A seconds pendulum is suspended from rof of a vehicle that is moving along a circular track of radius `(10)/(sqrt(3))m` with speed `10m//s`. Its period of oscillation will be `(g = 10m//s^(2))`

Options

A. `sqrt(2)`s
B. 2s
C. 1s
D. 0.5s

Answer

A

Explanation

`a = (v^(2))/(R) = (10^(2))/(((10)/(sqrt(3))))=10 sqrt(3) m//s^(2)`
`g_(eff) = sqrt(g^(2) + a^(2)) = sqrt(10^(2) + (10 sqrt(3))^(2))=20 m//s`
`(T^(1))/(T) = sqrt((g)/(g_(eff))) rArr (T^(1))/(2) = sqrt((10)/(20))`
`rArr T^(1) = sqrt(2)s`