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Question

A particle’s x-projection from circular motion is \( x = 10 \cos (2\pi t) \) (in m). What is its
maximum acceleration?

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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. Maximum acceleration: aₘₐₓ = ω² A . A = 10 m, ω = 2π s⁻¹ . aₘₐₓ = (2π)² × 10 ≈ 39.48 × 10 ≈ 394.8 m/s² . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 394.8

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