Any hint of how to represent a mass spring system with vibration damped?
I find damped system:
damping force. $\displaystyle Fd = -c \frac{dx}{dt}$
For the spring mass system with damping force through a fluid in the opposite direction of motion of the mass
$\displaystyle m \frac{dx^2}{dt^2} = -kx - c \frac{dx}{dt}$
$\displaystyle m \frac{dx^2}{dt^2} + c \frac{dx}{dt} + kx = 0$
Auxiliary equation
$\displaystyle mw^2 + cw + k = 0$
$\displaystyle w1 = \frac{-c + \sqrt{c^2-4mk}}{2m}$
$\displaystyle w2 = \frac{-c - \sqrt{c^2-4mk}}{2m}$
The system overdamped
$\displaystyle c^2 - 4mk > 0$
Then:
$\displaystyle x = C1e^{w1t} + C2e^{w2t}$
How do I determine C1 and C2?
Oh sorry, those are determined by initial conditions. You should have initial conditions that say something like $\displaystyle x(0) = f(x)$ and $\displaystyle \frac{dx(0)}{dt}=g(x).$ With these initial conditions you can solve for both $\displaystyle C_1$ and $\displaystyle C_2.$ They are basically constants of integration that define the family of solutions.
Well, $\displaystyle x(0)$ would be something given. Say you know the spring is extended a distance of $\displaystyle L$ at the initial time then we know $\displaystyle x(0) = L$ and therefore
$\displaystyle x(0) = C_1e^{w_1 \cdot 0} + C_2e^{w_2\cdot 0} = C_1 + C_2 = L.$
Because there are two unknowns we needed the second initial condition $\displaystyle x_t(0) = g(x).$ Say, $\displaystyle g(x) = 0$ then we can solve for $\displaystyle C_1,C_2$ using
$\displaystyle x_t(0) = w_1\cdot C_1 e^{w_1\cdot 0} + w_2 \cdot C_2 e^{w_2\cdot 0} = w_1\cdot C_1 + w_2 \cdot C_2 = 0$
Thus,
$\displaystyle C_1 = -\frac{w_2}{w_1} C_2$
then substituting this into $\displaystyle x(0) = C_1 + C_2 = L$
we have
$\displaystyle C_2 = \dfrac{w_1}{w_1 - w_2} L$
and
$\displaystyle C_1 = \dfrac{w_2}{w_2 - w_1} L$
Is a deduction of the vibration model.
My solution so far
Damped spring mass system
$\displaystyle mx'' = -kx - cx'$
$\displaystyle mx'' + cx' + kx = 0$
$\displaystyle x(t) = Acoswt + Bsinwt$
$\displaystyle x'(t) = -Awsinwt + Bwcoswt$
$\displaystyle x''(t) = -Aw^2coswt - Bw^2sinwt$
c is damping constant
Substituting I can not find: $\displaystyle mw^2 + cw + k =0$
But is the correct equation, or not ?
Okay, the way to solve this differential equation is by finding the roots to the equation
$\displaystyle m w^2 + cw + k = 0$
like you wanted to. This can be done using the quadratice formula, i.e.,
$\displaystyle w = \dfrac{-c \pm \sqrt{c^2 - 4mk}}{2m}$
Unfortunately, you do not no if $\displaystyle w \in \mathbb{R}, w \in \mathbb{C}$ or $\displaystyle w$ is a repeated root and each one of these cases leads to a different general solution. Therefore, you would have to solve the differential equation using each of these cases and then add all the solutions together.
If $\displaystyle w$ are repeated real roots then the solution is of the form
$\displaystyle x(t) = C_1 + C_2 t.$
If $\displaystyle w \in \mathbb{R}$ the solution is of the form
$\displaystyle x(t) = C_3 e^{w_1t} + C_4e^{w_2t}.$
Finally, if $\displaystyle w \in \mathbb{C}$ the solution is of the form
$\displaystyle x(t) = C_5 \cos w t + C_6 \sin w t.$
The constants $\displaystyle C_1,C_2,C_3,C_4,C_5,C_6$ are all determined by the initial conditions.
EDIT: I am sorry this keeps getting more and more complicated, but your questions are so general they have to have a general answer.
But I not find the equation $\displaystyle m w^2 + cw + k = 0$
$\displaystyle m(-Aw^2coswt - Bw^2sinwt) + c(-Awsinwt + Bwcoswt) + k(Acoswt + Bsinwt) = 0$
$\displaystyle (mw^2+cw+k)(-Acoswt -Bsinwt -Asinwt + Bcoswt + Acoswt + Bsinwt) = 0$
$\displaystyle (mw^2+cw+k)(-Asinwt+Bcoswt) = 0$
This equation not correct !!!
I guess I don't understand what you are asking. If we assume that $\displaystyle -A\sin wt + B \cos wt \ne 0$ this leads you to conclude the only time this is zero is when $\displaystyle mw^2 + cw + k = 0.$ Therefore, the we must solve for the roots of the characteristic equatioin $\displaystyle mw^2 + cw + k = 0.$ Again, I am not sure what you are having issues with, and so I am sorry I am not of much help.