Page 143 - Schaum's Outline of Differential Equations

126 SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS [CHAP. 14

The roots of the associated characteristic equation are H the general solution of the differential
equation is

At t = 0, 20 percent of the 10-in length of the cylinder, or 2 in, is out of the water. Using the results of Problem 14.19,
we know that the equilibrium position has 1.75 in above the water, so at t = 0, the cylinder is raised 1/4 in or 1/48 ft
above its equilibrium position. In the context of Fig. 14-3, jc(0) = 1/48 ft. The initial velocity is 5 ft/sec in the down-
ward or negative direction in the coordinate system of Fig. 14-3, so x(0) = - 5. Applying these initial conditions to
(_/), we find that

Equation (_/) becomes

14.21. Determine whether a cylinder of diameter 10 cm, height 15 cm, and weight 19.6 N can float in a deep
3
pool of water of weight density 980 dynes/cm .
Let h denote the length (in centimeters) of the submerged portion of the cylinder at equilibrium. With r = 5 cm
6
and mg = 19.6 N = 1.96 X 10 dynes, it follows from Eq. (14.9) that

Since this is more height than the cylinder possesses, the cylinder cannot displace sufficient water to float and will
sink to the bottom of the pool.

14.22. Determine whether a cylinder of diameter 10 cm, height 15 cm, and weight 19.6 N can float in a deep
3
pool of liquid having weight density 2450 dynes/cm .
Let h denote the length of the submerged portion of the cylinder at equilibrium. With r = 5 cm and mg = 19.6 N
6
= 1.96 X 10 dynes, it follows from Eq. (14.9) that

Thus, the cylinder will float with 15 - 10.2 = 4.8 cm of length above the liquid at equilibrium.

14.23. Determine an expression for the motion of the cylinder described in Problem 14.22 if it is released at
rest with 12 cm of its length fully submerged.
3
Here r = 5 cm, p = 2450 dynes/cm , m = 19.6/9.8 = 2 kg = 2000 g, and Eq. (14.10) becomes

The roots of the associated characteristic equation are the general solution of the differential
equation is

At t = 0, 12 cm of the length of the cylinder is submerged. Using the results of Problem 14.22, we know that the
equilibrium position has 10.2 cm submerged, so at t= 0, the cylinder is submerged 12 - 10.2 = 1.8 cm below its
equilibrium position. In the context of Fig. 14-3, x(Q) = —1.8 cm with a negative sign indicating that the equilibrium
line is submerged. The cylinder begins at rest, so its initial velocity is x(0) = 0. Applying these initial conditions to
(_/), we find that c 1 = —1.8 and c 2 = 0. Equation (1) becomes