For a 1D harmonic oscillator $L = \frac{1}{2}m\dot{x}^2 - \frac{1}{2}kx^2$, the E-L equation gives:

$\frac{\partial L}{\partial \dot{x}} = m\dot{x}$, $\frac{\partial L}{\partial x} = -kx$, so E-L: $m\ddot{x} + kx = 0$.

Correct answer: $m\ddot{x} + kx = 0$

Noether's theorem states that a cyclic coordinate $q_i$ (where $\partial L / \partial q_i = 0$) implies:

If $q_i$ is cyclic, E-L gives $\dot{p}_i = 0$, so $p_i$ is constant.

Correct answer: The conjugate momentum $p_i = \partial L/\partial \dot{q}_i$ is conserved

Write the Lagrangian for a simple pendulum (mass $m$, length $l$, angle $\theta$, no friction).

Kinetic energy $T = \frac{1}{2}ml^2\dot{\theta}^2$, potential $V = -mgl\cos\theta$.

Correct answer: $L = \frac{1}{2}ml^2\dot{\theta}^2 + mgl\cos\theta$