Spring Damping

1. Nov 17, 2003

DaveMan

so, f = ma, and f = -kx -bv.
rearrange to the form of a differential equation.
i am stuck when the next line simply states the general solution.
i have done this for first differential only, not second derivatives.

2. Nov 18, 2003

jamesrc

For the homogeneous solution to ma = -kx -bv, it is standard practice to find the characteristic equation:

First, rewrite into a standard form:

$\ddot{x} + \frac{b}{m}\dot{x} + \frac{k}{m}x$

Set
$\frac{k}{m} = \omega_n^2$

$\frac{b}{m} = 2\zeta\omega_n$

(the reason why should be clear by the end of the problem; natural frequency and damping ration are useful, meaningful quantities in the study of oscillations)

characteristic equation:

$s^2 + 2\zeta\omega_n s + \omega_n^2 = 0$

find the roots of the characteristic equation (it's just a quadratic in s), s 1,2 , so that the solution to the differential equation is written:

$x(t) = C_1 e^{s_1 t} + C_2 e^{s_2 t}$

using the Euler identity and some algebra, you end up with the solution:

$x(t) = A e^{-\zeta\omega_n t}\cos\left(\omega_d t + \phi \right)$

where the damped frequency $\omega_d = \omega_n \sqrt{1-\zeta^2}$ and the constants A and &phi; (magnitude and phase) are determined by the initial conditions. (You can solve it with a sine or cosine, you'll just end up with a different phase.) That's just the basics off the top of my head, but I hope that helps.

3. Dec 3, 2003

DaveMan

I dont understand how you get the charateristic equation.

4. Dec 3, 2003

HallsofIvy

Staff Emeritus
That's usually one of the first things you learn in an introductory differential equation course.

The linear homogeneous differential equation a y"+ by'+ cy= 0 has
"characteristic equation" ar2+ br+ c= 0.

More generally, you replace the nth derivative with rn.