Measure each rod angle from the downward vertical, as in the held diagram. The upper mass rises by

2 l (1 - cos θ_{1})

; the lower mass rises by

2 l (1 - cos θ_{1}) + l (1 - cos θ_{2})

. With the hanging equilibrium as the zero of gravitational potential energy,

U = m g l [4 (1 - cos θ_{1}) + (1 - cos θ_{2})] ≃ \frac{m g l}{2} (4 θ_{1}^{2} + θ_{2}^{2}) .

The retained expression is quadratic in the small displacements. Both masses contribute to the upper rod's gravitational term.

Solution

The two positions are

r_{1} = (2 l sin θ_{1}, - 2 l cos θ_{1})

and

r_{2} = r_{1} + (l sin θ_{2}, - l cos θ_{2})

. Differentiating and summing their kinetic energies gives the exact expression

K = \frac{m l ^{2}}{2} [8 \dot{θ}_{1}^{2} + \dot{θ}_{2}^{2} + 4 \dot{θ}_{1} \dot{θ}_{2} cos (θ_{1} - θ_{2})] .

In the linear small-oscillation approximation, set the cosine to one in this quadratic velocity expression:

K ≃ \frac{m l ^{2}}{2} (8 \dot{θ}_{1}^{2} + 4 \dot{θ}_{1} \dot{θ}_{2} + \dot{θ}_{2}^{2}) .

The cross term arises because the lower mass participates in motion generated by both rods.

A symmetric quadratic form has twice its off-diagonal entry as the coefficient of the mixed product. Comparing with the preceding potential energy and kinetic energy expressions gives

V = m g l (4001), T = m l^{2} (8221) .

Here

V

T

the mass matrix in angular coordinates; their units differ accordingly. The dimensionless mass-matrix determinant is

8 - 4 = 4 > 0

and its first diagonal entry is positive, so the kinetic energy is positive for every nonzero angular velocity.

The quadratic total mechanical energy is

E = \frac{1}{2} \dot{θ}^{T} T \dot{θ} + \frac{1}{2} θ^{T} V θ, \frac{d E}{d t} = \dot{θ}^{T} (T \ddot{θ} + V θ) .

For this frictionless conservative double pendulum, the Euler-Lagrange equations for

L = K - U

give

T \ddot{θ} + V θ = 0,

8 \ddot{θ}_{1} + 2 \ddot{θ}_{2} + (4 g / l) θ_{1} = 0

and

2 \ddot{θ}_{1} + \ddot{θ}_{2} + (g / l) θ_{2} = 0

. Substitution immediately verifies energy conservation.

Logical qualification. Energy conservation does not uniquely determine equations of motion: the scalar equation

d E / d t = 0

alone does not determine both equations. For example, replacing their right side by

G \dot{θ}

with any nonzero antisymmetric matrix

G

preserves the same energy because

\dot{θ}^{T} G \dot{θ} = 0

, while changing the dynamics. Such a gyroscopic force is absent from the specified pendulum. The conservative-force structure, supplied here by the Lagrangian, is the additional information needed to justify the displayed equations rather than infer two independent equations from one scalar identity.