- #1

DaveMan

- 9

- 0

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!

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- Thread starter DaveMan
- Start date

- #1

DaveMan

- 9

- 0

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!

- #2

jamesrc

Science Advisor

Gold Member

- 477

- 1

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

[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

DaveMan

- 9

- 0

I don't understand how you get the charateristic equation.

- #4

HallsofIvy

Science Advisor

Homework Helper

- 43,021

- 970

The linear homogeneous differential equation a y"+ by'+ cy= 0 has

"characteristic equation" ar

More generally, you replace the nth derivative with r

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