Euler-Lagrange Equations from d'Alembert

Derivation

Euler-Lagrange Equations from d'Alembert

D-038 Home PU-101 Threads force · energy · symmetry Depends on The Work-Energy Theorem, Conservative Forces and Potential Energy

Statement

For a system of N particles subject to ideal (workless) holonomic constraints and described by n independent generalized coordinates \( q_1,\dots,q_n \), d'Alembert's principle \( \sum_i \left(\mathbf{F}_i - m_i \ddot{\mathbf{r}}_i\right)\cdot \delta\mathbf{r}_i = 0 \) is exactly equivalent to the n Euler–Lagrange equations \[ \frac{d}{dt}\!\left(\frac{\partial L}{\partial \dot{q}_j}\right) - \frac{\partial L}{\partial q_j} = 0, \qquad L = T - V, \] provided the applied forces derive from a potential \( V=V(q,t) \) that is independent of the generalized velocities.

Why it matters

This is the bridge from vectorial (Newtonian) mechanics to analytical mechanics. Newton's second law is written per particle in an inertial frame with every constraint force explicit; d'Alembert's principle removes the constraint forces by projecting onto virtual displacements, and this derivation shows that the projection collapses into a single scalar function \( L \) whose stationarity governs the entire system. Once you have the Euler–Lagrange equations you never again write a constraint force you do not care about.

It also explains why the Lagrangian is \( T-V \) rather than something you must postulate: the combination falls out of the inertial term reorganizing into derivatives of the kinetic energy, with the potential entering through the generalized force. Hamilton's variational principle is the same statement integrated in time, so this route is the physical, force-based justification of the action principle — the Lagrangian formalism is a theorem of Newtonian mechanics for this class of systems, not an independent axiom.

Assumptions

Ideal (workless) constraint forces.If constraint forces do net virtual work — as with sliding friction — they survive in \( \sum_i \mathbf{C}_i\cdot\delta\mathbf{r}_i \) and cannot be dropped, so d'Alembert's principle in the reduced coordinates picks up extra terms and the bare Euler–Lagrange form fails.

Holonomic constraints.Only holonomic constraints \( f(q,t)=0 \) let the \( \delta q_j \) be independent, which is what allows setting each coefficient to zero. Non-holonomic constraints (general rolling, inequalities) require Lagrange multipliers and the bracket-splitting step is illegal.

Velocity-independent potential.If the applied forces cannot be written as \( Q_j = -\partial V/\partial q_j \) with \( V \) independent of \( \dot q \), the equations stop at \( \tfrac{d}{dt}(\partial T/\partial\dot q_j)-\partial T/\partial q_j = Q_j \); velocity-dependent forces (Lorentz force) need a generalized potential \( U(q,\dot q,t) \) and a modified \( L \).

Smooth invertible coordinate map \( \mathbf{r}_i(q,t) \).The cancellation-of-dots and commutation lemmas use \( C^2 \) smoothness and equality of mixed partials; at coordinate singularities (e.g. the polar origin) the Jacobian \( \partial\mathbf{r}_i/\partial q_j \) loses rank and the Euler–Lagrange form holds only on charts where it is full rank.

Derivation

\[ \sum_i \left(\mathbf{F}_i - m_i \ddot{\mathbf{r}}_i\right)\cdot \delta\mathbf{r}_i = 0 \]

d'Alembert's principle: constraint forces are ideal, so they drop out of the virtual-work sum; \( \mathbf{F}_i \) is the applied force on particle \( i \). A

\[ \mathbf{r}_i = \mathbf{r}_i(q_1,\dots,q_n,t), \qquad \delta\mathbf{r}_i = \sum_j \frac{\partial \mathbf{r}_i}{\partial q_j}\,\delta q_j \]

Holonomic constraints parametrize positions by the \( q_j \); a virtual displacement is taken at fixed time (\( \delta t = 0 \)), so the explicit-time term \( \partial\mathbf{r}_i/\partial t \) does not appear. B

\[ \sum_i \mathbf{F}_i\cdot\delta\mathbf{r}_i = \sum_j Q_j\,\delta q_j, \qquad Q_j \equiv \sum_i \mathbf{F}_i\cdot\frac{\partial \mathbf{r}_i}{\partial q_j} \]

Substitute step 2 and collect the coefficient of each \( \delta q_j \); this defines the generalized force \( Q_j \) conjugate to \( q_j \). B

\[ \sum_i m_i \ddot{\mathbf{r}}_i\cdot\delta\mathbf{r}_i = \sum_j \left(\sum_i m_i \ddot{\mathbf{r}}_i\cdot\frac{\partial \mathbf{r}_i}{\partial q_j}\right)\delta q_j \]

Same substitution of step 2 into the inertial term; the inner bracket must now be rewritten as energy derivatives. B

\[ m_i \ddot{\mathbf{r}}_i\cdot\frac{\partial \mathbf{r}_i}{\partial q_j} = \frac{d}{dt}\!\left(m_i \dot{\mathbf{r}}_i\cdot\frac{\partial \mathbf{r}_i}{\partial q_j}\right) - m_i \dot{\mathbf{r}}_i\cdot\frac{d}{dt}\frac{\partial \mathbf{r}_i}{\partial q_j} \]

Product rule run backwards to expose a total time derivative — the standard trick that will generate the \( \tfrac{d}{dt}(\partial T/\partial\dot q) \) structure. C

\[ \dot{\mathbf{r}}_i = \sum_k \frac{\partial \mathbf{r}_i}{\partial q_k}\dot q_k + \frac{\partial \mathbf{r}_i}{\partial t} \;\Rightarrow\; \frac{\partial \dot{\mathbf{r}}_i}{\partial \dot q_j} = \frac{\partial \mathbf{r}_i}{\partial q_j} \]

"Cancellation of the dots." The velocity is linear in the \( \dot q_k \) with coefficients depending only on \( (q,t) \), so differentiating w.r.t. \( \dot q_j \) picks out the \( j \)-th coefficient. C

\[ \frac{d}{dt}\frac{\partial \mathbf{r}_i}{\partial q_j} = \sum_k \frac{\partial^2 \mathbf{r}_i}{\partial q_j\,\partial q_k}\dot q_k + \frac{\partial^2 \mathbf{r}_i}{\partial q_j\,\partial t} = \frac{\partial \dot{\mathbf{r}}_i}{\partial q_j} \]

Equality of mixed partials (\( C^2 \) map) lets \( d/dt \) and \( \partial/\partial q_j \) commute; the right side is exactly \( \partial\dot{\mathbf r}_i/\partial q_j \). C

\[ \sum_i m_i \ddot{\mathbf{r}}_i\cdot\frac{\partial \mathbf{r}_i}{\partial q_j} = \frac{d}{dt}\!\left(\frac{\partial T}{\partial \dot q_j}\right) - \frac{\partial T}{\partial q_j}, \qquad T \equiv \sum_i \tfrac12 m_i \dot{\mathbf{r}}_i^{\,2} \]

Insert lemmas 6 and 7 into step 5: both dot products become derivatives of \( \tfrac12 m_i\dot{\mathbf r}_i^2 \) using \( \mathbf{v}\cdot\partial\mathbf{v}/\partial s = \partial(\tfrac12\mathbf{v}^2)/\partial s \); sum over \( i \). C

\[ \sum_j \left[\frac{d}{dt}\!\left(\frac{\partial T}{\partial \dot q_j}\right) - \frac{\partial T}{\partial q_j} - Q_j\right]\delta q_j = 0 \]

Combine steps 3, 4 and 8 back into d'Alembert's principle (step 1) and collect the common coefficient of each \( \delta q_j \). B

\[ \frac{d}{dt}\!\left(\frac{\partial T}{\partial \dot q_j}\right) - \frac{\partial T}{\partial q_j} = Q_j, \qquad j = 1,\dots,n \]

The \( q_j \) are independent (holonomic), so the \( \delta q_j \) are arbitrary; a sum of independent variations vanishing for all \( \delta q_j \) forces each bracket to vanish separately. C

\[ Q_j = -\frac{\partial V}{\partial q_j}, \qquad V = V(q,t), \quad \frac{\partial V}{\partial \dot q_j} = 0 \]

The applied forces are conservative (using the prior result that a conservative force is \( -\nabla V \)); the potential does not depend on the velocities. A

\[ \frac{d}{dt}\!\left(\frac{\partial L}{\partial \dot q_j}\right) - \frac{\partial L}{\partial q_j} = 0, \qquad L \equiv T - V \]

Move \( -\partial V/\partial q_j \) into the \( \partial/\partial q_j \) slot; since \( \partial V/\partial\dot q_j = 0 \), \( \partial L/\partial\dot q_j = \partial T/\partial\dot q_j \) and the time-derivative term is unchanged. The two potential-dependent pieces assemble into a single function \( L \). A

Result

\[ \boxed{\;\frac{d}{dt}\!\left(\frac{\partial L}{\partial \dot q_j}\right) - \frac{\partial L}{\partial q_j} = 0, \qquad L = T - V, \qquad j=1,\dots,n\;} \]

Reading. Each generalized coordinate obeys one second-order equation of motion: the rate of change of the generalized momentum \( p_j = \partial L/\partial\dot q_j \) equals the generalized force \( \partial L/\partial q_j \). All ideal constraint forces have been eliminated, and the dynamics are packed into the single scalar \( L \). If \( L \) is independent of a coordinate \( q_j \) (cyclic), then \( p_j \) is conserved — the seed of Noether's theorem.

Units check. \( L \) has units of energy, J. For a length-like \( q_j \) (m): \( \partial L/\partial\dot q_j \) has J/(m·s\(^{-1}\)) = kg·m·s\(^{-1}\) (momentum), its \( d/dt \) is kg·m·s\(^{-2}\) = N, matching \( \partial L/\partial q_j \) = J/m = N. For an angle \( q_j \) (dimensionless rad): \( \partial L/\partial\dot q_j \) is J·s (angular momentum), \( d/dt \) gives J = N·m (torque), matching \( \partial L/\partial q_j \) in J. The equation is dimensionally homogeneous as a generalized-force balance for any coordinate type.

Limiting cases

Single Cartesian coordinate, \( q=x \): \( L = \tfrac12 m\dot x^2 - V(x) \) gives \( m\ddot x = -dV/dx \), recovering Newton's second law exactly.
Free particle, \( V=0 \), \( q_j \) Cartesian: \( \tfrac{d}{dt}(m\dot q_j) = 0 \), momentum conserved (Newton's first law).
No explicit time dependence (\( \partial L/\partial t = 0 \)): the energy function \( h = \sum_j \dot q_j\,\partial L/\partial\dot q_j - L \) is conserved; for scleronomic \( T \) it equals \( T+V \).
Cyclic coordinate (\( \partial L/\partial q_j = 0 \)): the conjugate momentum \( p_j = \partial L/\partial\dot q_j \) is a constant of motion (linear momentum for translations, angular momentum for rotations).
Static equilibrium (\( \dot q = \ddot q = 0 \)): reduces to \( \partial V/\partial q_j = 0 \), the principle of virtual work / stationary potential energy.

Breaks when

Non-holonomic or non-ideal constraints. If constraints cannot be written as \( f(q,t)=0 \), the \( \delta q_j \) are not independent and step 10 fails; general rolling and sliding-with-friction systems need Lagrange multipliers or extra generalized forces. Dissipative forces (drag, kinetic friction) do virtual work and cannot be absorbed into any \( V \).
Velocity-dependent forces without a generalized potential. A force that is not \( -\partial V/\partial q_j \) for velocity-independent \( V \) stops the derivation at step 10; the magnetic Lorentz force is the borderline case that does admit a generalized potential \( U(q,\dot q) = e\phi - e\,\mathbf{A}\cdot\mathbf{v} \), so \( L=T-U \) works only there.
Relativistic or field regimes. The kinetic energy \( T = \sum_i \tfrac12 m_i\dot{\mathbf r}_i^2 \) and the fixed-particle picture assume \( v \ll c \); at relativistic speed \( L = -mc^2\sqrt{1-v^2/c^2} - V \) replaces \( \tfrac12 mv^2 \), and continuous fields require the Lagrangian-density Euler–Lagrange equations instead.
Non-inertial reference frame. The derivation starts from \( m_i\ddot{\mathbf r}_i \) in an inertial frame; if \( \mathbf r_i \) is measured in a rotating/accelerating frame the inertial term differs and Coriolis/centrifugal pseudo-forces must be included in \( V \) or \( Q_j \).

Failure modes

Treating \( q_j \) and \( \dot q_j \) as dependent while differentiating. In \( \partial L/\partial\dot q_j \) and \( \partial L/\partial q_j \) the coordinates and velocities are independent arguments; only the total time derivative in the first term imposes the \( q \)–\( \dot q \) link. Substituting the equation of motion too early gives wrong terms.
Forgetting the \( \partial T/\partial q_j \) term. Curvilinear coordinates make \( T \) depend on \( q_j \) (e.g. \( T = \tfrac12 m(\dot r^2 + r^2\dot\theta^2) \) in polar). Dropping \( \partial T/\partial q_j \) loses the centrifugal term and gives incorrect radial dynamics.
Sign error \( L = T+V \). Writing the Lagrangian as the total energy is the single most common slip; \( L \) is the difference, with the sign of \( V \) fixed by \( Q_j = -\partial V/\partial q_j \).
Applying \( L=T-V \) to friction. Expecting Euler–Lagrange to hold with dissipation present; friction must enter as a non-conservative \( Q_j \) (or a Rayleigh function), never inside \( L \).
Choosing dependent generalized coordinates. If the chosen \( q_j \) are not independent (a constraint left unused), step 10 is invalid and one over-counts the equations of motion.

Discussion

The physical heart of the derivation is that ideal constraint forces are geometrically perpendicular to the allowed virtual displacements: a bead confined to a wire feels a normal force that does no virtual work along the wire. d'Alembert's principle turns Newton's dynamical statement into a statement about the tangent space of the configuration manifold, and the projection \( \mathbf{F}_i\cdot(\partial\mathbf{r}_i/\partial q_j) \) is precisely the push-forward of forces onto that tangent space. Everything downstream is bookkeeping that recognizes the resulting quadratic-in-velocity expression as derivatives of kinetic energy.

Two small lemmas carry the entire technical load: cancellation of the dots (\( \partial\dot{\mathbf r}/\partial\dot q = \partial\mathbf r/\partial q \)) and commutation of the time and coordinate derivatives (\( \tfrac{d}{dt}\,\partial\mathbf r/\partial q = \partial\dot{\mathbf r}/\partial q \)). Both follow from the map \( \mathbf r(q,t) \) being smooth and from \( \dot{\mathbf r} \) being linear in the generalized velocities. This linearity is why kinetic energy is a homogeneous quadratic form \( T = \tfrac12\sum_{jk} M_{jk}(q)\dot q_j\dot q_k \) in scleronomic systems, and it is what makes \( \partial T/\partial\dot q_j \) the generalized momentum.

Conceptually, the result explains why Hamilton's principle \( \delta\int L\,dt = 0 \) works: the Euler–Lagrange equations are the stationarity conditions of the action, so a variational principle and Newton's laws coincide for this class of systems. Historically d'Alembert (1743) came first; the action principle is a later repackaging. Deriving Euler–Lagrange from d'Alembert therefore demystifies the action — it is not a new law of nature but a compact encoding of Newtonian dynamics after constraint forces are projected out.

The deeper geometric reading is that \( Q_j\,\delta q_j \) and \( \big(\tfrac{d}{dt}\partial T/\partial\dot q_j - \partial T/\partial q_j\big)\delta q_j \) are one-forms on configuration space; the equation of motion is their equality as covectors, independent of coordinate choice. The term \( \tfrac{d}{dt}(\partial T/\partial\dot q_j) - \partial T/\partial q_j \) is precisely the covariant acceleration \( g_{jk}(\ddot q^k + \Gamma^k_{lm}\dot q^l\dot q^m) \) associated with the kinetic-energy metric \( T = \tfrac12 g_{jk}(q)\dot q^j\dot q^k \); free motion (\( Q_j=0 \)) is geodesic motion on the configuration manifold. This is the seed of the Riemannian formulation of mechanics and of general relativity's geodesic equation \( \ddot x^\mu + \Gamma^\mu_{\alpha\beta}\dot x^\alpha\dot x^\beta = 0 \), itself an Euler–Lagrange equation for \( L = \tfrac12 g_{\mu\nu}\dot x^\mu\dot x^\nu \).

Common misconceptions. (1) The Lagrangian is not the total energy — \( L=T-V \), and only the Hamiltonian/energy function equals \( T+V \) (for scleronomic systems with quadratic \( T \)). (2) d'Alembert's principle is not "static equilibrium with an inertial force" tacked on; the inertial term \( -m_i\ddot{\mathbf r}_i \) is genuine dynamics, and the "equilibrium" holds only in the space of virtual displacements. (3) Cyclic does not mean absent from the motion — a cyclic coordinate still evolves; it is its momentum that is conserved.

Worked examples

Example 1 — Simple plane pendulum. A bob of mass \( m = 0.50\ \text{kg} \) on a rigid massless rod of length \( \ell = 1.00\ \text{m} \) swings in a vertical plane under gravity \( g = 9.81\ \text{m s}^{-2} \). Use the angle \( \theta \) from the downward vertical.

\[ x = \ell\sin\theta,\ y = -\ell\cos\theta \;\Rightarrow\; T = \tfrac12 m\ell^2\dot\theta^2,\quad V = -mg\ell\cos\theta \]

Rigid rod is the ideal constraint; speed is \( \ell\dot\theta \), potential measured from the pivot. B

\[ L = \tfrac12 m\ell^2\dot\theta^2 + mg\ell\cos\theta \]

Form \( L=T-V \). A

\[ \frac{\partial L}{\partial\dot\theta} = m\ell^2\dot\theta,\quad \frac{d}{dt}\frac{\partial L}{\partial\dot\theta} = m\ell^2\ddot\theta,\quad \frac{\partial L}{\partial\theta} = -mg\ell\sin\theta \]

Differentiate treating \( \theta,\dot\theta \) as independent, then take the time derivative. A

\[ m\ell^2\ddot\theta + mg\ell\sin\theta = 0 \;\Rightarrow\; \ddot\theta = -\frac{g}{\ell}\sin\theta \]

Euler–Lagrange equation; the mass and one power of \( \ell \) cancel and the rod tension never appears. A

\[ \theta = 30^\circ = 0.5236\ \text{rad}: \quad \ddot\theta = -\frac{9.81}{1.00}\times 0.5000 = -4.905\ \text{rad s}^{-2} \]

Insert numbers last; \( \sin 30^\circ = 0.5 \). A

\[ \ddot\theta = -\frac{g}{\ell}\sin\theta = -4.91\ \text{rad s}^{-2} \quad(\text{independent of } m) \]

Reading. The angular acceleration depends only on \( g,\ell,\theta \) — the classic mass-independence of pendulum motion, obtained without ever writing the rod tension. Small oscillations have \( \omega_0=\sqrt{g/\ell} \).

Example 2 — Atwood machine. Two masses \( m_1 = 3.0\ \text{kg} \) and \( m_2 = 2.0\ \text{kg} \) hang from an ideal massless inextensible string of length \( \ell \) over a frictionless massless pulley, \( g = 9.81\ \text{m s}^{-2} \). Let \( x \) be the length of string on the \( m_1 \) side.

\[ \dot x_1 = \dot x,\quad \dot x_2 = -\dot x \;\Rightarrow\; T = \tfrac12(m_1+m_2)\dot x^2 \]

Inextensible string is the ideal holonomic constraint tying both speeds to \( \dot x \); one coordinate suffices. B

\[ V = -m_1 g x - m_2 g(\ell - x) \;\Rightarrow\; L = \tfrac12(m_1+m_2)\dot x^2 + m_1 g x + m_2 g(\ell-x) \]

Heights: \( m_1 \) at depth \( x \), \( m_2 \) at depth \( \ell-x \); the constraint has eliminated the second coordinate. B

\[ \frac{d}{dt}\frac{\partial L}{\partial\dot x} = (m_1+m_2)\ddot x,\qquad \frac{\partial L}{\partial x} = (m_1-m_2)g \]

Euler–Lagrange pieces; the constant \( \ell \)-term drops and the string tension never appears (the eliminated constraint force). A

\[ (m_1+m_2)\ddot x = (m_1-m_2)g \;\Rightarrow\; \ddot x = \frac{m_1-m_2}{m_1+m_2}\,g \]

Solve for the acceleration; symbols before numbers. A

\[ \ddot x = \frac{3.0-2.0}{3.0+2.0}\times 9.81 = \frac{1}{5}\times 9.81 = 1.96\ \text{m s}^{-2} \]

Substitute numbers with units. A

\[ \ddot x = \frac{m_1-m_2}{m_1+m_2}\,g = 1.96\ \text{m s}^{-2}\ (m_1\text{ side downward}) \]

Reading. The heavier mass descends at \( \tfrac15 g \), set by the mass ratio. One generalized coordinate captured both masses and eliminated the tension entirely — the payoff of the Lagrangian route.

Problems

(Easy) A particle of mass \( m \) moves in one dimension in a potential \( V(x) \). Starting from \( L = \tfrac12 m\dot x^2 - V(x) \), obtain the equation of motion and identify it.
Solution
\( \partial L/\partial\dot x = m\dot x \Rightarrow \frac{d}{dt}(\cdot) = m\ddot x \); \( \partial L/\partial x = -dV/dx \). Euler–Lagrange: \( m\ddot x + dV/dx = 0 \), i.e. \( m\ddot x = -dV/dx \) — Newton's second law with conservative force \( F=-dV/dx \). Units: kg·m·s\(^{-2}\) = J/m = N.
(Easy) A force \( \mathbf{F} = (3,-2,0)\ \text{N} \) acts on a particle at position \( \mathbf{r} = (q, q^2, 0)\ \text{m} \), with \( q \) a length in metres. Find the generalized force \( Q \) and evaluate it at \( q=1\ \text{m} \).
Solution
\( \partial\mathbf{r}/\partial q = (1, 2q, 0) \). \( Q = \mathbf{F}\cdot\partial\mathbf{r}/\partial q = 3(1) + (-2)(2q) + 0 = 3 - 4q\ \text{N} \). At \( q=1 \): \( Q = 3-4 = -1\ \text{N} \). The negative sign means the generalized force opposes increasing \( q \) there.
(Medium) Using \( \ddot\theta = -(g/\ell)\sin\theta \) from Example 1, find the small-oscillation angular frequency and period for \( \ell = 0.25\ \text{m} \), \( g = 9.81\ \text{m s}^{-2} \).
Solution
Small angle \( \sin\theta\approx\theta \) gives \( \ddot\theta \approx -(g/\ell)\theta \), SHM with \( \omega_0=\sqrt{g/\ell} = \sqrt{9.81/0.25} = \sqrt{39.24} = 6.26\ \text{rad s}^{-1} \). Period \( T = 2\pi/\omega_0 = 2\pi\sqrt{\ell/g} = 2\pi(0.1597) = 1.00\ \text{s} \). Units: \( \sqrt{\text{m s}^{-2}/\text{m}} = \text{s}^{-1} \).
(Medium) A mass \( m = 0.50\ \text{kg} \) on a spring of constant \( k = 200\ \text{N m}^{-1} \) moves horizontally without friction. Write \( L \), derive the equation of motion, and find the frequency \( f \).
Solution
\( T = \tfrac12 m\dot x^2 \), \( V = \tfrac12 kx^2 \), \( L = \tfrac12 m\dot x^2 - \tfrac12 kx^2 \). \( \partial L/\partial\dot x = m\dot x \Rightarrow \frac{d}{dt}(\cdot) = m\ddot x \); \( \partial L/\partial x = -kx \). Euler–Lagrange: \( m\ddot x + kx = 0 \Rightarrow \ddot x = -(k/m)x \). \( \omega = \sqrt{k/m} = \sqrt{200/0.50} = \sqrt{400} = 20\ \text{rad s}^{-1} \). \( f = \omega/2\pi = 3.18\ \text{Hz} \). Units: \( \sqrt{\text{N m}^{-1}/\text{kg}} = \text{s}^{-1} \).
(Hard) A bead of mass \( m \) slides without friction on a wire shaped as the parabola \( y = \tfrac12 a x^2 \) in uniform gravity \( g \) (\( a \) in m\(^{-1}\)), with \( x \) the generalized coordinate. Derive the exact equation of motion and the small-oscillation frequency about \( x=0 \). Evaluate \( \omega \) for \( a = 2.0\ \text{m}^{-1} \), \( g = 9.81\ \text{m s}^{-2} \).
Solution
\( \mathbf{r} = (x, \tfrac12 a x^2) \), so \( \dot{\mathbf r} = (\dot x, a x\dot x) \) and \( T = \tfrac12 m\dot x^2(1 + a^2 x^2) \). \( V = mgy = \tfrac12 mga x^2 \). \( L = \tfrac12 m\dot x^2(1+a^2x^2) - \tfrac12 mgax^2 \). Then \( \partial L/\partial\dot x = m\dot x(1+a^2x^2) \Rightarrow \frac{d}{dt}(\cdot) = m(1+a^2x^2)\ddot x + 2ma^2x\dot x^2 \), and \( \partial L/\partial x = ma^2x\dot x^2 - mgax \). Euler–Lagrange: \( m(1+a^2x^2)\ddot x + 2ma^2x\dot x^2 - ma^2x\dot x^2 + mgax = 0 \), i.e. \[ (1+a^2x^2)\ddot x + a^2x\dot x^2 + gax = 0 \quad(\text{exact}). \] For small \( x \), drop quadratic-in-\( (x,\dot x) \) terms: \( \ddot x + gax \approx 0 \), so \( \omega = \sqrt{ga} = \sqrt{9.81\times 2.0} = \sqrt{19.62} = 4.43\ \text{rad s}^{-1} \). The position-dependent effective mass \( m(1+a^2x^2) \) shows the \( \partial T/\partial x \) term is essential — the classic trap.