A massless spring hangs from the ceiling with a small object attached to its low

Question

A massless spring hangs from the ceiling with a small object attached to its lower end. The object is initially held at rest in a position yi such that the spring is at its rest length. The object is then released from yi and oscillates up and down, with its lowest position being 23 cm below yi.

The Frequency of the oscillation is 1.45005

The Speed of the object is 0.432987599

An object of mass 500 g is attached to the first object, after which the system oscillates with half the original frequency. The mass of the first object is 166.666667 g.

How far below yi is the new equilibrium (rest) position with both objects attached to the spring?

Please show your work.

Please calculate a final answer.

Explanation / Answer

Then twice the amplitude = 23cm, and A = 11.5cm F = kx m·9.8m/s² = 0.115m·k k/m = 9.8/0.115s² = 85.21 /s² sqrt(k/m) = 9.23/s f = 9.23/2ps = 1.46 /s b) max Ep = ½k·(0.115m)² = 6.6125 e-3m²·k at y = 19cm, Ep = ½k(0.015m)² = 1.1e-4m²·k ?E = 1.4e-3m²·k This must have become kinetic energy, Ek = ½mv² Set them equal, and solve for v²: v² = 1.4e-3m²·2·k/m From above, we have k/m = 85.21/s² v = sqrt(1.4e-3m² · 2 · 85.21/s²) c) sqrt(k/(m + 0.525kg) )= ½sqrt(k/m) square both sides k/(m+0.525kg) = k/(4m) m + 0.525kg = 4m 3m = 0.525kg m = 0.175kg = 175g d) we've quadrupled the force on the spring; therefore we've quadrupled the displacement. 4 * 11.5c m = 46 cm answer

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