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Simple Harmonic motion of vertical spring.

  1. Dec 11, 2008 #1
    1. The problem statement, all variables and given/known data
    An object of unknown mass M is hung from a vertical spring of unknown spring constant k, and the object is observed to be at rest when the spring has extended by 14cm. The object is then given a slight push and executes SHM. Determine the period of this oscillation.


    2. Relevant equations
    T = 1/f
    w = (k/M)^(1/2) = 2pi*f
    F=kx

    3. The attempt at a solution
    The object is hanging, so it has a force of Mg pulling it down and kx pulling it up. It's in equilibrium at a displacement of 14cm (0.14m). So I set kx = Mg and solved for k:
    k = Mg / x

    I then plugged this into w = (k/M)^(1/2) so the M's cancel and I'm left with w = (g/x)^(1/2) = (9.8 / 0.14)^(1/2) = 70rad/s

    w = 2pi*f => f = w/(2pi) = 11.14 Hz
    T = 1/f = 0.089s

    This isn't the right answer according to the book, and I keep getting this kind of problem wrong. I'm not entirely sure what I'm doing wrong.
     
  2. jcsd
  3. Dec 11, 2008 #2

    rl.bhat

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    3. The attempt at a solution


    I then plugged this into w = (k/M)^(1/2) so the M's cancel and I'm left with w = (g/x)^(1/2) = (9.8 / 0.14)^(1/2) = 70rad/s


    Check the above calculation
     
  4. Dec 11, 2008 #3

    tiny-tim

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    Hi Kizaru! :smile:

    (have a square-root: √ and an omega: ω :wink:)
    Nooo … √70 :wink:
     
  5. Dec 11, 2008 #4
    Ah yes, turns out my mistake was ignoring the square root. Thanks. I made that same mistake for all of these types of problems hehe, thanks :) Embarassing, guess I should check for elementary mistakes before I look for physics mistakes :)
     
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