Mathematical Tools for Physics

4—Differential Equations 105

These are negative, and that’s what you should expect. There’s no damping and the springs provide restoring
forces that should give oscillations. That’s just what these imaginaryα’s provide.
When you examine the equationsax+by= 0andcx+dy= 0the condition that the determinant vanishes
is the condition that the two equations are really only one equation, and that the other is not independent of it;
it’s actually a multiple of the first. You still have to solve that equation forxandy. Here, I arbitrarily pick the
first of the equations ( 29 ) and find the relation betweenAandB.

α^21 =−

k 1

m

=⇒

(

k 1 +k 3 +m(−(k 1 /m))

)

A+

(

−k 3

)

B= 0 =⇒ B=A

α^22 =−

k 1 + 2k 3

m

=⇒

(

k 1 +k 3 +m(−(k 1 + 2k 3 /m))

)

A+

(

−k 3

)

B= 0 =⇒ B=−A

For the first case,α 1 =±iω 1 =±i

√

k 1 /m, there are two solutions to the original differential equations. These
are called ”normal modes.”
x 1 (t) =A 1 eiω^1 t
x 2 (t) =A 1 eiω^1 t

and

x 1 (t) =A 2 e−iω^1 t

x 2 (t) =A 2 e−iω^1 t

The other frequency has the corresponding solutions ( 27 )

x 1 (t) =A 3 eiω^2 t

x 2 (t) =−A 3 eiω^2 t

and

x 1 (t) =A 4 e−iω^2 t

x 2 (t) =−A 4 e−iω^2 t

The total solution to the differential equations is the sum of all four of these.

x 1 (t) =A 1 eiω^1 t+A 2 e−iω^1 t+A 3 eiω^2 t+A 4 e−iω^2 t
x 2 (t) =A 1 eiω^1 t+A 2 e−iω^1 t−A 3 eiω^2 t−A 4 e−iω^2 t (33)
The two second order differential equations have four arbitrary constants in their solution. You can specify
the initial values of two positions and of two velocities this way. As a specific example suppose that all initial
velocities are zero and that the first mass is pushed to coordinatex 0 and released.

x 1 (0) =x 0 =A 1 +A 2 +A 3 +A 4

x 2 (0) = 0 =A 1 +A 2 −A 3 −A 4

vx 1 (0) = 0 =iω 1 A 1 −iω 1 A 2 +iω 2 A 3 −iω 2 A 4

vx 2 (0) = 0 =iω 1 A 1 −iω 1 A 2 −iω 2 A 3 +iω 2 A 4 (34)