Post a New Question


posted by .

You hang mass m=0.22 kg on a spring with constant 12 N/m. You pull the mass down a distance of 0.04 m from its equilibrium position and release it. You measure the period to be 0.85 s.

If you instead pull the mass down 0.12 m from its equilibrium position and release it, what would the new period be?

  • Physics -

    I believe the period time will be tripled from the original time at 2.55s because the distance that the spring was puled down was 3 times the original length.

    Answer= 2.55 seconds

  • Physics -

    The period is
    T= 2π•sqrt(m/k) =
    =2•3.14•sqrt(0.22/12) =0.85 s.
    It doesn’t depend on „x“

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question