MHB Transfer function of a damped hanging mass

[Dustinsfl]

How do I find the transfer function of damped masses hanging?

I know that the transfer function is
\[
H(s) = \frac{\mathcal{L}\{y(t)\}}{\mathcal{L}\{x(t)\}}
\]
where \(u\) is the input which is a force and \(x_1\) is the output.

Given the following diagram (see below), how do I find the input and output functions?

http://imagizer.imageshack.us/v2/800x600q90/40/eh7q.png

 

[Dustinsfl]

The force of damping is
\[
F_d = b\frac{dx}{dt}
\]
so by Newton's Law, the system can be written as
\[
u - b(\dot{x}_1 + \dot{x}_2) = m(\ddot{x_1} + \ddot{x}_2).
\]
Is this Correct?

How do I separate out the input from the output?

 

[Dustinsfl]

I tried modeling the forces separately but not sure if this is wise either.

For the first mass, we have $-b\dot{x}_1 = m\ddot{x_1}$, correct?

Now the second is $-b(\dot{x}_1 + \dot{x}_2) + u = m\ddot{x}_2$, correct?

I should be able to take the Laplace transform of both and end up with what I need to construct the transfer function which is
$$
H(s) = \frac{X_1(s)}{U(s)}.
$$
However, I still have and $X_2(s)$. Are my equations of motion wrong? If not, what am I doing incorrectly?

If take the Laplace transform of both, we have
\[
X_1(s)(s^2m + bs) = 0
\]
and
\[
X_2(s)(s^2m + sb) + sbX_1(s) - U(s) = 0
\]
The initial conditions are zero since when finding the transfer function, the initial conditions are zero by definition. If this is correct, how do I get rid of \(X_2(s)\)?