Question

Being a punctual man, the lift operator of a skyscraper hung an exact pendulum clock on the lift wall to know the end of the working day. The lift moves with an upward and downward accelerations during the same time (according to a stationary clock), the magnitudes of the accelerations remaining unchanged.
Will the operator finish his working day in time, or will he work more (less) than required?

Answer 1

Correct Answer - A::B::C::D
`T_(0) = 2pisqrt((L)/(g))`
(Lift is stationary)
`T_(1) = 2pisqrt((L)/(g + a))`
(Lift is accelerated upward )
`T_(2) = 2pisqrt((L)/(g-a))`
(Lift is deacelerated upward)
Let x = total upward distance travelled
`rArr x/2 = 1/2 at^(2) rArr t = sqrt((x)/(a))`
`:.` for upward accelerated motion
`DeltaT = (t(T_(1))/(T_(2)) - t) xx 2 = 2(sqrt((x)/(a)) - 1) (sqrt((g)/(g+a)))"..."(i)`
& for upward decelerated motion
`Delta T = (t_(1)(T_(2))/(T_(0)) -t) xx 2`
`= 2(sqrt((x)/(a)) - 1) (sqrt((g)/(g-a)))`
`:. Delta T_("total") = 2(sqrt((x)/(a)))(sqrt((g)/(g-a)) - sqrt((g)/(g+a)))`
`= sqrt((4x)/(a))(sqrt((g)/(g-a))-sqrt((g)/(g+a)))`