Practice question
Question
A pendulum oscillates with a period of \( 1.5 \, \text{s} \) on Earth (\( g = 9.8 \, \text{m/s}^2 \)).
What is its length?
Explanation
**General equation of SHM** x = A sin(ωt + φ) or A cos(ωt + φ) includes amplitude A (m), angular frequency ω = √(k/m) (rad/s) for spring system, and initial phase φ (rad) setting t=0 position. Phase (ωt + φ) determines instantaneous state, phase difference Δφ governs interference of two SHM motions. T = 2π √((L/g)) . 1.5 = 2π √((L/9.8)) ⇒ (1.5/2π) = √((L/9.8)) . ((1.5/2 × 3.14))² = (L/9.8) ⇒ L = 9.8 × (0.238)² ≈ 0.56 m . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.56
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