Worked exercise — Two pendulums coupled by a wire

Problem

Two identical pendulums of length $\ell = 1$ m hang from the same horizontal wire. The torsional coupling of the wire provides a small constant $\kappa$ ($\kappa \ll K$). Find the two normal modes and describe the motion when only the first pendulum is displaced.

Solution

Symmetric mode (in phase). The two pendulums oscillate together, the wire does not twist: the frequency is that of the isolated pendulum, $\omega_+ = \sqrt{\frac{g}{\ell}} = \sqrt{\frac{9{,}81}{1}} \approx 3{,}13 \ \text{rad/s} \quad (T \approx 2{,}0 \ \text{s})$

Antisymmetric mode (out of phase). The pendulums oscillate in opposition, the wire twists and adds a small restoring effect: $\omega_- \approx \omega_+\left(1 + \frac{\kappa}{K}\right)$ slightly greater than $\omega_+$.

Motion with only one pendulum displaced. Displacing only the first and releasing it, the initial state is the superposition of the two modes. Since $\omega_-$ and $\omega_+$ differ only slightly, the modes slowly dephase: the energy migrates from the first pendulum to the second, then back to the first, with a beat period $\ev{T_b = \frac{2\pi}{\omega_- - \omega_+} \propto \frac{K}{\kappa}}$ the weaker the coupling, the longer the period. This is an experience that is easily reproducible in a school laboratory, and foreshadows the pair of nearby levels of quantum mechanics.

Collegamenti

Argomenti: Oscillations and simple harmonic motion Concetti: Normal modes · Beats · Simple pendulum Oggetti: Simple pendulum