A wire frame in the shape of an equilateral triangle is hinged at one vertex so that it can swing freely in a vertical plane, with the plane of the `Delta` always remaining vertical. The side of the frame is `1//sqrt(3)m`. The time period in seconds of small oscillations of the frame will be-

Options

A. `pi//sqrt(2)`
B. `pi//sqrt(3)`
C. `pi//sqrt(6)`
D. `pi//sqrt(5)`

Answer

D

Explanation

`OP = a sin 60^(@) = (sqrt(3)a)/(2)`
`OC = l = (2)/(3)(OP) = (a)/sqrt(3)`
`I = I_(0) = 2((ma^(2))/(3)) + m(OP)^(2) + (ma^(2))/(12)`
`= (2)/(3)ma^(2) + (3)/(4)ma^(2) + (1)/(12)ma^(2) = (3)/(2)ma^(2)`
`T = 2pi sqrt((I)/((3m))gl)`
`= 2pi sqrt (((3//2m a^(2)))/((3m)g((a)/(sqrt3))))`
`= 2pi sqrt((sqrt(3))/(2)(a)/(g))`
Putting `a = (1)/sqrt(3)m` and `g = 10m//s^(2)`
we get, `T = (pi)/sqrt(5)s`