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7 Lagrangian and Hamiltonian Mechanics
(b) Symmetric mode ω 2 =
k
2m
B = +2A
7.24 There are three coordinates x 1 , x 2 and x 3 , Fig. 7.26:
Fig. 7.26
T =
1
2
m ˙
x
2
1 +
1
2
M x
2
2 +
1
2
m ˙
x
2
3
(1)
V =
1
2
k
(x 2 − x 1 )
2
+ (x 3 − x 2 )
2
=
1
2
k
x
2
1 − 2x 1 x 2 + 2x
2
2 − 2x 2 x 3 + x
2
3
(2)
L =
1
2
m ˙
x
2
1 +
1
2
M ˙
x
2
2 +
1
2
m ˙
x
2
3 −
1
2
k
x
2
1 − 2x 1 x 2 + 2x
2
2 − 2x 2 x 3 + x
2
3
(3)
Lagrange’s equations
d
dt
∂ L
∂ ˙
x 1
−
∂ L
∂ x 1
= 0,
d
dt
∂ L
∂ ˙
x 2
−
∂ L
∂ x 2
= 0,
d
dt
∂ L
∂ ˙
x 3
−
∂ L
∂ x 3
= 0 (4)
yield
m ¨
x 1 + k(x 1 − x 2 ) = 0
( 5 )
M ¨
x 2 + k(−x 1 + 2x 2 − x 3 ) = 0
( 6 )
m ¨
x 3 + k(−x 2 + x 3 ) = 0
( 7 )
Let the harmonic solutions be
x 1 = A sin ωt, x 2 = B sin ωt, x 3 = C sin ωt
( 8 )
∴ ¨
x 1 = −Aω
2 sin ωt, ¨
x 2 = −Bω
2 sin ωt, ¨
x 3 = −Cω
2 sin ωt
( 9 )
Substituting (8) and (9) in (5), (6) and (7)
(k − mω
2
)A − k B = 0
(10)
− k A + (2k − Mω
2
)B − kC = 0
(11)
− k B + (k − mω
2
) C = 0
(12)