Spring Damping

  • Thread starter DaveMan
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  • #1
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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.
please help!
 

Answers and Replies

  • #2
jamesrc
Science Advisor
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For the homogeneous solution to ma = -kx -bv, it is standard practice to find the characteristic equation:

First, rewrite into a standard form:

[itex] \ddot{x} + \frac{b}{m}\dot{x} + \frac{k}{m}x [/itex]

Set
[itex] \frac{k}{m} = \omega_n^2 [/itex]

[itex] \frac{b}{m} = 2\zeta\omega_n [/itex]

(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:

[itex] s^2 + 2\zeta\omega_n s + \omega_n^2 = 0 [/itex]

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:

[itex] x(t) = C_1 e^{s_1 t} + C_2 e^{s_2 t} [/itex]

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

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

where the damped frequency [itex] \omega_d = \omega_n \sqrt{1-\zeta^2} [/itex] and the constants A and φ (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
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I dont understand how you get the charateristic equation.
 
  • #4
HallsofIvy
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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.
 

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