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Question

A simple pendulum has a frequency of \( 0.25 \, \text{Hz} \) on Earth (\( g = 9.8 \, \text{m/s}^2 \)).
What is its length?

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Explanation

**Angular frequency** ω = 2πf = 2π/T characterizes rapidity, independent of amplitude for SHM. Phase constant φ shifts sine/cosine, allowing any initial condition, e.g., x(0)=0 requires φ=0 for sine form. Displacement, velocity, acceleration share ω but differ in phase by 90° and 180°. Period: T = (1/v) = (1/0.25) = 4 s . T = 2π √((L/g)) ⇒ 4 = 2π √((L/9.8)) . √((L/9.8)) = (4/2π) ≈ 0.637 ⇒ (L/9.8) = (0.637)² ⇒ L ≈ 3.98 m . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 3.98 m follows, reflecting

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