Are the Forces Different in Parallel and Series Springs?

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SUMMARY

The discussion centers on the behavior of springs in series and parallel configurations, specifically regarding their displacement and force characteristics. In parallel springs, the displacement (x) is identical, but the forces differ, while in series, the forces are uniform, but the displacements vary. The correct answer to the homework question regarding arrangements that yield an equivalent spring constant of k is not option (b) but requires calculating the equivalent spring constants using the formulas for series and parallel configurations. The relevant equations are: for parallel springs, keq = k1 + k2, and for series springs, 1/keq = 1/k1 + 1/k2.

PREREQUISITES
  • Understanding of Hooke's Law
  • Knowledge of spring constants (k)
  • Familiarity with algebraic manipulation for equations
  • Concept of equivalent spring constants in series and parallel arrangements
NEXT STEPS
  • Calculate equivalent spring constants for various configurations of springs
  • Explore the implications of Hooke's Law in real-world applications
  • Study the effects of spring constants on mechanical systems
  • Learn about energy storage in springs and their applications in engineering
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Students studying physics, mechanical engineers, and anyone interested in understanding the principles of spring mechanics and their applications in various configurations.

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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