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

0.315758099469 m

Step-by-step 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* 9.81=1000* \pi(2.25* 10^(-2))^2x* 9.81+31x\\\Rightarrow 1.5* 9.81=46.60213x\\\Rightarrow x=(1.5* 9.81)/(46.60213)\\\Rightarrow x=0.315758099469\ m`

The length of the submerged cylinder is 0.315758099469 m