1. ## Differential equations

Any hint of how to represent a mass spring system with vibration damped?

2. You could search the net for spring dashpot.

3. Originally Posted by lvleph
You could search the net for spring dashpot.

I need to find the equation for damped system.

I found without damping:

$y(t) = y.coswt + \frac{v.}{w}sinwt$

4. I find damped system:

damping force. $Fd = -c \frac{dx}{dt}$

For the spring mass system with damping force through a fluid in the opposite direction of motion of the mass

$m \frac{dx^2}{dt^2} = -kx - c \frac{dx}{dt}$

$m \frac{dx^2}{dt^2} + c \frac{dx}{dt} + kx = 0$

Auxiliary equation

$mw^2 + cw + k = 0$

$w1 = \frac{-c + \sqrt{c^2-4mk}}{2m}$

$w2 = \frac{-c - \sqrt{c^2-4mk}}{2m}$

The system overdamped
$c^2 - 4mk > 0$

Then:

$x = C1e^{w1t} + C2e^{w2t}$

How do I determine C1 and C2?

5. That c is inherent to the system, i.e., it come from experimental data. Therefore, in a mathematics problem it should be given.

6. Originally Posted by lvleph
That c is inherent to the system, i.e., it come from experimental data. Therefore, in a mathematics problem it should be given.
C is inherent to the system.

I do not understand, C1 and C2 are what?

7. Oh sorry, those are determined by initial conditions. You should have initial conditions that say something like $x(0) = f(x)$ and $\frac{dx(0)}{dt}=g(x).$ With these initial conditions you can solve for both $C_1$ and $C_2.$ They are basically constants of integration that define the family of solutions.

8. Originally Posted by lvleph
Oh sorry, those are determined by initial conditions. You should have initial conditions that say something like $x(0) = f(x)$ and $\frac{dx(0)}{dt}=g(x).$ With these initial conditions you can solve for both $C_1$ and $C_2.$ They are basically constants of integration that define the family of solutions.

I understand what you said. But if x is function of time. $x(t)$

I do not know which determine the condition for x(0), ie t = 0

9. Well, $x(0)$ would be something given. Say you know the spring is extended a distance of $L$ at the initial time then we know $x(0) = L$ and therefore
$x(0) = C_1e^{w_1 \cdot 0} + C_2e^{w_2\cdot 0} = C_1 + C_2 = L.$
Because there are two unknowns we needed the second initial condition $x_t(0) = g(x).$ Say, $g(x) = 0$ then we can solve for $C_1,C_2$ using
$x_t(0) = w_1\cdot C_1 e^{w_1\cdot 0} + w_2 \cdot C_2 e^{w_2\cdot 0} = w_1\cdot C_1 + w_2 \cdot C_2 = 0$
Thus,
$C_1 = -\frac{w_2}{w_1} C_2$
then substituting this into $x(0) = C_1 + C_2 = L$
we have
$C_2 = \dfrac{w_1}{w_1 - w_2} L$
and
$C_1 = \dfrac{w_2}{w_2 - w_1} L$

10. Originally Posted by lvleph
Well, $x(0)$ would be something given. Say you know the spring is extended a distance of $L$ at the initial time then we know $x(0) = L$ and therefore
$x(0) = C_1e^{w_1 \cdot 0} + C_2e^{w_2\cdot 0} = C_1 + C_2 = L.$
Because there are two unknowns we needed the second initial condition $x_t(0) = g(x).$ Say, $g(x) = 0$ then we can solve for $C_1,C_2$ using
$x_t(0) = w_1\cdot C_1 e^{w_1\cdot 0} + w_2 \cdot C_2 e^{w_2\cdot 0} = w_1\cdot C_1 + w_2 \cdot C_2 = 0$
Thus,
$C_1 = -\frac{w_2}{w_1} C_2$
then substituting this into $x(0) = C_1 + C_2 = L$
we have
$C_2 = \dfrac{w_1}{w_1 - w_2} L$
and
$C_1 = \dfrac{w_2}{w_2 - w_1} L$

Sorry, it did not understand the second initial condition, which is g (x)?

11. Just like the initial condition $x(0) = f(x)$ is given $x_t(0) = g(x)$ will be given. They are defined by the problem. What are you trying to do? I will be of more help if I knew specifics.

12. Originally Posted by lvleph
Just like the initial condition $x(0) = f(x)$ is given $x_t(0) = g(x)$ will be given. They are defined by the problem. What are you trying to do? I will be of more help if I knew specifics.
Is a deduction of the vibration model.

My solution so far

Damped spring mass system

$mx'' = -kx - cx'$
$mx'' + cx' + kx = 0$

$x(t) = Acoswt + Bsinwt$
$x'(t) = -Awsinwt + Bwcoswt$
$x''(t) = -Aw^2coswt - Bw^2sinwt$
c is damping constant

Substituting I can not find: $mw^2 + cw + k =0$
But is the correct equation, or not ?

13. Okay, the way to solve this differential equation is by finding the roots to the equation
$m w^2 + cw + k = 0$
like you wanted to. This can be done using the quadratice formula, i.e.,
$w = \dfrac{-c \pm \sqrt{c^2 - 4mk}}{2m}$
Unfortunately, you do not no if $w \in \mathbb{R}, w \in \mathbb{C}$ or $w$ 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 $w$ are repeated real roots then the solution is of the form
$x(t) = C_1 + C_2 t.$
If $w \in \mathbb{R}$ the solution is of the form
$x(t) = C_3 e^{w_1t} + C_4e^{w_2t}.$
Finally, if $w \in \mathbb{C}$ the solution is of the form
$x(t) = C_5 \cos w t + C_6 \sin w t.$

The constants $C_1,C_2,C_3,C_4,C_5,C_6$ 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.

14. Originally Posted by lvleph
Okay, the way to solve this differential equation is by finding the roots to the equation
$m w^2 + cw + k = 0$
like you wanted to. This can be done using the quadratice formula, i.e.,
$w = \dfrac{-c \pm \sqrt{c^2 - 4mk}}{2m}$
Unfortunately, you do not no if $w \in \mathbb{R}, w \in \mathbb{C}$ or $w$ 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 $w$ are repeated real roots then the solution is of the form
$x(t) = C_1 + C_2 t.$
If $w \in \mathbb{R}$ the solution is of the form
$x(t) = C_3 e^{w_1t} + C_4e^{w_2t}.$
Finally, if $w \in \mathbb{C}$ the solution is of the form
$x(t) = C_5 \cos w t + C_6 \sin w t.$

The constants $C_1,C_2,C_3,C_4,C_5,C_6$ 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 $m w^2 + cw + k = 0$

$m(-Aw^2coswt - Bw^2sinwt) + c(-Awsinwt + Bwcoswt) + k(Acoswt + Bsinwt) = 0$
$(mw^2+cw+k)(-Acoswt -Bsinwt -Asinwt + Bcoswt + Acoswt + Bsinwt) = 0$
$(mw^2+cw+k)(-Asinwt+Bcoswt) = 0$
This equation not correct !!!

15. I guess I don't understand what you are asking. If we assume that $-A\sin wt + B \cos wt \ne 0$ this leads you to conclude the only time this is zero is when $mw^2 + cw + k = 0.$ Therefore, the we must solve for the roots of the characteristic equatioin $mw^2 + cw + k = 0.$ Again, I am not sure what you are having issues with, and so I am sorry I am not of much help.

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