A simple pendulum is suspnded from the ceilignof a car accelerating uniformly on a horizontal road.If the acceleration is `a_0` and the length of the pendulum is l, find the time period of small oscillations about the mean position.

Explanation

We shall work in the car frame. As it is accelerated with respect to the road, we shall have to apply a psuedo force ma0 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 peusdo force will add to zero in this position. Hence, resultant of mg and `ma_(0)` (say F = m `sqrt(g^(2)+a_(0)^(2))` has to be along the string.
`:. tan theta_(0) = (ma_(0))/(mg) = (a_(0))/(g)`
Now , suppose the string is further deflected by an angle `theta` as shown in figure. Now, restoring torque about point O can be given by `tau = lalpha `
(F sin `theta ) l = - ml^(2)alpha`
Substituting F and using sin `theta = 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^(2)= sqrt(g^(2)a_(0)^(2))/(l)`
This is an equation of simple harmonic motion with
time period.
`T = (2pi)/(omega)= 2pi(sqrt(l))/((g^(2)+a_(0)^(2))^(1//4))`