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Simple Harmonic Motion

  1. Jul 30, 2013 #1
    1. The problem statement, all variables and given/known data

    A spring is freely hanged on a ceiling. You attach a mass to the end of the spring and let the mass go. It falls down a distance of 49 cm and comes back to where it started. It contineous to oscillate in a simple harmonic motion going up and down - a total distance of 49 cm from top to bottom. What is the period of the simple harmonic oscillation?

    2. Relevant equations

    Newtons Second Law --->> Fr = Fg

    3. The attempt at a solution

    To find the period of the simple harmonic motion, my first objective was to find the angular frequency using the equation (k/m) = (g/Xm) which comes from newtons second law, however the answer im suppose to get is ∏/[itex]\sqrt{}10[/itex] but i got ∏[itex]\sqrt{}10[/itex] ---> please explain how it works
     
  2. jcsd
  3. Jul 30, 2013 #2

    ehild

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    Show your work, please. How did you get that result? What is Xm?

    ehild
     
  4. Jul 30, 2013 #3
    Given that the period of the armonic motion of the spring is given by
    $$T=2\pi\sqrt{\frac{m}{k}}$$
    the only thing you need to find is indeed ##k##. You did correctly finding the equilibrium situation; indeed in the middle of the motion (means @##x_{1/2}=24.5## cm) you have the gravitational force balancing the elastic one (the mass goes on because at that point it has a speed, but you do not care about this). Therefore you can find ##k## imposing
    $$kx_{1/2}=mg$$
    and the trick is done. Plugging numbers I indeed find ##\pi/\sqrt{10}##.
     
  5. Jul 30, 2013 #4
    Thanks tia89, ill provide the whole work out in the next question - im new to the forum and im not very good at inserting equations etc. Please understand ...
     
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