A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscillations, if two masses each of 'm' are attached at distance` 'L//2' ` from its centre on both sides, it reduces the oscillation frequency by `20%`. The value of ratio `m//M` is close to :

Options

A. 0.57
B. 0.17
C. 0.77
D. 0.37

Answer

D

Explanation

Moment of inertia of rod along the centre of rod

`I = (1)/(12) M (2L)^(2) = (1)/(2)ML^(2)`
Masses each of m. are attached ,br> `I. =(ML^(2))/(3) + 2(mL^(2))/(4) = L^(2)((M)/(3) + (m)/(2))`
Frequency of oscillation
`f = (1)/(2pi) sqrt((K)/(I)) rArr (f)/(f.) = (sqrt(I.))/(sqrt(I))`
By increasing mass, frequency is reduced to 80%
`because f. = 0.8f`
`(f)/(f.) = (sqrt(I.))/(sqrt(I)) rArr (1)/(0.8) = sqrt((L^(2)((M)/(3) + (m)/(2)))/((1)/(3)ML^(2)))`
`(m)/(M) = (9)/(24) = (3)/(8) = 0.375`