Question

A cone made of a material of relative density $(s=\frac{27}{64})$ and height H = 4m floats with its apex downward in water. Find the time period of vertical oscillation (approximately) if it is slightly displaced from the equilibrium position. $g=10m/s^2$ [A = base area]___

Answer 1

At equilibrium,
$S_s\left(\frac{1}{3}\pi R^{2}H\right)S_{\lambda }\frac{1}{3}\left(\pi r^{2}h\right)g\cdots \left(1\right)$
When displaced by a small amount x downwards,
$[S_1\frac{1}{3}\pi r^{2}xx=S_s\frac{1}{3}\pi r^{2}Hg]=ma$
$S_1\times \frac{1}{3}\pi r^{2}x=S_s\frac{1}{3}\pi r^{2}Ha$
$From-\left(1\right)$
$\frac{Hg}{h^2}X=Ha$
$a=\frac{g}{h^2}X\cdots \left(2\right)$
$Also,\frac{r}{h}=\frac{R}{H}thusfrom\left(1\right)\frac{H^3}{h^3}=\frac{S_1}{S_3}\Rightarrow \frac{H}{h}=\sqrt[3]{3\frac{64}{27}}=\frac{4}{3}-\cdots \left(3\right)$
$\left(3\right)in\left(2\right)a=\frac{16g}{9H^2}X$
$a=\frac{g}{9}X;sinceH=4m$
$T=2\pi \sqrt{\frac{9}{g}};g={\pi }^2$
$=6s$