A simple pendulum suspended from the ceilling of a stationary trolley has a length l. its period of oscillation is `2pisqrt(l//g)`. What will be its period of oscillation if the trolley moves forward with an acceleration f ?

Options

A. `2pisqrt(l/(f-g))`
B. `2pisqrt(l/(f+g))`
C. `2pisqrt(l/((f^(2)+g^(2))^(1//2)))`
D. `2pisqrt(l/(f^(2)-g^(2)))`

Answer

C

Explanation

`2pisqrt(l/((f^(2)+g^(2))^(1//2)))`