Mass-spring measurements and an energy cross-check
A classroom group studies a horizontal cart attached to a spring on an idealized low-friction track. For this calculation the spring is massless, the track is level, friction is neglected and the restoring law is linear over all displacements used. The cart mass is 0.20 kg and the spring stiffness is supplied as 50 N/m. The group displaces the cart by 0.030 m from equilibrium and releases it from rest. It then repeats the measurement with a total moving mass of 0.40 kg while keeping the same spring and amplitude. One student predicts that doubling mass doubles the period; another predicts that the cart has its largest acceleration as it passes equilibrium. The group must check both predictions using a dynamical calculation and an independent energy calculation.
1. Calculate the period in each run and evaluate the doubling claim.
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For the ideal model T = 2pi sqrt(m/k). With 0.20 kg, T = 2pi sqrt(0.20/50) = 0.3974 s. With 0.40 kg, T = 0.5620 s. The ratio is sqrt(2), approximately 1.414, so doubling the mass does not double the period. The same amplitude is irrelevant to the period only because the specified restoring law is linear.
2. Find total mechanical energy and maximum speed for the 0.20 kg cart. Explain the acceleration at equilibrium.
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Total energy is one half kA squared = 0.5(50)(0.030 squared) = 0.0225 J. At equilibrium that energy is kinetic, so maximum speed is sqrt(2E/m) = 0.4743 m/s. Acceleration is -kx/m, hence zero at equilibrium and greatest in magnitude at the turning points. Maximum speed and maximum acceleration do not occur at the same position.