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:

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