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Example: Two identical pendulums of mass 1 kg each are connected by a light spring with spring constant k_c = 2 N/m. Each pendulum alone would swing with frequency f_0 = 1 Hz (ω_0 = 2π rad/s). Find the two normal mode frequencies when they are coupled. Step 1: Write equations of motion. For small angles, each pendulum behaves like a mass on a spring. The equations are: 1) m ẍ_1 + k x_1 - k_c (x_2 - x_1) = 0 2) m ẍ_2 + k x_2 - k_c (x_1 - x_2) = 0 Step 2: Assume solutions of the form x_1 = A cos(ωt), x_2 = B cos(ωt). Plug into equations. Step 3: This leads to a matrix equation: [(k + k_c) - mω²]A - k_c B = 0, and similarly for B. Step 4: For nontrivial solutions, determinant must be zero. Solve: (k + k_c - mω²)² - k_c² = 0. Step 5: Solve for ω²: ω² = (k + 2k_c)/m or ω² = k/m. Step 6: Compute numeric values. With k = (m g / L) but given f_0 = 1 Hz, we know ω_0 = 2π rad/s, so k/m = ω_0² = (2π)² ≈ 39.5. Then ω²_1 = 39.5 + 4 = 43.5 → ω_1 ≈ 6.6 rad/s → f_1 ≈ 1.05 Hz. ω²_2 = 39.5 → ω_2 = 6.28 rad/s → f_2 = 1.00 Hz. So the two normal mode frequencies are about 1.00 Hz and 1.05 Hz.