Question 16

**Setting up the motion** Each pendulum has length L = 8 m and angular amplitude θ₀ with cos θ₀ = 0.9. By energy conservation, the maximum speed (at the lowest point) is v_max = √(2gL(1 − cos θ₀)) = √(2 × 10 × 8 × 0.1) = √16 = 4 m/s. Since T and R are pulled in opposite directions and released simultaneously, their motions are mirror images: when they pass the lowest point they are either both moving toward each other (both at 4 m/s) or both moving away from each other (both at 4 m/s), half a period apart. **Doppler shifts** Speed of sound c = 330 m/s, f₀ = 660 Hz, v_s = v_o = 4 m/s. Maximum frequency (approaching each other): f_max = f₀ (c + v_o)/(c − v_s) = 660 × 334/326 ≈ 676.2 Hz Minimum frequency (receding from each other): f_min = f₀ (c − v_o)/(c + v_s) = 660 × 326/334 ≈ 644.2 Hz **Maximum variation** Δf = f₀ [(334² − 326²)/(326 × 334)] = 660 × (8 × 660)/108884 ≈ 32.0 Hz

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