General relationship for springs connected in series

Click For Summary
SUMMARY

The effective force constant for two springs connected in series is calculated using the formula Keff = k1k2 / (k1 + k2). This relationship indicates that the effective spring constant is not simply the sum of the individual spring constants but rather a function of their product divided by their sum. The discussion emphasizes that while the same force acts on both springs, their extensions differ, leading to the derivation of the effective constant. Understanding this relationship is crucial for accurately modeling systems involving multiple springs in series.

PREREQUISITES
  • Understanding of Hooke's Law and spring constants
  • Familiarity with basic mechanics concepts
  • Knowledge of algebraic manipulation for deriving equations
  • Concept of effective resistance in electrical circuits for analogy
NEXT STEPS
  • Study the derivation of the effective spring constant for multiple springs in series
  • Explore the concept of springs in parallel and their effective constants
  • Learn about the applications of spring constants in mechanical systems
  • Investigate the relationship between force, extension, and energy stored in springs
USEFUL FOR

Students studying physics, mechanical engineers, and anyone involved in designing systems with spring mechanisms will benefit from this discussion.

Vandetah
Messages
13
Reaction score
0

Homework Statement



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?

Homework Equations


im not sure if i should use
hookes_law_equation_spring_force_constant.png


or is the formula for a series circuit also relevant?

The Attempt at a Solution

 
Physics news on Phys.org
Hi Vandetah! :smile:

When massless springs are connected in series, the forces exerted by each is the same. Can you derive a relation using this fact?
 
so if there's two springs then it would be
k1 = k2?
 
Vandetah said:
so if there's two springs then it would be
k1 = k2?

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?
 
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?
 
Vandetah said:
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?

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

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

Noo...Use the equations I wrote in my earlier posts to derive the relation for the effective force constant!
 
is there a difference between just a force constant and an effective one?
 
  • #10
Vandetah said:
is there a difference between just a force constant and an effective one?

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
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
Vandetah said:
so if i derive the equations u mentioned i will get

k1 = \frac{F}{x1}

and

k2 = \frac{F}{x2}


so k = k1+k2??

Noo :redface:

Use x = x1 + x2.
 
  • #13
Infinitum said:
Noo :redface:

Use x = x1 + x2.

could it be?

x = \frac{F K_{2}+ F K_{1}}{K_{1}K_{2}}
 
Last edited:
  • #14
Vandetah said:
could it be?

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

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

Take x = F/k.
 
  • #15
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
 

Attachments

  • springss.JPG
    springss.JPG
    10.4 KB · Views: 518
  • #16
lol that post above got me more confused :(
 
  • #17
Vandetah said:
lol that post above got me more confused :(

Ehild put all I was saying in one post.

Just use x=F/k in your last equation, where k is the effective constant. :smile:
 
  • #18
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
azizlwl said:
Keff =k1k2/k1+k2 like resistors in parallel

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

ehild
 

Similar threads

  • · Replies 56 ·
2
Replies
56
Views
4K
  • · Replies 17 ·
Replies
17
Views
1K
  • · Replies 2 ·
Replies
2
Views
4K
Replies
17
Views
2K
Replies
29
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 14 ·
Replies
14
Views
4K
Replies
6
Views
986
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K