A rod of mass `m` and length `L` is kept on a horizontal smooth surface. The rod is hinged at its one end `A`. There are two springs, perpendicular to the rod, kept in the same horizontal plane. The spring of spring constant `k` is rigidly attached to the lower end of rod and spring of spring constant `12k` just touches the rod at its midpoint `C`, as shown in the figure. If the rod is slightly displaced leftward and released then, the time period for small osciallation of the rod will be

Options

A. `pisqrt((m)/(3k))`
B. `(3pi)/(2)sqrt((m)/(3k))`
C. `pi[sqrt((m)/(3k))+sqrt((3g)/(2L))]`
D. `(3pi)/(2)[sqrt((m)/(3k))+4sqrt((3g)/(2L))]`

Explanation

When rod moves right word to the line `AB`
`tau_(A)=kLthetaLcostheta+12k(L/2)theta(L//2)costheta`
Hence, `theta to 0 rArr costheta=1`
`tau_(A)=[(4kL^(2)+12kL^(2))/(4)]thetarArr(mL^(2))/(3)alpha=((4k+12k)L^(2))/(4)theta`
`rArralpha=(3(4k+12k)theta)/(4m)=(12k)/(m)theta`
`rArr (d^(2)theta)/(dt^(2))+(12k)/(m)theta=0rArr`Equation of SHM
`omega= 2sqrt((3k)/(m))rArrT_(1)=pisqrt((m)/(3k))`

When the rod moves leftward of line `AB`, spring of spring constant `k_(2)` will no play any role in the motion of the rod, so
`(mL^(2)alpha)/(3)=kL^(2)thetarArralpha=(3k)/(m)thetarArr(d^2)theta)/(dt^(2))+(3k)/(m)theta=0`
`rArr T_(2)=pisqrt((m)/(3k))`
`T=(1)/(2)(T_(1)+T_(2))=(3pi)/(2)sqrt((m)/(3k))`
(##FIT_JEE_PHY_GMP_ASS_E01_281_S02##)