We shall work in the car frame. As it is accelerated with respect to the road, we shall have to apply a pseudo force `ma_(0)` on the bob of mass `m`.
For mean position, the acceleration of the bob with respect to the car should be zero. If `theta_(0)` be the angle made by the string with the vertical, the tension, weight and the pseudo force will add to zero in this position.
Hence, resultant of `mg` and `ma_(0)` (say `F = msqrt(g^(2) + a_(0)^(2))`) has to be along the sting.
`:. tan theta_(0) = (ma_(0))/(mg) = (a_(0))/(g)`
Now, suppose the string is further deflected by an angele `theta` as shown in figure.
Now, restoring torque can be given by
`(F sintheta)l = -ml^(2)alpha`
Substituting `F` and using `sintheta = theta`, for small `theta`.
`(msqrt(g^(2) + a_(0)^(2))) l theta = -ml^(2)alpha`
or, `alpha = - sqrt(g^(2) + a_(0)^(2))/(l)theta` so, `omega = sqrt(g^(2) + a_(0)^(2))/(l)`
This is an equation of simple harmonic motin with time period
`T = (2pi)/(omega) = 2pi(sqrt(l))/(g^(2) + a_(0)^(2))^(1//4)`
