SHM Math Explodes (Req: help finding Real parts)

  • Thread starter Thread starter malweth
  • Start date Start date
  • Tags Tags
    parts Shm
Click For Summary

Homework Help Overview

The discussion revolves around the mathematical treatment of simple harmonic motion (SHM) in the context of a damped system, specifically focusing on the ordinary differential equation (ODE) governing the motion of a mass attached to a spring and dashpot. The original poster is exploring the energy lost due to damping and is attempting to derive real values from the complex solutions of the ODE.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to substitute complex solutions back into the equation to extract real parts for energy calculations. They express confusion over the complexity of their results compared to examples in their reference materials. Some participants question the correctness of the original poster's approach to the solution and the handling of constants of integration.

Discussion Status

The discussion is active, with participants providing feedback on the original poster's attempts. There is acknowledgment of a missing constant of integration, and some participants suggest that the original poster is on the right track regarding the transition to real space for energy calculations.

Contextual Notes

The original poster notes that they are working from specific textbooks, which may influence their expectations of the solution's complexity. There is an indication of a potential misunderstanding regarding the integration constants in the context of a second-order ODE.

malweth
Messages
3
Reaction score
0

Homework Statement


I'm not attempting a specific problem, I'm just trying to find a correct way of doing the math behind simple harmonic motion ODEs...

The example problem I've given myself (based on the books I'm using) is a dashpot with constant [tex]c[/tex], spring with constant [tex]k[/tex] and a mass [tex]m[/tex].

Specifically, I'm trying to understand the energy lost by the dashpot, but I'm in the "Calculus" subforum for a reason:

Homework Equations



(variables [tex]/dt[/tex])
[tex]m\ddot x + c\dot x + kx = 0[/tex] - (Unforced, damped SHM).
which is eqivalent to:
[tex]\ddot x + \gamma\dot x + \omega_0^2 x = 0[/tex]

The solution is
[tex]x = \hat A e^{\alpha t}[/tex] and is differentiated & substituted back into get:

[tex]\alpha^2 + \gamma\alpha + \omega_0^2 = 0[/tex]

which is easy to solve for alpha:

[tex]\alpha = -\gamma / 2 \pm j\sqrt{\omega_0^2 - \gamma^2/4}[/tex]

The Attempt at a Solution


The problem I have is in substituting back in... at first it's not too bad, but I'm trying to get real values out... even with letting [tex]\omega_r = \sqrt{\omega_0^2 - \gamma^2/4}[/tex] it gets complicated quickly in my pen & paper scratches...

[tex]x = \hat A e^{-\gamma/2}(e^{j\omega_r t} + e^{-j\omega_r t})[/tex]

gets differentiated to find velocity (making it more complicated, but manageable), and then "de-Eulered" to find a real part. Am I doing something wrong? I don't see anything half as complicated in the books I'm using (primary book is "Fundamentals of Acoustics," by Kinsler and my additional resource is from Feynman's first volume on physics).

The books make it seem as though I'm doing something wrong, but I can't find how. Any help would be appreciated!
 
Physics news on Phys.org
malweth said:
[tex]x = \hat A e^{-\gamma/2}(e^{j\omega_r t} + e^{-j\omega_r t})[/tex]

This is wrong. You've lost the decaying exponential part. Also, you only have one constant of integration. In a 2nd order ODE you must have 2 constants of integration in your general solution.

It should be:

[tex]x(t)=e^{-\gamma t/2}\left(Ae^{j\omega_rt}+Be^{-j\omega_rt}\right)[/tex]
 
Thanks! I forgot about the other constant. The missing [tex]t[/tex] value in the exponent was a mistype.

Am I otherwise doing things correctly when trying to get back into real space (for Energy calculations)?
 
It looks to me like you are.
 
great! Thanks for the help.
 

Similar threads

Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
9
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 21 ·
Replies
21
Views
4K