Question

A block of wood is floating in water; it is depressed slightly and then released to oscillate up and down. Assume that the top and bottom of the block are parallel planes which remain horizontal during the oscillations and that the sides of the block are vertical. Show that the period of the motion (neglecting friction) is 2π ph/g, where h is the vertical height of the part of the block under water when it is floating at rest. Hint: Recall that the buoyant force is equal to the weight of displaced water.

Answer 1

Explanation:

Equilibrium position in y direction:

W = Fb (Weight of the block is equal to buoyant force)

m*g = V*p*g

V under water = A*h

hence,

m = A*h*p

Using Newton 2nd Law

$-m*\frac{d^2y}{dt^2} = Fb - W\\\\-m*\frac{d^2y}{dt^2} = p*g*(h+y)*A - A*h*p*g\\\\-A*h*p*\frac{d^2y}{dt^2} = y *p*A*g\\\\\frac{d^2y}{dt^2} + \frac{g}{h} * y =0$

Hence, T time period

T = 2*pi*sqrt ( h / g )