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Question

A simple pendulum has a length of \( 1.44 \, \text{m} \) and oscillates with \( g = 9.8 \, \text{m/s}^2
\). What is its period?

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Explanation

**Angular SHM** of pendulum results from restoring torque τ = -m g L sinθ ≈ -m g L θ for small θ, analogous to linear SHM with ω = √(g/L). This relation allows period determination from length and local gravity, illustrating gravitational influence on oscillation. Period: T = 2π √((L/g)) = 2π √((1.44/9.8)) ≈ 2 × 3.14 √(0.1469) ≈ 2.406 s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 2.406 s follows, reflecting SHM dependence on amplitude A, ω and system parameters.

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