1550078481-Ordinary_Differential_Equations__Roberts_

Applications of Linear Equations with Constant Coefficients 287

EXAMPLE 2 Equation of Motion of a Damped Mass on a Spring

A. l kg mass is attached to one end of a spring, the other end is attached to

a fixed support, and the system is allowed to come to rest. In the equilibrium

position the spring is stretched. 25 m. The mass is pulled down an additional

.5 m and released without imparting any velocity. If the damping constant of

the system is c = 1 kg/s, write the equation of motion of the system.

SOLUTION

Since the spring is stretched .25 m =£by a mass of .l kg= m, by Hooke's

law the spring constant k satisfies mg = k£. So solving for k, we have

k =mg/£= (.1 kg)(9.8 m/s

2

)/(.25 m) = 3. 92 kg-m^2 /s^2.

So the equation of motion of this spring-mass system satisfies the initial value

problem

my" + cy' + ky = O; y(O) = .5 m, y' (0) = 0 m/s

or
.ly" + y' + 3.92y = O; y(O) = .5 m , y' (0) = 0 m/s.
Using the computer software POLYRTS, we find the roots of the auxiliary
equation .lr^2 + r + 3.92 = 0 are r 1 = -5 + 3.7683i and r 2 = -5 - 3.7683i.
Since the roots are complex conjugate roots, the system is executing damped
oscillatory motion, and the equation of motion is

y(t) = Ae-^5 tsin(3.7683t+¢)

where A and ¢are constants which are to be determined to satisfy the initial

conditions. Differentiating the equation of motion, we see

y' (t) = Ae-^5 t[3.7683 cos (3.7683t + ¢) - 5 sin (3.7683t + ¢ )].

To satisfy the given initial conditions A and ¢must simultaneously satisfy

y(O) = A sin¢ = .5 m and y' (0) = A[3. 7683 cos¢ - 5 sin¢] = 0 m/s.

Dividing the second equation by the first, we see ¢ must satisfy

3.7683cot¢-5=0 or cot¢=5/3.7683=1.3269.

So ¢ = .6458 rads. Solving the equation A sin¢ = .5 m for A, we find A =

.5 m/ sin(.6458) = .831 m. Therefore, the equation of motion is

y(t) = .83le-^5 t sin (3. 7683t + .6458).