A thin rod of length (L) and area of cross - section (S) is pivoted at its lowest point (P) inside a stationary, homegeneous and non - viscous liquid. The rod is free to ratate in a vertical plane about a horizontal axis passing through (P). The density (d_1) of the material of the rod is smaller than the density (d_2) of the liquid. The rod is displaced by a small angle (theta) from its equilibrium position and then released. Show that the motion of the rod is simple harmonic and determine its angular frequency in terms of the given parameters.
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Answer

A, B, C, D

Explanation

Let theta be the angle made by the rod at any instant t.
(##JMA_CHMO_C10_029_S01##)
The volume of the rod `=LS`,
Weight of the rod `=LS d_(1)g`
Upthrust acting on rod `=LS d_(1)g`
Since, `d_(2)gtd_(1) (given). Therfore net force acting at the centre of mass of the rod at tilted position of
`(LS d_(2)g-LSd_(1)g)
Taking moment of force about (P)
`tau=(LSd_(2)d-LSd_(1)g)xxPN`
where `PN=perpendincular distance of line of action of force from (P)
:. `tau=LSg(d_(2)-d_(1))xxL/2sin theta`
when `theta` is small, `sin theta~~theta`
:. `t =L^(2)sg)/2(d_(2)-d_(1))theta`. Since, tau prop theta`
:. Motion is simpoe harmonic.
:. On comparing it with `tau=C theta`
`We get `C=(L^(2)Dg)/2(d_(2)-d_(1))`
rArr `I omega^(2)=(L^(2)Sg)/2(d_(2)-(d_(1))`
The moment of inertia I of the rod (P)`
`1/(12)ML^(2)+M(L/2)^(2)`
`I=1/3ML^(2)=1/3LSd_(1)L^(2)`
From (i) and (ii)`
`omega^(2)xxL^(3)/3Sd_(1) =(L^(2)Dg)/2(d_(2)-d_(1)`
`rArr omega=sqrt((3Sg(d_(2)-d_(1)))/(2LSd_(1) rArr omega=sqrt(3(d_(2)-d_(1))g)/(2d_(1)L)`.