2nd Order DE with undamped motion

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    2nd order Motion
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Homework Help Overview

The discussion revolves around solving a second-order differential equation related to undamped motion, specifically the initial value problem involving the equation u'' + u = 0.5cos(0.8t) with initial conditions u(0) = 0 and u'(0) = 0.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the general solution for forced vibration problems and express confusion regarding the interpretation of terms in the solution. Questions arise about identifying the homogeneous and particular solutions, as well as the implications of initial conditions on the overall solution.

Discussion Status

There is ongoing exploration of different interpretations of the solution components, with some participants suggesting that the initial conditions may lead to a solution that is entirely determined by the particular solution. Guidance is offered regarding the roles of the driving force and natural frequency in the context of the problem.

Contextual Notes

Participants note potential misinterpretations of the general solution format and the implications of having both initial displacement and velocity set to zero. The discussion includes references to specific equations and terms from course materials, indicating a reliance on established methods for solving such differential equations.

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Homework Statement


Solve the initial value problem
[tex]u\prime\prime+u=0.5cos (0.8t)\\[/tex]

[tex]u(0)=0 \ u\prime(0) = 0[/tex]

Homework Equations





[tex]u(t) = [A*cos (w_nt)+ B*sin (w_nt)] + \frac{F_0}{m(w^2_n-w^2)}[/tex][tex]\left\{\begin{array}{cl}<br /> sin(wt)\\<br /> cos(wt)<br /> \end{array}\right.[/tex]

The Attempt at a Solution



http://img503.imageshack.us/img503/5746/untitledhh8.th.jpg

Here is picture of the notes for the equation if it is unclear

http://img518.imageshack.us/img518/6043/untitled2bh1.th.jpg

[tex]F_0 = 0.5[/tex]
[tex]w_n=1[/tex]
[tex]w=0.8[/tex]
The professor gave the above general equation for solving forced vibration problems.
I can get the answer, but i get an extra two terms (the ones underlined with question mark).
 

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I think maybe I am interpreting the sin(wt) and cos(wt) in the brackets of the general solution wrong. They represent the imaginary and real parts respectively.
 
Last edited:
I'm not following your work all that well.

What is your homogeneous solution?

Which one is your particular solution?
 
I found the complimentary solution to be

I think its the particular part of the solution that is confusing me.

The notes say this (its also posted in the 2nd image, but image shack is acting up on display it )
[tex]u(t) = u_c(t)+u_p(t)[/tex]
where

[tex]u_c(t) = [A*cos (w_nt)+ B*sin (w_nt)][/tex] (complimentary)
and

[tex]u_p(t) = A\exp(iwt)[/tex] where [tex]\\A = \frac{F_0}{m(w^2_n-w^2)}[/tex]

so I found the following:
[tex]u_c(t)-1.38cos(t) -1.112sin (t) \\[/tex]
[tex]u_p(t) = \frac{0.5(cos 0.8t+sin0.8t)}{1(1^2-0.8^2)}[/tex]
 
Last edited:
Both the initial displacement and the initial velocity are zero?

If so, there will not be a homogeneous (or transient) solution, and the solution will be entirely determined by the particular (or steady-state) solution. The physical analogy means that your system is at rest, until somebody hits a switch that drives it at a particular frequency.

In that case, I have not memorized the formula you are using, but I think you are using it wrong. The F_0 is supposed to mean your driving force, the m is the mass (which is one), omega_n is the natural frequency (one), and omega is the driving frequency (0.8).

What I got with my own method, simply guessing that the particular would be of the form Acos(wt) + Bsin(wt), and a homogeneous solution of 0, was:

[tex]u(t) = \frac{0.5}{1-0.8^2} cos(0.8t)[/tex]
 
This is a sample problem ffrom my book. They just used the equation for the phenomena of beats, but I was trying to see if I could get the same answer by using the general solution in my first post. I think i know what I did wrong, the sin and cos term in the brackets of the equation I posted don't represent sin wt + cost wt, which i used in my solution.
 

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