320
7 Lagrangian and Hamiltonian Mechanics
The eigenfrequency equation is obtained by equating to zero the determinant formed by the coefficients of A and B:
(2k − mω 2 )
−k
−k
(k − mω 2 )
= 0
(11)
Expanding the determinant, we obtain
m
2
ω
4
− 3mkω
2
+ k
2
= 0
ω
2
=
3 ±
√
5
2
k
m
(12)
∴ ω 1 = 1.618
k
m
, ω 2 = 0.618
k
m
(13)
(c) Inserting ω = ω 1 in (10) we find B = 1.618 A. This corresponds to a
symmetric mode as both the amplitudes have the same sign.
Inserting ω = ω 2 in (10), we find B = −0.618 A. This corresponds
to asymmetric mode. These two modes of oscillation are depicted in
Fig. 7.24 with relative sizes and directions of displacement.
(a) Symmetric mode ω 1 = 1.618
k
m
(b) Asymmetric mode ω 2 = 0.618
k
m
Fig. 7.24
7.23 (a) Let x 1 and x 2 be the displacements of the beads of mass 2m and m,
respectively.
T =
1
2
(2m) ˙
x
2
1 +
1
2
(m) ˙
x
2
2
(1)
V =
1
2
· 2kx
2
1 +
1
2
k(x 2 − x 1 )
2
(2)
L = m
˙
x
2
1 +
1
2
˙
x
2
2
− k
3
2
x
2
1 − x 1 x 2 +
1
2
x
2
2
(3)
Lagrange’s equations are
d
dt
∂ L
∂ ˙
x 1
−
∂ L
∂ x 1
= 0,
d
dt
∂ L
∂ ˙
x 2
−
∂ L
∂ x 2
= 0
( 4 )