Being a punctual man, the lift operator of a skyscraper hung an exact pendulum clock on the lift
Answers
The period of a simple pendulum is inversely proportional.to the square root of the free fall acceleration:
Let the magnitude of the acceleration of the lift be a. Then the period of the pendulum for the lift moving upwards with an acceleration a will be
and for the lift moving downwards with the same acceleration
Obviously, the time measured by the pendulum clock moving upwards with the acceleration a is proportional to the ratio of the time tup of the upward uniformly accelerated motion to the period Tup:
The time measured by the pendulum clock moving downwards with the acceleration a is
By hypothesis, the times of the uniformly accelerated downward and upward motions are equal:
where t1 is the total time of accelerated motion of the lift. Therefore, the time measured by the pendulum clock during a working day is
Here to is the time of the uniform motion of the lift. The stationary pendulum clock would indicate
It can easily be seen that the inequality
is fulfilled. Indeed,
Hence it follows that on the average the pendulum clock in the lift lags behind: t' < t, and hence the operator works too much.