Question

A hollow spherical ball of radius $20\text{cm}$ floats in still water, with half of its volume submerged. Taking the density of water as $1000{\text{kg/m}}^3$, and the acceleration due to gravity as $10{\text{m/s}}^2$, the natural frequency of small oscillations of the ball, normal to the water surface is
(roundoff to 2 decimal places).

• A. $5.66\text{rad/s}$
• B. $4.86\text{rad/s}$
• C. $8.66\text{rad/s}$
• D. $9.66\text{rad/s}$

Answer 1

The correct option is C $8.66\text{rad/s}$

Given that:
Radius of ball, $R=20\text{cm}=0.20\text{m}$
Density of the water, $\rho =1000{\text{kg/m}}^3$
$g=10{\text{m/s}}^2$ $[V^{\prime }=\text{Volume displaced}=\frac{V_s}{2}]$
$mg=F_B=\rho V^{\prime }g$
$m=\rho .\frac{V_s}{2}$

When the ball is slightly displaced from the equilibrium position, the additional buoyant force exerts on the sphere is $f_B^{\prime }$
$m\frac{d^{2}x}{d{t}^2}+f_{\text{B extra}}^{\prime }=0$
$m\frac{d^{2}x}{d{t}^2}+V_{\text{extra}}\rho g=0$
$\rho \frac{V_s}{2}.\frac{d^{2}x}{d{t}^2}+(\pi R^2.x)\rho g=0$
$\frac{d^{2}x}{d{t}^2}\left(\frac{4}{3}\frac{\pi R^3}{2}\right)+\left(\pi R^{2}g\right)x=0$
$\Rightarrow \frac{d^{2}x}{d{t}^2}+\frac{3g}{2R}x=0$
${\omega }_n=\sqrt{\frac{3g}{2R}}=\sqrt{\frac{3\times 10}{2\times 0.20}}=\sqrt{\frac{30}{0.4}}$
${\omega }_n=\sqrt{\frac{300}{4}}=\sqrt{75}=8.66\text{rad/s}$

$\begin{array}{c}\text{Why this question?}\\ \text{Hint: Since a very small portion of the sphere dribbles in and out}\\ \text{of water, so it can be safely considered as a very thin cylinder.}\end{array}$