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Two springs, each with unstretched length 0.200 m but with

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 102CP Chapter 14

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 102CP

Two springs, each with unstretched length 0.200 m but with different force constants ?k?1 and ?k?2 are attached to opposite ends of a block with mass m on a level, frictionless surface. The outer ends of the springs are now attached to two pins ?P?1 and ?P?2, 0.100 in from the original positions of the ends of the springs (below Fig.). Let ?K?1 = 2.00 N/m, ?K?2 = 6.00 N/m, and ?m = 0.100 kg. (a) Find the length of each spring when the block is in its new equilibrium position after the springs have been attached to the pins. (b) Find the period of vibration of the block if it is slightly displaced from its new equilibrium position and released. Figure:

Step-by-Step Solution:

Solution 102CP Step 1: a) In the initial equilibrium position F = F 1 2 Where F = k x 1 1 1 2 x2 2 The force constants are k = 2.0N1 K =6.0N/m 2 So , F 1 0.1 =0.2N F = 6×0.1 =0.6N 2 we can write F =F -F 2 1 I F = 0.4N It can be expressed as 0.4N = (k + k ) x 1 2 From this equation we can find out x 0.4N X = (k1 + k2) 0.4N X = 8N/m = 0.05 m So it’s clear the box makes a movement of 0.05m from the initial equilibrium position. Total length on both sides of the box is 0.3m . The change in length of the first spring is = 0.30m - 0.05m =0.250m And the change in length of the second spring =0.30m +0.05m = 0.350m

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Chapter 14, Problem 102CP is Solved
Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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Two springs, each with unstretched length 0.200 m but with