Unformatted text preview: x2 .
∂x2 It now follows from Newton’s second law of motion that
Force = Mass × Acceleration,
and hence d2 x1
d2 x2
,
F2 = m 2 .
dt2
dt
Thus we obtain a secondorder system of diﬀerential equations,
F1 = m m d2 x1
= −k1 x1 + k2 (x2 − x1 ) = −(k1 + k2 )x1 + k2 x2 ,
dt2 d2 x2
= k2 (x1 − x2 ) − k3 x2 = k2 x1 − (k2 + k3 )x2 .
dt2
We can write this system in matrix form as
m m d2 x
= Ax,
dt2 where −(k1 + k2 )
k2
k2
−(k2 + k3 ) A= 50 . (2.14) Note that A is indeed a symmetric matrix. The potential energy is given by the
expression
1
x
V (x1 , x2 ) = − x1 x2 A 1 .
x2
2
Example. Let us consider the special case of the massspring system in which
m = k1 = k2 = k3 = 1,
so that the system (2.14) becomes
d2 x
=
dt2 −2
1 1
x.
−2 (2.15) To ﬁnd the eigenvalues, we must solve the characteristic equation
det −2 − λ
1 1
−2 − λ = (λ + 2)2 − 1 = 0, which yields
λ = −2 ± 1.
The eigenvalues in this case are
λ1 = −1, and λ2 = −3. The eigenspace corresponding to the eigenvalue −1 is
1
1 W−1 = {b ∈ R2 : (A + I )b = 0} = . . . = span . It follows from the argument i...
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 Winter '14
 Equations

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