Forced Damped Oscillation

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Homework Statement



A Force F(t) = F0(1 - e-at), where both F0 and a are constants, acts over a damped oscillator. In t = 0, the oscillator is in it's equilibrium position. The mass of the oscillator is m, the spring constant is k = 2ma2 and the damping constant is b = 2ma.

Find x(t)

Homework Equations



Well, the differential equation is: d2x / dt2 + 2adx/dt + 2a2x = F0/m - F0/m * e-at

Also, x(0) = 0; x'(0) = 0.

The Attempt at a Solution



First, I tried to find the solution to the homogeneous equation associated.
So I tried a solution of the kind x = ept, and the I found: p2 + 2ap + 2 = 0
p = (-2a [tex]\pm[/tex] [tex]\sqrt{-4a^2}[/tex]) / 2
p = -a [tex]\pm[/tex] ia

So, it's an under-dampeing case, wich the general solution is A*e-at*cos(at + [tex]\theta)[/tex].

Then I have to find the particular solutions, where xp1 = F0/m and xp2 = -F0/m * e-at.

xp1 is rather easy. C*xp1 = F0/m ; 2a2*xp1 = F0/m ; xp1 = F0/2ma2

Now I'm stuck at the other one. I tried solutions like x = C*ept and x = C*t2*ept (C = constant) and haven't got it right.

The answer is: x(t) = F0/ma2 * [[tex]\sqrt{2}[/tex]*e-at*cos(at + P/4) + 1 - 2e-at]

How do I find the other particular solution? What am I missing?
(By the way, should this be here or in the Calculus sub-forum?)
 
Last edited:

Answers and Replies

  • #2
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Now I'm stuck at the other one. I tried solutions like x = C*ept and x = C*t2*ept (C = constant) and haven't got it right.
...
How do I find the other particular solution? What am I missing?
(By the way, should this be here or in the Calculus sub-forum?)
Hi,

You're almost there. To xp2, if you select x = C*ept as trial function, the only case that will work is p=a.
 
  • #3
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I used the trial [itex]x_{p2} = C e^{-a t}[/itex] and got close to the correct answer.

[tex]x_{p2} = C e^{-a t}[/tex]
[tex]\dot{x}_{p2} = -a C e^{-a t}[/tex]
[tex]\ddot{x}_{p2} = a^2 C e^{_a t}[/tex]

Substituting:

[tex]a^2 C + 2 a^2 C - 2 a^2 C = \frac{-F_0}{m}[/tex]
[tex]C = \frac{F_0}{m a^2}[/tex]

So:

[tex]x(t) = A e^{-at}cos(at + \theta) + \frac{F_0}{2ma^2} - \frac{F_0}{ma^2} e^{-at}[/tex]

Applying the initial conditions:

[tex]x(0) = 0 = A cos(\theta) - \frac{F_0}{2ma^2}[/tex]
[tex]\dot{x}(0) = 0 = -a A cos(\theta) - a A sin(\theta) + \frac{F_0}{m a}[/tex]

Substituting [itex]A cos(\theta) = \frac{F_0}{2 m a^2}[/itex] on the second equation:

[tex]A sin(\theta) = \frac{F_0}{2 m a^2}[/tex]

[tex]A sin(\theta) = A cos(\theta)[/tex]
[tex]\theta = \frac{\pi}{4}[/tex]

Then:

[tex]A cos(\frac{\pi}{4}) = \frac{F_0}{2 m a^2}[/tex]

[tex]A = \frac{\sqrt{2} F_0} {2 m a^2}[/tex]

And, finally:

[tex]x(t) = \frac{F_0}{2 m a^2} [\sqrt{2} e^{-at} cos(a t + \frac{\pi}{4}) + 1 - 2 e^{-a t}][/tex]

Still can't see what's wrong, or whatever I'm missing...
 
Last edited:

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