4—Differential Equations 86The results of problem1.23and of section4.9tell you all about such equations. In particular,for the pair of equationsax+by= 0andcx+dy= 0, the only way to have a non-zero solution for
xandyis for the determinant of the coefficients to be zero: ad−bc= 0. Apply this result to the
problem at hand. EitherA= 0andB= 0with a trivial solutionorthe determinant is zero.
(k 1 +k 3 +m 1 α^2
)(
k 2 +k 3 +m 2 α^2
)
−(k 3
) 2
= 0 (4.49)
This is a quadratic equation forα^2 , and it determines the frequencies of the oscillation. Note the plural
in the word frequencies.
Equation (4.49) is just a quadratic, but it’s still messy. For a first example, try a special,
symmetric case:m 1 =m 2 =mandk 1 =k 2. There’s a lot less algebra.
(k 1 +k 3 +mα^2
) 2
−(k 3
) 2
= 0 (4.50)
You could use the quadratic formula on this, but why? It’s already set up to be factored.
(k 1 +k 3 +mα^2 −k 3 )(k 1 +k 3 +mα^2 +k 3 ) = 0
The product is zero, so one or the other factors is zero. These determine theαs.
α^21 =−
k 1
m
and α^22 =−
k 1 + 2k 3
m
(4.51)
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 just one equation, and that the other is not
independent of it; it is actually a multiple of the first. You must solve that equation forxandy. Here,
arbitrarily pick the first of the equations (4.48) 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
andx 1 (t) =A 2 e−iω^1 t
x 2 (t) =A 2 e−iω^1 t
The other frequency has the corresponding solutions
x 1 (t) =A 3 eiω^2 t
x 2 (t) =−A 3 eiω^2 t
andx 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.