A spring is hung from a ceiling

Question

A spring is hung from the ceiling. A 0.383 -kg block is then attached to the free end of the spring. When released from rest, the block drops 0.198 m before momentarily coming to rest, after which it moves back upward. (a) What is the spring constant of the spring? (b) Find the angular frequency of the block's vibrations.

I can't figure out the formula to use for finding the constant.

If you can help that would be appreciated.

Answer 1

There is the downward force of gravity on the block given by -mg = -0.383kg x 9.8m/s². At the moment it comes to rest, the restoring force on the spring = -kX = -k x 0.198m. So, when the block is momentarily at rest, the downward force of gravity must equal the upward force due to the spring -0.383 kg x 9.8 m/s² = -k x 0.198m. Solve for k.

Answer 2

(a) At first stop: mg = kx, so k = mg/x = 0.383kg·9.8m/s²/0.198m = 18.96 N/m;

(b) Period vibrations: T = 2π√m/k). Then frequenct f = 1/T = 1/(2π)√(k/m) = 1/(2π)√(0,198 N/m/9.383lg

= 0.1144 Hz