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**VonNemo19** Hi. So...

Suppose that an object weighing 4lbs stretches a spring 2 inches. The object is displaced an additional six inches below the equilibrium position and then released. The object is in a medium that exerts a damping force nemuerically equal to 2 times the instantaneous velocity. Determine the equation of motion.

So, Here I know that since F=ks by Hooke's Law, we can determine k as follows:

$\displaystyle 4=k(\frac{1}{6})\Rightarrow{k}=24$.

Now, since from newton's second law we know that $\displaystyle F=ma$, where $\displaystyle m=W/g=8/32=1/4$ and $\displaystyle a=\frac{d^2x}{dt^2}$ and we can equate this to the sum of the restoring force of the spring and the resistant force of the medium so that

$\displaystyle \frac{1}{4}\frac{d^2x}{dt^2}=-24x-2\frac{dx}{dt}$

Am I on the right track?