Differential equation - mass spring system So we have a mass spring system, which is supposed to be modelling a bridge, and the equation that the displacement of the bridge $x$ satisfies is given by $$M\ddot{x}+c\dot{x}+kx=0$$ where $M=4\times 10^5\text{kg}, c=5\times 10^4\text{kg/s}, k=10^7\text{kg} $, where $m$ is the mass, $k$ is the 'spring constant'and $c$ is the friction or level of damping or something, I'm not sure - this equation is pretty standard for this type of problem so you probably know what it means more than me anyway.
Now there is a part of the question where another factor is introduced, so that $x$ now satisfies the equation $$M\ddot{x}+c\dot{x}+kx=300\dot{N}.$$
Now my question is, is the damping level a constant value, so is it still $c$ or has it changed due to the new factor affecting $\dot{x}$ so would the damping now be $c-300N$? or does it just stay at $c$?
I think the main problem here is that I don't really know what is damping (what actually is damping )in general, so if anyone could just answer the question I asked above or say something about damping in general in these types of problems that would explain my confusion.
 A: Damping in this model is a force that resists motion and that is proportional to the instantaneous velocity,
$$
F_{\rm damping} = -c \frac{dx}{dt}
$$
This is a very typical force when you're immersed in a fluid. In this case $c$ is a constant, the larger $c$ the stronger is the force opposing the motion. So for instance between oil and water, oil has a larger $c$.
Now, there is also a restoring force in your model that is taken proportional to the deformation of the bridge
$$
F_{\rm restoring} = -k x
$$
And finally an external force (e.g. wind) $F_{\rm external}$. Applying Newton's second law you find
\begin{eqnarray}
 Ma = \sum_i F_i &=& F_{\rm restoring} + F_{\rm damping} + F_{\rm external} \\
M \ddot{x} + a\dot{x} + kx &=& F_{\rm external}
\end{eqnarray}
$c$ is still constant, regardless of the fact that the bridge is being externally forced to move!
A: $Mx''+cx'+kx=0$
$x = Ae^{-\frac c{2M}t}\sin\big(\phi + t\sqrt {\frac {k}{M} -(\frac{c}{2M})^2} \big) $
I hope I have that right, I eyeballed it.
Now we change the equation.
$Mx''+cx'+kx=300$
What happens to $x?$ $x$ adjusts by a constant.
$x = Ae^{-\frac c2t}\sin(\phi + t\sqrt {Mk -\frac {c^2}4} ) + K$
$x'$  and $x''$ are unchanged.
$kK = 300\\
K = \frac {300}{k}$
$x = Ae^{-\frac c2t}\sin(\phi + t\sqrt {Mk -\frac {c^2}4} ) + \frac {300}{k}$
