An ideal mass–spring oscillator of mass 0.34 kg has period 1.08 s and amplitude 0.054 m. At an instant when displacement x = +0.0324 m, it moves toward equilibrium. Find its velocity, acceleration, total energy and kinetic-energy fraction. Take positive displacement to the right and set spring potential energy zero at equilibrium. Explain the signs and why the speed is not proportional to the displacement.

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Velocity −0.2513 m s⁻¹; acceleration −1.097 m s⁻²; total energy 0.01678 J; kinetic fraction 0.64. The angular frequency is ω = 2π/T = 5.818 rad s⁻¹. The restoring acceleration is a = −ω²x = −1.097 m s⁻² because the positive displacement requires a leftward restoring force. Energy conservation gives ½mω²A² = ½mv² + ½mω²x². Hence the speed magnitude is ω√(A² − x²), which is a square-root dependence rather than a direct proportionality to x. Here x/A = 0.6, so |v| = 0.8ωA = 0.2513 m s⁻¹. The stated motion toward equilibrium from positive x selects the negative velocity. Total energy is ½mω²A² = 0.01678 J. The potential-energy fraction is x²/A² = 0.36, leaving a kinetic-energy fraction of 0.64. At the turning points x = ±A the same formula gives zero speed; at equilibrium it gives the maximum speed ωA. These limiting cases check the energy expression and illustrate the continual exchange between kinetic and elastic potential energy.

Speed and energy away from equilibrium · set 7

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