A spring id vertically, as shown below, and a 500g mass is hung from its lower end. The EQUILIBRIUM extension of the spring is observed to be 10.0cm.

Question

(I) What is the spring constant of the spring?
(II) If the is pulled down by 5.00cm and released, what is the frequency of the subsequent oscillations.

Answer 1

(i) The spring constant is
k = M*g/deflection
= 0.500 kg*9.8 m/s^2/0.10 m = 49 N/m

(ii) The oscillation frequency is
[1/(2 pi)]*sqrt(k/M)
and is independent of the deflection amplitude.