A rectangular block of mass `m` and area of cross a small vertical displacement from equilibrium it undergoes oscillation with a time period `T` then

Options

A. `T prop sqrt(p)`
B. `T prop (1)/(sqrt(A))`
C. `T prop (1)/(p)`
D. `T prop (1)/(sqrt(m))`

Answer

B

Explanation

Force applied on the body will be equal to upthust for verticle oscilation
Let block is displeced through `x` then the weight of displaced water or upbrus (upwards)
`= - Axrhog`
wihere A is area of cross section of the block and `p`I s density The must be equal to force `(= ma)` applied where `m` is mass of the block and a is acceleration
`:. ma = -Arhog`
or `a =m - (Arhog)/(m) x = - omega^(2)x`
This is the equation of simple harmonic motion time period of oscilation
`T = (2pi)/(omega) = 2pi sqrt((m)/(Arhog)) rArr T prop (1)/(sqrt(g))`