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 pseudo force `ma_0` on the bob of mass m.
For mean positon the acceleration of the bob with respect to the car should be zero. It theta be the angle made by the string with vertical the tension, weight and the pseudo force will add to zero in this position.

Suppose at some instant during oscillation, the stirng is further deflected by an angle alpha so that the displacement of the bob is x. Taking te components perpendicular to the string.
component of T=0
Component of `mg=mgsin(alpha+theta)and
component `of ma_0=-ma_0cos (alpha+theta)`
Thus the resultant component F
`=m[gsin(alphas+theta)-a_0cos(alpha+thet)]`
Expanding the sine and cosine and putting `cosalpha=1, siN/Alpha=alpha=x/l` we get
`F=m[gsintheta-a_0costheta+(gcostheta+a_0sintheta)x/l]`.....i
At x=0 the force F on the bob should be zero as this is the mean position. Thus by i
`0=m[gsintheta-a_0costheta]`...........ii
giving `tantheta=(a_0)/(sqrt(a_0^2+g^2))`........iii
` costheta=g/(sqrt(a_0^2+g^2))`.......iv
Putting ii, iii and iv in i, `F=msqrt(g^2+a_0^2) x/l`
or `F=momega^2x, where omega^2=(sqrt(g^2+a_0^2))/l`
This is an equation of simle harmonic motion withh time period
`t=(2pi)/omega=2pi sqrtl/((g^2+a_0^2)^(1/4))`
An easy working rule may be found out as follows. In the mean position the tension the weight and the pseudo force balance.
From ure the tension is
`T=sqrt((ma_0)^2+(mg)^2)`
`T/m=sqrt(a_0^2+g^2)`

this plays the role of effective g. Thus the time period is
`t=2pisqrt(l/(T/m))=2pi sqtl/([g^2+a_0^2]^(1/4))`