H (Classical Mechanics)

The lower mass has coordinates

(x + l sin θ, - l cos θ)

. Its squared speed is

\overset{x}{˙}^{2} + 2 l cos θ \overset{x}{˙} \dot{θ} + l^{2} \dot{θ}^{2}

. Adding the bead's kinetic energy gives the kinetic part of the supplied Lagrangian. Equal spring prestresses give spring energy

k x^{2}

up to a constant: the two squared extensions are proportional to

(a + x - ℓ_{0})^{2} + (a - x - ℓ_{0})^{2}

. The gravitational potential is

- m_{2} g l cos θ

. Thus

L = T - V

has exactly the stated form, with constant energy offsets discarded.

Since

L

has no explicit time dependence, the conserved energy is

E = \frac{1}{2} (m_{1} + m_{2}) \overset{x}{˙}^{2} + m_{2} l cos θ \overset{x}{˙} \dot{θ} + \frac{1}{2} m_{2} l^{2} \dot{θ}^{2} + k x^{2} - m_{2} g l cos θ .

(m_{1} + m_{2}) \overset{x}{¨} + m_{2} l cos θ \ddot{θ} - m_{2} l sin θ \dot{θ}^{2} + 2 k x m_{2} l cos θ \overset{x}{¨} + m_{2} l^{2} \ddot{θ} + m_{2} g l sin θ = 0, = 0.

At equilibrium,

x = 0

and

sin θ = 0

. There are two geometrically distinct positions,

(x, θ) = (0, 0), (0, π)

2 π

. The downward one has positive potential Hessian

diag (2 k, m_{2} g l)

and is stable; the inverted one has a negative angular curvature and is unstable.

Keeping only quadratic terms about the downward equilibrium gives

L_{2} = \frac{1}{2} (m_{1} + m_{2}) \overset{x}{˙}^{2} + m_{2} l \overset{x}{˙} \dot{θ} + \frac{1}{2} m_{2} l^{2} \dot{θ}^{2} - k x^{2} - \frac{1}{2} m_{2} g l θ^{2} .

M = (2111)

and

K = (2001)

det (K - ω^{2} M) = 0

ω^{4} - 4 ω^{2} + 2 = 0

. Therefore the normal angular frequencies are

ω_{-} = 2 - 2, ω_{+} = 2 + 2 .

 Ancestors (10)

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