7—Operators and Matrices 164Example: Coupled Oscillators
Another example drawn from physics: Two masses are connected to a set of springs and fastened
between two rigid walls. This is a problem that appeared in chapter 4, Eq. (4.45).
m 1 d^2 x 1 /dt^2 =−k 1 x 1 −k 3 (x 1 −x 2 ), and m 2 d^2 x 2 /dt^2 =−k 2 x 2 −k 3 (x 2 −x 1 )
The exponential form of the solution was
x 1 (t) =Aeiωt, x 2 (t) =Beiωt
The algebraic equations that you get by substituting these into the differential equations are apair of linear equations forAandB, Eq. (4.47). In matrix form these equations are, after rearranging
some minus signs,
(
k 1 +k 3 −k 3
−k 3 k 2 +k 3
)(
A
B
)
=ω^2
(
m 1 0
0 m 2
)(
A
B
)
You can make it look more like the previous example with some further arrangement
[(k 1 +k 3 −k 3
−k 3 k 2 +k 3
)
−ω^2
(
m 1 0
0 m 2
)](
A
B
)
=
(
0
0
)
The matrix on the left side maps the column matrix to zero. That can happen only if the matrix has
zero determinant (or the column matrix is zero). If you write out the determinant of this 2 × 2 matrix
you have a quadratic equation inω^2. It’s simple but messy, so rather than looking first at the general
case, look at a special case with more symmetry. Takem 1 =m 2 =mandk 1 =k 2.
det[(
k 1 +k 3 −k 3
−k 3 k 1 +k 3
)
−ω^2 m
(
1 0
0 1
)]
= 0 =
(
k 1 +k 3 −mω^2
) 2
−k^23
This is now so simple that you don’t even need the quadratic formula; it factors directly.
(k 1 +k 3 −mω^2 −k 3
)(
k 1 +k 3 −mω^2 +k 3
)
= 0
The only way that the product of two numbers is zero is if one of the numbers is zero, so either
k 1 −mω^2 = 0 or k 1 + 2k 3 −mω^2 = 0
This determines two possible frequencies of oscillation.
ω 1 =
√
k 1
m
and ω 2 =
√
k 1 + 2k 3
m
You’re not done yet; these are just the eigenvalues. You still have to find the eigenvectors andthengo
back to apply them to the original problem. This isF~=m~aafter all. Look back to section4.10for
the development of the solutions.