the equation to it, I am given some points would you like them?

Answer 4

Can you append it as an image?

Answer 5

no but it looks like what i sent basically

Answer 6

Ok, so you have to decide what kind of periodicity you have, choose a normal form with some variables, like for example \(f(t) = at\sin(bt+c)\) and then insert the known values for t and f(t) to find a,b and c.

Answer 7

okay can I take a sec to type the points I am given i think once you see the points i have you can be more specific?

Answer 8

Ok, we can try that. In the mean time I'm gonna grab something to eat ;-)

Answer 9

ok one sec typing them real fast

Answer 10

(-.75,0), (-.5,-.7906), (-.25, -.8274), (0,0), (.25, .94868), (.75,0), (1, -1.039), (1.25, -1.088), (1.5,0), (1.75, 1.1915), (2, 1.2471), (2.25,0)

Answer 11

there you go :)

Answer 12

hopefully it helps

Answer 13

You should be able to see, that your data has a periodicity of \(1.5\).

Answer 14

yes I see that, so what is the next step that is where I am lost :)

Answer 15

sorry not sure why that posted 3 times

Answer 16

You assume that the maxima behave like a linear function \(a+bt\) and then you get \(f(t) = (a+bt)\sin(\frac{2π}{1.5}t)\) And you only have to insert some value, to see what a and b are.

Answer 17

what do I plug in to solve for a and b..well your b value would be whater 2pi/1.5 is correct? so I just need "a"?

Answer 18

No, \(\frac{2π}{1.5}\) is for the periodicity and \(b\) is the slope of the maxima. That's something different. Just insert two non-trivial value pairs and solve the resulting system of linear equations.

Answer 19

what do you mean by two non-trivial pairs can you look at my data and give me examples of what you mean

Answer 20

You can choose any of them. But if \(f(t) = 0\) you will only have \(0 = 0\), that's what I mean by trivial.

Answer 21

oh so can i do 2.25,0 and .75,0?

Answer 22

Just try and see what you get.

Answer 23

ok hold on so do i solve for a or b first or does it not matter

Answer 24

is "t" x?

Answer 25

In physics you often have pairs of values denoted by \((t, f(t))\). But you can of course also call them \((x, f(x))\). And solving the system of linear equations does not demand anything on the order.

Answer 26

ok can you walk me through sorry im just a little confused with all these variables..my point of choice is (.75,0) so 0=[a+b(.75)]sin(2pi/1.5)(.75)?

Answer 27

I bet you can simplify that equation. You plugged in the values correctly.

Answer 28

ok i think it simplifies to 0=a+.75bsin(pi) ??

Answer 29

where does the b go

Answer 30

ok wait sin(pi)=0 so that just leave sme with 0=a+.75b right?

Answer 31

leaves me *

Answer 32

I read that wrong, you need parentheses: \(0 = (a+0.75b)\sin(π)\)
So now you can insert the value of \(\sin\).

Answer 33

what do you mean sin(pi) is 0?

Answer 34

Yes. That's what I tried to tell you above, but now you found out yourself. The resulting equation \(0=0\) tells you nothing about a and b, so you have to choose another point. Apart from that the way is good.

Answer 35

hmm well i guess it can't be any other x-intercept so in other words (2.25,0) right?

Answer 36

Yes :-)

Answer 37

ok lets try .5, .94868?

Answer 38

or would 2, 1.2471 be easier

Answer 39

that would give us 1.2471=(a+2b)sin(pi)

Answer 40

is that better?

Answer 41

except that it's \(\sin(\frac{4π}{1.5})\), ;-)

Answer 42

why

Answer 43

?

Answer 44

....i thought it was 2pi/1.5t

Answer 45

?

Answer 46

you there?

Answer 47

\[\frac{2π}{1.5}\cdot 2 = \frac{4π}{1.5}\]

Answer 48

oh okay, so sin(4pi/1.5) = .8660 so i have now: 1.2471=(a+2b)(.8660) so that becomes 1.2471=.8660a+1.732b?

Answer 49

:/

Answer 50

Yeah, do that for another point and you can calculate a and b.

Answer 51

ok im doing (1, -1.039) and that came out to...-1.039=(a+b)(-.8660) so -1.039=-.8660a-.8660b?

Answer 52

looks good.

Answer 53

hm i dont think that worked?

Answer 54

If you divide the factor on the other side you get equations \(a+b = C_1\) and \(a+2b = C_2\) with some numbers \(C_i\). Then you can solve for a and b.

Answer 55

??

Answer 56

so divide one equation by the other?

Answer 57

1.2471=.866a + 1.732b/-1.039=-.866a-.866b i got...-1.200= -a-2b?

Answer 58

:/

Answer 59

You get \[1.2471=(a+2b)\cdot 0.8660 ⇒ a+2b = \frac{1.2471}{0.866} =: C_1\] and \[-1.039=(a+b)\cdot (-0.8660) ⇒ a+b = \frac{1.039}{0.866} =: C_2\] So then \[b = (a+2b) - (a+b) = C_1 - C_2 = \frac{0.2081}{0.866}\] and thus \[a = (a+b) - b = C_2 - (C_1 - C_2) = 2C_2 - C_1 = \frac{0.8309}{0.866}\] if I calculated it correctlyj

Answer 60

ok hold on so whats a and b then

Answer 61

It's standing right there... :-(

Answer 62

the point was (1, -1.039) so a +b = -1.039/.866 then right? as c2?

Answer 63

Just look what I wrote there. You'll see I took your equations and ended it.

Answer 64

what do you mean ended it? the point was (1, -1.039) you have it as positive 1.039

Answer 65

in that case a = .959 and b=.866 so y=.959sin(.866x)? the points im getting dont match up with the ones im given

Answer 66

Sorry, I'm to tired to do this discussion with you now. Just read thoroughly what I wrote. That equation is certainly not what results from our ansatz.

Answer 67

oh wait okay, its y=(.9594+.2403x)sin(2pi/1.5x)?

Answer 68

with that i get (1.75, 1.1951) not (1.75, 1.1915)? :/ hm i wonder if its close enough though

Answer 69

All the calculations and also the given values where only approximations, so you should not expect getting exactly the same values back. Just check that all of the points are close enough.

Answer 70

is (-.25, -.7788) close enough its supposed to be -.8274

Answer 71

WAIT i think we might have made a small error..a=(a+b)−b=C2−(C1−C2)=2C2−C1=0.83090.866 how is c2-(c1-c2) = 2c2-c1?

Answer 72

i think that last line might be wrong slightly