Question

A cylindrical rod of length $l=2m$ and density $\frac{\rho }{2}$ floats vertically in a liquid of density $\rho$ as shown in figure (a).
(a) Show that it performs SHM when pulled slightly up and released and find its time period. Neglect change in liquid level.
(b) Find the time taken by the rod to completely immerse when released from position shown in figure (b). Assume that it remains vertical throughout its motion.(take$g={\pi }^{2}m/s^2$)

Answer 1

The unbalanced force as the rod is shifted in vertical direction by a distance x is

$F=Ax\rho g$

Also, acceleration will be,

$a=\frac{-F}{m}=-\frac{Ax\rho g}{Al\frac{\rho }{2}}=-\frac{2g}{l}x$

$a=-10x=-{\omega }^{2}x$

$\omega =\sqrt{10}{s}^{-1}$

Thus, time period is

$T=\frac{2\pi }{\omega }=\frac{2\pi }{\sqrt{10}}=2s$

Equilibrium position of the rod is x (assume)

$A\rho xg=A\frac{\rho }{2}lg$

$x=1m$ length is immersed inside. Thus, time taken to immerse completely, that is $\frac{1}{2}$ length is 1s.

Answer is 2s and 1s.