# Spring with damping force numerically equal to the velocity

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.

• "offering a damping force numerically equal to the velocity" Aug 17, 2019 at 13:11
• @cesareo yes that's why I would assume the coefficient is just one. but in the answer, it says it is 16. Aug 17, 2019 at 17:56

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 $$2 \ddot x + \dot x + 128 x = 0 , \qquad x(0) = \tfrac13.$$