A rod of mass `m` and length `l` is pivoted at a point `O` in a car whose acceleration towards left is `a_(0)`. The rod is free to oscillate in a vertical plane. In the equilibrium state the rod remains horizontal then other end is suspended by a spring of stiffness `k`. The time period of oscillations is..........

Options

A. `2pi sqrt((lm)/(ma_(0)+kl))`
B. `2pi sqrt((2lm)/(ma_(0)+kl))`
C. `2pi sqrt((lm)/(3ma_(0)+4kl))`
D. `2pi sqrt((2lm)/(3ma_(0)+6kl))`

Answer

D

Explanation

Let `e` be the initial elongation in the spring at equilibrium position `:.` about 'O'
`tau = mg ((l)/(2))-kel = 0 rArr e = (mg)/(2k) …………(1)`
let the angular displacement be `theta`:

`:. Tau_("restoring") = mg ((l)/(2) cos theta) +ma_(0) ((l)/(2) sin theta)`
`-k (e-l sin theta)l cos theta = I alpha = (ml^(2))/(3)alpha`
`rArr alpha = ?`
`=w^(2) theta :. T = (2pi)/(w)`