Question 3.8.1: A mass weighing 2 pounds stretches a spring 6 inches. At t =......
A mass weighing 2 pounds stretches a spring 6 inches. At t = 0 the mass is released from a point 8 inches below the equilibrium position with an upward velocity of \frac{4}{3} ft/s. Determine the equation of free motion.
Because we are using the engineering system of units, the measurements given in terms of inches must be converted into feet: 6 in = \frac{1}{2} \mathrm{ft} ; 8 in = \frac{2}{3} \mathrm{ft}. In addition, we must convert the units of weight given in pounds into units of mass. From m=W/g we have m=\frac{2}{32}=\frac{1}{16} slug. Also, from Hooke’s law 2=k\left(\frac{1}{2}\right) implies that the spring constant is k = 4 lb/ft. Hence (1) gives
\frac{1}{16} \frac{d^2 x}{d t^2}=-4 x \quad \text { or } \quad \frac{d^2 x}{d t^2}+64 x=0
The initial displacement and initial velocity are x(0)=\frac{2}{3}, x^{\prime}(0)=-\frac{4}{3} where the negative sign in the last condition is a consequence of the fact that the mass is given an initial velocity in the negative, or upward, direction.
Now \omega^2=64 or \omega=8 , so that the general solution of the differential equation is
x(t)=c_1 \cos 8 t+c_2 \sin 8 t (4)
Applying the initial conditions to x(t) and x'(t) gives c_1=\frac{2}{3} \text { and } c_2=-\frac{1}{6} Thus the equation
of motion is
x(t)=\frac{2}{3} \cos 8 t-\frac{1}{6} \sin 8 t . (5)