Question 3.8.5: A 16-pound weight is attached to a 5-foot-long spring. At eq......

The elongation of the spring after the weight is attached is 8.2 – 5 = 3.2 ft so it follows from Hooke’s law that 16 = k(3.2) or k = 5 lb/ft. In addition, $m=\frac{16}{32}=\frac{1}{2}$ slug so that the differential equation is given by

$\frac{1}{2} \frac{d^2 x}{d t^2}=-5 x-\frac{d x}{d t} \quad \text { or } \quad \frac{d^2 x}{d t^2}+2 \frac{d x}{d t}+10 x=0 .$         (20)

Proceeding, we find that the roots of $m^2+2 m+10=0$ are  $m_1=-1+3 i$ and $m_2=-1-3 i$, which then implies the system is underdamped and

$x(t)=e^{-t}\left(c_1 \cos 3 t+c_2 \sin 3 t\right) .$         (21)

Finally, the initial conditions x(0) = -2 and x'(0) = 0 yield $c_1=-2$ and $c_2=-\frac{2}{3},$ so the equation of motion is

$x(t)=e^{-t}\left(-2 \cos 3 t-\frac{2}{3} \sin 3 t\right) .$         (22)