Linear Algebra, 2020a

Topic: Coupled Oscillators 485
we also discussed there are two cases: when the twist imparted by the spring’s
motion is in the same direction as the twist given by the rotational oscillation
and when they are opposed. In either case to get a normal mode the peaks must
coincide.
x(t) =A 1 cos(ωt + φ) x(t) =A 1 cos(ωt + φ)
θ(t) =A 2 cos(ωt + φ) θ(t) =A 2 cos(ωt +(φ + π))
We will work through the left-hand case, leaving the other as an exercise.
We want to find which ω’s are possible. Take the second derivatives
d 2 x(t)
dt
=−A 1 ω 2 cos(ωt + φ)
and plug into the equations of motion (∗∗).
d 2 θ(t)
dt
=−A 2 ω 2 cos(ωt + φ)
m · (−A 1 ω 2 cos(ωt + φ)) + k · (A 1 cos(ωt + φ)) + ɛ 2 · (A 2 cos(ωt + φ)) = 0
I · (−A 2 ω 2 cos(ωt + φ)) + κ · (A 2 cos(ωt + φ)) + ɛ 2 · (A 1 cos(ωt + φ)) = 0
Factor out cos(ωt + φ) and divide through by m.
( k
m − ω2) · A 1 +
ɛ
2m · A 2 = 0
(κ
I − ω2) · A 2 +
ɛ
2m · A 1 = 0
We are assuming that k/m = ω 2 x and replace κ/I = ω 2 θ
are equal, and writing
ω 2 0
for that number. Make the substitution and restate it as a matrix equation.
(
)( ) ( )
ω 2 0 − ω2 ɛ/2m A 1 0
ɛ/2I ω 2 0 − =
ω2 A 2 0
Obviously this system has the trivial solution A 1 = 0, A 2 = 0, for the case
where the mass is at rest. We want to know for which frequencies ω this system
has a nontrivial solution.
(
ω 2 0
ɛ/2I ω 2 0
)(
ɛ/2m
)
A 1
A 2
( )
= ω 2 A 1
A 2
The normal mode angular frequencies ω are the eigenvalues of the matrix.
To calculate it take the determinant and set it to zero.
∣ ω2 0 − ω2 ɛ/2m ∣∣∣∣
∣ ɛ/2I ω 2 0 − = 0 =⇒ ω 4 −(2ω 2 0) ω 2 +(ω 4
ω2 0 − ɛ2
4mI )=0