A $~2$-pound weight is hung on a spring and stretches it $~\frac{1}{2}~$ foot. The mass spring system is then put into motion in a medium offering a damping force numerically equal to the velocity. If the mass is pulled down $~4~$ inches from equilibrium and released, write the initial value problem describing the position $~x(t)~$. Find the equation of motion.
Answer: $$x'' + 16x' + 32x = 0$$
Here is my attempt. The formula is $$mx'' + cx' + kx = 0~.$$
So I have solved for $~k~$ which is $~\frac{\text{Force}}{x} = \frac{32\times 2}{0.5} = 128$ pdl/ft
Then so far I have $$2x'' + 128x~.$$
The problem is that I don't know how to get the dampening coefficient $~(c)~$.
According to the question, I would assume it's just is $~1~$.
But that does not seem to be correct.
I presume that the spring is vertical w.r.t. the gravity field. At equilibrium, the sum of the forces is zero. The spring's stiffness constant $k$ satisfies $$ k = \frac{m g}{\Delta \ell} = \frac{2\times 32}{0.5} = 128\; \text{pdl/ft}, $$ where $m = 2$ lb is the mass, $g = 32$ pdl/lb is the g-force, and $\Delta \ell = 0.5$ ft is the spring's length variation at equilibrium. Out of equilibrium, the relative elongation $x$ in ft satisfies $$ m \ddot x + c\dot x + k x = 0 , $$ where a damping force $c\dot x$ numerically equal to the velocity $\dot x$ has been added -- that is, $c=1$ lb/s. Note that the gravity force does not appear here (it has been eliminated by subtracting the equilibrium equation). Finally, we end up with the initial-value problem $$ 2 \ddot x + \dot x + 128 x = 0 , \qquad x(0) = \tfrac13. $$