Page 144 - Schaum's Outline of Differential Equations

CHAP. 14] SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS 127

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14.24. A solid cylinder partially submerged in water having weight density 62.5 Ib/ft , with its axis vertical, oscil-
lates up and down within a period of 0.6 sec. Determine the diameter of the cylinder if it weighs 2 Ib.
With p = 62.5 Ib/ft 3 and m = 2/32 slugs, Eq. (14.10) becomes

which has as its general solution

Its circular frequency is its natural frequency is its period is
T= !//= 1/8.92r. We are given 0.6 = T= 1/8.92r, thus r = 0.187 ft = 2.24 in with a diameter of 4.48 in.

14.25. A prism whose cross section is an equilateral triangle with sides of length / floats in a pool of liquid of
weight density p with its height parallel to the vertical axis. The prism is set in motion by displacing it
from its equilibrium position (see Fig. 14-4) and giving it an initial velocity. Determine the differential
equation governing the subsequent motion of this prism.
Equilibrium occurs when the buoyant force of the displaced liquid equals the force of gravity on the body. The
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area of an equilateral triangle with sides of length / is A = v3/ /4. For the prism depicted in Fig. 14-4, with h units
of height submerged at equilibrium, the volume of water displaced at equilibrium is -J^fhl 4, providing a buoyant
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force of v3/ Ap / 4. By Archimedes' principle, this buoyant force at equilibrium must equal the weight of the prism
mg; hence,

We arbitrarily take the upward direction to be the positive x-direction. If the prism is raised out of the water by
x(t) units, as shown in Fig. 14-4, then it is no longer in equilibrium. The downward or negative force on such a body
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remains mg but the buoyant or positive force is reduced to ^J3l [h - x(t)]p 14. It now follows from Newton's second
law that

Substituting (1) into this last equation, we simplify it to

Fig. 14.4