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Frequency of Oscillation, with mass and k value given

  1. Dec 5, 2014 #1
    1. The problem statement, all variables and given/known data

    A mass m = 2.0 kg is attached to a spring having a force constant k = 990 N/m as in the figure. The mass is displaced from its equilibrium position and released. Its frequency of oscillation (in Hz) is approximately _____ .

    upload_2014-12-5_17-52-8.png

    2. Relevant equations
    [itex] \omega=\sqrt\frac{k}{m} [/itex]
    [itex] T=\frac{2\pi}{\omega} [/itex] [itex] f=\frac{\omega}{2\pi} [/itex]



    3. The attempt at a solution

    m= 2.0 kg
    k= 990 N/m
    f = ?
    omega = sqrt((990N/m)/(2.0kg)) = 22.25 rad/s
    f= (22.25 rad/s)/(2pi) = 3.5 Hz

    The answer to the problem, is 4.0 Hz. However, I don't believe I should have rounded up. I believe I have made an error in how to solve the problem.

    Have I gone wrong in choosing the formulas to use? or
    Should I have taken something else into account based on the mention of the equilibrium statement? and if so, what does that imply?
    I recall doing another type of spring problem where it went past its point of equilibrium, but we were using velocity and its angle and solving for v using
    dy/dt = -A(omega)sin(omega(t)+phi), however I was reviewing the slides before this problem, and it doesn't seem to address the equilibrium.

    I thank you for your assistance very much. I'm studying for my final on Monday and I'm trying to be prepared.
     
  2. jcsd
  3. Dec 5, 2014 #2

    gneill

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    Staff: Mentor

    Hi LaLaLina, Welcome to Physics Forums.

    Your solution method and result look fine. It's possible that at some point someone changed a parameter in the problem in order to make it a "new" problem, but didn't update the answer key.
     
  4. Dec 5, 2014 #3

    BruceW

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    Homework Helper

    your calculation looks fine to me. 3.5 Hz is correct to 2 significant figures and 4 Hz is correct to 1 significant figure. Although, it is weird that your teacher (or professor or book) wrote 4.0 Hz, since this would usually imply 2 significant figures of precision... maybe it is just an error in the book like gneill says
     
  5. Dec 5, 2014 #4
    Thank you so much. I got an email back from him, and he said I did wonderful. That it was supposed to be an approximation. So, there in lies the answer I guess. I on the other hand am sitting here worrying my brain out over knowing how to do a problem correctly with this final coming up that is 30% of my grade, so I wanted to make SURE I knew what I was doing.

    Thank you both for your feedback.I feel much better now.
     
  6. Dec 5, 2014 #5

    gneill

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    Staff: Mentor

    Glad we could help. That's what Physics Forums is all about :)
     
  7. Dec 5, 2014 #6
    Oh my! I just now noticed the T.A.R.D.I.S.
    fellow Whovian in the house!

    Thanks again.
     
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