Simple Pendulum: Lagrangian Dynamics & Large-Angle Non-Linearity
Aim: to formulate the simple pendulum from the Lagrangian \(L = T - V\), explore the breakdown of the small-angle approximation \(\sin\theta \approx \theta\), observe exact elliptic period expansion \(T(\theta_0)\), and measure local gravitational acceleration \(g\) using high-precision photogate timing.
Virtual Apparatus & Optical Photogate
Phase Space Portrait \((\theta, \dot{\theta})\)
Non-Linear Period Expansion \(T/T_0\) vs \(\theta_0\)
Observation Table (\(T^2\) vs \(l\) Fit to Measure \(g\))
Lagrangian Formulation of the Pendulum
Consider a bob of mass \(m\) suspended by a massless rigid cord of length \(l\) in a uniform gravitational field \(g\). Choosing the angular displacement \(\theta\) from the downward vertical as the single generalized coordinate:
$$ T = \frac{1}{2} m v^2 = \frac{1}{2} m l^2 \dot{\theta}^2, \qquad V = m g l (1 - \cos\theta) $$The Lagrangian function is \(L(\theta, \dot{\theta}) = T - V\):
$$ L = \frac{1}{2} m l^2 \dot{\theta}^2 - m g l (1 - \cos\theta) $$Applying the Euler–Lagrange equation \(\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{\theta}}\right) - \frac{\partial L}{\partial\theta} = Q_{nc}\):
$$ m l^2 \ddot{\theta} + m g l \sin\theta = -\gamma \dot{\theta} \implies \ddot{\theta} + \frac{g}{l}\sin\theta + \frac{\gamma}{m l^2}\dot{\theta} = 0 $$In the small-angle limit (\(\theta \ll 1\)), \(\sin\theta \approx \theta\), yielding simple harmonic motion with natural frequency \(\omega_0 = \sqrt{g/l}\) and period \(T_0 = 2\pi\sqrt{l/g}\).
Exact Elliptic Integral Solution & Large-Angle Non-Linearity
For finite release amplitude \(\theta_0\), the restore torque is strictly weaker than linear (\(\sin\theta < \theta\)), causing the period to lengthen. Integrating conservation of energy \(\frac{1}{2}m l^2\dot{\theta}^2 + m g l(1-\cos\theta) = m g l(1-\cos\theta_0)\):
$$ T(\theta_0) = 4 \sqrt{\frac{l}{2g}} \int_0^{\theta_0} \frac{d\theta}{\sqrt{\cos\theta - \cos\theta_0}} = \frac{2}{\pi} T_0 K\left(\sin\frac{\theta_0}{2}\right) $$where \(K(k) = \int_0^{\pi/2} \frac{d\phi}{\sqrt{1 - k^2\sin^2\phi}}\) is the complete elliptic integral of the first kind. Expanding in powers of \(\theta_0\):
$$ T(\theta_0) = T_0 \left(1 + \frac{1}{16}\theta_0^2 + \frac{11}{3072}\theta_0^4 + \frac{173}{737280}\theta_0^6 + \dots\right) $$Viva Questions
Why does the period increase at large angles?
Because \(\sin\theta < \theta\). The restoring torque \(-\frac{g}{l}\sin\theta\) is weaker than a true linear Hookean spring at large angles, so the pendulum moves more sluggishly near its turnarounds.
What happens to the period as \(\theta_0 \to 180^\circ\) (\(\pi\) rad)?
The elliptic modulus \(k = \sin(\theta_0/2) \to 1\), and \(K(1) \to \infty\). The period diverges logarithmically as \(T \approx \frac{2}{\pi}T_0 \ln\left(\frac{8}{\pi - \theta_0}\right)\), since an inverted pendulum is an unstable equilibrium point.