A spring and a block?

Question

A spring is hung from the ceiling. A 0.482-kg block is then attached to the free end of the spring. When released from rest, the block drops 0.222 m before momentarily coming to rest.
a) What is the spring constant of the spring?
b) Find the angular frequency of the block's vibrations

Answer 1

According the Hooke's Law, the force required to stretch a spring is directly proportional to the amount it stretches.

The spring stops when the energy exerted in pulling it back is the same as the potential energy caused by gravity. The potential energy of a spring is (1/2)*k*x^2, and the potential energy by gravity is m*g*h. Set the two equal to each other, and you get (1/2)*k*x^2 = m*g*h. Solve for k to get k = 2*m*g*h/(x^2) = 2*0.482kg*9.80m/s^2*0.222m/(0.222m)^2 = 42.6 N/m.

The angular frequency of the spring is simply sqrt(k/m). So, you get: w = sqrt(42.6N/m / 0.482kg) = 9.40 rad/s.

Answer 2

a) k = 0.482*9.81/0.111 = 42.6 N/m
b) ω = √(k/m) = √(42.6 N/m/0.482 kg) = 9.4 rad/s