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(A) Given that the density of the cone $\rho_C$ is half of the density of the water $\rho_W$,so $\rho_C = \frac{1}{2} \rho_W$.Let $h$ be the height of

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$\frac{1}{2 \pi} \sqrt{\frac{6 g}{H} \frac{1}{4^{\frac{1}{3}}}}$

Vedclass · Answer

(A) Given that the density of the cone $ho_C$ is half of the density of the water $ho_W$,so $ho_C = \frac{1}{2} ho_W$.Let $h$ be the height of the cone immersed in water at equilibrium. For floating,the weight of the cone equals the buoyant force:$\frac{1}{3} \pi R^2 H ho_C = \frac{1}{3} \pi r^2 h ho_W$Since $\frac{r}{h} = 	an 30^{\circ} = \frac{1}{\sqrt{3}}$ and $\frac{R}{H} = \frac{1}{\sqrt{3}}$,we have $r = \frac{h}{\sqrt{3}}$ and $R = \frac{H}{\sqrt{3}}$.Substituting these into the equilibrium equation:$\frac{1}{3} \pi (\frac{H}{\sqrt{3}})^2 H ho_C = \frac{1}{3} \pi (\frac{h}{\sqrt{3}})^2 h ho_W$$\frac{H^3}{3} ho_C = \frac{h^3}{3} ho_W \Rightarrow \frac{h^3}{H^3} = \frac{ho_C}{ho_W} = \frac{1}{2} \Rightarrow h = H (\frac{1}{2})^{\frac{1}{3}}$.When the cone is displaced by a small distance $\delta$ downward,the additional buoyant force is $F = \pi r^2 ho_W g \delta$.The restoring force constant is $k = \pi r^2 ho_W g$.The mass of the cone is $M = \frac{1}{3} \pi R^2 H ho_C = \frac{1}{3} \pi (\frac{H}{\sqrt{3}})^2 H (\frac{1}{2} ho_W) = \frac{1}{18} \pi H^3 ho_W$.The frequency of $SHM$ is $f = \frac{1}{2 \pi} \sqrt{\frac{k}{M}}$.$k = \pi (\frac{h}{\sqrt{3}})^2 ho_W g = \frac{\pi h^2 ho_W g}{3} = \frac{\pi H^2 (1/2)^{2/3} ho_W g}{3}$.$f = \frac{1}{2 \pi} \sqrt{\frac{\pi H^2 (1/2)^{2/3} ho_W g / 3}{\pi H^3 ho_W / 18}} = \frac{1}{2 \pi} \sqrt{\frac{6 g (1/2)^{2/3}}{H}} = \frac{1}{2 \pi} \sqrt{\frac{6 g}{H} \frac{1}{4^{1/3}}}$.

$A$ cone with half the density of water is floating in water as shown in the figure. It is depressed down by a small distance $\delta (\ll H)$ and released. The frequency of simple harmonic oscillations of the cone is

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