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Simple Spring Help

  1. Mar 19, 2007 #1
    1. The problem statement, all variables and given/known data
    A mass of 3.15 kg stretches a vertical spring 0.290 m. If the spring is stretched an additional 0.197 m and released, how
    long does it take to reach the (new) equilibrium position again?

    2. Relevant equations

    F=-kx
    T=2(pi)(m/k)^.5

    3. The attempt at a solution

    First I solved for k using F=kx where F is the weight of the mass

    mg=kx

    Then I solved for the period, T...

    T=2(pi)(m/k)^.5

    and the time required to reach equilibrium would be half a period so T/2 right?
     
    Last edited: Mar 19, 2007
  2. jcsd
  3. Mar 19, 2007 #2

    Doc Al

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    Staff: Mentor

    No. Trace out the motion of one complete period, which is the time the mass takes to return to its starting position and velocity. Hint: The initial position is y = -A; what other positions does it pass through during one period?
     
  4. Mar 19, 2007 #3
    It'd go from -A, through its equilibrium position, to A, back through equilibrium, then to –A and repeat.

    Did I at least solve for “k” correctly?

    Also, why care about amplitude? The period has been derived and proved to be independent of amplitude.

    The time required for the mass to go from state to the exact same state again (same position and velocity) is one period.

    So instead of doing T/2 it’s just T right?
     
  5. Mar 19, 2007 #4

    Doc Al

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    Exactly right. (Think about how long it takes for each of these position changes in terms of period.)

    Yes. (Assuming you used the correct displacement.)

    This is true.

    Also true.

    No. See my first parenthetical remark above.
     
  6. Mar 19, 2007 #5
    Each position change takes T/4, and there are 4 of them...

    The time required for it to go from some initial velocity and position back to the exact same velocity and position is one period.

    How is that not what the problem is asking for?
     
    Last edited by a moderator: Mar 19, 2007
  7. Mar 19, 2007 #6

    Doc Al

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    Right.

    Still true (but not what was asked).

    Let's reread the problem:

    You tell me: Starting position = ? End position = ?
     
  8. Mar 19, 2007 #7
    I'm assuming when the problem states "new" equilibrium position that they mean the .290+0.197 position, which would be the starting and ending position, and it would take a period to go from start to finish.
     
  9. Mar 19, 2007 #8

    Doc Al

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    The ".290+0.197 position" is the initial stretched postion, not the equilibrium position. Remember that the spring is stretched and released--so the initial position is what I called y = -A.
     
  10. Mar 19, 2007 #9
    Ohhhh. So it's 3/4T?
     
  11. Mar 19, 2007 #10
    Also, so you think when they say "new" equilibrium position they mean the .29?
     
  12. Mar 19, 2007 #11

    Doc Al

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    Nope. Review what you wrote (about successive positions) in posts #3 and 5.
     
  13. Mar 19, 2007 #12

    Doc Al

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    Yes, that's what they mean. (As opposed to the unstretched equilibrium position of the spring before the mass was added.)
     
  14. Mar 19, 2007 #13
    Please tell me it's 1/4T then...
     
  15. Mar 19, 2007 #14

    Doc Al

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    Finally.... :wink:
     
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