A cone with half the density of water is floating in water as shown in figure. It is depressed down by a…

Given that the density of the cone is half of the density of the water. If the height of the cone that is immersed in water at equilibrium is $h$, then for floating

$$\begin{gathered} \frac{1}{3}\pi R^{2}H{\rho }_C=\frac{1}{3}\pi r^{2}h{\rho }_W \\ \Rightarrow \frac{r^{2}h}{R^{2}H}=\frac{{\rho }_C}{{\rho }_W}=\frac{1}{2} \\ \Rightarrow \frac{h^3}{H^3}=\frac{1}{2} \\ \Rightarrow h={\left(\frac{1}{2}\right)}^{\frac{1}{3}}H \end{gathered}$$

Now 

$$\begin{gathered} \tan 30°=\frac{r}{h} \\ \Rightarrow r=\frac{h}{\sqrt{3}} \end{gathered}$$

If the cone is displaced by $dh$ downward, extra force due to buoyancy on the cone will be,

$F=\left(\pi r^2{\rho }_{W}g\right)dh$

The mass of the cone is

$M=\frac{1}{3}\pi R^{2}H{\rho }_C$

Therefore, the frequency of the SHM will be,

$$\begin{gathered} f=\frac{1}{2\pi }\sqrt{\frac{K}{M}}=\frac{1}{2\pi }\sqrt{\frac{\pi r^2{\rho }_{W}g}{\frac{1}{3}\pi R^{2}H{\rho }_C}}=\frac{1}{2\pi }\sqrt{\frac{3\times 2g}{H}\times \frac{h^2}{H^2}} \\ \Rightarrow f=\frac{1}{2\pi }\sqrt{\frac{6g}{H}\frac{1}{4^{{\displaystyle \frac{1}{3}}}}} \end{gathered}$$