1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Equilibrium position

  1. Nov 8, 2003 #1
    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.

  2. jcsd
  3. Nov 8, 2003 #2


    User Avatar

    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.
  4. Nov 8, 2003 #3
    so the answer should be T/4?
  5. Nov 8, 2003 #4


    User Avatar
    Staff Emeritus
    Science Advisor

    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).
  6. Nov 9, 2003 #5
    so t should equal to .628s/4 = .157 s
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?