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Question

A particle in SHM has \( x = 4 \cos (3t) \) (in m). What is its maximum acceleration?

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Explanation

**Energy in SHM** interconverts between kinetic K = ½ m v² = ½ m ω² (A² - x²) and potential U = ½ k x² = ½ m ω² x², total E = K + U = ½ k A² = ½ m ω² A² constant, independent of time. At mean position x=0, E = K_max = ½ m ω² A², at extremes x=±A, E = U_max = ½ k A². Maximum acceleration: aₘₐₓ = ω² A . A = 4 m, ω = 3 s⁻¹ . aₘₐₓ = 3² × 4 = 9 × 4 = 36 m/s² . Applying x = A cos(ωt

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