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Unformatted text preview: · · 1
0
0 · · · −2
xn We can write this as where x= 1
V (x1 , . . . , xn ) = − k xT Ax,
2
x1
x2
x3
·
xn , A= −2
1
0
·
1 1
−2
1
·
0 0
1
−2
·
0 ···
···
···
···
··· or equivalently as
n n 1
V (x1 , . . . , xn ) = − k
aij xi xj ,
2 i=1 j =1
54 1
0
0
·
−2 , where aij denotes the (i, j )component of the matrix A.
Just as in the preceding section, the force acting on the ith cart can be
calculated as minus the derivative of the potential energy with respect to the
position of the ith cart, the other carts being held ﬁxed. Thus for example,
n F1 = − n n ∂V
1
1
=k
a1j xj + k
ai1 xi = k
a1j xj ,
∂x1
2 j =1
2 i=1
j =1 the last step obtained by using the symmetry of A. In general, we obtain the
result:
n
∂V
Fi = −
=k
aij xj ,
∂xi
j =1
which could be rewritten in matrix form as
F = kAx. (2.16) On the other hand, by Newton’s second law of motion,
m d2 x
= F.
dt2 Substitution into (2.16) yields
m d2 x
= kAx
dt2 or d2 x
k
= Ax,
dt2
m (2.17) where A is...
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 Winter '14
 Equations

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