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A spring-block system on a frictional plane

  1. Mar 16, 2007 #1
    Maybe I should have posted this question in 'elementary physics' but the question seemed rather difficult to me(although involving elementary concepts) so i am posting it at this forum.Here goes.

    1. The problem statement, all variables and given/known data

    A block of mass m is attached to a spring of spring constant k.the other end of the spring is attached to a wall.The mass rests on a surface with friction coefficient [tex]\mu[/tex].Now the mass is given a displacement [tex]x_{0}[/tex] from its equilibrium position.Calculate x at any subsequent time.

    2. Relevant equations

    The forces acting on the block are(taking the equilibrium position as origin)
    force due to the spring:
    [tex]\\F_{spring} = -kxi\hat \\
    [/tex]

    [tex]
    \\ F_{friction} = \mu mgi\hat \\ \mbox{ when the block is moving in the -i direction}
    [/tex]

    [tex]
    F_{friction} = -\mu mgi\hat \\ \mbox{ when the block is moving in +i direction}
    [/tex]

    3. The attempt at a solution

    The equations i obtain are
    [tex] m\ddot x +kx = \mu mg\ \mbox{when moving towards -i} \\
    [/tex]
    [tex]
    \\ \mbox{and} m\ddot x + kx = -\mu mg\ \mbox{when moving towards +i}
    [/tex]
    Solving them i get,for the first half cycle:
    [tex]
    \\x = (x_{0} - \frac{\mu mg}{k})\cos \frac{k}{m}t + \frac{\mu mg}{k}\\
    [/tex]
    [tex]

    \mbox{At the end of the first half cycle,the position of the block is,then } x'_{0} = -x_{o} + 2\frac{\mu mg}{k}
    [/tex]
    [tex]
    \\ \mbox{Now,if i take t= 0 at} x= x'_{0} \ \mbox{ ,i get,from solving the second differential equation}[/tex]
    [tex]
    \\x = -( x'_{o}+ \frac{\mu mg}{k}) \cos \frac{k}{m}t - \frac{\mu mg}{k}
    [/tex]
    [tex]
    \mbox{At the end of this half-cycle the position is }
    [/tex]
    [tex]
    x_{0} - 4\frac{\mu mg}{k}
    [/tex]

    So i'm getting the result that for each half cycle the distance traversed goes down by [tex] 2\frac{\mu mg}{k}[/tex].I get the same reult from work-energy theorem.
    is my method correct?if yes,is there a more elegant way of doing it?is there at least an elegant way of writing the answer?In my answer i'm putting t=0 at the start of each half-cycle,so i can't boil it down to less than 2 equations.

    P.S:this is te first time i'm using latex.
     
    Last edited: Mar 16, 2007
  2. jcsd
  3. Mar 16, 2007 #2
    this looks all right.must've edited it 20 times or so! how does one start a new line in latex here? \\ doesn't seem to work.
     
    Last edited: Mar 16, 2007
  4. Mar 16, 2007 #3

    Mentz114

    User Avatar
    Gold Member

    This might help,

    http://en.wikipedia.org/wiki/Harmonic_oscillator

    most of the damped SHO models have a damping force proportional to velocity, different from yours.
     
  5. Mar 16, 2007 #4
    thanx mentz,but that doesn't help.
    yep,this is different from your usual damped oscillator,which is why i'm finding it difficult(and interesting)!!
     
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