The Pendulum and the Bridge
A civil engineering team is designing a pedestrian bridge in Yerevan. The bridge deck is supported by cables that can oscillate like pendulums under wind or foot traffic. The team models each cable as a simple pendulum of length L = 10 m. They know that the period of a simple pendulum is given by T = 2π√(L/g), where g ≈ 9.8 m/s². However, they are concerned about resonance: if the frequency of footfalls or wind gusts matches the natural frequency of the cables, large oscillations could occur. The average walking pace is about 2 steps per second. The team also considers adding a damping system to reduce oscillations. They test a prototype cable with a damping coefficient that reduces amplitude by 50% over 5 seconds.
1. Calculate the natural frequency of the 10 m cable. Determine whether walking at 2 steps per second could cause resonance. Justify your answer with calculations.
Read the worked answer
The period T = 2π√(L/g) = 2π√(10/9.8) ≈ 2π√(1.0204) ≈ 2π(1.010) ≈ 6.35 s. The natural frequency f = 1/T ≈ 1/6.35 ≈ 0.157 Hz. Walking at 2 steps per second corresponds to a frequency of 2 Hz (or 1 Hz if considering one footfall per cycle, but typically 2 Hz for both feet). The natural frequency (0.157 Hz) is much lower than 2 Hz, so direct resonance from walking is unlikely. However, if the cable length were shorter, say 0.25 m, the natural frequency would be higher. For L=10 m, the frequency is too low to match walking. But wind gusts could have lower frequencies (e.g., 0.1–0.2 Hz), so resonance with wind is possible. Thus, walking at 2 steps/s is not a concern, but wind-induced resonance should be considered.
2. The damping system reduces amplitude by 50% every 5 seconds. Calculate the decay constant λ (in s⁻¹) assuming exponential decay A(t) = A₀e^(-λt). Then, estimate the time required for the amplitude to drop to 10% of its initial value.
Read the worked answer
Given A(5) = 0.5 A₀ = A₀ e^(-5λ). So e^(-5λ) = 0.5 → -5λ = ln(0.5) = -0.693 → λ = 0.693/5 ≈ 0.1386 s⁻¹. For A(t) = 0.1 A₀: e^(-λt) = 0.1 → -λt = ln(0.1) = -2.3026 → t = 2.3026/0.1386 ≈ 16.6 s. So it takes about 16.6 seconds for the amplitude to drop to 10%.