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Example 1

A mass of one gram is suspended from a vertical spring with spring constant k = 100, as in Figure 14.7.1. At time t = 0, the mass is at position y(0) = 2 cm and has velocity y'(0) = 50 cm/sec. Find the equation of motion of the mass. It is understood that there is no damping, and the origin is at the point where the spring is at its natural length.

14_differential_equations-225.gif

Figure 14.7.1

In this problem there are two forces, the force of the spring and the force of gravity. The force of gravity is a constant and is equal to mg dynes, always

in the downward direction. The system is described by a second order linear differential equation with a constant forcing term,

my" + ky = -mg.

In this case k = 100, m = 1, and g = 980, so the differential equation is

(4)

y" + 100y= -980.

To solve the problem, we first find some particular solution of the differential equation (4), then use Theorem 1 to find the general solution, and finally substitute to find the particular solution for the given initial values y(0) = 2 and y'(0) = 50.

Since the forcing term is a constant f(t) = -980, we guess that the differential equation (4) has a particular solution, which is a constant. In this example it is easy to see by inspection that the constant function

14_differential_equations-226.gif

is a particular solution of the differential equation (4). By the method of the preceding section, the characteristic polynomial z2 + 100 has roots ± i10, and the corresponding homogeneous differential equation x" + 100x = 0 has the general solution

A cos (10t) + B sin(10t).

According to Theorem 1, the general solution of the original differential equation (4) is the sum of the particular solution of the original equation and the general solution of the homogeneous equation. Thus the general solution of equation (4) is

y(t) = A cos (10t) + B sin (10t) - 9.8.

Use the initial value y(0) = 2 to find A.

2 = A cos (0) + B sin (0) - 9.8 = A - 9.8, A = 11.8.

Now compute y'(t), and substitute the given initial value y'(0) = 50 to find B.

y'(t) = - 10A sin (10t) + 10B cos (10t).

50 = - 10A sin (0) + 10B cos (0) = 105, B = 5.0.

The required particular solution is thus

y(t) = 11.8 cos (10t) + 5.0 sin(10t) - 9.8,

shown in Figure 14.7.2.

14_differential_equations-227.gif

Figure 14.7.2 Example 1


Last Update: 2006-11-16