A simple pendulum of length L is suspended from the roof of a train. If the train moves in a horizontal direction with an acceleration 'a' then the period of the simple pendulum is given by

Options

A. `T= 2pi sqrt((l)/(g))`
B. `T=2pisqrt((l)/(g+a))`
C. `T=2pisqrt((l)/((g^(2)+a^(2))))`
D. `T=2pisqrt((l)/(g-a))`

Answer

C

Explanation

Resul"tan"t acceleration `=sqrt(g^(2)+a^(2))`
`therefore T= 2pi sqrt((l)/((g^(2)+a^(2)))`