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