Find Spring Constant: Prove 1/K_eff = 1/k_1 + 1/k_2

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SUMMARY

The discussion focuses on proving the relationship for the effective spring constant, represented as 1/K_eff = 1/k_1 + 1/k_2, for a system with two springs of constants k1 and k2. Participants clarify that the forces on both springs are equal, leading to the conclusion that the resultant force can be expressed as F_r = -K_eff * (x1 + x2). The key takeaway is that the effective spring constant can be derived by considering the total displacement of the system.

PREREQUISITES
  • Understanding of Hooke's Law (F = -kx)
  • Basic knowledge of spring systems and their configurations
  • Familiarity with algebraic manipulation of equations
  • Concept of effective spring constant in parallel spring systems
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  • Study the derivation of effective spring constants in parallel spring configurations
  • Learn about the implications of Hooke's Law in complex mechanical systems
  • Explore examples of spring systems in physics problems
  • Investigate the role of displacement in determining forces in spring systems
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Students studying physics, particularly those focusing on mechanics and spring systems, as well as educators looking for clear explanations of effective spring constants.

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Finding the given spring constant

Homework Statement




Consider the system represented on the figure, constituted by the mass m and two springs of constant k1 and k2.
(The image is attached)
Prove that:

\frac{1}{K_{eff}} = \frac{1}{k_{1}} + \frac{1}{k_{2}}

Homework Equations



F = -kx

The Attempt at a Solution



I don't know where to begin. I have to consider two different displacements: One for the first spring, and one for the second.
I think, but I am not sure, that I can consider both forces are equal.
So:

F1 = -k_{1} x_{1}

F2 = -k_{2} x_{2}

and

1)

F1 + F2 = 0

and a resultant force =

F_{r} = -k_{eff} x

2)

x = x1 + x2

I tried making a system with equations 2 and 1, but I am getting nowhere. Can anyone help?
 

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The force on both springs are equal i.e. F1=F2

Now the force on one spring is also equal to keff(x1+x2).

I think you can now find keff
 
rock.freak667 said:
Now the force on one spring is also equal to keff(x1+x2).

can you explain why is that so, please?
 
Fanta said:
can you explain why is that so, please?

The force on the spring should be the same throughout.
 
Like this i found it easy to do, thanks.

but isn't that the force for both springs combined?
I mean, you have the total displacement, and the Keff.

Or I can choose any of the springs, say F1 = Keff(x1+x2) ?
 
Last edited:
Fanta said:
Like this i found it easy to do, thanks.

but isn't that the force for both springs combined?
I mean, you have the total displacement, and the Keff.

Or I can choose any of the springs, say F1 = Keff(x1+x2) ?

You can choose any spring and it should work out.
 

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