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Motion of a mass on a spring suspended vertically

  1. Aug 4, 2005 #1
    Okay I have part of this problem right but am not sure how to proceed on the last part.

    A 50 g spring is made from a new steel/titanium alloy. Engineers determine that if a 500 g mass is hung from the spring it oscillates with a period of exactly 1.00 seconds. What is the spring's constant?
    Equation used is = system mass = hanging mass + (1/3) spring mass
    system mass = 500 + (1/3)50 = 516.6666667
    Now I have to use the equation Period^2 = 4pi^2 system mass/spring constant.
    I have to solve for mass and don't know how to turn this last equation around.
     
  2. jcsd
  3. Aug 4, 2005 #2

    Doc Al

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    Staff: Mentor

    You start with this:
    [tex]T^2 = \frac{4 \pi^2 m}{k}[/tex]

    To solve for the spring constant, try this: First multiply both sides by [itex]k[/itex], then divide both sides by [itex]T^2[/itex]. That will isolate k.
     
  4. Aug 4, 2005 #3
    So then that equation should be k = 4pi^2/T^2?

    4pi^2(516.6666667)/1.00^2 = 20397.8243 is the spring constant?
     
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