Differential equations

**Apprentice123** · November 30th 2009, 06:34 AM

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

**lvleph** · November 30th 2009, 06:57 AM

You could search the net for spring dashpot.

**Apprentice123** · November 30th 2009, 07:07 AM

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:

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

**Apprentice123** · November 30th 2009, 07:54 AM

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?

**lvleph** · November 30th 2009, 08:50 AM

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

**Apprentice123** · November 30th 2009, 09:34 AM

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?

**lvleph** · November 30th 2009, 12:50 PM

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.

**Apprentice123** · November 30th 2009, 01:30 PM

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

**lvleph** · November 30th 2009, 02:06 PM

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
$C_1 = \dfrac{w_2}{w_2 - w_1} L$

**Apprentice123** · November 30th 2009, 02:17 PM

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)?

**lvleph** · November 30th 2009, 02:21 PM

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.

**Apprentice123** · November 30th 2009, 02:37 PM

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 ?

**lvleph** · November 30th 2009, 02:53 PM

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.

**Apprentice123** · December 1st 2009, 05:57 AM

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 !!!

**lvleph** · December 1st 2009, 06:05 AM

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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