Posted by **Julia** on Thursday, January 24, 2013 at 9:24am.

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 -
**Joseph**, Thursday, January 24, 2013 at 11:34am
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 -
**Elena**, Thursday, January 24, 2013 at 12:07pm
The period is

T= 2π•sqrt(m/k) =

=2•3.14•sqrt(0.22/12) =0.85 s.

It doesn’t depend on „x“

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