# Forced Damped Oscillation

## 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 $$\pm$$ $$\sqrt{-4a^2}$$) / 2
p = -a $$\pm$$ ia

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

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 * [$$\sqrt{2}$$*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:

Related Introductory Physics Homework Help News on Phys.org
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.

I used the trial $x_{p2} = C e^{-a t}$ and got close to the correct answer.

$$x_{p2} = C e^{-a t}$$
$$\dot{x}_{p2} = -a C e^{-a t}$$
$$\ddot{x}_{p2} = a^2 C e^{_a t}$$

Substituting:

$$a^2 C + 2 a^2 C - 2 a^2 C = \frac{-F_0}{m}$$
$$C = \frac{F_0}{m a^2}$$

So:

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

Applying the initial conditions:

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

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

$$A sin(\theta) = \frac{F_0}{2 m a^2}$$

$$A sin(\theta) = A cos(\theta)$$
$$\theta = \frac{\pi}{4}$$

Then:

$$A cos(\frac{\pi}{4}) = \frac{F_0}{2 m a^2}$$

$$A = \frac{\sqrt{2} F_0} {2 m a^2}$$

And, finally:

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

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

Last edited: