Question 3.8.5: A 16-pound weight is attached to a 5-foot-long spring. At eq......
A 16-pound weight is attached to a 5-foot-long spring. At equilibrium the spring measures 8.2 feet. If the weight is pushed up and released from rest at a point 2 feet above the equilibrium position, find the displacements x(t) if it is further known that the surrounding medium offers a resistance numerically equal to the instantaneous velocity.
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)