Solving ODE Oscillations: Find T for Buoy & Train Round Trip

[djh101]

1. A spherical buoy of radius r floats half-submerged in water. If it is depressed slightly, a restoring force equal to the weight of the displaced water presses it upward; and if it is then released, it will bob up and down. Find the period of oscillation if the friction of the water is neglected.

T = 2π/ω
y'' + ω²y = 0
F = my'' = .5V_waterρ_waterg
y = cos(ωt) & y'' = ω²cos(ωt)

So ω needs to be found, which can be done by setting t = 0 and a_max = ω² = .5Vρg.
.5V = 2/3πr³ and ρ = 1, so ω² = 2/3πr³g.
All that is left is to take the square root and divide from 2 to get T. However, the book gives the answer 2π√(2r/3g)s.
Summary: $2\pi\sqrt{\frac{3}{2\pi r^{3}g}} \neq 2\pi\sqrt{\frac{2r}{3g}}$


2. Suppose that a straight tunnel is drilled through the Earth between any two points on the surface. If tracks are laid, then a train placed in the tunnel at one end will roll through the Earth under its own weight, stop at the other end, and return. Estimate the value of the time required to complete one round trip.

y'' = g = GM/y²
M = 4/3 πR³ρ
y'' = 4/3 πGρy, which gives ω² = 4/3 πGρ. The book, however, gives T = 2π√(R/g).
Summary: $2\pi\sqrt{\frac{3}{4πGρ}} \neq 2\pi\sqrt{\frac{R}{g}}$

Am I over thinking these? The answers in the book are pretty simple, but I'm having a little bit of a hard time figuring out where they came from.

 

[mighty2000]

it is important to always critically evaluate and question the results and answers presented to you. It is possible that the book has simplified the equations or assumptions for the sake of clarity or ease of understanding. It is also possible that there may be a mistake in the book's answer, and it is always important to double check your calculations and assumptions. Don't be afraid to seek clarification or ask for help if needed. Remember, science is all about questioning and seeking the truth.