Page 137 - Schaum's Outline of Differential Equations

120 SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS [CHAP. 14

14.4. Determine the circular frequency, natural frequency, and period for the simple harmonic motion
described in Problem 14.1.
Circular frequency:
Natural frequency:
Period:

14.5. A 10-kg mass is attached to a spring, stretching it 0.7 m from its natural length. The mass is started in
motion from the equilibrium position with an initial velocity of 1 ml sec in the upward direction. Find
the subsequent motion, if the force due to air resistance is -90i N.
2
Taking g = 9.8 m/sec , we have w = mg = 98 N and k = w/l = 140 N/m. Furthermore, a = 90 and F(t) = 0 (there
is no external force). Equation (14.1) becomes

The roots of the associated characteristic equation are Xj = -2 and ^ = -7, which are real and distinct; hence this
problem is an example of overdamped motion. The solution of (1) is

The initial conditions are x(0) = 0 (the mass starts at the equilibrium position) and i(0) = -1 (the initial velocity is
2
7
in the negative direction). Applying these conditions, we find that c 1 = — c 2 = --|, so that x = ^(e" ' - e~ '). Note that
x —> 0 as t —> °°; thus, the motion is transient.

14.6. A mass of 1/4 slug is attached to a spring, whereupon the spring is stretched 1.28 ft from its natural length.
The mass is started in motion from the equilibrium position with an initial velocity of 4 ft/sec in the down-
ward direction. Find the subsequent motion of the mass if the force due to air resistance is -2ilb.
Here m=l/4, a = 2, F(t) = Q (there is no external force), and, from Hooke's law, k = mgll
= (l/4)(32)/1.28 = 6.25. Equation (14.1) becomes

The roots of the associated characteristic equation are Xj = -4 + O and ^ = -4 - 13, which are complex conjugates;
hence this problem is an example of oscillatory damped motion. The solution of (1) is

The initial conditions are x(0) = 0 and i(0) = 4. Applying these conditions, we find that Cj = 0 and c 2 = |; thus,
4
x = |e~ ' sin3t. Since x —> 0 as t —> m, the motion is transient.

14.7. A mass of 1/4 slug is attached to a spring having a spring constant of 1 Ib/ft. The mass is started in motion
by initially displacing it 2 ft in the downward direction and giving it an initial velocity of 2 ft/sec in the
upward direction. Find the subsequent motion of the mass, if the force due to air resistance is -lilb.
Here m = 1/4, a = 1, k =1, and F(t) = 0. Equation (14.1) becomes

The roots of the associated characteristic equation are A^ = A^ = -2, which are equal; hence this problem is an example
of critically damped motion. The solution of (1) is

The initial conditions are x(0) = 2 and i(0) = -2 (the initial velocity is in the negative direction). Applying these
conditions, we find that Cj = c 2 = 2. Thus,

Since x —> 0 as t —> °°, the motion is transient.