All Problems

Mechanical Oscillations

Problem 4.21

A pendulum clock is mounted in an elevator car which starts going up with a constant acceleration \(w,\) with \(w < g .\) 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?

Reveal Answer
t=2hw1+η−1−η1−1−η where η=w/g . t=\sqrt{\frac{2 h}{w}} \frac{\sqrt{1+\eta}-\sqrt{1-\eta}}{1-\sqrt{1-\eta}} \text { where } \eta=w / g \text { . }
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