Solution: Coupled Pendulums with Slack Cord
The motion consists of two distinct phases:
- Phase 1 (Moving Apart): The bobs move outward from the equilibrium position. The cord stretches, providing a restoring force \(2kx\). The effective restoring force is \(F = -(mg/l + 2k)x\).
- Phase 2 (Moving Together): The bobs move inward past the equilibrium line. The cord goes slack (tension = 0). The restoring force is gravity only: \(F = -(mg/l)x\).
Time Period Calculation
The total period is the sum of the half-periods of these two phases.
$$ \omega_1 = \sqrt{\frac{g}{l} + \frac{2k}{m}} \quad \implies \quad t_1 = \frac{\pi}{\omega_1} $$ $$ \omega_2 = \sqrt{\frac{g}{l}} \quad \implies \quad t_2 = \frac{\pi}{\omega_2} $$
$$ T = \pi \left[ \left( \frac{g}{l} \right)^{-1/2} + \left( \frac{g}{l} + \frac{2k}{m} \right)^{-1/2} \right] $$