Mathematical Tools for Physics

4—Differential Equations 104

When you plug this into the differential equations for the masses, all the factors ofeαtcancel, just the way it
happens in the one variable case.

m 1 α^2 A=−k 1 A−k 3 (A−B), and m 2 α^2 B=−k 2 B−k 3 (B−A) (28)

Rearrange these to put them into a neater form.

(

k 1 +k 3 +m 1 α^2

)

A+

(

−k 3

)

B= 0

(

−k 3 )A+

(

k 2 +k 3 +m 2 α^2

)

B= 0 (29)

The results of problem1.23and of section4.7tell 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 forxandyis for the
determinant of the coefficients to be zero:ad−bc= 0. Apply this result to the problem at hand. EitherA= 0
andB= 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 (30)

This is a quadratic equation forα^2 , and it determines the frequencies of the oscillation. Note the plural in the
word frequencies.
Equation ( 30 ) is only 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 (31)

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 determineα.

α^21 =−

k 1

m

and α^22 =−

k 1 + 2k 3

m

(32)