Question

A particle of mass \(m\) moves under a potential \(V(x)\). The Lagrangian of the system is given by \(L = \tfrac{1}{2}m\dot{x}^2 - V(x)\). According to Lagrange's equation of motion, which of the following is the correct equation of motion for the particle?

• A. \(m\ddot{x} + V(x) = 0\)
• B. \(m\ddot{x} = -\dfrac{dV(x)}{dx}\)
• C. \(m\ddot{x} = \dfrac{dV(x)}{dx}\)
• D. \(m\ddot{x} + V(x) = 0\)

Hint:

Plug \(L = \tfrac{1}{2}m\dot{x}^2 - V(x)\) into \(\frac{d}{dt}(\partial L/\partial\dot{x}) - \partial L/\partial x = 0\); this is just \(F = ma\) with \(F = -dV/dx\).

Answer 1

The Correct Option is B

Step 1: The Euler-Lagrange equation for the coordinate \(x\) is
\[\frac{d}{dt}\!\left(\frac{\partial L}{\partial \dot{x}}\right) - \frac{\partial L}{\partial x} = 0\]
Step 2: With \(L = \tfrac{1}{2}m\dot{x}^2 - V(x)\), the momentum term is
\[\frac{\partial L}{\partial \dot{x}} = m\dot{x} \quad\Rightarrow\quad \frac{d}{dt}\!\left(\frac{\partial L}{\partial \dot{x}}\right) = m\ddot{x}\]
Step 3: The coordinate term is
\[\frac{\partial L}{\partial x} = -\frac{dV}{dx}\]
Step 4: Substitute both into the Euler-Lagrange equation:
\[m\ddot{x} - \left(-\frac{dV}{dx}\right) = 0 \quad\Rightarrow\quad m\ddot{x} = -\frac{dV}{dx}\]
\[\boxed{m\ddot{x} = -\frac{dV(x)}{dx}}\]