2nd order dampened harmonic oscillator (Shot in the dark)

[zaurus]

I have the following equation subject to y(0)=0 and y'(0) = 0

my'' + b y' + k y = C

I have done an experiment where I measured force at given depth for a dampened harmonic oscillator. Is it possible to use the force I measured to solve for displacement and then back out coefficient b for the dampened (y') term?

For instance, measured force = 1000 N

Fspring = k y = E A y / L so for a given spring... properties and length i can solve for the change in spring length, y.


The solution to my'' + b y' + k y = C is of the form y(t) = e^Mt cos (Bt) + e^Nt sin (Bt) + some particular solution (for the C)...or something like that

Since the sine wave dies out over time (due to dampening) for every y value there should be ONE specific corresponding value of time. So basically what I am saying, knowing y can I find a time and then back out b, the coefficient for my y' term?

 

[Hootenanny]

I have the following equation subject to y(0)=0 and y'(0) = 0

my'' + b y' + k y = C

I have done an experiment where I measured force at given depth for a dampened harmonic oscillator. Is it possible to use the force I measured to solve for displacement and then back out coefficient b for the dampened (y') term?

For instance, measured force = 1000 N

Fspring = k y = E A y / L so for a given spring... properties and length i can solve for the change in spring length, y. The solution to my'' + b y' + k y = C is of the form y(t) = e^Mt cos (Bt) + e^Nt sin (Bt) + some particular solution (for the C)...or something like that

The form of the solution will depend on the damping ratio of the system:
\xi = \frac{b}{2\sqrt{mk}}
If \xi = 1 then the solution will be the product of a polynomial and exponential in t. If \xi > 1 then the solution will be purely exponential (real). And if \xi< 1 then the solution will be the product of a trigonometric and exponential function in t.

Since the sine wave dies out over time (due to dampening) for every y value there should be ONE specific corresponding value of time.

This is only true for positive \xi, i.e. for the cases of critical and over-damping. For the case of under damping where the solutions are of the form:
y\left(t\right) = e^{-\xi\omega t}\left\{A\sin\left(\omega^\prime t\right) + B\cos\left(\omega^\prime t\right)\right\}
the solution has the same y value for multiple t values.

 

[Dr.D]

You say, "I have done an experiment where I measured force at given depth for a dampened harmonic oscillator." What sort of osciallator do you have, what does depth refer to, and what force did you measure?

You also said, "what I am saying, knowing y can I find a time and then back out b, the coefficient for my y' term?" How do you know y? What is y? Why would you think that knowing a value for the displacement you could then find a corresponding time? If this oscillator goes through many cycles, and this particular displacement value is repeated many, many times, how would you think that you are going to be able to determine a corresponding value of the time t?

I do hope that you can clarify some of your ideas for me. I really cannot follow you.

 

[zaurus]

The oscillator is a cable with a mass on it being lowered into water. I measured the tension in the cable at a specific depth in the water (at a given depth the mass system resonates...something I want to model). I was thinking that I might be able to find the damping coefficient by solving the ODE backwards...

Since I know tension in the cable, I was going to try to relate it to a spring
T = k y = (E A / L) * y

Solving backwards, y = LT / EA

I then thought maybe I can just assume an over or critically damped case where I have a unique displacement for a given time value.
I thought I could then plug these values into the solution for the assumed solution...back solve for the damping coefficient b.

I could then use this damping coefficient to semi understand the behavior of similar systems (with slightly different masses and spring constants).

Hootenanny brought up the reason why this won't work...I didn't think it all the way through. :(