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: Identify the physical pieces of the Lagrangian. The kinetic energy is $T = \tfrac{1}{2}m\dot{x}^2$ and the potential energy is $V(x)$, so $L = T - V$ is the standard mechanical form.
Step 2: For such a system the Lagrange equation reduces to Newton's second law, because the generalized force is minus the gradient of the potential:
$$F = -\frac{dV}{dx}.$$
Step 3: Newton's law states $F = m\ddot{x}$. Equating the two expressions for the force:
$$m\ddot{x} = -\frac{dV}{dx}.$$
Step 4: Notice the sign: the particle is pushed toward decreasing potential energy, which is why the derivative carries a minus sign. This eliminates the $+V(x)=0$ options (dimensionally wrong) and the $+\dfrac{dV}{dx}$ option (wrong sign).
\[\boxed{m\ddot{x} = -\frac{dV(x)}{dx}}\]