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F = MA 2009 #16 (Simple Question about Springs)

  1. Jan 29, 2013 #1
    1. The problem statement, all variables and given/known data
    https://aapt.org/physicsteam/2010/upload/2009_F-maSolutions.pdf
    #16


    2. Relevant equations
    ω = √(k/m)


    3. The attempt at a solution
    So in this problem, I thought that given two identical masses on each side (that exert let's say Force F each), the net tension is not 2F, but F. This the force that is resisted is F = - kx, not twice that. I thought that there was nothing special about k, but apparently I'm wrong. Also, I thought given two objects, the mass is 2m; yielding the following answer:
    √(k/2m) - The wrong answer.

    The correct one is :
    √(2k/m)
     
  2. jcsd
  3. Jan 29, 2013 #2

    TSny

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    Imagine the blocks sitting at rest with the spring unstretched so each block is at its equilibrium position. Think about how you would start the system oscillating so that each mass oscillates about its equilibrium position with the midpoint of the spring always staying at rest.

    If at some moment of the oscillation the block on the right is a distance x to the right of its equilibrium position, where is the other block in relation to its equilibrium position? How much is the spring stretched in terms of x? How would you write the force that the spring exerts on the right-hand block in terms of x?
     
  4. Jan 29, 2013 #3
    Well if A is the amplitude, I would think that as one block is X to the right, the other would be A-X to the left. The spring is stretched by A-X + X = A? The force that the spring exerts on the right-hand block I'm not sure how to write.
     
  5. Jan 29, 2013 #4

    TSny

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    If one block has moved 0.1 m from its equilbrium position, how far has the other block moved from its equilibrium position? How much has the spring stretched?
     
  6. Jan 29, 2013 #5
    The other block compresses .1m?
     
  7. Jan 29, 2013 #6

    TSny

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    Yes, for the .1 m. Not sure about the interpretation of "compresses". Consider the figure below which shows the equilibrium position of each mass. If at some instant block B happens to be displaced a distance x to the right of its equilibrium position as shown, then where is block A at this instant in relation to its equilibrium position?
     

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  8. Jan 29, 2013 #7
    To the right of its EP.
     
  9. Jan 29, 2013 #8

    TSny

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    No. If both blocks move to the right by the same distance, then the whole system simply displaces to the right with no change in the amount of stretch of the spring. That would not correspond to oscillatory motion of the system.
     
  10. Jan 29, 2013 #9
    Whoops. Where can I study oscillatory motion with two masses - my text is no help.

    Must it move x to the left?
     
  11. Jan 29, 2013 #10

    TSny

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    Yes, it must move to the left.

    Imagine starting the system oscillating by simultaneously pulling block A to the left a certain distance and block B to the right the same distance and letting them go at the same instant. Can you visualize what the system will do after that?
     
  12. Jan 29, 2013 #11
    They will oscillate between getting relatively close to relatively far away
     
  13. Jan 29, 2013 #12

    TSny

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    So, if block B has displaced a distance x to the right of its equilibrium position, block A will have displaced a distance x to the left of its equilibrium position. How much has the spring stretched from its natural length? (Express in terms of x.)
     
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