Question

When a force of 20.0N is applied to a spring, it elongates 0.20m. Determine the period of oscillation of a 4.0kg object suspended from this spring

Answer 1

So what we can do is apply the Hooke's law wich states that
F = -kx ( P.S the -ve sign means opposite in direction ) 
Also we will need to determine the spring's constant with the formula:
k = F / x 
Where F = the force ( = 20 N ) 
x = the displacement of the end of the spring from it's position ( = 0.20 m ) 
k = the spring's constant ( = unknown ) 
So this would be: k = 20 / 0.20 = 100 N/m 
The period of oscillation of 4 kg : T = 2 * pi * square root m / k 
T = 2 * pi * square root 4 / 100 
T = 1.256 seconds
Hope it helps