The period of small oscillations of a simple pendulum of length `l` if its point of suspension `O` moves `a` with a constant acceleration `alpha = alpha_(1)overset(wedge)(i) - alpha_(2)overset(wedge)(j)` with respect to earth is (`overset(wedge)(i)` and `overset(wedge)(j)` are unit vectors in horizontal in horizontal and vertically upward directions respectively)

Options

A. `T=2pisqrt((l)/({(g-alpha_(2))^(2) + alpha_(1)^(2)}^(1//2)))`
B. `T=2pisqrt((l)/({(g-alpha_(1))^(2) + alpha_(2)^(2)}^(1//2)))`
C. `T=2pisqrt((l)/(g))`
D. `T=2pisqrt((l)/((g^(2) + alpha_(1)^(2))^(1//2)))`

Answer

A

Explanation

Point `O` is moving as shown
So `acc`. Of particle `w.r.t O`
`= (-alpha_(1)hat(i) + (alpha_(2) - g)hat(j))`
So `g_(eff**) = sqrt(alpha_(1)^(2) + (g - alpha_(2)^(2))`
So time period
`= 2pisqrt((l)/((alpha_(1)^(2) + (g - alpha_(2))^(2))^(1//2)))`