Are the Forces Different in Parallel and Series Springs?

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Homework Help Overview

The discussion revolves around the behavior of springs in different configurations—specifically, how they function in series versus parallel arrangements. The original poster presents a question regarding the equivalent spring constant when using four springs, each with a given spring constant.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the differences in displacement and force for springs in series and parallel. The original poster expresses confusion about the correct answer to a multiple-choice question regarding spring arrangements. Others suggest performing algebraic calculations to determine the equivalent spring constants for various configurations.

Discussion Status

The discussion is ongoing, with participants providing insights into the relationships between spring constants in different arrangements. Some guidance has been offered regarding the equations for calculating equivalent spring constants, but no consensus has been reached on the correct answer to the original question.

Contextual Notes

The original poster requests a simple confirmation of their answer, indicating a desire for validation rather than a detailed explanation. The problem is framed within the context of homework, suggesting constraints on the level of assistance expected.

posto002
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Homework Statement
You are given four springs, each with a spring constant k. What arrangements using all
four springs will result in an equivalent spring constant of k?
(a) All four springs in series.
(b) All four springs in parallel.
(c) Three springs in parallel which are then connected in series with the fourth spring.
(d) Two springs in parallel then connected in series with the other two springs which themselves
are in parallel.
(e) No arrangement of the four springs will result in an equivalent spring constant of k
Relevant Equations
Hooke's law
I know that in parallel springs, x (the displacement of the spring) is the same for both springs. However, the forces resulting for each string are different. For springs in a series, x may be different, but the force is the same on each string. I got the answer b, seeing how the weight would be evenly distributed between the four springs. A simple 'yes' or 'no' on whether I answered the question right will suffice. Thank you for any help!
 
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posto002 said:
Homework Statement: You are given four springs, each with a spring constant k. What arrangements using all
four springs will result in an equivalent spring constant of k?
(a) All four springs in series.
(b) All four springs in parallel.
(c) Three springs in parallel which are then connected in series with the fourth spring.
(d) Two springs in parallel then connected in series with the other two springs which themselves
are in parallel.
(e) No arrangement of the four springs will result in an equivalent spring constant of k
Homework Equations: Hooke's law

I know that in parallel springs, x (the displacement of the spring) is the same for both springs. However, the forces resulting for each string are different. For springs in a series, x may be different, but the force is the same on each string. I got the answer b, seeing how the weight would be evenly distributed between the four springs. A simple 'yes' or 'no' on whether I answered the question right will suffice. Thank you for any help!
Do the algebra. If one spring is displaced by x, what is the displacement of the others? What net force will result?
 
Oh! Right, thank you! That completely went over my head!
 
No b) is not the correct answer.

You can find the correct answer if you know that when two springs are connected in parallel, then the equivalent total ##k_{eq}## is ##k_{eq}=k_1+k_2## while if they are connected in series then the equivalent ##k_{eq}## is ##\frac{1}{k_{eq}}=\frac{1}{k_1}+\frac{1}{k_2}##.

Using these basic equations you can calculate the ##K_{eq}## for each of the configurations (a) to (d) and see which turns out to be equal to just ##k##.
 

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