Question

A spring with spring constant 31 N/m is attached to the ceiling, and a 4.5-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:

0.315758099469 m

Explanation:

m = Mass of cylinder = 1.5 kg

$\rho$ = Density of water = 1000 kg/m³

V = Volume = Ah

A = Area = $\pi r^2$

k = Spring constant = 31 N/m

x = Displacement

g = Acceleration due to gravity = 9.81 m/s²

Here the forces are conserved

Weight of cylinder = Buoyant force + Spring force

$mg=\rho Vg+kx\\\Rightarrow mg=\rho Ahg+kx\\\Rightarrow 1.5\times 9.81=1000\times \pi(2.25\times 10^{-2})^2x\times 9.81+31x\\\Rightarrow 1.5\times 9.81=46.60213x\\\Rightarrow x=\dfrac{1.5\times 9.81}{46.60213}\\\Rightarrow x=0.315758099469\ m$

The length of the submerged cylinder is 0.315758099469 m