2nd Order DE with undamped motion

In summary: Instead they represent two different scenario's. In my case (cos (wt)), the sin term in the brackets is zero and vice versa.In summary, the conversation discusses solving a forced vibration problem using the general equation for forced vibration problems. The professor gave the equation u(t) = [A*cos (w_nt)+ B*sin (w_nt)] + \frac{F_0}{m(w^2_n-w^2)} \left\{\begin{array}{cl} sin(wt)\\ cos(wt) \end{array}\right. to solve the problem, but the student is having trouble understanding the particular part of the solution. After discussing the initial conditions and the physical analogy of the problem, it is determined that
  • #1
teknodude
157
0

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}
sin(wt)\\
cos(wt)
\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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  • #2
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.
 
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  • #3
I'm not following your work all that well.

What is your homogeneous solution?

Which one is your particular solution?
 
  • #4
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]
 
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  • #5
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]
 
  • #6
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.
 

1. What is a 2nd order differential equation with undamped motion?

A 2nd order differential equation with undamped motion is a mathematical model that describes the motion of a system that is not subject to any damping forces, such as friction or air resistance. It is usually in the form of m x'' + kx' + cx = 0, where m is the mass of the system, x is the displacement, k is the spring constant, and c is the damping coefficient.

2. What are the solutions to a 2nd order DE with undamped motion?

The solutions to a 2nd order DE with undamped motion depend on the initial conditions of the system. If the initial displacement and velocity values are known, then the general solution can be found using the characteristic equation: ar² + br + c = 0, where r is the damping ratio. The solutions can be classified as underdamped, critically damped, or overdamped.

3. How does the damping ratio affect the behavior of a 2nd order DE with undamped motion?

The damping ratio, denoted by r, is a measure of the system's response to an external force. It is defined as the ratio of the actual damping coefficient to the critical damping coefficient. A system is considered underdamped when r < 1, critically damped when r = 1, and overdamped when r > 1. The damping ratio affects the amplitude, frequency, and decay rate of the system's oscillations.

4. How do initial conditions affect the solution of a 2nd order DE with undamped motion?

The initial conditions, specifically the initial displacement and velocity values, determine the values of the constants in the general solution of the 2nd order DE with undamped motion. This means that different initial conditions will result in different solutions, and therefore, different behaviors of the system. The initial conditions also affect the system's natural frequency and damping ratio.

5. What are some real-life examples of systems described by a 2nd order DE with undamped motion?

Some common examples of systems that can be described by a 2nd order DE with undamped motion include a mass-spring system, a pendulum, and a vibrating guitar string. Other examples include a swinging door, a diving board, and a car's suspension system. In general, any system that exhibits oscillatory behavior without any damping forces can be described by a 2nd order DE with undamped motion.

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