How is a simple harmonic motion affected by resistive forces? In this case, a spring block system is placed on rough horizontal surface. How to derive the block's displacement equation?

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How is a simple harmonic motion affected by resistive forces? In this case, a spring block system is placed on rough horizontal surface. How to derive the block&#039;s displacement equation?

Kyan Mcdonald · Accepted Answer

Let us consider a block on a surface, under the action of a restoring force   −  k  x and a static-sliding friction force. For simplicity we consider the case where the block is initially at rest, i.e. its initial velocity is zero,             x      ˙        =  0First of all, if       |    x      |    &amp;lt;  μ  N      /    k no motion will occur, since the static friction force balances the restoring force. If       |    x      |    &amp;gt;  μ  N      /    k the motion will occur, governed by the Newton&#039;s equation  m            x      ¨        =  ±  μ  N  −  k  x  ,where the sign in front of the friction force depends on the direction of the block motion. formally, this could be written as  m            x      ¨        =  −  sign  (            x      ˙        )  μ  N  −  k  x  ,where  sign  (            x      ˙        )  =      {                            +          1          ,           if                                     x              ˙                                &amp;gt;          0          ,                                      −          1          ,           if                                     x              ˙                                &amp;lt;          0                          .As was mentioned in the beginning, it is easier to solve this problem in piecewise manner:if       x    0    &amp;gt;  μ  N      /    k, the motion is governed by equation  m            x      ¨        =  μ  N  −  k  x  ,which is the equation of an oscillator under the action of a constant force, with equilibrium position       x          e      q        =  μ  N      /    k, and the amplitude   A  =      x    0    −  μ  N      /    k. It will swing through its equilibrium position and stop at the point       x    1    =      x          e      q        −  A  =  2  μ  N      /    k  −      x    0  if       x    1    &amp;lt;  −  μ  N      /    k, the oscillator will swing back. Its velocity is now positive, and motion is now governed by equation  m            x      ¨        =  −  μ  N  −  k  x  ,which is the equation of an oscillator under the action of a constant force   −  μ  N, with equilibrium position       x          e      q        =  −  μ  N      /    k, and amplitude   A  =      |        x    1    −      x          e      q            |    =  −  μ  N      /    k  −      x    1   The oscillator will thus stop at       x    2    =  −  μ  N      /    k  +  A  =  −  2  μ  N      /    k  −      x    1    =      x    0    −  4  μ  N      /    kWe can continue reasoning in this way and arrive at the following recursive solution:      x          2      n      +      1        =  2  μ  N      /    k  −      x          2      n        ,    x      2    n    +    2    =  −      x          2      n      +      1        −  2  μ  N      /    k  .The solution of this equations for the stopping points is      x          2      n        =      x    0    −            4      n      μ      N        k    ,        x          2      n      +      1        =            2      (      2      n      +      1      )      μ      N        k    −      x    0    ,while       |        x          i            |    &amp;gt;  μ  N      /    kWith some patience this solution could be generalized to the case of arbitrary initial conditions.

Resistive forces on Simple Harmonic motion.

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