# Springs,Hooke's Law

1. Oct 28, 2012

### MilenaMath

A spring with spring constant k and of unstretched length L is cut in half and both parts are put side by side.
Determine the total spring constant of the parallel springs.
Can anyone explain me the solution to this.

2. Oct 28, 2012

### DrChicken

Since the springs are in parallel, they will both have the same displacement. Knowing this and examining the equation for the restoring force of the spring you can find that:

F = kx

F = k1x + k2x = (k1+k2)x

Keeping the second equation in the same "form" as the first,

k = k1 + k2

This is more of an introductory physics question. Make sure to read the stickies before posting.

3. Oct 28, 2012

### MilenaMath

But the correct answer is 4K,I did the same thing,but the professor says 4K,I cant find the mistake

4. Oct 28, 2012

### Redbelly98

Staff Emeritus
What is the spring constant of half of the original spring?

5. Oct 28, 2012

### MilenaMath

My solution was: I assumed the F=kx,k is the spring constant of the initial spring,then I said $k_1= F/(x/2)$ so is $k_2$ and thus $k_1=k_2=2k => k_t=4k$
But he says my mistake is that I cant say $k_1= F/(x/2)$ but the rest is ok

6. Oct 28, 2012

### PhanthomJay

Your answer is correct but so is the professor's comment. You need to explain more clearly, in terms of the spring's unstretched lengths, not its displacement, why cutting a spring in half doubles its stiffness.

7. Oct 29, 2012

### MilenaMath

I found out my mistake: So the right way to proceed is We apply force $F_x$ to the spring and let's assume it stretches to length $x_n-L$ then we have $F_x=k(x_n-L)$ .Then we cut the spring into half and both of them would stretch to$x/2$ length when $F_x$ force is applied but both of them will have untretched lengths equal to $L/2$ so we have $F_x=k_1(\frac{x}{2}-\frac{L}{2})$ and $F_x=k_2(x/2-L/2)$ we can't say anything about force constants $k_1$ and $k_2$ yet. but combining last 2 equations and knowing it is equal to the equation at the beginning we find that total force(spring) constant is equal to 4k.

8. Oct 29, 2012

### PhanthomJay

This is a bit of a 'circular' proof in that you assume that the spring will stretch only 1/2 as much when it it is cut in half, but this is what you are trying to prove.
The deflection of the spring is directly proportional to its length and force, and the amount of its deflection is dependent on its geometric and elastic properties. Since you halve its length, you halve its deflection under a given force, and thus double its stiffness k. Then you put the cut springs in parallel, so each sees just half the force, which effectively quadruples the effective equivalent stiffness.