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If a pendulum-driven clock gains 5.00 s/day, what fractional change in pendulum length must be made for it to keep perfect time?
$l_{2}=1.0001157 \cdot l_{1}$
Physics 101 Mechanics
Physics 103
Chapter 16
Oscillatory Motion and Waves
Periodic Motion
Wave Optics
University of Michigan - Ann Arbor
Hope College
University of Sheffield
McMaster University
Lectures
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you know that the period the equation would be to pi multiplied by the length of the pendulum al, divided by G, the acceleration due to gravity. And so we can then say that the period is directly proportional to the square root of the length of the pendulum. And so we can say that here, we're going to say that Tisa bow, which would essentially be t long divide by teeth of s, would be equaling the square root of the length long length short. And so we can then say that here one minus length, it's a pal over at length. Some s, uh this would be essentially equaling. This would be the period long. So there are 86,400 seconds in one day, and then it is always five seconds late. This would be divided by the length some short or this would be the length correct. Essentially. And this would be 86,400 seconds quantity squared. And so we can then say that here this would be equaling one minus l over l. Whether we can say l sub divided by Alsup s, this is equaling one minus 1.16 And so we can say that here, the answer would be we'd have to increase the length by a factor. Ah, 0.116 And so we can say that where we can increase the length by a percentage of 0.116 percent. And so this would be our final answer. That is the end of the solution. Thank you for one.
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