A second pendulum is attached to roof of a car that is moving with acceleration `10m//s^(2)` on a straight road. Its frequency of oscillation is

Options

A. `(1)/(2)`Hz
B. 1 Hz
C. `(1)/(sqrt(2))`Hz
D. `2^(-3//4)` Hz

Answer

D

Explanation

`g_(eff) = sqrt(g^(2) + a^(2)) = sqrt(10^(2) + 10^(2)) = 10 sqrt(2)`
`(T^(1))/(T) = sqrt((g)/(g_(eff))) = (T^(1))/(2) = sqrt((10)/(10 sqrt(2)))`
`rArr T^(1) = (2)/(sqrt(2)) = 2^(3//4)" "rArr f = (1)/(T^(1)) = 2^(-3//4)`