Question

A spring with spring constant 40 N/m is attached to the ceiling, and a 5.1-cm-diameter, 1.5 kg metal cylinder is attached to its lower end. The cylinder is held so that the spring is neither stretched nor compressed, then a tank of water is placed underneath with the surface of the water just touching the bottom of the cylinder. When released, the cylinder will oscillate a few times but, damped by the water, quickly reach an equilibrium position.
When in equilibrium, what length of the cylinder is submerged?

Answer 1

Answer: 24.5 cm

Explanation:

Given

Force constant of spring, k = 40 N/m

Diameter of spring, d = 5.1 cm = 0.051 m

Mass of cylinder, m = 1.5 kg

Let us assume that the cylinder is hanging in such a way that the circular end is parallel with the water. Also, we assume that the tank water level is not materially affected by the displacement of the cylinder while the cylinder sinks. The water is fresh and as we all know, the density of water is 1000 kg/m³

To solve this, we assume x to be the spring extension and it's equivalent sinking distance(in meters). We then apply the formula,

mg = kx + ρgAx

mg = x(k + ρgA)

x = mg / (k + ρgA), where

A = πd²/4

A = (3.142 * 0.051²)/4

A = 0.0082 / 4

A = 0.00205 m²

x = 1.5 * 9.81 / [40 + (1000 * 9.81 * 0.00205)]

x = 14.715 / (40 + 20.1105)

x = 14.715 / 60.1105

x = 0.245 m

or 24.5 cm of stretch or sinking