CHALLENGE · OSCILLATORS EVERYWHERE

COUPLED PENDULUM NORMAL MODES

Two identical pendulums (length L = 0.50 m, bob mass m = 0.25 kg) are connected by a weak coupling spring with constant k_c = 0.80 N/m. Find: (a) the natural frequency ω₀ of each uncoupled pendulum, (b) the coupling frequency ω_C = √(2k_c/m), (c) the two normal-mode angular frequencies ω₁ (in-phase) and ω₂ (anti-phase), (d) the beat period T_beat. Then, given the initial condition θ₁(0) = 0.1 rad and θ₂(0) = 0, both at rest, find θ₁ and θ₂ at t = 3 s.

Linked equations:x'' + \omega^2 x = 0\omega_1 = \omega_0,\quad \omega_2 = \sqrt{\omega_0^2 + \omega_C^2}
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Step-by-step solution

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Step 1

Each pendulum has the same uncoupled equation θ″ + (g/L)θ = 0. What is its angular frequency ω₀?

Hint

ω₀ = √(g/L). This is exactly the same universal oscillator pattern — θ plays the role of x, and g/L plays the role of k/m.

Step 2

In the in-phase normal mode, both bobs swing together at the same angle. The coupling spring never stretches. What is ω₁?

Step 3

In the anti-phase normal mode, the bobs swing in opposite directions. The spring is always stretched or compressed by 2θ. What is ω₂?

Step 4

The beat period arises from the two modes going in and out of phase. Express T_beat in terms of ω₁ and ω₂.

Solution walkthrough
Two identical pendulums coupled by a spring are the textbook example of a system with normal modes. Rather than tracking complicated mutual forces, we look for special initial conditions — normal modes — where every part of the system oscillates at a single frequency. Mode 1 (in-phase): both bobs swing together. The spring never stretches, so it might as well not be there, and ω₁ = ω₀ = √(g/L) ≈ 4.43 rad/s. Mode 2 (anti-phase): bobs swing in opposite directions. The spring stretches by 2θ and applies a restoring force 2k_c·θ to each bob. This doubles the spring's contribution, giving an extra ω_C² = 2k_c/m = 6.4 rad²/s² to the frequency equation. So ω₂ = √(ω₀² + ω_C²) ≈ 4.82 rad/s. Any motion is a superposition of these modes. When only pendulum 1 is displaced, both modes are excited with equal amplitude, and because ω₁ ≠ ω₂ they drift in and out of phase with beat frequency (ω₂−ω₁)/2π ≈ 0.062 Hz, giving T_beat ≈ 16 s. Over that period, all the energy migrates from pendulum 1 to pendulum 2 and back. Lissajous observed exactly this kind of coupled oscillation in 1857 using tuning forks and a mirror.
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