A mass $m$ on a frictionless spring with stiffness $k$, at position $q$
A block of mass $m$ slides without friction on a horizontal surface, attached to a wall by a spring with stiffness $k$. The coordinate $q$ is the displacement of the block from the position where the spring is relaxed, and the spring pulls the block back with the force $-k q$ (Hooke’s law). By Newton’s second law, the motion of the block obeys
\[m \ddot{q} = -k q.\]Terms
| Term | This example | Why |
|---|---|---|
| lagrangian | $L = \frac{1}{2} m \dot{q}^2 - \frac{1}{2} k q^2$ | The Lagrangian of a mass on a frictionless spring is $L = T - V$, with the kinetic energy $T = \frac{1}{2} m \dot{q}^2$ and the potential energy $V = \frac{1}{2} k q^2$. The Euler–Lagrange equation is $m \ddot{q} = -k q$. |
| conjugate momentum | $p = m \dot{q}$ | The conjugate momentum of a mass on a frictionless spring is $p = \partial L / \partial \dot{q} = m \dot{q}$, which is the ordinary momentum. |
| hamiltonian | $H = p^2 / (2 m) + \frac{1}{2} k q^2$ | The Hamiltonian of a mass on a frictionless spring is $H = p \dot{q} - L$ with $\dot{q} = p / m$, which gives the total energy $p^2 / (2 m) + \frac{1}{2} k q^2$. |
| conservative system | ✓ | A mass on a frictionless spring is a conservative system, because the spring force $-k q$ is the negative derivative of the potential energy $V(q) = \frac{1}{2} k q^2$. |