1550078481-Ordinary_Differential_Equations__Roberts_

298 Ordinary Differential Equations

spring-mass system satisfy the linear system of differential equations

(28a)

(28b)

Solving equation (28a) for Y2, we get

(29)

Differentiating twice, we find

(4) k

11 _ ffi1Y1 + (l + ____'.:_) II

(30) Y2 - k2 k2 Y1·

Substituting these last two expressions for y 2 and y~ into equation (28b ), we
obtain the following single fourth-order differential equation for y 1

(4) k

m2m1 Y1 ( · 1 ) /1 11 ( )

k

2

+ m2 1 + k

2

Y1 = -m1Y1 - k1 + k2 Y1 + k2Y1

or multiplying by k2 and rearranging

(31) m1m2yi

4

) + [m2(k1 + k2) + m1k2]y{ + k1k2Y1 = 0.

Exercise 1. a. For m 1 = .2 kg, m2 = .7 kg, k 1 = 5 kg-m^2 /s^2 , and k 2 =
11 kg-m^2 /s^2 find the general solution of equation (31). (HINT: Use POLYRTS
or your computer software to find the roots of the auxiliary equation associated
with equation (31).)

b. Use equation (29) and the answer to part a. to find y 2 (t).
c. Find the solution to the initial value problem consisting of the system
of two first-order differential equations (28) and the initial conditions:

Y1(0) = .1 m, y~(O) = .3 m/s, Y2(0) = - .15 m, and y~(O) = .4 m/s.

Another Coupled Spring-Mass System A second coupled spring-mass
system which consists of two masses, m 1 and m 2 , connected to two fixed
supports by three springs which have spring constants k 1 , k 2 , and k 3 is shown
in Figure 6.8. Neglecting the effects of damping, the system of differential
equations which describes the displacements y 1 and Y2 of masses m 1 and m2,
respectively, from their equilibrium positions is

(32a)

(32b)