# Mass-spring system with damping

• Dec 18th 2009, 09:36 AM
Apprentice123
Mass-spring system with damping
The differential equation system is

$\displaystyle m \frac{d^2x}{dt^2} + \gamma \frac{dx}{dt} + kx(t) = F_o cos(wt)$

To solve this differential equation requires a particular solution xp(t) and two fundamental solutions of the homogeneous equation corresponding X1(t) and X2(t)

$\displaystyle x(t) = xp(t) + C1x1(t) + C2x2(t)$

I find X1(t) and X2(t) for the following cases

Overdamped:

$\displaystyle x(t) = C1e^{r1t} + C2e^{r2t}$

$\displaystyle r1 = \frac{ - \gamma + \sqrt{( \gamma )^2 -4mk}}{2m}$
$\displaystyle r2 = \frac{ - \gamma - \sqrt{( \gamma )^2 -4mk}}{2m}$

Underdamped

$\displaystyle x(t) = e^{-(\frac{c}{2m})t} (C1cos(rt) + C2sin(rt))$

r -> imaginary part
$\displaystyle r = \frac{ \sqrt{( \gamma )^2 -4mk}}{2m}$

How do I find the particular solution xp(t) ?
• Dec 19th 2009, 12:19 AM
CaptainBlack
Quote:

Originally Posted by Apprentice123
The differential equation system is

$\displaystyle m \frac{d^2x}{dt^2} + \gamma \frac{dx}{dt} + kx(t) = F_o cos(wt)$

To solve this differential equation requires a particular solution xp(t) and two fundamental solutions of the homogeneous equation corresponding X1(t) and X2(t)

$\displaystyle x(t) = xp(t) + C1x1(t) + C2x2(t)$

I find X1(t) and X2(t) for the following cases

Overdamped:

$\displaystyle x(t) = C1e^{r1t} + C2e^{r2t}$

$\displaystyle r1 = \frac{ - \gamma + \sqrt{( \gamma )^2 -4mk}}{2m}$
$\displaystyle r2 = \frac{ - \gamma - \sqrt{( \gamma )^2 -4mk}}{2m}$

Underdamped

$\displaystyle x(t) = e^{-(\frac{c}{2m})t} (C1cos(rt) + C2sin(rt))$

r -> imaginary part
$\displaystyle r = \frac{ \sqrt{( \gamma )^2 -4mk}}{2m}$

How do I find the particular solution xp(t) ?

Use a trial solution for the particular solution of the form:

$\displaystyle x_p(t)=A\cos(\omega t) +B \sin(\omega t)$

CB
• Dec 19th 2009, 04:49 AM
HallsofIvy
Quote:

Originally Posted by CaptainBlack
Use a trial solution for the particular solution of the form:

$\displaystyle x_p(t)=A\cos(\omega t) +B \sin(\omega t)$

CB

As long as $\displaystyle \omega\ne \frac{ \sqrt{( \gamma )^2 -4mk}}{2m}$.

If $\displaystyle \omega$ is equal to that you will need to try
$\displaystyle x_p(t)= A t cos(\omega t)+ B t sin(\omega t)$
• Dec 19th 2009, 07:04 AM
Apprentice123
The damped

$\displaystyle m \frac{d^2x}{dt^2} + \gamma \frac{dx}{dt} + kx(t) = 0$

The characteristic equation

$\displaystyle mr^2 + \gamma r + k = 0$

$\displaystyle r1 = \frac{ - \gamma + \sqrt{( \gamma )^2 -4mk}}{2m}$
$\displaystyle r2 = \frac{ - \gamma - \sqrt{( \gamma )^2 -4mk}}{2m}$

In overdamped
$\displaystyle ( \gamma )^2 -4mk > 0$

What are the steps now to find the solution:
$\displaystyle X(t) = C1e^{r1t} + C2^{r2t}$
???

And how do I get the equations in the case of mechanical vibrations: Beats and resonance ???
• Dec 19th 2009, 02:31 PM
mr fantastic
Quote:

Originally Posted by Apprentice123
The damped

$\displaystyle m \frac{d^2x}{dt^2} + \gamma \frac{dx}{dt} + kx(t) = 0$

The characteristic equation

$\displaystyle mr^2 + \gamma r + k = 0$

$\displaystyle r1 = \frac{ - \gamma + \sqrt{( \gamma )^2 -4mk}}{2m}$
$\displaystyle r2 = \frac{ - \gamma - \sqrt{( \gamma )^2 -4mk}}{2m}$

In overdamped
$\displaystyle ( \gamma )^2 -4mk > 0$

What are the steps now to find the solution:
$\displaystyle X(t) = C1e^{r1t} + C2^{r2t}$
???

And how do I get the equations in the case of mechanical vibrations: Beats and resonance ???

You have found the homogenous solution. Now you have to add to it the particular solution. You have already been told how to find the particular solution. All this stuff will be found in any decent textbook that covers second order differential equations with constant coefficients (go to your school library) and I'm certain that Google will turn up pages and pages of links.

Your problem is attempting abstract application questions like these before developing a sufficient grasp of the basic theory in a sufficient number and variety of concrete cases. You are advised to go back and review that theory.