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Mechanical Oscillation

  1. Jul 9, 2010 #1
    A 4.0kg block extends a spring 16cm from its unstretched position.The block is removed and a 0.50Kg body is hung from the same spring.If the spring is stretched and released,the period of oscillation is :


    My Work :

    T = 2pi/w

    T = 2pi*sqrt{Mass Totall/K}

    I need to find K .

    By applying law of conservaton of energy :

    EMi = EMf

    0.5K(16*10^-2)^2 = -mgh

    h = 16*10^-2 ? ?

    Any help please.
    Thank you
  2. jcsd
  3. Jul 9, 2010 #2

    Filip Larsen

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    Gold Member

    Try analyze which forces that are acting on the first mass when it hangs still extending the spring 16 cm. Think about what the sum of all these forces must be and how that will allow you to find k.
  4. Jul 9, 2010 #3


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    Homework Helper

    The very first sentence of the problem statement contains enough information to calculate the spring constant. From that you should be able to calculate the frequency and therfore, period using the 2nd mass.
  5. Jul 9, 2010 #4
    So at equilibrium

    mg = k*(16*10^-2)

    Am I correct ?!
  6. Jul 9, 2010 #5

    Filip Larsen

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    Gold Member

    Yes, that is correct.
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