# General relationship for springs connected in series

1. Jul 8, 2012

### Vandetah

1. The problem statement, all variables and given/known data

What is the effective force constant of 2 springs which are connected in series? What is the general relationship for n springs connected in series?

2. Relevant equations
im not sure if i should use

or is the formula for a series circuit also relevant?

3. The attempt at a solution

2. Jul 8, 2012

### Infinitum

Hi Vandetah!

When massless springs are connected in series, the forces exerted by each is the same. Can you derive a relation using this fact?

3. Jul 8, 2012

### Vandetah

so if theres two springs then it would be
k1 = k2?

4. Jul 8, 2012

### Infinitum

Not quite. Even though their forces are same, their extension will be different. For two strings, you have,

$$x_1 = \frac{F}{k_1}$$

$$x_2 = \frac{F}{k_2}$$

And, $x= x_1 + x_2$

Do you see a relation between the spring constants now?

5. Jul 8, 2012

### Vandetah

u might laugh at this but, does it mean that two constant springs connected to each other create a signle force constant and therefore act as if it were a single spring?

6. Jul 8, 2012

### Infinitum

No, they produce a single force. Not the same force constant. That should be clear from the above equations

7. Jul 8, 2012

### Vandetah

so that is the general relationship, but what about the effective force constant? Can i use the equation i mentioned in the op?

8. Jul 8, 2012

### Infinitum

Noo...Use the equations I wrote in my earlier posts to derive the relation for the effective force constant!

9. Jul 8, 2012

### Vandetah

is there a difference between just a force constant and an effective one?

10. Jul 8, 2012

### Infinitum

Well, force constant is for a simple, single spring. The effective constant is the effective spring you replace a given setup with, so that you get the same force exerted by that single spring which the whole setup would exert.

11. Jul 8, 2012

### Vandetah

so if i derive the equations u mentioned i will get

k1 = $\frac{F}{x1}$

and

k2 = $\frac{F}{x2}$

so k = k1+k2??

12. Jul 8, 2012

### Infinitum

Noo

Use x = x1 + x2.

13. Jul 8, 2012

### Vandetah

could it be?

x = $\frac{F K_{2}+ F K_{1}}{K_{1}K_{2}}$

Last edited: Jul 8, 2012
14. Jul 8, 2012

### Infinitum

Yes, but you need to find the effective constant, and not the final extension

Take x = F/k.

15. Jul 8, 2012

### ehild

Imagine that you measure the extension of both springs applying the same force to each. You get ΔL1 with the first spring and ΔL2 with the other spring. If you connect the springs, the same force acts on each - you pull the second one with F, it stretches by ΔL2, the second spring pulls the first one with the same force, it extends by ΔL1, so the connected springs stretch by the sum ΔL1+ΔL2.

You put the connected springs in a black box to hide the point where they are connected. You say it is a spring inside and ask somebody to measure the spring constant, applying force F. The person measures the change of length ΔL, and calculates the spring constant k=F/ΔL. That is the effective spring constant of the device inside the black box.

You know that ΔL=ΔL1+ΔL2, and also that ΔL1=F/k1, ΔL2=F/k2, so you can express the effective spring constant with the individual ones.

ehild

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16. Jul 8, 2012

### Vandetah

lol that post above got me more confused :(

17. Jul 9, 2012

### Infinitum

Ehild put all I was saying in one post.

Just use x=F/k in your last equation, where k is the effective constant.

18. Jul 9, 2012

### azizlwl

1. Let 2 bodies connected with inextensible string.
If 1st body is pulled by a force and remained at rest, EQUAL force also exerted on 2nd object.

Thus spring#1 and spring#2 extended with equal force, x1 and x2

F=-k1x1=-k2x2 .....(1)

2. Total extension is x1+x2 due to F.
F=Keff(x1+x2)

Subt. (1) in (2)
-k2x2=-Keff((k2/k1)x2+x2)
k2=Keff((k2/k1)+1)

Keff =k1k2/k1+k2 like resistors in parallel

19. Jul 9, 2012

### ehild

You mean that Keff=k1k2/k1+k2, that is Keff=2k2? It is wrong.
Or you just forgot the parentheses...

ehild