rubber band
I found delta L to be .002 m as well.
Here's what I did.
Fc=m*(dL+1)*w^2
Fs=k*dL
.5 N=k*.01 m
k=50 N/m
Fc=Fs
m*(dL+1)*w^2=k*dL
m*w^2=k*dL  m*w^2*dL
m*w^2=dL*(km*w^2)
dL=(m*w^2)/(km*w^2)
When you plug in the values (m=.1 kg; w=1 rps; k=50 N/m) the answer is dL=.1/49.9
So dL = .020 m
