EXAM · THE SIMPLE PENDULUM
GRAVITY FROM TWO PENDULUMS
A physicist measures two pendulums at the same location. Pendulum 1 has length L₁ = 0.50 m and measured period T₁ = 1.4187 s. Pendulum 2 has length L₂ = 2.00 m and measured period T₂ = 2.8375 s. From each measurement determine g independently, then average the two estimates and compute the fractional discrepancy between them.
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Step-by-step solution
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Step 1
Use ω₁² = g / L₁ to find g₁ = L₁ × ω₁². What is g₁ in m/s²?
Hint
The small-angle pendulum gives ω² = g/L, so g = L × ω².
Step 2
Find g₂ = L₂ × ω₂². What is g₂ in m/s²?
Step 3
Average the two estimates: g_avg = (g₁ + g₂) / 2. What is g_avg in m/s²?
Solution walkthrough
This problem mirrors a real laboratory technique for measuring local gravitational acceleration. Two independent pendulum experiments reduce random error and reveal whether systematic effects are present.
Step 1 — ω₁: ω₁ = 2π/T₁ = 2π/1.4187 ≈ 4.4288 rad/s.
Step 2 — g₁: g₁ = L₁ × ω₁² = 0.5 × (4.4288)² = 0.5 × 19.6145 ≈ 9.8073 m/s².
Step 3 — ω₂: ω₂ = 2π/T₂ = 2π/2.8375 ≈ 2.2143 rad/s.
Step 4 — g₂: g₂ = L₂ × ω₂² = 2.0 × (2.2143)² = 2.0 × 4.9033 ≈ 9.8066 m/s².
Step 5 — g_avg: g_avg = (9.8073 + 9.8066)/2 ≈ 9.8069 m/s².
Step 6 — discrepancy: |9.8073 − 9.8066|/9.8069 ≈ 0.0000705 ≈ 0.007%. The two estimates agree to better than one part in ten thousand.
Forward check: T₁_pred = 2π√(0.5/9.8069) ≈ 1.4187 s and T₂_pred = 2π√(2.0/9.8069) ≈ 2.8375 s — recovering the measured values to four decimal places.
This level of consistency would indicate an extremely well-controlled experiment. In practice, air resistance, string elasticity, and finite bob size introduce systematic shifts that would spread g₁ and g₂ further apart. The technique is still used in modern geodesy to map local g variations across Earth's surface.
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