Problem Set 3

# Problem Set 3 - 4.17 Eective Spring Constant First consider...

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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( ωt + φ ) for ω 2 = 2 k/m and constants A and φ 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( ω prime t + φ prime ) where ω prime 2 = k/m . Note the original A is given as d , and φ can be determined from the fact x (0) = d/ 2 d cos φ = d/ 2 φ = 2 π 3 1

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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 φ prime by continuity of x - sin(2 π/ 3) = - A prime ω prime sin φ prime by continuity of ˙ x substitute in for the frequencies and obtain d/ 2 = A prime cos φ prime d radicalbig 3 / 2 = A prime sin φ prime Square both equations and sum, to see d 2 4 + 3 d 2 2 = A prime 2 (cos 2 φ prime + sin 2 φ prime ) A prime = 7 2 d Substitute this back in to equation for continuity of position...
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