Does Cutting a Spring in Half Change Its Spring Constant K?

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SUMMARY

Cutting a spring in half does change its spring constant, denoted as "k," specifically for helically wound compression or extension springs. When a spring is cut, the number of active coils is halved, which results in the spring constant doubling. The equation governing this relationship is k = d^4 * G / (8 * D^3 * N), where "d" is the wire diameter, "G" is the modulus of rigidity, "D" is the mean diameter of the coil, and "N" is the number of active coils. Thus, the spring constant is a function of the material properties and coil configuration, not the overall length of the spring.

PREREQUISITES
  • Understanding of spring mechanics and Hooke's Law
  • Familiarity with the equation for spring constant k = d^4 * G / (8 * D^3 * N)
  • Knowledge of material properties, specifically modulus of rigidity
  • Basic concepts of helically wound springs
NEXT STEPS
  • Explore the relationship between spring constant and material properties in-depth
  • Learn about different types of springs and their applications
  • Investigate the effects of coil diameter and wire diameter on spring performance
  • Study advanced topics in mechanical vibrations and oscillations involving springs
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Students and professionals in physics, mechanical engineering, and materials science who are interested in understanding the behavior of springs and their mechanical properties.

physicsstudent12
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if you cut a spring with a constant "k" in half, does the new spring's "k" change?
 
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What do you think would happen and why?

:wink:
 
I'm not sure, I was thinking that the "k" would double, because by cutting the spring in half you are reducing the spring's displacement.
 
When you cut it in two you are not changing the displacement at all. That is, if by the displacement you mean how much you are compressing it.
 
okay, so now I'm really confused.
My other idea about the question was that maybe the "k" doesn't change at all, because it is a constant, so it is not effected by the change in the size of the spring.
 
The units for the constant are in F/L where the L refers to the length of compression. Does it say anything about the spring's actual length?
 
Hey, Fred Garvin, thanks for replying, but there's nothing about the actual length of the spring. I don't need an exact numerical value, I just need to know if the spring constant changes, and if so, how does it change.
 
I think I found out the answer to my question. I think that k would be doubled, because the force of the spring is the same, and the length of compression is reduced. Can anybody tell me if that's right?
 
No

The spring constant does not change.

:smile: :smile:
 
  • #10
physicsstudent12 said:
Hey, Fred Garvin, thanks for replying, but there's nothing about the actual length of the spring. I don't need an exact numerical value, I just need to know if the spring constant changes, and if so, how does it change.
What I was trying to get at was that the spring constant is a function of material and the way it is wound. It is not a function of the overall length of the spring. The only thing that cutting a spring in half does is halves the total distance you can compress it before you reach the spring's solid height.
 
  • #11
I'm not sure, I was thinking that the "k" would double, because by cutting the spring in half you are reducing the spring's displacement.
The tacit assumption I believe you're making is:
- This is in reference to a helically wound compression or extension spring.
- The spring is literally cut in half so that the material, wire diameter and coil diameter stay the same. Only the number of active coils gets reduced by 1/2.

If those are your assumptions, then you are correct. The k will double. The equation for spring constant for such a spring is:

k = d^4 * G / (8 * D^3 * N)

Where d = wire diameter
G = Modulus of rigidity
D = mean diameter of coil
N = Number of active coils
 
  • #12
Thanks for that Q_Goest.

I always had thought it was just a function of the material and the way it was wound.

Sorry PhysicsStudent12 you were right all along :wink:
 

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