§ 37. Coupled Pendulums: Normal Modes & Energy Beats (Grade 11 · Uni)
Periodic exchange of mechanical energy between two pendulums coupled by a spring. Any motion of a linear coupled system is a superposition of its fundamental normal modes.
Theory Summary & Scientific Mechanism
Two identical pendulums of mass m and length L coupled by a horizontal spring of stiffness k form a canonical two-degree-of-freedom coupled oscillator: mx₁ = -(mg)/(L)x₁ - k(x₁ - x₂) and mx₂ = -(mg)/(L)x₂ - k(x₂ - x₁).
Transforming to normal coordinates q₊ = x₁ + x₂ and q₋ = x₁ - x₂ diagonalizes the equations into two independent normal modes: 1) the in-phase symmetric mode (x₁ = x₂), where the coupling spring remains unstretched and ω₁ = √((g)/(L)); 2) the out-of-phase antisymmetric mode (x₁ = -x₂), where the spring stretches by 2x₁ and raises the frequency to ω₂ = √((g)/(L) + (2k)/(m)).
Displacing only pendulum 1 at t = 0 excites an equal superposition of both normal modes: x₁(t) = Acos ≤ft((ω₂ - ω₁)/(2)t )cos (ωt) and x₂(t) = Asin ≤ft((ω₂ - ω₁)/(2)t )sin (ωt). This produces energy beats: with beat period T_beat = (2π)/(ω₂ - ω₁), pendulum 1 comes to a complete halt while transferring 100% of its energy to pendulum 2, and then back again!
Everyday Analogy
Hang two identical swings from a flexible beam or link two pendulums with a light spring! Push only the first pendulum: within seconds it comes to a dead stop after handing 100% of its energy to the second, and then the energy flows right back!
Interactive Question & Derivation
How does the period of complete energy transfer between the two pendulums (beat period T_b = (2π)/(ω₂ - ω₁)) change if the stiffness of the coupling spring κ is increased?
- A: The beat period T_b decreases: stronger elastic coupling widens the frequency splitting ω₂ - ω₁, so energy transfers faster between the pendulums
- B: The beat period T_b increases proportionally to κ
- C: The rate of energy exchange depends only on pendulum length L and not on the spring
Study Hint: Energy beats arise from the difference between symmetric and antisymmetric normal-mode frequencies (ω_beat = ω₂ - ω₁). What happens to the energy transfer speed when spring stiffness k_c increases?
Proof & Explanation: The symmetric (in-phase) mode oscillates at ω₁ = √(g/L) (spring unstretched), while the antisymmetric (out-of-phase) mode oscillates at ω₂ = √(g/L + 2κ/m). Increasing spring stiffness κ increases the frequency difference Δω = ω₂ - ω₁, shortening the beat period T_b = 2π / Δω.
Key Equations & 3D Simulation Parameters
x₁(t) ∝ cos ≤ft((ω₂ - ω₁)/(2)t )cos (ωt), Δω = ω₂ - ω₁
- Pendulum suspension length L: 1.6 … 3.6 m (default: 2.6 m)
- Coupling spring stiffness κ: 0.4 … 5 N/m (default: 1.8 N/m)
- Mass of each bob m: 0.8 … 3 kg (default: 1.5 kg)
- Mode (0 = beats, 1 = in-phase, −1 = anti-phase): -1 … 1 (default: 0)