Question

A wooden cube (density of wood $'d'$) of side $'\ell'$ floats in a liquid of density $'\rho'$ with its upper and lower surfaces horizontal. If the cube is pushed slightly down and released, it performs simple harmonic motion of period $'T'$. Then, $'T'$ is equal to :

• A. $2\pi\sqrt{\frac{\ell d}{\rho g}}$
• B. $2\pi\sqrt{\frac{\ell\rho}{dg}}$
• C. $2\pi\sqrt{\frac{\ell d}{\left(\rho-d\right)g}}$
• D. $2\pi\sqrt{\frac{\ell\rho}{\left(\rho-d\right)g}}$

Answer 1

Answer: (A)

$2\pi\sqrt{\frac{\ell d}{\rho g}}$

Answer 2

At equilibrium $F_{b} = mg$ $\rho A\ell_{0}g = dA\ell g \quad...........\left(i\right)$ Restoring force, $F = mg - F_{b}'$ $F = mg - \rho A\left(\ell_{0} + x\right)g$ $dA\ell a = dA\ell g - \rho A\ell_{0}g - \rho gAx$ $a = \frac{\rho g}{d\ell}x$ $\omega = \sqrt{\frac{\rho g}{d\ell}}$ $T = 2\pi\sqrt{\frac{\ell d}{\rho g}}\quad...........\left(i\right)$