A rod of mass `M` and length `L` is hinged at its one end and carries a particle of mass `m` at its lower end. A spring of force constant `k_(1)` is installed at distance a from the hinge and another of force constant `k_(2)` at `B` distance `b` as shown in the figure. If the whcich arrangement rests on a smoth horizontal table top, the frequency of vibration is :

Options

A. `A. (1)/(2pi)sqrt((k_(1)a^(2)+k_(2)b^(2))/(L^(2)(m + (M)/(3))))`
B. `B. (1)/(2pi)sqrt((k_(2)+k_(1))/(M+m))`
C. `C. (1)/(2pi)sqrt((k_(2)+k_(1)(a_(2))/(b^(2)))/(4(M)/(3)+m))`
D. `D. (1)/(2pi)sqrt((k_(1)+(k_(2)b_(2))/(a^(2)))/((4)/(3)m+M))`

Answer

A

Explanation

For small angular displacement `(theta)`
Net torque on body `= Ialpha`
`= (k_(1)asintheta)a + (k_(2)b sintheta)b = (mL^(2) + (ML^(2))/(3))alpha`
For small `theta rArr a = (k_(1)a^(2) + k_(2)b^(2))/(mL^(2) + (ML^(2))/(3)) theta rArr` frequency `= (1)/(2pi)sqrt((k_(1)a^(2) + k_(2)b^(2))/(L^(2)(m + M//3)))`