Mass on a Spring: Solving for k and T with 100g and 1kg

[Akibarika]

Homework Statement

a steel spring that it extends by 10cm in equilibrium when you attach the upper end of the spring to a fixed support and hang a weight of 100g at the lower end.

And I found k and T, so how about if I change the weight to 1kg at the lower end and do the same thing?

Homework Equations

F = -kx

The Attempt at a Solution

I found they actually same.
when 100g
mg=kx
k=9.81

T=2pi*sqrt(m/k) = 2*pi*sqrt(0.1/9.81) = ...

when 1kg
mg=kx
k = 98.1
T=2pi*sqrt(m/k) = 2*pi*sqrt(1/98.1) = ...

same!?

I think I did wrong, who can help me?
THANK YOU!

 

[Doc Al]

Why do you think the spring constant changes? If you are using the same spring, then you'll have the same spring constant. (You cannot assume that the spring extends by the same amount with a heavier weight hanging from it. If you do, then of course you'll get the same answer since m/k will be the same.)