Question

A cylindrical object of outer diameter 20 cm and mass the 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, d = 20

$\Rightarrow r=\frac{d}{2}$ = 10 cm

When depressed downward the net unbalanced force will cause SHM. Let x $\to$ displacement of the block from the equilibrium position. So,

Driving force = U = V ${\rho }_{w}g$

$\Rightarrow ma=\pi r^2\left(X\right)\times {\rho }_{w}g$

a= $\frac{\pi r^2{\rho }_{w}9x}{2\times {10}^3}$

[because m = 2 kg = $2\times {10}^3$ g]

T= $2\pi \sqrt{\frac{\text{displacement}}{\text{Acceleration}}}$

= $2\pi \sqrt{\frac{\left(x\right)\times 2\times {10}^3}{\left(\pi r^2{\rho }_{w}g\right(x\left)\right)}}$

= $2\pi \sqrt{\frac{2\times {10}^3}{\pi \times (10{)}^2\times 1\times 10}}$

= $2\pi \sqrt{\frac{2}{\pi \times 10}}$ = 0.5 sec.