285lect22 - 1 Lecture 22 Example 1 Two pendulums are of...

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1 Lecture 22 Example 1 Two pendulums are of length L 1 and L 2 , and located at the respected distances R 1 and R 2 from the center of Earth, have periods p 1 and p 2 . Prove that p 1 p 2 = R 1 R 2 p L 1 p L 2 : Solution. & 00 + g L & = 0 ) ! 0 = r g L : Because g = G M Earth R 2 ; where R is the distance from the mass to the center of Earth, we get ! 0 = r g L = r G M Earth R 2 L = 1 R r GM Earth L . Hence, p 1 = 2 ± ! 01 = 2 ± 1 R 1 q GM Earth L 1 = 2 ±R 1 p L 1 p GM Earth ; p 2 = 2 ± ! 02 = 2 ± 1 R 2 q GM Earth L 2 = 2 ±R 2 p L 2 p GM Earth : So, p 1 p 2 = 2 &R 1 p L 1 p GM Earth 2 &R 2 p L 2 p GM Earth = R 1 R 2 p L 1 p L 2 . Example 2 A certain pendulum keeps a perfect time in Paris, where radius of the earth is R = 3956 mi . But this clock loses 2 min 40 sec per day at a location on the equator. Find the equatorial bulge of the earth. Solution. If the pendulum in the clock execute n cycles per day (86,400sec) in Paris, then its period is p 1 = 86 ; 400 n sec : 1
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At the equatorial location, it takes 24h 2min and 40 sec=86,560sec for the same number of cycles. Hence, p 2 = 86 ; 560 n :
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This note was uploaded on 01/26/2012 for the course MATH 285 taught by Professor Staff during the Fall '08 term at University of Illinois, Urbana Champaign.

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285lect22 - 1 Lecture 22 Example 1 Two pendulums are of...

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