Understanding the Effective Spring Constant When Cutting a Spring in Half

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When a spring is cut in half, the effective spring constant doubles because each half behaves as a stiffer spring. This is due to the fact that while the tension remains constant, the stretch for each half is reduced to half, leading to a spring constant that is twice the original. The discussion highlights that if a spring of length 1 meter and spring constant k is divided into n equal lengths, each segment will have a spring constant of nk. However, it is noted that the actual spring constant also depends on material properties, cross-sectional area, and the number of turns per unit length. Understanding these principles is essential for accurate spring design and application.
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I should probably know this being a college grad w/ 4 years of mechanics, but can anyone give me a clear explanation why the effective spring constant of a spring doubles when you cut the spring in half? I understand physically smaller springs are stiffer but is there a proof someone can give? thanks
 
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Consider this: Stretch the spring a distance X. The tension is the same throughout the spring. Now consider the spring as consisting of two springs (each half as big) hooked together. They exert the same force, but each is only stretched X/2. So the spring constant of each must be twice the original. Make sense?
 
So is this to say that if I manufacture springs originally with a length of 1 meter and spring constant k, then if i cut the thing into n equal lengths, the spring constant of each length will be nk?

Obviously I would assume this is only a very basic high level description of a spring and the actual constant is a function ofthe material, cross sectional area, and numbers of turns per unit lenght?
 
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