Question 3.8.4: An 8-pound weight stretches a spring 2 feet. Assuming that a......
An 8-pound weight stretches a spring 2 feet. Assuming that a damping force numerically equal to two times the instantaneous velocity acts on the system, determine the equation of motion if the weight is released from the equilibrium position with an upward velocity of 3 ft/s .
From Hooke’s law we see that 8 = k(2) gives k = 4 lb/ft and that W = mg gives m=\frac{8}{32}=\frac{1}{4} slug. The differential equation of motion is then
\frac{1}{4} \frac{d^2 x}{d t^2}=-4 x-2 \frac{d x}{d t} \quad \text { or } \quad \frac{d^2 x}{d t^2}+8 \frac{d x}{d t}+16 x=0 . (17)
The auxiliary equation for (17) is m^2+8 m+16=(m+4)^2=0 so that m_1=m_2=-4.
Hence the system is critically damped and
x(t)=c_1 e^{-4 t}+c_2 t e^{-4 t} (18)
Applying the initial conditions x(0) = 0 and x'(0) = 3, we find, in turn, that c_1=0 and c_2=-3. Thus the equation of motion is
x(t)=-3 t e^{-4 t} (19)
To graph x(t) we proceed as in Example 3. From x^{\prime}(t)=-3 e^{-4 t}(1-4 t) x'(t) = 0 when t=\frac{1}{4} The corresponding extreme displacement is x\left(\frac{1}{4}\right)=-3\left(\frac{1}{4}\right) e^{-1}=-0.276 ft. As shown in FIGURE 3.8.10, we interpret this value to mean that the weight reaches a maximum height of 0.276 foot above the equilibrium position.
