# Cutting a spring in half and then in 1/4 and 3/4

1. Oct 20, 2014

### Dustinsfl

1. The problem statement, all variables and given/known data
A helical spring of stiffness k is cut into two halves and a mass m is connected to the two halves. The natural time period of this system is found to be 0.5 sec. If an identical spring is cut so that one part is one-fourth and the other part is three-fourths of the original length, and the mass m is connected to the two parts, what would be the natural period of the system?

2. Relevant equations

3. The attempt at a solution
Attached is the solution. Why is k2 = 4k? I understand the solution up until that point.

2. Oct 20, 2014

### BiGyElLoWhAt

3. Oct 20, 2014

### Dustinsfl

4. Oct 20, 2014

### BiGyElLoWhAt

Oh wait, I think I misunderstood the question.

F=kx => k =F/x and if we halve x we double k.

So basically if you take a spring, with some number N coils and stretch it by some length x, each winding or coil stretches by a factor of x/N. Now halve N (you're cutting the spring in half). Stretch the spring by the same length, each coil now stretches by a factor of 2x/N, effectively doubling the spring constant. (It takes 2x as much force to stretch or compress the spring by the same amount)

5. Oct 20, 2014

### BiGyElLoWhAt

6. Aug 29, 2017

### killercrew

Why did the text bookuse series of spring and not parallell springs Keq=k1+K2?

7. Aug 29, 2017

### haruspex

Not sure which step you are referring to.
The working has two stages. First, we need to find the effective k of each section of the cut spring. E.g. for the 1/4:3/4 split:
We can think of the uncut spring as these two sections in series:
$\frac 1k=\frac 1{k_{\frac 14}}+\frac 1{k_{\frac 34}}$.
Using the argument in post #4, we know that $k_{\frac 14}=4k$. Substituting that leads to $k_{\frac 34}=\frac 43k$.
In the second stage, we consider inserting the mass between the two sections. At the mass, for a given displacement, the two forces add. This is the same as for springs in parallel, so we use $k_{net}=k_{\frac 14}+k_{\frac 34}$.