A pendulum clock is mounted in an elevator car which starts goint up with a constant acceleration `w`, with `w lt g`. At a height `h` the acceleration of the car reverses, its magnitude remaining constant. How soon after the start of the motion will the clock show the right time again ?

Explanation

Let the downward acceleration of the elevator car has continued for time `t^(')` then the sought time
`t=sqrt((2h)/(w))+t^(')`, where obviously `sqrt((2h)/(w))` is th etime of upward acceleration fo the elevator. One shoule note that if the point of suspension of a mathematical pendulum moves with an acceleration `vec(w)`, then the time period of the pendulum becomes
`2pi sqrt((l)/(|vec(g)-vec(w)|))`
In this problem the time period of the pendulum while it is moving upward with acceleration `w` becomes
`2pisqrt((l)/(g+w))` and its time period while the elevator moves downward with the same magnitude of acceleration becomes
`2pisqrt((l)/(g-w))`
As the time of upward acceleration equals `sqrt((2h)/(w))`, the total number of oscillations during this time equals
`(sqrt(2h//w))/(2pisqrt(l//(g+w)))`
Thus the indicated time `=(sqrt(2h//w))/(2pisqrt(l//(g+w))). 2pisqrt(l//w)=sqrt(2h//w)sqrt((g+w)//g)`
Similarly the indicated time for the time interval `t^(')`
`(t^('))/(2pisqrt(l//(g-w)))2pisqrt(l//g)=t^(')sqrt((g-w)//g)`
we demand that
`sqrt(2h//w)sqrt((g+w)//g)+t^(')sqrt((g-w)//g)=sqrt(2h//w)+t^(')`
or, `t^(')sqrt(2h//w)(sqrt(g+w)-sqrt(g))/(sqrt(g)-sqrt(g-w))`
`=sqrt((2h)/(w))(sqrt(1+beta)-sqrt(1-beta))/(1-sqrt(1-beta))`, where` beta=w//g`