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Unformatted text preview: 4.17) Effective Spring Constant First consider two springs with constants k 1 , k 2 in parallel. When they are stretched from their equilibirum length by a distance x , the force is F x = k 1 x k 2 x = k eff x so we see k eff = k 1 + k 2 Now consider two springs in series, stretched by a total distance x . Note that the springs are not necessarily both stretched x/ 2. If the first spring is stretched by x 1 , the second is stretched by x 2 = x x 1 . These distances must be such that the tension in the springs is constant along their length, otherwise there will be a net force at the point where the two springs connect. k 1 x 1 = k 2 ( x x 1 ) x 1 = k 2 k 1 + k 2 x So the total force is F x = k 1 x 1 = k 2 x 2 = k 2 k 1 k 1 + k 2 x = k eff x So we have k eff = k 1 k 2 k 1 + k 2 4.19) Removing a Spring Initially, the equation of motion for the mass is m x = 2 kx which has solution x = A cos( t + ) for 2 = 2 k/m and constants A and determined from initial conditions. If one spring is removed, the equation of motion will become m x = kx which we can solve with x = A prime cos( prime t + prime ) where prime 2 = k/m . Note the original A is given as d , and can be determined from the fact x (0) = d/ 2 d cos = d/ 2 = 2 3 1 Now determine the new constants, knowing that even though the force on the particle changes, its position and velocity must remain continuous. d/ 2 = A prime cos prime by continuity of x d sin(2 / 3) = A prime prime sin prime by continuity of x substitute in for the frequencies and obtain d/ 2 = A prime cos prime d radicalbig 3 / 2 = A prime sin prime Square both equations and sum, to see d 2 4 + 3 d 2 2 = A prime 2 (cos 2 prime + sin 2 prime ) A...
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This note was uploaded on 01/22/2012 for the course PHYS 3201 taught by Professor Deheer during the Spring '10 term at Georgia Institute of Technology.
 Spring '10
 deHeer
 mechanics, Force

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