- #1

Wavefunction

- 99

- 4

## Homework Statement

The point of suspension of a simple pendulum of length [itex] l [/itex] and mass [itex] m [/itex] is constrained to move on

a parabola [itex] z = ax^2 [/itex] in the vertical plane. Derive a Hamiltonian governing the motion of the

pendulum and its point of suspension. This is a two-dimensional problem.

## Homework Equations

eq(1) [itex] H=T+U [/itex]

## The Attempt at a Solution

[itex] T=\frac{m}{2}[\dot{x}^2+\dot{z}^2] [/itex]

[itex] U=mgz [/itex]

eq(2) [itex] H = \frac{m}{2}[\dot{x}^2+\dot{z}^2]+mgz [/itex]

Applying the constraint [itex] z=ax^2 [/itex], [itex] \dot{z} = 2ax\dot{x} [/itex]

[itex] H = \frac{m}{2}[\dot{x}^2+4a^2x^2\dot{x}^2]+mgax^2 [/itex]

The resulting Hamilton's equations I got were:

eq(3) [itex] \dot{p_x} = -(4ma^2x\dot{x}^2+2mgax) = -\frac{∂H}{∂x} [/itex]

eq(4) [itex] \dot{x} = \frac{p_x}{m(1+4a^2x^2)} = \frac{∂H}{∂p_x} [/itex]

and [itex] p_x = \frac{∂L}{∂\dot{x}} = m(1+4a^2x^2)\dot{x} [/itex]

This problem seemed uncharacteristically easy, so I definitely appreciate anybody who checks it over (: Thank you in advance!