A 0.400 kg cart attached to a horizontal spring of constant 64.0 N m⁻¹ oscillates with amplitude 0.0800 m on a low-friction track. Find the angular frequency, period, total energy, and speed when x = 0.0500 m. Starting at x = +A at t = 0, determine the first time the cart reaches x = 0.0500 m. Describe how modest damping would change the position–time graph without changing the equilibrium position.

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ω = 12.6 rad s⁻¹, T = 0.497 s, E = 0.2048 J, v = 0.790 m s⁻¹, and the first arrival is at 0.0709 s. Damping produces a decaying envelope and slightly longer period. For an ideal mass–spring oscillator, ω = √(k/m) = √(64.0/0.400) = 12.65 rad s⁻¹ and T = 2π/ω = 0.497 s. At a turning point all energy is elastic, so E = ½kA² = ½(64.0)(0.0800)² = 0.2048 J. Energy conservation at x gives ½mv² = ½k(A²−x²). Thus v = ω√(A²−x²) = 0.790 m s⁻¹; the first passage is toward negative x. With x = A cosωt, cosωt = 0.0500/0.0800 = 0.625. Therefore t = cos⁻¹(0.625)/12.65 = 0.0709 s. Damping removes energy each cycle, so successive extrema lie inside a decreasing envelope. The oscillation becomes slightly slower for modest damping, while the fixed spring's force still vanishes at the same equilibrium location.

Energy and timing in a spring oscillator

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