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.

  • 1
    $\begingroup$ "offering a damping force numerically equal to the velocity" $\endgroup$
    – Cesareo
    Aug 17, 2019 at 13:11
  • $\begingroup$ @cesareo yes that's why I would assume the coefficient is just one. but in the answer, it says it is 16. $\endgroup$
    – Meraj Haq
    Aug 17, 2019 at 17:56

1 Answer 1


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

  • $\begingroup$ thank you for your answer, but if you look at the answer given by the question it does not seem to be correct. the correct differential equation is supposed to be 2x¨+16x˙+32x=0 $\endgroup$
    – Meraj Haq
    Aug 17, 2019 at 17:54
  • $\begingroup$ the question was given by my professor along with the answer $\endgroup$
    – Meraj Haq
    Aug 17, 2019 at 23:48
  • 1
    $\begingroup$ @MerajHaq ok, then if you get the answer some day (maybe by your prof him/herself), then just post it as an answer here. $\endgroup$
    – EditPiAf
    Aug 18, 2019 at 10:26

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