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Finding the period with mass attached to two springs

  1. Apr 11, 2010 #1
    1. The problem statement, all variables and given/known data
    A mass m slides on a frictionless horizontal surface, connected to two springs. If the springs have force constants k1 and k2, show that the simple harmonic sliding motion has period: T = 2pi*sq root(m(k1+k2)/(k1k2)
    There is a diagram and the springs are connected horizontally to each other and then attached to the mass.


    2. Relevant equations
    T=1/f
    f=1/2pi*sq root(k/m)


    3. The attempt at a solution
    I figured out that for one spring T=2pi*sq root(m/k1), but i dont understand how to get k1k2 at the bottom of the equation that i need to show.
     
  2. jcsd
  3. Apr 11, 2010 #2

    rl.bhat

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    Homework Helper

    When the two springs of spring constant k1 and k2 are connected in series, they behave like a single spring of equivalent spring constant k = k1*k2/(k1+k2).
     
  4. Apr 15, 2010 #3
    This is my first post, so I hope this is okay to post! Kindly let me know if it is not.

    I just did this in my homework this past weekend!!

    starting with Hooke's law, F=kx

    for both k values:

    F=k1x1
    F=k2x2

    so

    x1=F/k1
    x2=F/k2

    distance traveled is x = x1+x2

    x1 + x2 = F/k1 + F/k2

    get common denominator of k1k2

    x = (Fk2 + Fk1)/ k1k2

    factor out F

    x = F (k2 + k1) / k1k2

    divide by F

    x/F = (k2 + k1)/k1k2

    now you have your k value that you can plug into your formula.
     
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