Losing contact with an oscillating platform
● VERBATIM — examined word-for-wordAn object resting on a vertically oscillating plate stays in contact only while the plate's downward acceleration ≤ g (normal force N ≥ 0). Contact is lost at the highest point, where downward acceleration is largest, when ω²x₀ exceeds g. Condition to stay in contact: ω²x₀ ≤ g. N = 0 at the moment a = g.
just loses contact: ω²x₀ = gWhat this actually means
An object riding a vertically oscillating platform is held by two forces only: weight down and the normal force up. The platform can push but never pull.
Trouble arrives at the top of the motion, where the platform accelerates DOWNWARDS at its maximum ω²x₀. Gravity can supply at most g of downward acceleration; if the platform demands more, the object cannot follow and contact breaks with N = 0.
So the condition to stay in contact is ω²x₀ ≤ g, and 'just loses contact' means ω²x₀ = g at the highest point. Solve that for the limiting frequency or amplitude.
Prove it — watch it be true
- Read a_max = ω²x₀ off the a–t graph: it occurs at the extremes
- Compare it with g = 9.81 m s⁻²
- If the top-of-motion downward acceleration beats g, a passenger there would lift off