Page 138 - Schaum's Outline of Differential Equations

CHAP. 14] SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS 121

14.8. Show that the types of motions that result from free damped problems are completely determined by the
2
quantity a — 4 km.
For free damped motions F(t) = 0 and Eq. (14.1) becomes

The roots of the associated characteristic equation are

2
2
2
If a - 4 km > 0, the roots are real and distinct; if a - 4 km = 0, the roots are equal; if a - 4 km < 0, the roots
are complex conjugates. The corresponding motions are, respectively, overdamped, critically damped, and oscillatory
damped. Since the real parts of both roots are always negative, the resulting motion in all three cases is transient.
(For overdamped motion, we need only note that whereas for the other two cases the real parts are
both-a/2m.)

14.9. A 10-kg mass is attached to a spring having a spring constant of 140 N/m. The mass is started in motion
from the equilibrium position with an initial velocity of 1 m/sec in the upward direction and with an
applied external force F(t) = 5 sin t. Find the subsequent motion of the mass if the force due to air
resistance is -90iN.
Here m = 10, k = 140, a = 90, and F(t) = 5 sin t. The equation of motion, (14.1), becomes

The general solution to the associated homogeneous equation x + 9x + 14x = 0 is (see Problem 14.5)

Using the method of undetermined coefficients (see Chapter 11), we find

The general solution of (_/) is therefore

Applying the initial conditions, x(0) = 0 and i(0) = -1, we obtain

Note that the exponential terms, which come from x h and hence represent an associated free overdamped motion,
quickly die out. These terms are the transient part of the solution. The terms coming from x p, however, do not die
out as t —> °°; they are the steady-state part of the solution.

14.10. A 128-lb weight is attached to a spring having a spring constant of 64 Ib/ft. The weight is started in
motion with no initial velocity by displacing it 6 in above the equilibrium position and by simultaneously
applying to the weight an external force F(t) = 8 sin 4t. Assuming no air resistance, find the subsequent
motion of the weight.
Here m = 4, k = 64, a = 0, and F(t) = 8 sin 4t; hence, Eq. (14.1) becomes