Equilibrium position

  • Thread starter timtng
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  • #1
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A load of mass .2 g is hanging from a light spring whose elastic constant is 20 N/m. The load is pulled down .1 m from its equilibrium position and released.

How long is required for the load to reach its equilibrium position?

T=2pi(square root(m/k))=.628s

Please verify to see if I did it correctly.

Thanks
 

Answers and Replies

  • #2
FZ+
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I'm not looking carefully, but you seem to be wrong. It asks the time taken to reach the initial equilibrium position, while you seem to have given the periodic time for an oscillation in SHM, which would be 4*t.
 
  • #3
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so the answer should be T/4?
 
  • #4
HallsofIvy
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Yes. The "T" you give is the time to go up to the maximum height, then back down to the initial position: 1 cycle. The weight will take exactly 1/4 of that time to go back to the equilibrium point (1/2 T to reach the highest point, 3/4 T to pass the equilibrim point again and then at T back to the initial point).
 
  • #5
25
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so t should equal to .628s/4 = .157 s
 

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