# Swapping springs in series

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1. Oct 8, 2014

### user977784078

1. The problem statement, all variables and given/known data
Suppose you have 2 springs in series, one is linear and one is non-linear. Initially the linear spring is at the top position and the non-linear is on bottom. Does swapping the springs affect the system?

2. Relevant equations
1/k_eq = 1/k1 + 1/k2
This equation doesn't apply though since we have a non-linear spring.
x_tot = x1 + x2
3. The attempt at a solution
I would say that it does not affect the system because x_tot in both cases will be equal, hence, F will be equal. x_tot = x1 + x2 = x2 + x1

Is this correct?

2. Oct 8, 2014

### BvU

Dear user, welcome to PF :)

If you remember where the relevant equation came from, you see that there they added the extensions: xtotal = x1+ x2 and substituted F/k1 and F/k2, respectively. So each of the springs feels the same F. That is also the case if the spring is not ideal and x is some other kind of function of F. So F isn't equal because the x_tot is equal (that is not a given; in fact that's what the exercise asks you to show!). But x_tot is equal because the F that cause the xi is the same for each spring involved, threfore the xi are equal, and yes, x1 + x2 = x2 + x1

3. Oct 8, 2014

### OldEngr63

The same force acts through both springs (because they are in series). Each spring sees a particular deflection across that spring; the total deflection of the series pair is the sum of these deflections. The order in which the load gets to the spring is immaterial; they will experience the same relative deflections in either case. As BvU said (in different notation), d1 + d2 = d2 + d1

4. Oct 9, 2014

### user977784078

Thank you. I understand now.