Question

A cylindrical object of outer diameter 20 cm and mass 2 kg floats in water with its axis vertical. If it is slightly depressed and then released, find the time period of the resulting simple harmonic motion of the object.

Answer 1

Given:
Outer diameter of the cylindrical object, d = 20 cm
∴ Radius$,r=\frac{d}{2}=10\mathrm{cm}$
Mass of the object, m = 2 kg
When the object is depressed into water, the net unbalanced force causes simple harmonic motion.
Let x be the displacement of the block from the equilibrium position.
Now,
$$\begin{gathered} \mathrm{Driving}\mathrm{force}=U=V{\rho }_{\mathrm{w}}g \\ \mathrm{Here}, \\ U\mathrm{is}\mathrm{the}\mathrm{upward}\mathrm{thrust}. \\ V\mathit{}\mathrm{is}\mathrm{the}\mathrm{volume}. \\ {\rho }_{\mathrm{w}}\mathit{}\mathrm{is}\mathrm{the}\mathrm{density}\mathrm{of}\mathrm{water}. \\ \mathrm{Thus},\mathrm{we}\mathrm{have}: \\ ma=\mathrm{\pi }{r}^2\left(x\right)\times {\rho }_{\mathrm{w}}g \\ a\mathit{=}\frac{\pi r^{\mathit{2}}{\rho }_w\mathit{9}x}{\mathit{2}\mathit{\times }{\mathit{10}}^{\mathit{3}}} \\ \mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{Displacement}}{\mathrm{Acceleration}}} \\ =2\mathrm{\pi }\sqrt{\frac{\left(x\right)\times 2\times {10}^3}{\left(\mathrm{\pi }{r}^2{\mathrm{\rho }}_{w}g\left(x\right)\right)}} \\ =2\mathrm{\pi }\sqrt{\frac{2\times {10}^3}{\mathrm{\pi }\times (10{)}^2\times 1\times 10}} \\ =2\mathrm{\pi }\sqrt{\frac{2}{\mathrm{\pi }\times 10}}=0.5\mathrm{s} \end{gathered}$$
Therefore, the required time period is 0.5 second.